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+\section{Reductio ad Absurdum/Variant 1}
+Tags: Reductio ad Absurdum
+
+\begin{theorem}
+: $\neg p \implies \bot \vdash p$
+\end{theorem}<|endoftext|>
+\section{Reductio ad Absurdum/Variant 2}
+Tags: Reductio ad Absurdum
+
+\begin{theorem}
+:$\neg p \implies \paren {q \land \neg q} \vdash p$
+\end{theorem}<|endoftext|>
+\section{Proof by Contradiction/Variant 1}
+Tags: Proof by Contradiction
+
+\begin{theorem}
+:$\left({p \vdash \left({q \land \neg q}\right)}\right) \vdash \neg p$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\left({p \vdash \left({q \land \neg q}\right)}\right) \vdash \neg p}}
+{{Premise|1|p \vdash \left({q \land \neg q}\right)}}
+{{Assumption|2|p}}
+{{SequentIntro|3|1, 2|q \land \neg q|1, 2|[[Definition:By Hypothesis|by hypothesis]]}}
+{{Simplification|4|1, 2|q|3|1}}
+{{Simplification|5|1, 2|\neg q|3|2}}
+{{NonContradiction|6|1, 2|4|5}}
+{{Contradiction|7|1|\neg p|2|6}}
+{{EndTableau}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Proof by Contradiction/Variant 2}
+Tags: Proof by Contradiction
+
+\begin{theorem}
+==== [[Proof by Contradiction/Variant 2/Formulation 1|Formulation 1]] ====
+{{:Proof by Contradiction/Variant 2/Formulation 1}}
+==== [[Proof by Contradiction/Variant 2/Formulation 2|Formulation 2]] ====
+{{:Proof by Contradiction/Variant 2/Formulation 2}}
+\end{theorem}<|endoftext|>
+\section{Proof by Contradiction/Variant 3}
+Tags: Proof by Contradiction
+
+\begin{theorem}
+==== [[Proof by Contradiction/Variant 3/Formulation 1|Formulation 1]] ====
+{{:Proof by Contradiction/Variant 3/Formulation 1}}
+==== [[Proof by Contradiction/Variant 3/Formulation 2|Formulation 2]] ====
+{{:Proof by Contradiction/Variant 3/Formulation 2}}
+\end{theorem}<|endoftext|>
+\section{Equivalence of Definitions of Topological Group}
+Tags: Topological Groups
+
+\begin{theorem}
+Let $\left({G, \odot}\right)$ be a [[Definition:Group|group]].
+On its [[Definition:Underlying Set|underlying set]] $G$, let $\left({G, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+{{TFAE|def = Topological Group}}
+\end{theorem}
+
+\begin{proof}
+=== Definition 1 implies Definition 2 ===
+Let $\left({G, \odot, \tau}\right)$ be a [[Definition:Topological Group/Definition 1|topological group by Definition 1]].
+Let $\phi: \left({G, \tau}\right) \to \left({G, \tau}\right)$ be the [[Definition:Mapping|mapping]] defined as:
+:$\forall x \in G: \phi \left({x}\right) = x^{-1}$
+By definition:
+: $\odot: \left({G, \tau}\right) \times \left({G, \tau}\right) \to \left({G, \tau}\right)$ is a [[Definition:Continuous Mapping (Topology)|continuous mapping]]
+: $\phi: \left({G, \tau}\right) \to \left({G, \tau}\right)$ is a [[Definition:Continuous Mapping (Topology)|continuous mapping]]
+Let $\phi': G \times G \to G \times G$ be defined by
+:$\phi' \left({x, y}\right) = \left({x, \phi \left({y}\right) }\right)$.
+Let $\phi_1: G \times G \to G$ be defined as
+:$\phi_1 \left({x, y}\right) = x$.
+Let $\phi_2: G \times G \to G$ be defined as
+:$\phi_2 \left({x, y}\right) = y^{-1}$.
+Then for arbitrary open set $V \subset G$, both sets
+:$\phi_{1}^{-1} \left(V\right) = V \times G$
+and
+:$\phi_{2}^{-1} \left(V\right) = G \times \phi^{-1}\left(V\right)$
+are open in $G \times G$, thus $\phi_1$ and $\phi_2$ are [[Definition:Continuous Mapping (Topology)|continuous]].
+Since
+:$\phi' \left({x,y}\right) = \left( {\phi_1 \left(x)\right), \phi_2 \left(y\right)} \right)$,
+by [[Continuous Mapping to Topological Product]], $\phi'$ is [[Definition:Continuous Mapping (Topology)|continuous]].
+Let the [[Definition:Mapping|mapping]] $\psi: \left({G, \tau}\right) \times \left({G, \tau}\right) \to \left({G, \tau}\right)$ be defined as:
+:$\psi = \odot \circ \phi'$
+where $\circ$ denotes [[Definition:Composition of Mappings|composition of mappings]].
+By [[Composite of Continuous Mappings is Continuous]], $\psi$ is [[Definition:Continuous Mapping (Topology)|continuous]].
+Then:
+{{begin-eqn}}
+{{eqn | l = \psi \left({x, y}\right)
+      | r = \left({\odot \circ \phi'}\right) \left({x, y}\right)
+      | c = Definition of $\psi$
+}}
+{{eqn | r = \odot \left({\phi' \left({x, y}\right)}\right)
+      | c = Definition of [[Definition:Composition of Mappings|Composition of Mappings]]
+}}
+{{eqn | r = \odot \left({x, \phi \left({y}\right)}\right)
+      | c = Definition of $\phi'$
+}}
+{{eqn | r = \odot \left({x, y^{-1} }\right)
+      | c = Definition of $\phi$
+}}
+{{eqn | r = x \odot y^{-1}
+      | c = Definition of $\odot$
+}}
+{{end-eqn}}
+demonstrating that:
+:$\psi: \left({x, y}\right) = x \odot y^{-1}$
+Thus $\left({G, \odot, \tau}\right)$ is a [[Definition:Topological Group/Definition 2|topological group by Definition 2]].
+{{qed|lemma}}
+=== Definition 2 implies Definition 1 ===
+Let $\left({G, \odot, \tau}\right)$ be a [[Definition:Topological Group/Definition 2|topological group by Definition 2]].
+Let $e$ be the [[Definition:Identity Element|identity]] of $G$.
+Let the [[Definition:Mapping|mapping]] $\psi: \left({G, \tau}\right) \times \left({G, \tau}\right) \to \left({G, \tau}\right)$ be defined as:
+:$\forall \left({x, y}\right) \in G \times G: \psi: \left({x, y}\right) = x \odot y^{-1}$
+By definition of $\left({G, \odot, \tau}\right)$, $\psi$ is [[Definition:Continuous Mapping (Topology)|continuous]].
+By [[Continuous Mapping to Topological Product]] $\psi$ is [[Definition:Continuous Mapping (Topology)|continuous]] in each variable.
+Let $\phi: \left({G, \tau}\right) \to \left({G, \tau}\right)$ be the [[Definition:Mapping|mapping]] defined as:
+:$\forall x \in G: \phi \left({x}\right) = x^{-1}$
+Since $\phi \left({x}\right) = \psi \left({e, x}\right)$, it follows that $\phi$ is [[Definition:Continuous Mapping (Topology)|continuous]].
+Let $\phi': G \times G \to G \times G$ be defined by:
+:$\forall \left({x, y}\right) \in G \times G: \phi' \left({x, y}\right) = \left({x, \phi \left({y}\right) }\right)$
+Then $\phi'$ is [[Definition:Continuous Mapping (Topology)|continuous]].
+{{explain|$\phi'$ is [[Definition:Continuous Mapping (Topology)|continuous]]}}
+$\odot$ is the [[Definition:Composition of Mappings|composition]] of $\psi$ with $\phi'$.
+Thus by [[Composite of Continuous Mappings is Continuous]], $\odot$ is [[Definition:Continuous Mapping (Topology)|continuous]].
+Thus $\left({G, \odot, \tau}\right)$ is a [[Definition:Topological Group/Definition 1|topological group by Definition 1]].
+{{qed}}
+[[Category:Topological Groups]]
+i2rz3w2mxd19tziymyu26fajxkfnxu7
+\end{proof}<|endoftext|>
+\section{Implication is Left Distributive over Conjunction/Forward Implication/Formulation 1/Proof}
+Tags: Implication is Left Distributive over Conjunction
+
+\begin{theorem}
+: $p \implies \left({q \land r}\right) \vdash \left({p \implies q}\right) \land \left({p \implies r}\right)$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|p \implies \left({q \land r}\right) \vdash \left({p \implies q}\right) \land \left({p \implies r}\right)}}
+{{Premise|1|p \implies \left({q \land r}\right)}}
+{{Assumption|2|p}}
+{{ModusPonens|3|1, 2|q \land r|1|2}}
+{{Simplification|4|1, 2|q|3|1}}
+{{Simplification|5|1, 2|r|3|2}}
+{{Implication|6|1|p \implies q|2|4}}
+{{Implication|7|1|p \implies r|2|5}}
+{{Conjunction|8|1|\left({p \implies q}\right) \land \left({p \implies r}\right)|6|7}}
+{{EndTableau}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Implication is Left Distributive over Conjunction/Reverse Implication/Formulation 1/Proof}
+Tags: Implication is Left Distributive over Conjunction
+
+\begin{theorem}
+: $\left({p \implies q}\right) \land \left({p \implies r}\right) \vdash p \implies \left({q \land r}\right)$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\left({p \implies q}\right) \land \left({p \implies r}\right) \vdash p \implies \left({q \land r}\right)}}
+{{Premise|1|\left({p \implies q}\right) \land \left({p \implies r}\right)}}
+{{SequentIntro|2|1|\left({p \land p}\right) \implies \left({q \land r}\right)|1|[[Praeclarum Theorema]]}}
+{{Assumption|3|p}}
+{{Idempotence|4|3|p \land p|3|Conjunction}}
+{{ModusPonens|5|1, 3|q \land r|2|4}}
+{{Implication|6|1|p \implies \left({q \land r}\right)|3|5}}
+{{EndTableau}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Implication is Left Distributive over Conjunction/Forward Implication/Formulation 1}
+Tags: Implication is Left Distributive over Conjunction
+
+\begin{theorem}
+: $p \implies \left({q \land r}\right) \vdash \left({p \implies q}\right) \land \left({p \implies r}\right)$
+\end{theorem}<|endoftext|>
+\section{Implication is Left Distributive over Conjunction/Reverse Implication/Formulation 1}
+Tags: Implication is Left Distributive over Conjunction
+
+\begin{theorem}
+: $\left({p \implies q}\right) \land \left({p \implies r}\right) \vdash p \implies \left({q \land r}\right)$
+\end{theorem}<|endoftext|>
+\section{Implication is Left Distributive over Conjunction/Formulation 1/Proof 1}
+Tags: Implication is Left Distributive over Conjunction
+
+\begin{theorem}
+: $p \implies \left({q \land r}\right) \dashv \vdash \left({p \implies q}\right) \land \left({p \implies r}\right)$
+\end{theorem}
+
+\begin{proof}
+=== [[Implication is Left Distributive over Conjunction/Forward Implication/Formulation 1/Proof|Proof of Forward Implication]] ===
+{{:Implication is Left Distributive over Conjunction/Forward Implication/Formulation 1/Proof}}
+=== [[Implication is Left Distributive over Conjunction/Reverse Implication/Formulation 1/Proof|Proof of Reverse Implication]] ===
+{{:Implication is Left Distributive over Conjunction/Reverse Implication/Formulation 1/Proof}}
+\end{proof}<|endoftext|>
+\section{Implication is Left Distributive over Conjunction/Formulation 1}
+Tags: Implication is Left Distributive over Conjunction
+
+\begin{theorem}
+: $p \implies \left({q \land r}\right) \dashv \vdash \left({p \implies q}\right) \land \left({p \implies r}\right)$
+\end{theorem}<|endoftext|>
+\section{Implication is Left Distributive over Conjunction/Forward Implication/Formulation 2/Proof}
+Tags: Implication is Left Distributive over Conjunction
+
+\begin{theorem}
+: $\vdash \left({p \implies \left({q \land r}\right)}\right) \implies \left({\left({p \implies q}\right) \land \left({p \implies r}\right)}\right)$
+\end{theorem}
+
+\begin{proof}
+Let us use the following abbreviations
+{{begin-eqn}}
+{{eqn | l=\phi
+      | o=\text{ for }
+      | r=p \implies \left({q \land r}\right)
+      | c=
+}}
+{{eqn | l=\psi
+      | o=\text{ for }
+      | r=\left({p \implies q}\right) \land \left({p \implies r}\right)
+      | c=
+}}
+{{end-eqn}}
+{{BeginTableau|\left({p \implies \left({q \land r}\right)}\right) \implies \left({\left({p \implies q}\right) \land \left({p \implies r}\right)}\right)}}
+{{Assumption|1|\phi}}
+{{SequentIntro|2|1|\psi|1|[[Implication is Left Distributive over Conjunction/Forward Implication/Formulation 1|Implication is Left Distributive over Conjunction: Formulation 1]]}}
+{{Implication|3||\phi \implies \psi|1|2}}
+{{EndTableau}}
+Expanding the abbreviations leads us back to:
+: $\left({p \implies \left({q \land r}\right)}\right) \implies \left({\left({p \implies q}\right) \land \left({p \implies r}\right)}\right)$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Implication is Left Distributive over Conjunction/Reverse Implication/Formulation 2/Proof}
+Tags: Implication is Left Distributive over Conjunction
+
+\begin{theorem}
+: $\vdash \left({\left({p \implies q}\right) \land \left({p \implies r}\right)}\right) \implies \left({p \implies \left({q \land r}\right)}\right)$
+\end{theorem}
+
+\begin{proof}
+Let us use the following abbreviations
+{{begin-eqn}}
+{{eqn | l=\phi
+      | o=\text{ for }
+      | r=\left({p \implies q}\right) \land \left({p \implies r}\right)
+      | c=
+}}
+{{eqn | l=\psi
+      | o=\text{ for }
+      | r=p \implies \left({q \land r}\right)
+      | c=
+}}
+{{end-eqn}}
+{{BeginTableau|\left({\left({p \implies q}\right) \land \left({p \implies r}\right)}\right) \implies \left({p \implies \left({q \land r}\right)}\right)}}
+{{Assumption|1|\phi}}
+{{SequentIntro|2|1|\psi|1|[[Implication is Left Distributive over Conjunction/Reverse Implication/Formulation 1|Implication is Left Distributive over Conjunction: Formulation 1]]}}
+{{Implication|3||\phi \implies \psi|1|2}}
+{{EndTableau}}
+Expanding the abbreviations leads us back to:
+: $\left({\left({p \implies q}\right) \land \left({p \implies r}\right)}\right) \implies \left({p \implies \left({q \land r}\right)}\right)$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Implication is Left Distributive over Conjunction/Forward Implication/Formulation 2}
+Tags: Implication is Left Distributive over Conjunction
+
+\begin{theorem}
+: $\vdash \left({p \implies \left({q \land r}\right)}\right) \implies \left({\left({p \implies q}\right) \land \left({p \implies r}\right)}\right)$
+\end{theorem}<|endoftext|>
+\section{Implication is Left Distributive over Conjunction/Reverse Implication/Formulation 2}
+Tags: Implication is Left Distributive over Conjunction
+
+\begin{theorem}
+: $\vdash \left({\left({p \implies q}\right) \land \left({p \implies r}\right)}\right) \implies \left({p \implies \left({q \land r}\right)}\right)$
+\end{theorem}<|endoftext|>
+\section{Implication is Left Distributive over Conjunction/Formulation 2/Proof 1}
+Tags: Implication is Left Distributive over Conjunction
+
+\begin{theorem}
+: $\vdash \left({p \implies \left({q \land r}\right)}\right) \iff \left({\left({p \implies q}\right) \land \left({p \implies r}\right)}\right)$
+\end{theorem}
+
+\begin{proof}
+=== [[Implication is Left Distributive over Conjunction/Forward Implication/Formulation 2/Proof|Proof of Forward Implication]] ===
+{{:Implication is Left Distributive over Conjunction/Forward Implication/Formulation 2/Proof}}
+=== [[Implication is Left Distributive over Conjunction/Reverse Implication/Formulation 2/Proof|Proof of Reverse Implication]] ===
+{{:Implication is Left Distributive over Conjunction/Reverse Implication/Formulation 2/Proof}}
+\end{proof}<|endoftext|>
+\section{Implication is Left Distributive over Conjunction/Formulation 2}
+Tags: Implication is Left Distributive over Conjunction
+
+\begin{theorem}
+:$\vdash \paren {p \implies \paren {q \land r} } \iff \paren {\paren {p \implies q} \land \paren {p \implies r} }$
+\end{theorem}<|endoftext|>
+\section{Criterion for Ring with Unity to be Topological Ring}
+Tags: 
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring with Unity|ring with unity]].
+Let $\tau$ be a [[Definition:Topology|topology]] over $R$.
+Suppose that $+$ and $\circ$ are $\tau$-[[Definition:Continuous Mapping|continuous mappings]].
+Then $\struct {R, +, \circ, \tau}$ is a [[Definition:Topological Ring|topological ring]].
+\end{theorem}
+
+\begin{proof}
+As we presume $\circ$ to be [[Definition:Continuous Mapping|continuous]], we need only prove that $\struct {R, +, \tau}$ is a [[Definition:Topological Group|topological group]].
+As we presume $+$ to be [[Definition:Continuous Mapping|continuous]], we need only show that negation is [[Definition:Continuous Mapping|continuous]].
+As $\struct {R, \circ}$ is a [[Definition:Semigroup|semigroup]] and $\circ$ is [[Definition:Continuous Mapping|continuous]]:
+:$\struct{R, \circ, \tau}$ is a [[Definition:Topological Group|topological semigroup]].
+From [[Identity Mapping is Homeomorphism]], the [[Definition:Identity Mapping|identity mapping]] $I_R : \struct {R, \tau} \to \struct {R, \tau}$ is [[Definition:Continuous|continuous]].
+ 
+From [[Multiple Rule for Continuous Mappings to Topological Semigroup]], the mapping $\paren{- 1_R} \circ I_R : R \to R$ defined by:
+:$\forall b \in R : \map {\paren {\paren {-1_R} \circ I_R} } b = \paren {-1_R} \circ b$
+is [[Definition:Continuous|continuous]].
+From [[Product with Ring Negative]], for each $b \in R : -b = \paren {-1_R} \circ b$.
+Hence [[Definition:Ring Negative|negation]] is [[Definition:Continuous Mapping|continuous]].
+{{qed}}
+i43n6rn55jasvjiw2e43wgnc6882ov7
+\end{proof}<|endoftext|>
+\section{Union of Relations is Relation}
+Tags: Relation Theory
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Set|sets]].
+Let $\mathcal F$ be a family of [[Definition:Relation|relations]] from $S$ to $T$.
+Let $\displaystyle \mathcal R = \bigcup \mathcal F$, the [[Definition:Union of Set of Sets|union]] of all the elements of $\mathcal F$.
+ 
+Then $\mathcal R$ is a [[Definition:Relation|relation]] from $S$ to $T$.
+{{expand|Binary case}}
+\end{theorem}
+
+\begin{proof}
+By the definition of a [[Definition:Relation|relation]] from $S$ to $T$, each element of $\mathcal F$ is a [[Definition:Subset|subset]] of $S \times T$.
+By [[Union of Subsets is Subset/Set of Sets|Union of Subsets is Subset: Set of Sets]]:
+: $\mathcal R \subseteq S \times T$
+Therefore, by the definition of a [[Definition:Relation|relation]] from $S$ to $T$, $\mathcal R$ is a relation from $S$ to $T$.
+{{qed}}
+[[Category:Relation Theory]]
+82xv7p14zwlckbngevugib4r8b1ufv4
+\end{proof}<|endoftext|>
+\section{Combination Theorem for Complex Derivatives/Quotient Rule}
+Tags: Complex Analysis
+
+\begin{theorem}
+For all $z \in D$ with $\map g z \ne 0$:
+:$\map {\paren {\dfrac f g}'} z = \dfrac {\map {f'} z \map g z - \map f z \map {g'} z} {\paren {\map g z}^2}$
+\end{theorem}
+
+\begin{proof}
+Denote the [[Definition:Open Ball|open ball]] of $0$ with [[Definition:Radius of Open Ball|radius]] $r \in \R_{>0}$ as $\map {B_r} 0$.
+Let $z \in D \setminus \set {x \in D: \map g z = 0}$.
+By the [[Alternative Differentiability Condition]], it follows that there exists $r_0 \in \R_{>0}$ such that for all $h \in \map {B_{r_0} } 0 \setminus \set 0$:
+:$\map f {z + h} = \map f z + h \paren {\map {f'} z + \map {\epsilon_f} h}$
+:$\map g {z + h} = \map g z + h \paren {\map {g'} z + \map {\epsilon_g} h}$
+where $\epsilon_f, \epsilon_g: \map {B_{r_0} } 0 \setminus \set 0 \to \C$ are [[Definition:Continuous Complex Function|continuous functions]] that [[Definition:Convergent Function|converge]] to $0$ as $h$ tends to $0$.
+From [[Complex-Differentiable Function is Continuous]], it follows that $g$ is [[Definition:Continuous Complex Function|continuous]] at $z$.
+Then there exists $r_1 \in \R_{>0}$ such that for all $h \in \map {B_{r_1} } 0$, we have $\size {\map g {z + h} - \map g z} < \size {\map g z}$.
+Then by [[Backwards Form of Triangle Inequality]]:
+:$\map g {z + h} \ne 0$ for all $h \in \map {B_{r_1} } 0$
+Put $r = \min \set {r_0, r_1}$.
+Then for all $h \in B_r \setminus \set 0$:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\map f {z + h} } {\map g {z + h} }
+      | r = \dfrac {\map f z} {\map g z} + \dfrac {\map f {z + h} - \map f z} {\map g {z + h} } - \map f z \dfrac {\map g {z + h} - \map g z} {\map g z \map g {z + h} }
+      | c = adding and subtracting the same quantity
+}}
+{{eqn | r = \dfrac {\map f z} {\map g z} + \dfrac {h \paren {\map {f'} z + \map {\epsilon_f} h} } {\map g z + h \paren {\map {g'} z + \map {\epsilon_g} h} } - \dfrac {\map f z h \paren {\map {g'} z \map {\epsilon_g} h} } {\map g z \paren {\map g z + h \paren {\map {g'} z + \map {\epsilon_g} h} } }
+}}
+{{eqn | r = \dfrac {\map f z} {\map g z} + h \dfrac {\map {f'} z \map g z + \map g z \map {\epsilon_f} h - \map f z \map {g'} z - \map f z \map {\epsilon_g} h} {\paren {\map g z}^2 + \map {\epsilon_0} h}
+      | c = with $\map {\epsilon_0} h = h \map {g'} z \paren {\map {g'} z + \map {\epsilon_g} h}$
+}}
+{{eqn | r = \dfrac {\map f z} {\map g z} + h \paren {\dfrac {\map {f'} z \map g z - \map f z \map {g'} z} { {\paren {\map g z}^2 } } + \map \epsilon h}
+}}
+{{end-eqn}}
+where $\epsilon: \map {B_r} 0 \setminus \set 0 \to \C$ is defined by:
+:$\map \epsilon h = \dfrac {\map {f'} z \map g z - \map f z \map {g'} z} {\paren {\map g z}^2 + \map {\epsilon_0} h} - \dfrac {\map {f'} z \map g z - \map f z \map {g'} z} {\paren {\map g z}^2 } + \dfrac {\map g z \map {\epsilon_f} h - \map f z \map {\epsilon_g} h} {\paren {\map g z}^2 + \map {\epsilon_0} h}$
+From [[Combination Theorem for Continuous Functions]], it follows that $\epsilon$ is continuous.
+From [[Combination Theorem for Limits of Functions]], it follows that:
+{{begin-eqn}}
+{{eqn | l = \lim_{h \mathop \to 0} \map \epsilon h 
+      | r = \lim_{h \mathop \to 0} \dfrac {\map {f'} z \map g z - \map f z \map {g'} z} {\paren {\map g z}^2 + \map {\epsilon_0} h} - \dfrac {\map {f'} z \map g z - \map f z \map {g'} z} {\paren {\map g z}^2 } + \lim_{h \mathop \to 0} \dfrac {\map g z \map {\epsilon_f} h - \map f z \map {\epsilon_g} h} {\paren {\map g z}^2 + \map {\epsilon_0} h}
+}}
+{{eqn | r = \dfrac {\map {f'} z \map g z - \map f z \map {g'} z} {\paren {\map g z}^2} - \dfrac {\map {f'} z \map g z - \map f z \map {g'} z} {\paren {\map g z}^2 } + \map g z \lim_{h \mathop \to 0} \dfrac {\map {\epsilon_f} h} {\paren {\map g z}^2 + \map {\epsilon_0} h} - \map f z \lim_{h \mathop \to 0} \dfrac {\map {\epsilon_g} h} {\paren {\map g z}^2 + \map {\epsilon_0} h}
+}}
+{{eqn | r = 0
+}}
+{{end-eqn}}
+Then the [[Alternative Differentiability Condition]] shows that:
+:$\map {\paren {\dfrac f g}'} z = \dfrac {\map {f'} z \map g z - \map f z \map {g'} z} {\paren {\map g z}^2}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complex-Differentiable Function is Continuous}
+Tags: Complex Analysis, Continuous Mappings
+
+\begin{theorem}
+Let $f: D \to \C$ be a [[Definition:Complex Function|complex function]], where $D \subseteq \C$ is an [[Definition:Open Set (Complex Analysis)|open set]].
+Suppose that $f$ is [[Definition:Complex-Differentiable Function|complex-differentiable]] at $z \in D$.
+Then $f$ is [[Definition:Continuous Complex Function|continuous]] at $z$.
+\end{theorem}
+
+\begin{proof}
+By the [[Alternative Differentiability Condition]], it follows that there exists $r \in \R_{>0}$ such that for all $h \in B_r \left({0}\right) \setminus \left\{ {0}\right\}$:
+:$\displaystyle \lim_{h \to 0} f\left({z + h}\right) = f \left({z}\right) + h \left({f \left({z}\right) + \epsilon \left({h}\right) }\right)$
+It follows from [[Definition:Limit of Function|definition of limit]] that for all $\epsilon \in \R_{>0}$, there exists $\delta \in \R_{>0}$ with this property:
+If $\left\vert{h - 0}\right\vert < \delta$, then $\left\vert{f \left({z+h}\right) - f \left({z}\right)}\right\vert < \epsilon$.
+Put $z' = z+h$, so $\left\vert{z' - z}\right\vert = \left\vert{h}\right\vert$.
+It follows that:
+If $\left\vert{z' - z}\right\vert < \delta$, then $\left\vert{f \left({z'}\right) - f \left({z}\right)}\right\vert < \epsilon$.
+Then $f$ is [[Definition:Continuous Complex Function|continuous]] at $z$ by definition.
+{{qed}}
+[[Category:Complex Analysis]]
+[[Category:Continuous Mappings]]
+iuzm878f6b7h15nhuq8shtj2nvj1snw
+\end{proof}<|endoftext|>
+\section{Proof by Cases/Formulation 1/Forward Implication/Proof 1}
+Tags: Proof by Cases
+
+\begin{theorem}
+: $\left({p \implies r}\right) \land \left({q \implies r}\right) \vdash \left({p \lor q}\right) \implies r$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\left({p \implies r}\right) \land \left({q \implies r}\right) \vdash \left({p \lor q}\right) \implies r}}
+{{Premise|1|\left({p \implies r}\right) \land \left({q \implies r}\right)}}
+{{Simplification|2|1|p \implies r|1|1}}
+{{Simplification|3|1|q \implies r|1|2}}
+{{SequentIntro|4|1|p \lor q \implies r \lor r|2, 3|[[Constructive Dilemma]]}}
+{{Assumption|5|p \lor q}}
+{{ModusPonens|6|1, 5|r \lor r|4|5}}
+{{Idempotence|7|1, 5|r|6|Disjunction}}
+{{Implication|8|1|p \lor q \implies r|5|7}}
+{{EndTableau}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Closed Ball in Euclidean Space is Compact}
+Tags: Euclidean Space
+
+\begin{theorem}
+Let $x \in \R_n$ be a point in the [[Definition:Euclidean Space|Euclidean space]] $\R^n$.
+Let $\epsilon \in \R_{>0}$.
+Then the [[Definition:Closed Ball|closed $\epsilon$-ball]] $\map {B_\epsilon^-} x$ is [[Definition:Compact Subspace|compact]].
+\end{theorem}
+
+\begin{proof}
+From [[Closed Ball is Closed in Metric Space]], it follows that $\map {B_\epsilon^-} x$ is [[Definition:Closed Set/Metric Space|closed]] in $\R^n$.
+For all $a \in \map {B_\epsilon^-} x$ we have $\map d {x, a} \le \epsilon$, where $d$ denotes the [[Definition:Euclidean Metric/Real Vector Space|Euclidean metric]].
+Then $\map {B_\epsilon^-} x$ is [[Definition:Bounded Metric Space|bounded]] in $\R^n$.
+From the [[Heine-Borel Theorem/Euclidean Space|Heine-Borel Theorem]], it follows that $B_\epsilon^- \left({x}\right)$ is [[Definition:Compact Subspace|compact]].
+{{qed}}
+[[Category:Euclidean Space]]
+o4quf79inub19hlagn3qnzxwn1iwyuh
+\end{proof}<|endoftext|>
+\section{Isometric Image of Cauchy Sequence is Cauchy Sequence}
+Tags: Isometries, Cauchy Sequences
+
+\begin{theorem}
+Let $\struct {S_1, d_1}$ and $\struct {S_2, d_2}$ be [[Definition:Metric Space|metric spaces]].
+Let $f: S_1 \to S_2$ be an [[Definition:Isometry (Metric Spaces)|isometry]].
+Let $\sequence {x_n}$ be a [[Definition:Cauchy Sequence|Cauchy sequence]] in $S_1$.
+Let $\sequence {y_n} = \sequence {\map f {x_n} }$ be the image of $\sequence {x_n}$ under $f$.
+Then $\sequence {y_n}$ is a [[Definition:Cauchy Sequence|Cauchy sequence]].
+\end{theorem}
+
+\begin{proof}
+Let $\epsilon \in \R_{>0}$.
+By the definition of [[Definition:Cauchy Sequence|Cauchy sequence]], there is an $N \in \R$ such that:
+:$\paren{m > N} \land \paren {n > N} \implies \map {d_1} {x_m, x_n} < \epsilon$
+Since $f$ is an [[Definition:Isometry (Metric Spaces)|isometry]], $\map {d_2} {y_m , y_n} = \map {d_1} {x_m, x_n}$ for all $m$ and $n$.
+Thus:
+:$\paren {m > N} \land \paren {n > N} \implies \map {d_2} {y_m, y_n} < \epsilon$
+Hence $\sequence {y_n}$ is a [[Definition:Cauchy Sequence|Cauchy sequence]].
+{{qed}}
+[[Category:Isometries]]
+[[Category:Cauchy Sequences]]
+fsicej092lpjzhzgi3ez1wi0h3sxij4
+\end{proof}<|endoftext|>
+\section{Union of Subsets is Subset}
+Tags: Set Union, Subsets, Union of Subsets is Subset
+
+\begin{theorem}
+Let $S_1$, $S_2$, and $T$ be [[Definition:Set|sets]].
+Let $S_1$ and $S_2$ both be [[Definition:Subset|subsets]] of $T$.
+Then:
+:$S_1 \cup S_2 \subseteq T$
+That is:
+:$\paren {S_1 \subseteq T} \land \paren {S_2 \subseteq T} \implies \paren {S_1 \cup S_2} \subseteq T$
+\end{theorem}
+
+\begin{proof}
+Let:
+:$\paren {S_1 \subseteq T} \land \paren {S_2 \subseteq T}$
+Then:
+{{begin-eqn}}
+{{eqn | l = S_1 \cup S_2
+      | o = \subseteq
+      | r = T \cup T
+      | c = [[Set Union Preserves Subsets]]
+}}
+{{eqn | ll= \leadsto
+      | l = S_1 \cup S_2
+      | o = \subseteq
+      | r = T
+      | c = [[Union is Idempotent]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $x \in S_1 \cup S_2$.
+By the definition of [[Definition:Set Union|union]], either $x \in S_1$ or $x \in S_2$.
+[[Definition:By Hypothesis|By hypothesis]], $S_1 \subseteq T$ and $S_2 \subseteq T$.
+By definition of [[Definition:Subset|subset]]:
+:$x \in S_1 \implies x \in T$
+:$x \in S_2 \implies x \in T$
+By [[Proof by Cases]] it follows that $x \in T$.
+Hence the result by definition of [[Definition:Subset|subset]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Proof by Cases/Formulation 1/Forward Implication}
+Tags: Proof by Cases
+
+\begin{theorem}
+: $\left({p \implies r}\right) \land \left({q \implies r}\right) \vdash \left({p \lor q}\right) \implies r$
+\end{theorem}<|endoftext|>
+\section{Proof by Cases/Formulation 1}
+Tags: Proof by Cases
+
+\begin{theorem}
+: $\left({p \implies r}\right) \land \left({q \implies r}\right) \dashv \vdash \left({p \lor q}\right) \implies r$
+\end{theorem}<|endoftext|>
+\section{Distance between Closed Sets in Euclidean Space}
+Tags: Euclidean Space
+
+\begin{theorem}
+Let $S, T \subseteq \R^n$ be [[Definition:Closed Set of Metric Space|closed]], [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subsets]] of the [[Definition:Real Euclidean Space|real Euclidean space]] $R^n$.
+Suppose that $S$ is [[Definition:Bounded Metric Space|bounded]], and $S$ and $T$ are [[Definition:Disjoint Sets|disjoint]].
+Then there exists $x \in S$ and $y \in T$ such that:
+:$\map d {x, y} = \map d {S, T} > 0$
+where:
+:$d$ denotes the [[Definition:Euclidean Metric/Real Vector Space|Euclidean metric]]
+:$\map d {S, T}$ is the [[Definition:Distance from Subset|distance]] between $S$ and $T$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Distance from Subset|distance from subset]], we can for all $n \in \N$ find $x_n \in S, y_n \in T$ such that:
+:$\map d {S, T} \le \map d {x_n, y_n} < \map d {S, T} + \dfrac 1 n$
+so:
+:$\displaystyle \lim_{n \mathop \to \infty} \map d {x_n, y_n} = \map d {S, T}$
+By definition of [[Definition:Bounded Metric Space|bounded space]], there exists $a \in S$ and $K \in \R$ such that for all $x \in S$, we have $\map d {x, a} \le K$.
+It follows that $\sequence {x_n}$ is a [[Definition:Bounded Sequence|bounded sequence]].
+Then $\sequence {y_n}$ is also a [[Definition:Bounded Sequence|bounded sequence]], as:
+{{begin-eqn}}
+{{eqn | l = \map d {y_n, a}
+      | o = \le
+      | r = \map d {y_n, x_n} + \map d {x_n, a}
+      | c = [[Triangle Inequality for Vectors in Euclidean Space]]
+}}
+{{eqn | o = <
+      | r = \map d {S, T} + \dfrac 1 n + K
+}}
+{{eqn | o = \le
+      | r = \map d {S, T} + 1 + K
+}}
+{{end-eqn}}
+The [[Definition:Sequence|sequence]] $\sequence {\tuple {x_n, y_n} }$ in $\R^{2 n}$ is also bounded, as:
+{{begin-eqn}}
+{{eqn | l = \map d {\tuple {x_n, y_n}, \tuple {a, a} }
+      | r = \paren {\sum_{i \mathop = 1}^n \paren {x_i - a}^2 + \sum_{i \mathop = 1}^n \paren {y_i - a}^2}^{1 / 2}
+      | c = {{Defof|Euclidean Metric/Real Vector Space}}
+}}
+{{eqn | o = \le
+      | r = \sum_{i \mathop = 1}^n \paren {x_i - a}^2 + \sum_{i \mathop = 1}^n \paren {y_i - a}^2
+      | c = by [[Minkowski's Inequality for Sums]]
+}}
+{{eqn | o = \le
+      | r = 2 K + \map d {S, T} + 1 
+}}
+{{end-eqn}}
+From [[Bounded Sequence in Euclidean Space has Convergent Subsequence]], it follows that $\sequence {\tuple {x_n, y_n} }$ has a [[Definition:Subsequence|subsequence]] $\sequence {\tuple {x_{n_r}, y_{n_r} } }_{r \mathop \in N}$ that converges to a [[Definition:Limit of Sequence (Metric Space)|limit]] $\tuple {x, y} \in \R^{2n}$.
+Then $\displaystyle \lim_{r \mathop \to \infty} x_{n_r} = x$, and $\displaystyle \lim_{r \mathop \to \infty} y_{n_r} = y$.
+From [[Closed Set iff Contains all its Limit Points]], it follows that $x \in S$, and $y \in T$.
+Then from [[Distance Function of Metric Space is Continuous]]:
+:$\displaystyle \lim_{r \mathop \to \infty} \map d {x_{n_r}, y_{n_r} } = \map d {x, y}$
+As a [[Convergent Sequence in Metric Space has Unique Limit]], we have:
+:$\displaystyle \map d {x, y} = \lim_{r \mathop \to \infty} \map d {x_{n_r}, y_{n_r} } = \lim_{n \mathop \to \infty} \map d {x_n, y_n} = \map d {S, T}$
+As $S$ and $T$ are [[Definition:Disjoint Sets|disjoint]], it follows that $x \ne y$.
+Hence:
+:$0 < \map d {x, y} = \map d {S, T}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complement in Distributive Lattice is Unique}
+Tags: Distributive Lattices, Bounded Lattices
+
+\begin{theorem}
+Let $\left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Bounded Lattice|bounded]] [[Definition:Distributive Lattice|distributive lattice]].
+Then every $a \in S$ admits at most one [[Definition:Complement (Lattice Theory)|complement]].
+\end{theorem}
+
+\begin{proof}
+Let $a \in S$, and suppose that $b, c \in S$ are [[Definition:Complement (Lattice Theory)|complements]] for $a$.
+Then:
+{{begin-eqn}}
+{{eqn | l = b
+      | r = \top \wedge b
+      | c = $\top$ is the [[Definition:Identity Element|identity]] for $\wedge$
+}}
+{{eqn | r = \left({c \vee a}\right) \wedge b
+      | c = $c$ is a [[Definition:Complement (Lattice Theory)|complement]] for $a$
+}}
+{{eqn | r = \left({c \wedge b}\right) \vee \left({a \wedge b}\right)
+      | c = $S$ is a [[Definition:Distributive Lattice|distributive lattice]]
+}}
+{{eqn | r = \left({c \wedge b}\right) \vee \bot
+      | c = $b$ is a [[Definition:Complement (Lattice Theory)|complement]] for $a$
+}}
+{{eqn | r = c \wedge b
+      | c = $\bot$ is the [[Definition:Identity Element|identity]] for $\vee$
+}}
+{{end-eqn}}
+Interchanging $c$ and $b$ in the above gives that $c = c \wedge b$ as well.
+Hence $b = c$, as desired.
+{{qed}}
+[[Category:Distributive Lattices]]
+[[Category:Bounded Lattices]]
+kkeqa6hvsdk9ooj7vsfw4h28b6uszjc
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Top}
+Tags: Lattice Theory
+
+\begin{theorem}
+{{TFAE|def = Top (Lattice Theory)|view = $\top$ (top)|context = Lattice Theory}}
+Let $\left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Lattice|lattice]].
+\end{theorem}
+
+\begin{proof}
+By definition, $\top$ is the [[Definition:Greatest Element|greatest element]] of $S$ {{iff}} for all $a \in S$:
+:$a \preceq \top$
+By [[Ordering in terms of Meet]], this is equivalent to:
+:$a \wedge \top = a$
+If this equality holds for all $a \in S$, then by definition $\top$ is an [[Definition:Identity Element|identity]] for $\wedge$.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Bottom}
+Tags: Lattice Theory
+
+\begin{theorem}
+Let $\left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Lattice|lattice]].
+Let $\bot$ be a [[Definition:Bottom (Lattice Theory)|bottom]] of $\left({S, \vee, \wedge, \preceq}\right)$.
+{{TFAE|def = Bottom (Lattice Theory)|view = Bottom|context = Lattice Theory}}
+\end{theorem}
+
+\begin{proof}
+By definition, $\bot$ is the [[Definition:Smallest Element|smallest element]] of $S$ {{iff}} for all $a \in S$:
+:$\bot \preceq a$
+By [[Ordering in terms of Join]], this is equivalent to:
+:$a \vee \bot = a$
+If this equality holds for all $a \in S$, then by definition $\bot$ is an [[Definition:Identity Element|identity]] for $\vee$.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Top is Unique}
+Tags: Lattice Theory
+
+\begin{theorem}
+Let $\struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Lattice|lattice]].
+Then $S$ has at most one [[Definition:Top (Lattice Theory)|top]].
+\end{theorem}
+
+\begin{proof}
+By definition, a [[Definition:Top (Lattice Theory)/Definition 1|top]] for $S$ is a [[Definition:Greatest Element|greatest element]].
+The result follows from [[Greatest Element is Unique]].
+{{qed}}
+[[Category:Lattice Theory]]
+683yerk2ng1obdpxb8ac23vneubyuzt
+\end{proof}<|endoftext|>
+\section{Bottom is Unique}
+Tags: Lattice Theory
+
+\begin{theorem}
+Let $\left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Lattice|lattice]].
+Then $S$ has at most one [[Definition:Bottom (Lattice Theory)|bottom]].
+\end{theorem}
+
+\begin{proof}
+By definition, a [[Definition:Bottom (Lattice Theory)/Definition 1|bottom]] for $S$ is a [[Definition:Smallest Element|smallest element]].
+The result follows from [[Smallest Element is Unique]].
+{{qed}}
+[[Category:Lattice Theory]]
+g46t8oi3wi4enzdn1w1f57fg1p6zcp2
+\end{proof}<|endoftext|>
+\section{Isometry is Homeomorphism of Induced Topologies}
+Tags: Isometries, Homeomorphisms
+
+\begin{theorem}
+Let $\struct {S_1, d_1}$ and $\struct {S_2, d_2}$ be [[Definition:Metric Space|metric spaces]] or [[Definition:Pseudometric Space|pseudometric spaces]].
+Let $f: S_1 \to S_2$ be an [[Definition:Isometry (Metric Spaces)|isometry]] from $\struct {S_1, d_1}$ to $\struct {S_2, d_2}$.
+Let $\tau_1$ and $\tau_2$ be the [[Definition:Topology Induced by Metric|topologies induced]] on $S_1$ and $S_2$ by the [[Definition:Metric|metrics]] $d_1$ and $d_2$, respectively.
+Then $f$ is a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]] from $\struct {S_1, \tau_1}$ to $\struct {S_2, \tau_2}$.
+\end{theorem}
+
+\begin{proof}
+By the definition of an [[Definition:Isometry (Metric Spaces)|isometry]], $f$ is [[Definition:Bijective|bijective]].
+By [[Continuous Mapping is Continuous on Induced Topological Spaces]], $f$ is a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]] from $\struct {S_1, \tau_1}$ to $\struct {S_2, \tau_2}$.
+{{explain|It needs to be established that $f$ is continuous on the original metric spaces in the first place.}}
+By [[Inverse of Isometry of Metric Spaces is Isometry]], $f^{-1}$ is an [[Definition:Isometry (Metric Spaces)|isometry]].
+By [[Continuous Mapping is Continuous on Induced Topological Spaces]] once more, $f^{-1}$ is a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]] from $\struct {S_2, \tau_2}$ to $\struct {S_1, \tau_1}$.
+Thus $f$ is a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]] from $\struct {S_1, \tau_1}$ to $\struct {S_2, \tau_2}$.
+{{qed}}
+[[Category:Isometries]]
+[[Category:Homeomorphisms]]
+gdeqivac4rd0i7tmb3hgiu4auqz8vfi
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Distributive Lattice}
+Tags: Distributive Lattices
+
+\begin{theorem}
+{{TFAE|def = Distributive Lattice}}
+Let $\left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Lattice|lattice]].
+{{begin-axiom}}
+{{axiom | n = 1
+        | q = \forall x, y, z \in S
+        | m = x \wedge \left({y \vee z}\right) = \left({x \wedge y}\right) \vee \left({x \wedge z}\right)
+}}
+{{axiom | n = 1'
+        | q = \forall x, y, z \in S
+        | m = \left({x \vee y}\right) \wedge z = \left({x \wedge z}\right) \vee \left({y \wedge z}\right)
+}}
+{{axiom | n = 2
+        | q = \forall x, y, z \in S
+        | m = x \vee \left({y \wedge z}\right) = \left({x \vee y}\right) \wedge \left({x \vee z}\right)
+}}
+{{axiom | n = 2'
+        | q = \forall x, y, z \in S
+        | m = \left({x \wedge y}\right) \vee z = \left({x \vee z}\right) \wedge \left({x \vee y}\right)
+}}
+{{end-axiom}}
+\end{theorem}
+
+\begin{proof}
+=== 1 is equivalent to 1' ===
+By applying [[Meet is Commutative]] several times, we have:
+{{begin-eqn}}
+{{eqn | l = x \wedge \left({y \vee z}\right)
+      | r = \left({x \wedge y}\right) \vee \left({x \wedge z}\right)
+}}
+{{eqn | ll= \iff
+      | l = \left({y \vee z}\right) \wedge x
+      | r = \left({y \wedge x}\right) \vee \left({z \wedge x}\right)
+}}
+{{end-eqn}}
+which (after renaming variables as appropriate) establishes the equivalence.
+{{qed|lemma}}
+=== 2 is equivalent to 2' ===
+By applying [[Join is Commutative]] several times, we have:
+{{begin-eqn}}
+{{eqn | l = x \vee \left({y \wedge z}\right)
+      | r = \left({x \vee y}\right) \wedge \left({x \vee z}\right)
+}}
+{{eqn | ll= \iff
+      | l = \left({y \wedge z}\right) \vee x
+      | r = \left({y \vee x}\right) \wedge \left({z \vee x}\right)
+}}
+{{end-eqn}}
+which (after renaming variables as appropriate) establishes the equivalence.
+{{qed|lemma}}
+=== 1 implies 2 ===
+Suppose that $(1)$ holds, and hence $(1')$ as well.
+{{begin-eqn}}
+{{eqn | l = \left({x \vee y}\right) \wedge \left({x \vee z}\right)
+      | r = \left({\left({x \vee y}\right) \wedge x}\right) \vee \left({\left({x \vee y}\right) \wedge z}\right)
+      | c = by $(1)$
+}}
+{{eqn | r = x \vee \left({\left({x \vee y}\right) \wedge z}\right)
+      | c = $\wedge$ [[Definition:Absorb|absorbs]] $\vee$
+}}
+{{eqn | r = x \vee \left({\left({x \wedge z}\right) \vee \left({y \wedge z}\right)}\right)
+      | c = by $(1')$
+}}
+{{eqn | r = \left({x \vee \left({x \wedge z}\right)}\right) \vee \left({y \wedge z}\right)
+      | c = $\vee$ is [[Definition:Associative|associative]]
+}}
+{{eqn | r = x \vee \left({y \wedge z}\right)
+      | c = $\vee$ [[Definition:Absorb|absorbs]] $\wedge$
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== 2 implies 1 ===
+By inspection, aided by [[Dual Pairs (Order Theory)]], we see that $(2)$ is [[Definition:Dual Statement (Order Theory)|dual]] to $(1)$.
+Thus by [[Duality Principle (Order Theory)/Global Duality|Global Duality]], $(2)$ implies $(1)$ as soon as $(1)$ implies $(2)$.
+That direction was already established above.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Distance-Preserving Surjection is Isometry of Metric Spaces}
+Tags: Isometries
+
+\begin{theorem}
+Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $\phi: M_1 \to M_2$ be a [[Definition:Surjective|surjective]] [[Definition:Distance-Preserving Mapping|distance-preserving mapping]].
+That is:
+:$\forall a, b \in M_1: d_1 \tuple {a, b} = d_2 \tuple {\map \phi a, \map \phi b}$
+Then $\phi$ is an [[Definition:Isometry (Metric Spaces)|isometry]].
+\end{theorem}
+
+\begin{proof}
+The premises satisfy all elements of the definition of [[Definition:Isometry (Metric Spaces)|isometry]] except for [[Definition:Bijective|bijectivity]].
+As we presume $\phi$ to be [[Definition:Surjection|surjective]] we need only show that it is [[Definition:Injective|injective]].
+By [[Distance-Preserving Mapping is Injection of Metric Spaces]] then $\phi$ is [[Definition:Injective|injective]].
+{{qed}}
+[[Category:Isometries]]
+a6j4t4aomj1hrk6vjl69i65zqgq2xe2
+\end{proof}<|endoftext|>
+\section{Existence of Positive Root of Positive Real Number}
+Tags: Roots of Numbers, Real Numbers, Existence of Positive Root of Positive Real Number
+
+\begin{theorem}
+Let $x \in \R$ be a [[Definition:Real Number|real number]] such that $x > 0$.
+Let $n \in \Z$ be an [[Definition:Integer|integer]] such that $n \ne 0$.
+Then there exists a $y \in \R: y \ge 0$ such that $y^n = x$.
+\end{theorem}
+
+\begin{proof}
+=== [[Existence of Positive Root of Positive Real Number/Positive Exponent|Positive Exponent]] ===
+{{:Existence of Positive Root of Positive Real Number/Positive Exponent|Positive Exponent}}
+=== [[Existence of Positive Root of Positive Real Number/Negative Exponent|Negative Exponent]] ===
+{{:Existence of Positive Root of Positive Real Number/Negative Exponent|Negative Exponent}}
+\end{proof}<|endoftext|>
+\section{Proof by Cases/Formulation 2}
+Tags: Proof by Cases
+
+\begin{theorem}
+:$\vdash \paren {\paren {p \implies r} \land \paren {q \implies r} } \iff \paren {\paren {p \lor q} \implies r}$
+\end{theorem}<|endoftext|>
+\section{Proof by Cases/Formulation 2/Forward Implication}
+Tags: Proof by Cases
+
+\begin{theorem}
+: $\vdash \left({\left({p \implies r}\right) \land \left({q \implies r}\right)}\right) \implies \left({\left({p \lor q}\right) \implies r}\right)$
+\end{theorem}<|endoftext|>
+\section{Proof by Cases/Formulation 2/Reverse Implication}
+Tags: Proof by Cases
+
+\begin{theorem}
+: $\vdash \left({\left({p \lor q}\right) \implies r}\right) \implies \left({\left({p \implies r}\right) \land \left({q \implies r}\right)}\right)$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\vdash \left({\left({p \lor q}\right) \implies r}\right) \implies \left({\left({p \implies r}\right) \land \left({q \implies r}\right)}\right)}}
+{{Assumption|1|\left({p \lor q}\right) \implies r}}
+{{SequentIntro|2|1|\left({p \implies r}\right) \land \left({q \implies r}\right)|1|[[Proof by Cases/Formulation 1/Reverse Implication|Proof by Cases: Formulation 1: Reverse Implication]]}}
+{{Implication|3||\left({\left({p \lor q}\right) \implies r}\right) \implies \left({\left({p \implies r}\right) \land \left({q \implies r}\right)}\right)|1|2}}
+{{EndTableau}}
+{{qed}}
+[[Category:Proof by Cases]]
+pcm05869vx9skltwgpdk05y9lobkkvg
+\end{proof}<|endoftext|>
+\section{Set is Subset of Union/Set of Sets}
+Tags: Set Union, Subsets
+
+\begin{theorem}
+Let $\mathbb S$ be a [[Definition:Set of Sets|set of sets]].
+Then:
+: $\displaystyle \forall T \in \mathbb S: T \subseteq \bigcup \mathbb S$
+\end{theorem}
+
+\begin{proof}
+Let $T$ be any element of $\mathbb S$.
+We wish to show that $T \subseteq S$.
+Let $x \in T$.
+Then: 
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \in
+      | r = T
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | o = \in
+      | r = \bigcup \mathbb S
+      | c = {{Defof|Set Union}}
+}}
+{{end-eqn}}
+Since this holds for each $x \in T$:
+{{begin-eqn}}
+{{eqn | l = T
+      | o = \subseteq
+      | r = \bigcup \mathbb S
+      | c = {{Defof|Subset}}
+}}
+{{end-eqn}}
+As $T$ was arbitrary, it follows that:
+:$\forall T \in \mathbb S: T \subseteq \bigcup \mathbb S$
+{{qed}}
+[[Category:Set Union]]
+[[Category:Subsets]]
+3xl8aw1m4ibiplvh3jxz4ufmiey50ib
+\end{proof}<|endoftext|>
+\section{Proof by Cases/Formulation 3}
+Tags: Proof by Cases
+
+\begin{theorem}
+: $\vdash \left({\left({p \lor q}\right) \land \left({p \implies r}\right) \land \left({q \implies r}\right)}\right) \implies r$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\left({\left({p \lor q}\right) \land \left({p \implies r}\right) \land \left({q \implies r}\right)}\right) \implies r}}
+{{TheoremIntro|1|\left({\left({p \lor r}\right) \land \left({p \implies q}\right) \land \left({r \implies s}\right)}\right) \implies \left({q \lor s}\right)|[[Constructive Dilemma/Formulation 2|Constructive Dilemma: Formulation 2]]}}
+{{Substitution|2||\left({\left({p \lor q}\right) \land \left({p \implies r}\right) \land \left({q \implies r}\right)}\right) \implies \left({r \lor r}\right)|1|r|q|q|r|s|r}}
+{{Assumption|3|\left({p \lor q}\right) \land \left({p \implies r}\right) \land \left({q \implies r}\right)}}
+{{ModusPonens|4|3|r \lor r|2|3}}
+{{Idempotence|5|3|r|4|Disjunction}}
+{{Implication|6||\left({\left({p \lor q}\right) \land \left({p \implies r}\right) \land \left({q \implies r}\right)}\right) \implies r|3|5}}
+{{EndTableau}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Properties of Relation Compatible with Group Operation}
+Tags: Compatible Relations, Group Theory
+
+\begin{theorem}
+Let $\left({G,\circ}\right)$ be a [[Definition:Group|group]] with [[Definition:Identity Element|identity element]] $e$.
+Let $\mathcal R$ be a [[Definition:Endorelation|endorelation]] on $G$ which is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
+Let $x,y,z \in G$.
+Then the following hold:
+\end{theorem}<|endoftext|>
+\section{Existence of Positive Root of Positive Real Number/Positive Exponent}
+Tags: Existence of Positive Root of Positive Real Number
+
+\begin{theorem}
+Let $x \in \R$ be a [[Definition:Real Number|real number]] such that $x > 0$.
+Let $n \in \Z$ be an [[Definition:Integer|integer]] such that $n > 0$.
+Then there exists a $y \in \R: y \ge 0$ such that $y^n = x$.
+\end{theorem}<|endoftext|>
+\section{Existence of Positive Root of Positive Real Number/Negative Exponent}
+Tags: Existence of Positive Root of Positive Real Number
+
+\begin{theorem}
+Let $x \in \R$ be a [[Definition:Real Number|real number]] such that $x > 0$.
+Let $n \in \Z$ be an [[Definition:Integer|integer]] such that $n < 0$.
+Then there exists a $y \in \R: y \ge 0$ such that $y^n = x$.
+\end{theorem}
+
+\begin{proof}
+Let $m = -n$.
+Then $m > 0$.
+Let $g$ be the [[Definition:Real Function|real function]] defined on $\hointr 0 \to$ defined by:
+:$\map g y = y^m$
+Since $x \ge 0$:
+:$\dfrac 1 x \ge 0$
+By [[Existence of Positive Root of Positive Real Number/Positive Exponent|Existence of Positive Root of Positive Real Number: Positive Exponent]] there is a $y > 0$ such that:
+:$\map g y = \dfrac 1 x$
+It follows from the definition of [[Definition:Power (Algebra)|power]] that:
+{{begin-eqn}}
+{{eqn | l = \map f y
+      | r = \dfrac 1 {\map g y}
+}}
+{{eqn | r = \dfrac 1 {1 / x}
+}}
+{{eqn | r = x
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Existence of Positive Root of Positive Real Number]]
+2ioe7yjmlsdu080btfmgrb6l8e72o6z
+\end{proof}<|endoftext|>
+\section{Power of Ring Negative}
+Tags: Ring Theory, Powers
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $n \in \N_{>0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
+Let $x \in R$.
+Then:
+:If $n$ is [[Definition:Even Integer|even]]:
+:::$\map {\circ^n} {-x} = \map {\circ^n} x$
+:If $n$ is [[Definition:Odd Integer|odd]]:
+:::$\map {\circ^n} {-x} = -\map {\circ^n} x$
+\end{theorem}
+
+\begin{proof}
+First, suppose that $n$ is [[Definition:Even Integer|even]].
+Then for some $m \in \N_{>0}$:
+: $n = 2 m = m + m$
+Thus since $\circ$ is [[Definition:Associative|associative]]:
+:$\displaystyle \map {\circ^n} {-x} = \prod_{i \mathop = 1}^m \paren {-x} \circ \paren {-x}$
+By [[Product of Ring Negatives]]:
+:$\paren {-x} \circ \paren {-x} = x \circ x = \map {\circ^2} x$
+Thus:
+:$\displaystyle \map {\circ^n} {-x} = \prod_{i \mathop = 1}^m \map {\circ^2} x$
+By [[Definition:Associative|associativity]]:
+:$\map {\circ^n} {-x} = \map {\circ^{2 m} } x = \map {\circ^n} x$
+{{qed|lemma}}
+Now suppose instead that $n$ is [[Definition:Odd Integer|odd]].
+If $n = 1$, then:
+:$\map {\circ^n} {-x} = -x = -\map {\circ^n} x$
+Otherwise:
+{{begin-eqn}}
+{{eqn | l = \map {\circ^n} {-x}
+      | r = \paren {-x} \circ \paren {\map {\circ^{n - 1} } {-x} }
+}}
+{{eqn | r = \paren {-x} \circ \paren {\map {\circ^{n - 1} } x}
+      | c = $n-1$ is a [[Definition:Strictly Positive Integer|strictly positive]] [[Definition:Even Integer|even integer]], so the above case applies
+}}
+{{eqn | r = -\paren {x \circ \paren {\map {\circ^{n - 1} } x} }
+      | c = [[Product with Ring Negative]]
+}}
+{{eqn | r = -\map {\circ^n} x
+      | c = Definition of $\map {\circ^n} x$
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Ring Theory]]
+[[Category:Powers]]
+qnscwtcvihfvvopzzazedbfb0gbh64w
+\end{proof}<|endoftext|>
+\section{Paving Lemma}
+Tags: Complex Analysis, Euclidean Space, Named Theorems
+
+\begin{theorem}
+Let $S$ be an [[Definition:Open Set (Metric Space)|open]] [[Definition:Subset|subset]] of the [[Definition:Euclidean Space|Euclidean space]] $\R^m$ or the [[Definition:Complex Number|set of complex numbers]] $\C$.
+Let $\gamma: \closedint a b \to S$ be a [[Definition:Path (Topology)|path]] in $S$.
+Then there exists $K \in \R_{>0}$ such that:
+:For all $\epsilon \in \openint 0 K$, there exists a [[Definition:Normal Subdivision|normal subdivision]] $\set {x_0, x_1, \ldots, x_{n-1}, x_n}$ of the [[Definition:Closed Real Interval|closed interval]] $\closedint a b$ such that:
+::$\displaystyle \bigcup_{i \mathop = 0}^n \map {B_\epsilon} {\map \gamma {x_i} } \subseteq S$
+:and for all $i \in \set {0, 1, \ldots, n - 1}$:
+::$\map \gamma {\closedint {x_i} {x_{i + 1} } } \subseteq \map {B_\epsilon} {\map \gamma {x_i} }$
+Here, $\map {B_\epsilon} {\map \gamma {x_i} }$ denotes the [[Definition:Open Ball|open ball]] of $\map \gamma {x_i}$ with [[Definition:Radius of Open Ball|radius]] $\epsilon$.
+\end{theorem}
+
+\begin{proof}
+=== Finding the constant ===
+First, suppose that $S$ is a [[Definition:Subset|subset]] of $\R^m$.
+From [[Closed Real Interval is Compact in Metric Space]], it follows that $\closedint a b$ is [[Definition:Compact (Real Analysis)|compact]].
+Then [[Continuous Image of Compact Space is Compact]] shows that $\map \gamma {\closedint {x_i} {x_{i + 1} } }$ is compact.
+From the [[Heine-Borel Theorem/Euclidean Space|Heine-Borel Theorem]], it follows that $\map \gamma {\closedint {x_i} {x_{i + 1} } }$ is [[Definition:Bounded Metric Space|bounded]] and [[Definition:Closed Set (Metric Space)|closed]].
+Suppose that $S \ne \R^m$.
+As $\R^m \setminus S$ is [[Definition:Closed Set (Metric Space)|closed]], it follows from [[Distance between Closed Sets in Euclidean Space]] that:
+:$\map d {\map \gamma {\closedint a b }, \R^m \setminus S} > 0$
+Put $K = \map d {\map \gamma {\closedint a b }, \R^m \setminus S}$, the [[Definition:Distance from Subset|distance]] between $\map \gamma {\closedint a b }$ and $\R^m \setminus S$.
+If instead $S = \R^m$, then $K$ may be any [[Definition:Strictly Positive Real Number|strictly positive real number]].
+=== Finding the subdivision ===
+Let $\epsilon \in \openint 0 K$.
+From the [[Heine-Cantor Theorem]], it follows that $\gamma$ is [[Definition:Uniformly Continuous Real Function|uniformly continuous]].
+Then there exists $\delta \in \R_{>0}$ such that:
+:$\forall y, z \in \closedint a b : \size {y - z} < \delta$ implies $\map d {\map \gamma y, \map \gamma z } < \epsilon$
+where $d$ denotes the [[Definition:Euclidean Metric/Real Vector Space|Euclidean metric]].
+Choose $n \in \N$ such that $\dfrac {b - a} n < \delta$, and put $x_i = a + i \dfrac {b - a} n$ for all $i \in \set {0, 1, \ldots, n}$.
+Then $\set {x_0, x_1, \ldots, x_n}$ is a [[Definition:Normal Subdivision|normal subdivision]] of $\closedint a b$.
+It follows that $\map {B_\epsilon} {\map \gamma {x_i} }$ and $\R^m \setminus S$ are [[Definition:Disjoint Sets|disjoint]], as either:
+:$\epsilon < \map d {\map \gamma {\closedint a b }, \R^m \setminus S}$
+or:
+:$\R \setminus S = \varnothing$
+Then: 
+:$\displaystyle \bigcup_{i \mathop = 0}^n \map {B_\epsilon} {\map \gamma {x_i} } \subseteq S$
+Let $x \in \closedint {x_i} {x_{i+1} }$.
+Then:
+:$\size {x - x_i} \le \dfrac{b - a} n < \delta$
+As the uniorm continuity condition applies, it follows that:
+:$\map \gamma {\closedint {x_i} {x_{i + 1} } } \subseteq \map {B_\epsilon} {\map \gamma {x_i} }$
+{{qed|lemma}}
+=== Complex plane case ===
+Suppose that $S$ is a [[Definition:Subset|subset]] of $\C$.
+This theorem only uses the properties of $\C$ that depends on the [[Definition:Metric|metric]].
+From [[Complex Plane is Metric Space]], it follows that $\C$ and $\R^2$ are [[Definition:Homeomorphic Metric Spaces|homeomorphic]].
+It follows that the proof that applies for $\R^2$ also applies for $\C$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Proof by Cases/Formulation 1/Reverse Implication}
+Tags: Proof by Cases
+
+\begin{theorem}
+: $\left({p \lor q}\right) \implies r \vdash \left({p \implies r}\right) \land \left({q \implies r}\right)$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\left({p \lor q}\right) \implies r \vdash \left({p \implies r}\right) \land \left({q \implies r}\right)}}
+{{Premise|1|\left({p \lor q}\right) \implies r}}
+{{Assumption|2|p}}
+{{Addition|3|2|p \lor q|2|1}}
+{{ModusPonens|4|1, 2|r|1|3}}
+{{Implication|5|1|p \implies r|2|4}}
+{{Assumption|6|q}}
+{{Addition|7|6|p \lor q|6|2}}
+{{ModusPonens|8|1, 6|r|1|7}}
+{{Implication|9|1|q \implies r|6|8}}
+{{Conjunction|10|1|\left({p \implies r}\right) \land \left({q \implies r}\right)|5|9}}
+{{EndTableau}}
+{{qed}}
+[[Category:Proof by Cases]]
+mmig5x5s1otdg7m1su7k3gl33tj9fq7
+\end{proof}<|endoftext|>
+\section{Praeclarum Theorema/Formulation 1/Proof 2}
+Tags: Truth Table Proofs, Praeclarum Theorema
+
+\begin{theorem}
+:$\paren {p \implies q} \land \paren {r \implies s} \vdash \paren {p \land r} \implies \paren {q \land s}$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]] to the proposition.
+As can be seen for all [[Definition:Boolean Interpretation|boolean interpretations]] by inspection, where the [[Definition:Truth Value|truth value]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] on the {{LHS}} is $T$, that under the one on the {{RHS}} is also $T$:
+$\begin{array}{|ccccccc||ccccccc|} \hline
+(p & \implies & q) & \land & (r & \implies & s) & (p & \land & r) & \implies & (q & \land & s) \\
+\hline
+F & T & F & T & F & T & F & F & F & F & T & F & F & F \\
+F & T & F & T & F & T & T & F & F & F & T & F & F & T \\
+F & T & F & F & T & F & F & F & F & T & T & F & F & F \\
+F & T & F & T & T & T & T & F & F & T & T & F & F & T \\
+F & T & T & T & F & T & F & F & F & F & T & T & F & F \\
+F & T & T & T & F & T & T & F & F & F & T & T & T & T \\
+F & T & T & F & T & F & F & F & F & T & T & T & F & F \\
+F & T & T & T & T & T & T & F & F & T & T & T & T & T \\
+T & F & F & F & F & T & F & T & F & F & T & F & F & F \\
+T & F & F & F & F & T & T & T & F & F & T & F & F & T \\
+T & F & F & F & T & F & F & T & T & T & F & F & F & F \\
+T & F & F & F & T & T & T & T & T & T & F & F & F & T \\
+T & T & T & T & F & T & F & T & F & F & T & T & F & F \\
+T & T & T & T & F & T & T & T & F & F & T & T & T & T \\
+T & T & T & F & T & F & F & T & T & T & F & T & F & F \\
+T & T & T & T & T & T & T & T & T & T & T & T & T & T \\
+\hline
+\end{array}$
+Hence the result.
+Note that the two [[Definition:Propositional Formula|formulas]] are not [[Definition:Logical Equivalence|equivalent]], as the relevant columns do not match exactly.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Praeclarum Theorema/Formulation 1}
+Tags: Praeclarum Theorema
+
+\begin{theorem}
+: $\left({p \implies q}\right) \land \left({r \implies s}\right) \vdash \left({p \land r}\right) \implies \left({q \land s}\right)$
+\end{theorem}<|endoftext|>
+\section{Praeclarum Theorema/Formulation 2}
+Tags: Praeclarum Theorema
+
+\begin{theorem}
+: $\vdash \left({\left({p \implies q}\right) \land \left({r \implies s}\right)}\right) \implies \left({\left({p \land r}\right) \implies \left({q \land s}\right)}\right)$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\vdash \left({\left({p \implies q}\right) \land \left({r \implies s}\right)}\right) \implies \left({\left({p \land r}\right) \implies \left({q \land s}\right)}\right)}}
+{{Assumption|1|\left({\left({p \implies q}\right) \land \left({r \implies s}\right)}\right)}}
+{{SequentIntro|2|1|\left({p \land r}\right) \implies \left({q \land s}\right)|1|[[Praeclarum Theorema/Formulation 1|Praeclarum Theorema: Formulation 1]]}}
+{{Implication|3|1|\left({\left({p \implies q}\right) \land \left({r \implies s}\right)}\right) \implies \left({\left({p \land r}\right) \implies \left({q \land s}\right)}\right)|1|2}}
+{{EndTableau}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Strictly Positive Integer Power Function Strictly Succeeds Each Element}
+Tags: Ring Theory
+
+\begin{theorem}
+Let $\struct {R, +, \circ, \le}$ be an [[Definition:Ordered Ring|ordered ring]] [[Definition:Ring with Unity|with unity]].
+Let $\struct {R, \le}$ be a [[Definition:Directed Set|directed set]] with no [[Definition:Upper Bound of Set|upper bound]].
+Let $n \in \N_{>0}$.
+Let $f: R \to R$ be defined by:
+:$\forall x \in R: \map f x = \circ^n x$
+Then the [[Definition:Image of Mapping|image]] of $f$ has [[Definition:Element|elements]] [[Definition:Strictly Succeed|strictly succeeding]] each [[Definition:Element|elements]] of $R$.
+\end{theorem}
+
+\begin{proof}
+Let $b \in R$.
+By [[Directed Set has Strict Successors iff Unbounded Above]]:
+:$\exists c \in R: b < c$
+:$\exists d \in R: 1 < d$
+By the definition of a [[Definition:Directed Set|directed set]]:
+:$\exists e \in R: d \le e, c \le e$
+$\struct {R, +, \circ, \le}$ is an [[Definition:Ordered Ring|ordered ring]], so $\le$ is by definition a [[Definition:Transitive Relation|transitive relation]].
+Hence by [[Definition:Transitive Relation|transitivity]]:
+:$b < e$
+and:
+:$1 < e$
+By [[Strictly Positive Power of Strictly Positive Element Greater than One Succeeds Element]]:
+:$e \le \map f e$
+Thus by [[Definition:Transitive Relation|transitivity]]:
+:$b < \map f e$
+{{qed}}
+[[Category:Ring Theory]]
+413wwed1qe7xtz89l3kxirtjmke6ll7
+\end{proof}<|endoftext|>
+\section{Constructive Dilemma/Formulation 1/Proof 2}
+Tags: Truth Table Proofs, Constructive Dilemma
+
+\begin{theorem}
+: $p \implies q, r \implies s \vdash p \lor r \implies q \lor s$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]] to the proposition.
+As can be seen for all [[Definition:Boolean Interpretation|boolean interpretations]] by inspection, where the [[Definition:Truth Value|truth value]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] on the {{LHS}} is $T$, that under the one on the {{RHS}} is also $T$:
+: $\begin{array}{|ccccccc||ccccccc|} \hline
+(p & \implies & q) & \land & (r & \implies & s) & (p & \lor & r) & \implies & (q & \lor & s) \\
+\hline
+F & T & F & T & F & T & F & F & F & F & T & F & F & F \\
+F & T & F & T & F & T & T & F & F & F & T & F & T & T \\
+F & T & F & F & T & F & F & F & T & T & F & F & F & F \\
+F & T & F & T & T & T & T & F & T & T & T & F & T & T \\
+F & T & T & T & F & T & F & F & F & F & T & T & T & F \\
+F & T & T & T & F & T & T & F & F & F & T & T & T & T \\
+F & T & T & F & T & F & F & F & T & T & T & T & T & F \\
+F & T & T & T & T & T & T & F & T & T & T & T & T & T \\
+T & F & F & F & F & T & F & T & T & F & F & F & F & F \\
+T & F & F & F & F & T & T & T & T & F & T & F & T & T \\
+T & F & F & F & T & F & F & T & T & T & F & F & F & F \\
+T & F & F & F & T & T & T & T & T & T & T & F & T & T \\
+T & T & T & T & F & T & F & T & T & F & T & T & T & F \\
+T & T & T & T & F & T & T & T & T & F & T & T & T & T \\
+T & T & T & F & T & F & F & T & T & T & T & T & T & F \\
+T & T & T & T & T & T & T & T & T & T & T & T & T & T \\
+\hline
+\end{array}$
+Hence the result.
+Note that the two [[Definition:Propositional Formula|formulas]] are not [[Definition:Logical Equivalence|equivalent]], as the relevant columns do not match exactly.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Constructive Dilemma/Formulation 1}
+Tags: Constructive Dilemma
+
+\begin{theorem}
+: $p \implies q, r \implies s \vdash p \lor r \implies q \lor s$
+\end{theorem}<|endoftext|>
+\section{Constructive Dilemma/Formulation 2}
+Tags: Constructive Dilemma
+
+\begin{theorem}
+: $\vdash \paren {\paren {p \lor r} \land \paren {p \implies q} \land \paren {r \implies s} } \implies \paren {q \lor s}$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau  |\vdash \paren {\paren {p \lor r} \land \paren {p \implies q} \land \paren {r \implies s} } \implies \paren {q \lor s} }}
+{{Assumption    |1|\paren {\paren {p \lor r} \land \paren {p \implies q} \land \paren {r \implies s} } }}
+{{Simplification|2|1|\paren {p \implies q} \land \paren {r \implies s}|1|2|Cutting corners: should use Associativity first}}
+{{SequentIntro  |3|1|\paren {p \lor r} \implies \paren {q \lor s}|2|[[Constructive Dilemma/Formulation 1|Constructive Dilemma: Formulation 1]]}}
+{{Simplification|4|1|p \lor r|1|1}}
+{{ModusPonens   |5|1|q \lor s|3|4}}
+{{Implication   |6| |\paren {\paren {p \lor r} \land \paren {p \implies q} \land \paren {r \implies s} } \implies \paren {q \lor s}|1|5}}
+{{EndTableau}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Odd Power Function is Strictly Increasing/Real Numbers}
+Tags: Real Analysis, Real Numbers, Odd Power Function is Strictly Increasing
+
+\begin{theorem}
+Let $n \in \Z_{> 0}$ be an [[Definition:Odd Integer|odd]] [[Definition:Positive Integer|positive integer]].
+Let $f_n: \R \to \R$ be the [[Definition:Real Function|real function]] defined as:
+:$\map {f_n} x = x^n$
+Then $f_n$ is [[Definition:Strictly Increasing Real Function|strictly increasing]].
+\end{theorem}
+
+\begin{proof}
+From the [[Power Rule for Derivatives]]:
+:$\map {D_x} {x^n} = n x^{n - 1}$
+As $n$ is [[Definition:Odd Integer|odd]], $n - 1$ is [[Definition:Even Integer|even]].
+Thus by [[Even Power is Non-Negative]]:
+:$\map {D_x} {x^n} \ge 0$
+for all $x$.
+From [[Derivative of Monotone Function]], it follows that $f_n$ is [[Definition:Increasing Real Function|increasing]] over the whole of $\R$.
+The only place where $\map {D_x} {x^n} = 0$ is at $x = 0$.
+Everywhere else, $f_n$ is [[Definition:Strictly Increasing Real Function|strictly increasing]].
+By [[Sign of Odd Power]]:
+:$\map {f_n} x < 0 = \map {f_n} 0$ when $x < 0$ and
+:$\map {f_n} 0 = 0 < \map {f_n} x$ when $0 < x$.
+Thus $f_n$ is [[Definition:Strictly Increasing Real Function|strictly increasing]] on $\R$.
+{{qed}}
+[[Category:Real Analysis]]
+[[Category:Real Numbers]]
+[[Category:Odd Power Function is Strictly Increasing]]
+3ztw3ty74w9697jql4779mtobh2zy2y
+\end{proof}<|endoftext|>
+\section{Clavius's Law/Formulation 1}
+Tags: Clavius's Law
+
+\begin{theorem}
+:$\neg p \implies p \vdash p$
+\end{theorem}<|endoftext|>
+\section{Clavius's Law/Formulation 2}
+Tags: Clavius's Law
+
+\begin{theorem}
+:$\vdash \left({\neg p \implies p}\right) \implies  p$
+\end{theorem}<|endoftext|>
+\section{Equivalent Matrices may not be Similar}
+Tags: Matrix Algebra
+
+\begin{theorem}
+If two [[Definition:Square Matrix|square matrices of order $n > 1$]] over a [[Definition:Ring with Unity|ring with unity]] $R$ are [[Definition:Matrix Equivalence|equivalent]], they are not necessarily [[Definition:Matrix Similarity|similar]].
+\end{theorem}
+
+\begin{proof}
+As a counterexample, let $\mathbf A = \mathbf I_n$ be the [[Definition:Unit Matrix|unit matrix]] of order $n > 1$.
+Let $\mathbf B$ be any [[Definition:Invertible Matrix|invertible matrix]] over $R$ of order $n$ that is different from the [[Definition:Unit Matrix|unit matrix]].
+Then [[Definition:Matrix Equivalence|$\mathbf A \equiv \mathbf B$]], as:
+:$\mathbf I_n^{-1} \mathbf A \mathbf B = \mathbf I_n^{-1} \mathbf I_n \mathbf B = \mathbf B$.
+If $\mathbf P$ is an invertible [[Definition:Square Matrix|square matrix]] of order $n$, then:
+:$\mathbf P^{-1} \mathbf A \mathbf P = \mathbf P^{-1} \mathbf P = \mathbf I_n \ne \mathbf B$
+Hence, $\mathbf A$ is not [[Definition:Matrix Similarity|similar]] to $\mathbf B$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Odd Power Function is Strictly Increasing/General Result}
+Tags: Totally Ordered Rings, Odd Power Function is Strictly Increasing
+
+\begin{theorem}
+Let $\struct {R, +, \circ, \le}$ be a [[Definition:Totally Ordered Ring|totally ordered ring]].
+Let $n$ be an [[Definition:Odd Integer|odd]] [[Definition:Positive Integer|positive integer]].
+Let $f: R \to R$ be the [[Definition:Mapping|mapping]] defined by:
+:$\map f x = \map {\circ^n} x$
+Then $f$ is [[Definition:Strictly Increasing Mapping|strictly increasing]] on $R$.
+\end{theorem}
+
+\begin{proof}
+{{proofread}}
+Let $x, y \in R$ such that $0 < x < y$.
+By [[Power Function is Strictly Increasing on Positive Elements]]:
+:$\map f x < \map f y$
+Suppose that $x < y < 0$.
+By [[Properties of Ordered Ring]]:
+:$0 < -y < -x$
+By [[Power Function is Strictly Increasing on Positive Elements]] (applied to $-y$ and $-x$):
+:$0 < \map f {-y} < \map f {-x}$
+By [[Power of Ring Negative]]:
+:$\map f {-x} = -\map f x$
+:$\map f {-y} = -\map f y$
+Thus:
+:$0 < -\map f y < -\map f x$
+By [[Properties of Ordered Ring]]:
+:$\map f x < \map f y$
+By [[Sign of Odd Power]]:
+:$\map f x < 0 = \map f 0$ when $x < 0$
+:$\map f 0 = 0 < \map f x$ when $0 < x$
+Thus we have shown that $f$ is [[Definition:Strictly Increasing Mapping|strictly increasing]] on the [[Definition:Positivity Property|positive elements]] and the [[Definition:Negativity Property|negative elements]], and across [[Definition:Ring Zero|zero]].
+{{qed}}
+[[Category:Totally Ordered Rings]]
+[[Category:Odd Power Function is Strictly Increasing]]
+eh5d5yt5qq4k9gyuhnvli3v2uu8bfkm
+\end{proof}<|endoftext|>
+\section{Number of Bijective Restrictions}
+Tags: Bijections, Restrictions
+
+\begin{theorem}
+Let $f: S \to T$ be a [[Definition:Surjection|surjection]].
+Let $B$ be the [[Definition:Set|set]] of all [[Definition:Bijection|bijective]] [[Definition:Restriction of Mapping|restrictions]] of $f$.
+Then the [[Definition:Cardinality|cardinality]] of $B$ is:
+:$\displaystyle \card {\prod_{i \mathop \in I} \family {S / \mathcal R_f}_i}$
+where $S / \mathcal R_f$ denotes the [[Definition:Quotient Set|quotient set]] of the [[Definition:Equivalence Relation Induced by Mapping|induced equivalence]] of $f$ [[Definition:Indexing Set|indexed]] by $I$.
+\end{theorem}
+
+\begin{proof}
+{{proof wanted}}
+[[Category:Bijections]]
+[[Category:Restrictions]]
+rqgjekbhpxqzomvv9j2bt8q2j8ntpuw
+\end{proof}<|endoftext|>
+\section{Number of Injective Restrictions}
+Tags: Injections
+
+\begin{theorem}
+Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
+Let $Q$ be the [[Definition:Set|set]] of all [[Definition:Injection|injective]] [[Definition:Restriction of Mapping|restrictions]] of $f$.
+Then the [[Definition:Cardinality|cardinality]] of $Q$ is:
+:$\displaystyle \card {\prod_{i \mathop \in I} \prod_{j \mathop \in J_i} \family {\family {\powerset {S / \mathcal R_f} }_i}_j}$
+where:
+:$\mathcal P$ denotes [[Definition:Power Set|power set]]
+:$S / \mathcal R_f$ denotes [[Definition:Quotient Set|quotient set]] of the [[Definition:Equivalence Relation Induced by Mapping|induced equivalence]] of $f$.
+\end{theorem}
+
+\begin{proof}
+{{proof wanted}}
+[[Category:Injections]]
+r3x8kwe0p8kidy3iviauws20slgiqnl
+\end{proof}<|endoftext|>
+\section{Box Topology on Finite Product Space is Tychonoff Topology}
+Tags: Box Topology, Tychonoff Topology
+
+\begin{theorem}
+Let $n \in \N$.
+For all $k \in \set {1, \ldots, n}$, let $T_k = \struct {X_k, \tau_k}$ be [[Definition:Topological Space|topological spaces]].
+Let $\displaystyle X = \prod_{k \mathop = 1}^n X_k$ be the [[Definition:Finite Cartesian Product|cartesian product]] of $X_1, \ldots, X_n$.
+Then the [[Definition:Box Topology|box topology]] and the [[Definition:Tychonoff Topology|Tychonoff topology]] on $X$ are identical.
+\end{theorem}
+
+\begin{proof}
+Denote the [[Definition:Tychonoff Topology|Tychonoff topology]] on $X$ as $\tau$, and the [[Definition:Box Topology|box topology]] on $X$ as $\tau'$.
+Suppose that $U \in \tau'$.
+Then there exists an [[Definition:Indexed Set|index set]] $I$ such that:
+:$\displaystyle U = \bigcup_{i \mathop \in I} \struct {U_{i, 1} \times U_{i, 2} \times \cdots \times U_{i, n} }$
+where $U_{i, k} \in \tau_k$ for all $i \in I, k \in \set {1, \ldots, n}$.
+When $\pr_k: X \to X_k$ denotes the $k$th [[Definition:Projection on Family of Sets|projection]] on $X$, we have:
+:$\pr_k^{-1} \sqbrk {U_{i, k} } = X_1 \times X_2 \times \cdots \times X_{k - 1} \times U_{i, k} \times X_{k + 1} \times \cdots \times X_n$
+It follows that:
+:$\displaystyle U = \bigcup_{i \mathop \in I} \struct {U_{i, 1} \times U_{i, 2} \times \cdots \times U_{i, n} } = \bigcup_{i \mathop \in I} \bigcap_{k \mathop = 1}^{n_i} \pr_k^{-1} \sqbrk {U_{i, k} }$
+By [[Definition:Tychonoff Topology|definition of Tychonoff topology]], $U \in \tau$, so $\tau' \subseteq \tau$.
+{{qed|lemma}}
+Suppose that $U \in \tau$.
+Then there exists an [[Definition:Indexing Set|index set]] $I$, so $U$ can be expressed as:
+$U = \displaystyle \bigcup_{i \mathop \in I} \bigcap_{l \mathop = 1}^{m_i} \pr_{j_l}^{-1} \sqbrk {U_{i, l} }$
+where $m_i \in \N$, $j_1, j_2, \ldots, j_{m_i} \in \set {1, \ldots, n}$, and $U_{i, l} \in \tau_{j_l}$ for all $i \in I, l \in \set {1, \ldots, m_i}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = U
+      | r = \bigcup_{i \mathop \in I} \bigcap_{l \mathop = 1}^{m_i} \pr_{j_l}^{-1} \sqbrk {U_{i, l} }
+}}
+{{eqn | r = \bigcup_{i \mathop \in I} \bigcap_{k \mathop = 1}^n \bigcap_{l: k \mathop = j_l} \pr_k^{-1} \sqbrk {U_{i, l} }
+}}
+{{eqn | r = \bigcup_{i \mathop \in I} \bigcap_{k \mathop = 1}^n \struct {X_1 \times X_2 \times \cdots \times X_{k - 1} \times \bigcap_{l: k \mathop = j_l} U_{i, l} \times X_{k + 1} \times \cdots \times X_n}
+}}
+{{end-eqn}}
+As $\displaystyle \bigcap_{l: k \mathop = j_l} U_{i, l} \in \tau_k$, it follows that:
+:$\displaystyle \struct {X_1 \times X_2 \times \cdots \times X_{k - 1} \times \bigcap_{l: k \mathop = j_l} U_{i, l} \times X_{k + 1} \times \cdots \times X_n} \in \tau'$
+By [[Definition:Topology|definition of topology]], it follows that $U \in \tau'$, so $\tau \subseteq \tau'$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Strictly Positive Integer Power Function is Unbounded Above/General Case}
+Tags: Totally Ordered Rings, Rings with Unity
+
+\begin{theorem}
+Let $\struct {R, +, \circ, \le}$ be a [[Definition:Totally Ordered Ring|totally ordered ring]] [[Definition:Ring with Unity|with unity]].
+Suppose that $R$ has no [[Definition:Upper Bound of Set|upper bound]].
+Let $n \in \N_{>0}$.
+Let $f: R \to R$ be defined by:
+:$\map f x = \circ^n x$
+Then the [[Definition:Image of Mapping|image of $f$]] is [[Definition:Unbounded Above Set|unbounded above]] in $R$.
+\end{theorem}
+
+\begin{proof}
+Let $1_R$ be the [[Definition:Unity of Ring|unity]] of $R$.
+Let $b \in R$.
+We will show that $b$ is not an [[Definition:Upper Bound of Set|upper bound]] of the [[Definition:Image of Mapping|image]] of $f$.
+Since $R$ is [[Definition:Totally Ordered Ring|totally ordered]] and [[Definition:Unbounded Above Set|unbounded above]], there is an element $c \in R$ such that $b < c$ and $1_R < c$.
+By [[Strictly Positive Power of Strictly Positive Element Greater than One Succeeds Element]], $c \le \map f c$.
+Thus by [[Definition:Ordering|transitivity of ordering]], $b < \map f c$.
+{{qed}}
+[[Category:Totally Ordered Rings]]
+[[Category:Rings with Unity]]
+ourpo7kby1306f7ch5iz4zo25u4c3vb
+\end{proof}<|endoftext|>
+\section{Strictly Positive Integer Power Function is Unbounded Above}
+Tags: Real Analysis
+
+\begin{theorem}
+Let $\R$ be the [[Definition:Real Number|real numbers]] with the [[Definition:Usual Ordering|usual ordering]].
+Let $n \in \N_{>0}$.
+Let $f: \R \to \R$ be defined by:
+:$\map f x = x^n$
+Then $f$ is [[Definition:Unbounded Above Real-Valued Function|unbounded above]].
+\end{theorem}
+
+\begin{proof}
+If $n = 1$, then $f$ is the [[Definition:Identity Mapping|identity function]].
+By the [[Archimedean Principle]], the [[Definition:Real Number|real numbers]] are [[Definition:Unbounded Above Subset of Real Numbers|unbounded above]].
+Thus by definition of the [[Definition:Identity Mapping|identity function]]:
+$f$ is [[Definition:Unbounded Above Real-Valued Function|unbounded above]].
+{{qed|lemma}}
+Let $n \ge 2$.
+{{AimForCont}} that $f$ is [[Definition:Bounded Above Subset of Real Numbers|bounded above]] by $b \in \R$.
+{{WLOG}} suppose that $b > 0$.
+Then by the definition of an [[Definition:Upper Bound of Subset of Real Numbers|upper bound]]:
+:$\forall x \in \R: x^n \le b$
+Let $x > b$.
+Then:
+:$\dfrac {x^n - b} {x - b} \le 0$
+By the [[Mean Value Theorem]], there exists a point $p$ between $b$ and $x$ such that:
+:$\map {f'} p = \dfrac {x^n - b} {x - b}$
+By [[Power Rule for Derivatives/Natural Number Index|Derivative of Power]]:
+:$\map {f'} p = n p^{n - 1}$
+Therefore $n p^{n - 1} \le 0$.
+By [[Power of Strictly Positive Real Number is Strictly Positive/Positive Integer|Power of Strictly Positive Real Number is Positive]] and [[Positive Real Numbers Closed under Multiplication]] it follows that:
+:$p > 0 \implies n p^{n - 1} > 0$
+But this is impossible because it was previously established that $p \ge b > 0$.
+From this [[Proof by Contradiction|contradiction]] it follows that there can be no such $b$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ring Product preserves Inequalities on Positive Elements}
+Tags: Ordered Rings, Inequalities
+
+\begin{theorem}
+Let $\struct {R, +, \circ, \le}$ be an [[Definition:Ordered Ring|ordered ring]].
+Let $x, y, z, w \in R$.
+Let $0 < x < y$ and $0 < z < w$.
+Then:
+:$0 < z \circ x < w \circ y$
+\end{theorem}
+
+\begin{proof}
+By [[Properties of Ordered Ring]] $(6)$:
+:$z \circ x < z \circ y$
+:$z \circ y < w \circ y$
+Then by [[Definition:Transitive Relation|transitivity]] of $\circ$:
+:$z \circ x < w \circ y$
+Also by [[Properties of Ordered Ring]] $(6)$:
+:$z \circ 0 < z \circ x$
+Hence by [[Ring Product with Zero]]:
+:$0 < z \circ x$
+{{qed}}
+[[Category:Ordered Rings]]
+[[Category:Inequalities]]
+e0clz004vppggt5grtn0ls0bmjjxvfu
+\end{proof}<|endoftext|>
+\section{Power Function is Strictly Increasing on Positive Elements}
+Tags: Ordered Rings, Proofs by Induction
+
+\begin{theorem}
+Let $\struct {R, +, \circ, \le}$ be an [[Definition:Ordered Ring|ordered ring]].
+Let $x, y \in R$.
+Let $n \in \N_{>0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
+Let $0 < x < y$.
+Then:
+:$0 < \map {\circ^n} x < \map {\circ^n} y$
+\end{theorem}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]]:
+For all $n \in \N_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$0 < \map {\circ^n} x < \map {\circ^n} y$
+=== Basis for the Induction ===
+$\map P 1$ is the case:
+:$0 < \map {\circ^1} x < \map {\circ^1} y$
+which is just:
+:$0 < x < y$
+Thus $\map P 1$ is seen to hold.
+This is the [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
+So this is the [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$0 < \map {\circ^k} x < \map {\circ^k} y$
+from which it is to be shown that:
+:$0 < \map {\circ^{k + 1}} x < \map {\circ^{k + 1}} y$
+=== Induction Step ===
+This is the [[Definition:Induction Step|induction step]]:
+We have:
+:$0 < x < y$
+:$0 < \map {\circ^k} x < \map {\circ^k} y$
+By [[Ring Product preserves Inequalities on Positive Elements]]:
+:$0 < \map {\circ^k} x \circ x < \map {\circ^k} y \circ y$
+Hence:
+:$0 < \map {\circ^{k + 1}} x < \map {\circ^{k + 1}} y$
+So $\map P k \implies \map P {k + 1}$ and thus it follows by the [[Principle of Mathematical Induction]] that:
+:$\forall n \in \N_{> 0}: 0 < \map {\circ^n} x < \map {\circ^n} y$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Biconditional is Associative/Formulation 1}
+Tags: Biconditional is Associative
+
+\begin{theorem}
+: $p \iff \paren {q \iff r} \dashv \vdash \paren {p \iff q} \iff r$
+\end{theorem}
+
+\begin{proof}
+Proof of associativity by [[Definition:Natural Deduction|natural deduction]] is just too tedious to be considered.
+We apply the [[Method of Truth Tables]].
+As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
+$\begin{array}{|ccccc||ccccc|} \hline
+p & \iff & (q & \iff & r) & (p & \iff & q) & \iff & r \\
+\hline
+F & F & F & T & F & F & T & F & F & F \\
+F & T & F & F & T & F & T & F & T & T \\
+F & T & T & F & F & F & F & T & T & F \\
+F & F & T & T & T & F & F & T & F & T \\
+T & T & F & T & F & T & F & F & T & F \\
+T & F & F & F & T & T & F & F & F & T \\
+T & F & T & F & F & T & T & T & F & F \\
+T & T & T & T & T & T & T & T & T & T \\
+\hline
+\end{array}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Biconditional is Associative/Formulation 2}
+Tags: Biconditional is Associative
+
+\begin{theorem}
+:$\vdash \paren {p \iff \paren {q \iff r} } \iff \paren {\paren {p \iff q} \iff r}$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\vdash \paren {p \iff \paren {q \iff r} } \iff \paren {\paren {p \iff q} \iff r} }}
+{{Assumption|1|p \iff \paren {q \iff r} }}
+{{SequentIntro|2|1|\paren {p \iff q} \iff r|1|[[Biconditional is Associative/Formulation 1|Biconditional is Associative: Formulation 1]]}}
+{{Implication|3||\paren {p \iff \paren {q \iff r} } \implies \paren {\paren {p \iff q} \iff r}|1|2}}
+{{Assumption|4|\paren {p \iff q} \iff r}}
+{{SequentIntro|5|4|p \iff \paren {q \iff r}|4|[[Biconditional is Associative/Formulation 1|Biconditional is Associative: Formulation 1]]}}
+{{Implication|6||\paren {\paren {p \iff q} \iff r} \implies \paren {p \iff \paren {q \iff r} }|4|5}}
+{{BiconditionalIntro|7||\paren {p \iff \paren {q \iff r} } \iff \paren {\paren {p \iff q} \iff r}|3|6}}
+{{EndTableau|qed}}
+\end{proof}<|endoftext|>
+\section{Open Ball is Convex Set}
+Tags: Vector Spaces, Open Balls
+
+\begin{theorem}
+Let $V$ be a [[Definition:Normed Vector Space|normed vector space]] with [[Definition:Norm on Vector Space|norm]] $\left\Vert{\cdot}\right\Vert$ over $\R$ or $\C$.
+An [[Definition:Open Ball|open ball]] in the [[Definition:Metric Induced by Norm|metric induced by $\left\Vert{\cdot}\right\Vert$]] is a [[Definition:Convex Set (Vector Space)|convex set]].
+\end{theorem}
+
+\begin{proof}
+Let $v \in V$ and $\epsilon \in \R_{>0}$.
+Denote the [[Definition:Open Ball|open $\epsilon$-ball]] of $v$ as $B_\epsilon \left({v}\right)$.
+Let $x, y \in B_\epsilon \left({v}\right)$.
+Then $x + t \left({y - x}\right)$ lies on [[Definition:Convex Set (Vector Space)|line segment]] joining $x$ and $y$ for all $t \in \left[{0 \,.\,.\, 1}\right]$.
+The distance between $x + t \left({y - x}\right)$ and $v$ is:
+{{begin-eqn}}
+{{eqn |l= \left\Vert{x + t \left({y - x}\right) - v}\right\Vert
+      |r= \left\Vert{\left({1 - t}\right) \left({x - v}\right) + t \left({y - v}\right) }\right\Vert
+}}
+{{eqn |o= \le
+      |r= \left\Vert{\left({1 - t}\right) \left({x - v}\right) }\right\Vert + \left\Vert{t \left({y - v}\right) }\right\Vert
+      |c= by [[Definition:Metric Space/Triangle Inequality|triangle inequality]]
+}}
+{{eqn |r= \left({1 - t}\right) \left\Vert{x - v}\right\Vert + t \left\Vert{y-v}\right\Vert
+      |c= by [[Definition:Norm on Vector Space|norm axiom $(N2)$]]
+}}
+{{eqn |o= <
+      |r= \left({1 - t}\right) \epsilon + t \epsilon
+      |c= as $x, y \in B_\epsilon \left({v}\right)$
+}}
+{{eqn |r= \epsilon
+}}
+{{end-eqn}}
+Hence, $x + t \left({y - x}\right) \in B_\epsilon \left({v}\right)$.
+By [[Definition:Convex Set (Vector Space)|definition of convex set]], $B_\epsilon \left({v}\right)$ is a convex set.
+{{qed}}
+[[Category:Vector Spaces]]
+[[Category:Open Balls]]
+8lwxjlnq0kh1w3z7tvrti5n8wiaigx4
+\end{proof}<|endoftext|>
+\section{Pseudometric Space is Metric Space iff Kolmogorov}
+Tags: Metric Spaces, Pseudometric Spaces, T0 Spaces
+
+\begin{theorem}
+Let $M = \struct {S, d}$ be a [[Definition:Pseudometric Space|pseudometric space]].
+Let $T = \struct {S, \tau}$ be the [[Definition:Topological Space|topological space]] over $S$ [[Pseudometric induces Topology|induced by $d$]].
+Then $M$ is a [[Definition:Metric Space|metric space]] {{iff}} $T$ is a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov} space]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $M$ be a [[Definition:Metric Space|metric space]].
+From [[Metric Space is Hausdorff]], $M$ is a [[Definition:Hausdorff Space|Hausdorff space]].
+From:
+:[[T2 Space is T1 Space|$T_2$ (Hausdorff) Space is $T_1$ Space]]
+and:
+:[[T1 Space is T0 Space|$T_1$ Space is $T_0$ (Kolmogorov) Space]]
+it follows that $M$ is a [[Definition:Kolmogorov Space|Kolmogorov space]].
+{{qed|lemma}}
+=== Sufficient Condition ===
+Suppose now that $M$ is a [[Definition:Kolmogorov Space|Kolmogorov space]].
+Let $a, b \in S$ such that $a \ne b$.
+Then $\exists U \in \tau$ such that:
+:$\paren {a \in U \land b \notin U} \lor \paren {b \in U \land a \notin U}$
+{{WLOG}}, suppose that $a \in U$ and $b \notin U$.
+Then by the definition of the [[Pseudometric induces Topology|induced topology]]:
+:$\exists \epsilon \in \R_{>0}: \map {B_\epsilon} a \subset U$
+where $\map {B_\epsilon} a$ denotes the [[Definition:Open Ball|open $\epsilon$-ball]] of $a$ in $M$.
+Since $b \notin U$:
+:$\map d {a, b} \ge \epsilon$
+Since $\epsilon > 0$:
+:$\map d {a, b} > 0$
+Since this holds for any pair of [[Definition:Distinct|distinct]] points, $M$ is a [[Definition:Metric Space|metric space]].
+{{qed}}
+[[Category:Metric Spaces]]
+[[Category:Pseudometric Spaces]]
+[[Category:T0 Spaces]]
+rwglezgrcr8npj1mxld4jz80qzfdi6b
+\end{proof}<|endoftext|>
+\section{Sequence Converges to Point Relative to Metric iff it Converges Relative to Induced Topology}
+Tags: Metric Spaces
+
+\begin{theorem}
+Let $M = \struct {S, d}$ be a [[Definition:Metric Space|metric space]] or a [[Definition:Pseudometric Space|pseudometric space]].
+Let $T = \struct {S, \tau}$ be the [[Definition:Topology Induced by Metric|topological space induced by $d$]].
+Let $\sequence {x_n}$ be a [[Definition:Infinite Sequence|infinite sequence]] in $S$.
+Let $l \in S$.
+Then $\sequence {x_n}$ [[Definition:Convergent Sequence (Metric Space)|converges]] to $l$ relative to $d$ {{iff}} $\sequence {x_n}$ [[Definition:Convergent Sequence (Topology)|converges]] to $l$ relative to $\tau$.
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Suppose that $\sequence {x_n}$ [[Definition:Convergent Sequence (Metric Space)|converges]] to $l$ relative to $d$.
+When $\map {B_\epsilon} l$ denotes the [[Definition:Open Ball|open $\epsilon$-ball]] of $l$, this means:
+:$\forall \epsilon \in \R_{>0}: \exists N \in \R: n > N \implies x_n \in \map {B_\epsilon} l$
+Let $U \in \tau$ with $l \in U$.
+By definition of [[Definition:Topology Induced by Metric|induced topology]], there exists $\epsilon_0 \in \R_{>0}$ such that:
+:$\map {B_{\epsilon_0} } l \subseteq U$
+Then there exists $N_0 \in \R$ such that for all $n > N_0$:
+:$x_n \in \map {B_{\epsilon_0} } l \subseteq U$
+Hence, $\sequence {x_n}$ [[Definition:Convergent Sequence (Topology)|converges]] to $l$ in the [[Definition:Topology Induced by Metric|induced topology]] $\tau$.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Suppose that $\sequence {x_n}$ [[Definition:Convergent Sequence (Topology)|converges]] to $l$ in $\tau$.
+Let $\epsilon \in \R_{>0}$.
+Then $l \in \map {B_\epsilon} l$, and $\map {B_\epsilon} l \in \tau$.
+Then there exists $N \in \N$ such that for all $n > N$, we have $x_n \in \map {B_\epsilon} l$.
+Hence, $\sequence {x_n}$ [[Definition:Convergent Sequence (Metric Space)|converges]] to $l$ relative to $d$.
+{{qed}}
+[[Category:Metric Spaces]]
+4gt7042v97zx44p62ujy1c5l8nh60wp
+\end{proof}<|endoftext|>
+\section{Isometry Preserves Sequence Convergence}
+Tags: Isometries, Sequences, Isometry Preserves Sequence Convergence
+
+\begin{theorem}
+Let $M_1 = \struct {S_1, d_1}$ and $M_2 = \struct {S_2, d_2}$ both be [[Definition:Metric Space|metric spaces]] or [[Definition:Pseudometric Space|pseudometric spaces]].
+Let $\phi: S_1 \to S_2$ be an [[Definition:Isometry (Metric Spaces)|isometry]].
+Let $\sequence {x_n}$ be an [[Definition:Infinite Sequence|infinite sequence]] in $S_1$.
+Suppose that $\sequence {x_n}$ [[Definition:Convergent Sequence (Metric Space)|converges]] to a point $p \in S_1$.
+Then $\sequence {\map \phi {x_n}}$ [[Definition:Convergent Sequence (Metric Space)|converges]] to $\map \phi p$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | o =
+      | r = \lim_{n \mathop \to \infty} \map {d_2} {\map \phi {x_n}, \map \phi p}
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \map {d_1} {x_n, p}
+      | c = {{Defof|Isometry (Metric Spaces)|index = 1}}
+}}
+{{eqn | r = 0
+      | c = {{Defof|Convergent Sequence|subdef = Metric Space|index = 3}}
+}}
+{{end-eqn}}
+Hence, by definition, $\sequence {\map \phi {x_n}}$ [[Definition:Convergent Sequence/Metric Space/Definition 3|converges]] to $\map \phi p$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{proof wanted|proof from the fact that an isometry is a homeomorphism}}
+\end{proof}<|endoftext|>
+\section{Connected Domain is Connected by Staircase Contours}
+Tags: Complex Analysis
+
+\begin{theorem}
+Let $D \subseteq \C$ be an [[Definition:Open Set (Complex Analysis)|open set]].
+Then $D$ is a [[Definition:Connected Domain (Complex Analysis)|connected domain]] {{iff}}:
+:for all $z, w \in \C$, there exists a [[Definition:Staircase Contour|staircase contour]] in $D$ with [[Definition:Start Point of Contour (Complex Plane)|start point]] $z$ and [[Definition:End Point of Contour (Complex Plane)|end point]] $w$.
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Suppose $D$ is a [[Definition:Connected Domain (Complex Analysis)|connected domain]].
+If $z, w \in D$, there exists a [[Definition:Path (Topology)|path]] $\gamma: \left[{0 \,.\,.\, 1}\right] \to D$ with $\gamma \left({0}\right) = z$ and $\gamma \left({0}\right) = w$.
+From the [[Paving Lemma]], it follows that there exist $\epsilon \in \R_{>0}$ and a [[Definition:Subdivision (Real Analysis)|subdivision]] $\left\{ {x_0, x_1, \ldots, x_n }\right\}$ of $\left[{0 \,.\,.\, 1}\right]$ such that:
+:$\displaystyle \bigcup_{k \mathop = 0}^n B_\epsilon \left({\gamma \left({x_k}\right) }\right) \subseteq D$
+and for all $k \in \left\{ {0, 1, \ldots, n - 1}\right\}$:
+:$\gamma \left({\left[{x_k \,.\,.\, x_{k + 1} }\right] }\right) \subseteq B_\epsilon \left({\gamma \left({x_k}\right) }\right)$
+where $B_\epsilon \left({\gamma \left({x_k}\right) }\right)$ is the [[Definition:Open Ball|open $\epsilon$-ball]] about $\gamma \left({x_k}\right)$.
+Let $\operatorname{Re} \left({x_{k + 1} - x_k}\right)$ denote the [[Definition:Real Part|real part]] of $x_{k + 1} - x_k$, and $\operatorname{Im} \left({x_{k + 1} - x_k}\right) $ denote the [[Definition:Imaginary Part|imaginary part]] of $x_{k + 1} - x_k$.
+For all $k \in \left\{ {0, 1, \ldots, n - 1}\right\}$, define the [[Definition:Smooth Path (Complex Analysis)|smooth paths]] $\gamma_{2k+1}, \gamma_{2k+2}: \gamma: \left[{0 \,.\,.\, 1}\right] \to D$ by:
+:$\gamma_{2k+1} \left({t}\right) = x_k + t \left({\operatorname{Re} \left({x_{k + 1} - x_k}\right) }\right)$
+:$\gamma_{2k+2} \left({t}\right) = x_k + \operatorname{Re} \left({x_{k + 1} - x_k}\right) + i t \left({\operatorname{Im} \left({x_{k + 1} - x_k}\right) }\right)$
+These are the [[Definition:Convex Set (Vector Space)|line segments]] connecting $x_k$, $x_k + \operatorname{Re} \left({x_{k + 1} - x_k}\right)$, and $x_{k + 1}$.
+It follows that $x_k + \operatorname{Re} \left({x_{k + 1} - x_k}\right) \in B_\epsilon \left({\gamma \left({x_k}\right) }\right)$, as:
+{{begin-eqn}}
+{{eqn | l = \left\vert{\operatorname{Re} \left({x_{k + 1} - x_k}\right) }\right\vert
+      | o = \le 
+      | r = \left\vert{x_{k + 1} - x_k}\right\vert
+      | c = [[Modulus Larger than Real Part]]
+}}
+{{eqn | o = <
+      | r = \epsilon
+}}
+{{end-eqn}}
+Then [[Open Ball is Convex Set]] shows that $\gamma_{2k+1} \left({t}\right), \gamma_{2k+2} \left({t}\right) \in B_\epsilon \left({\gamma \left({x_k}\right) }\right) \subseteq D$ for all $t \in \left[{0 \,.\,.\, 1}\right]$.
+It follows that the [[Definition:Image of Mapping|images]] of $\gamma_{2k+1}$ and $\gamma_{2k+2}$ are [[Definition:Subset|subsets]] of $D$. 
+Define $C_k$ as the [[Definition:Directed Smooth Curve (Complex Plane)|directed smooth curve]] that is [[Definition:Parameterization of Directed Smooth Curve (Complex Plane)|parameterized]] by $\gamma_k$.
+Thus $C_{2 k - 1}$ has [[Definition:Start Point of Contour (Complex Plane)|start point]] $x_{k - 1}$, and $C_{2 k}$ has [[Definition:End Point of Contour (Complex Plane)|end point]] $x_k$.
+Define $C$ as the [[Definition:Concatenation of Contours (Complex Plane)|concatenation]] of $C_1, \ldots, C_{2n}$.
+Then $C$ is a [[Definition:Staircase Contour|staircase contour]] in $D$ with [[Definition:Start Point of Contour (Complex Plane)|start point]] $z$ and [[Definition:End Point of Contour (Complex Plane)|end point]] $w$.
+:[[File:ConnectedDomainStaircase.png]]
+Illustration of the open balls inside the connected domain $D$.
+The path $\gamma$ between $w$ and $z$ is grey, and the constructed staircase contour $C$ is red.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Suppose that for all $z, w \in \C$, there exists a [[Definition:Staircase Contour|staircase contour]] in $D$ with [[Definition:Start Point of Contour (Complex Plane)|start point]] $z$ and [[Definition:End Point of Contour (Complex Plane)|end point]] $w$.
+Let $\gamma: \left[{a \,.\,.\, b}\right] \to D$ be a [[Definition:Parameterization of Contour (Complex Plane)|parameterization]] of $C$, where $\left[{a \,.\,.\, b}\right]$ is a [[Definition:Closed Real Interval|closed interval]].
+Then $\gamma$ is a [[Definition:Path (Topology)|path]] in $D$ with $\gamma \left({a}\right) = z$ and $\gamma \left({b}\right) = w$.
+Define $\gamma_0: \left[{0 \,.\,.\, 1}\right] \to D$ by $\gamma_0 \left({t}\right) = \gamma \left({a + t \left({b - a}\right) }\right)$.
+Then $\gamma_0$ is also a [[Definition:Path (Topology)|path]] in $D$ with $\gamma_0 \left({0}\right) = z$ and $\gamma_0 \left({1}\right) = w$.
+By [[Definition:Path-Connected Metric Subspace|definition of path-connected]], it follows that $D$ is path-connected.
+Hence, $D$ is a [[Definition:Connected Domain (Complex Analysis)|connected domain]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Dedekind Completeness is Self-Dual}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
+Then $\struct {S, \preceq}$ is [[Definition:Dedekind Complete|Dedekind complete]] {{iff}} every [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $S$ that is [[Definition:Bounded Below Set|bounded below]] admits an [[Definition:Infimum of Set|infimum]] in $S$.
+That is, an [[Definition:Ordered Set|ordered set]] is [[Definition:Dedekind Complete|Dedekind complete]] {{iff}} its [[Definition:Dual Ordered Set|dual]] is [[Definition:Dedekind Complete|Dedekind complete]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $\struct {S, \preceq}$ be [[Definition:Dedekind Complete|Dedekind complete]].
+Let $A \subseteq S$ be [[Definition:Non-Empty Set|non-empty]] and [[Definition:Bounded Below Set|bounded below]].
+Let $B \subseteq S$ be [[Definition:Set|set]] of all [[Definition:Lower Bound of Set|lower bounds]] for $A$.
+Then every [[Definition:Element|element]] of $A$ is an [[Definition:Upper Bound of Set|upper bound]] for $B$.
+Therefore, $B$ is [[Definition:Non-Empty Set|non-empty]] and [[Definition:Bounded Above Set|bounded above]].
+By the definition of [[Definition:Dedekind Complete|Dedekind completeness]], $B$ admits a [[Definition:Supremum of Set|supremum]] $x \in S$.
+By the definition of [[Definition:Supremum of Set|supremum]], it follows that every [[Definition:Element|element]] of $A$ [[Definition:Succeed|succeeds]] $x$.
+That is, $x$ is a [[Definition:Lower Bound of Set|lower bound]] for $A$.
+If $y \in S$ is a [[Definition:Lower Bound of Set|lower bound]] for $A$, then $y \in B$, and so $y \preceq x$.
+Hence, $x$ is the [[Definition:Infimum of Set|infimum]] of $A$.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Follows directly from the [[Dedekind Completeness is Self-Dual#Necessary Condition|Necessary Condition]] and [[Dual of Dual Ordering]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Biconditional is Transitive/Formulation 1}
+Tags: Biconditional is Transitive, Truth Table Proofs
+
+\begin{theorem}
+:$p \iff q, q \iff r \vdash p \iff r$
+\end{theorem}<|endoftext|>
+\section{Biconditional is Transitive/Formulation 2}
+Tags: Biconditional is Transitive
+
+\begin{theorem}
+: $\vdash \left({\left({p \iff q}\right) \land \left({q \iff r}\right)}\right) \implies \left({p \iff r}\right)$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\vdash \left({\left({p \iff q}\right) \land \left({q \iff r}\right)}\right) \implies \left({p \iff r}\right)}}
+{{Assumption|1|\left({p \iff q}\right) \land \left({q \iff r}\right)}}
+{{Simplification|2|1|p \iff q|1|1}}
+{{Simplification|3|1|q \iff r|1|2}}
+{{SequentIntro|4|1|p \iff r|2, 3|[[Biconditional is Transitive/Formulation 1|Biconditional is Transitive: Formulation 1]]}}
+{{Implication|5||\left({\left({p \iff q}\right) \land \left({q \iff r}\right)}\right) \implies \left({p \iff r}\right)|1|4}}
+{{EndTableau}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Biconditional as Disjunction of Conjunctions/Formulation 1/Forward Implication}
+Tags: Biconditional as Disjunction of Conjunctions
+
+\begin{theorem}
+: $p \iff q \vdash \left({p \land q}\right) \lor \left({\neg p \land \neg q}\right)$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|p \iff q \vdash \left({p \land q}\right) \lor \left({\neg p \land \neg q}\right)}}
+{{Premise|1|p \iff q}}
+{{BiconditionalElimination|2|1|p \implies q|1|1}}
+{{BiconditionalElimination|3|1|q \implies p|1|2}}
+{{ExcludedMiddle|4|p \lor \neg p}}
+{{Assumption|5|p}}
+{{ModusPonens|6|1, 5|q|2|5}}
+{{Conjunction|7|1, 5|p \land q|5|6}}
+{{Addition|8|1, 5|\left({p \land q}\right) \lor \left({\neg p \land \neg q}\right)|7|1}}
+{{Assumption|9|\neg p}}
+{{ModusTollens|10|1, 9|\neg q|3|9}}
+{{Conjunction|11|1, 9|\neg p \land \neg q|9|10}}
+{{Addition|12|1, 9|\left({p \land q}\right) \lor \left({\neg p \land \neg q}\right)|11|2}}
+{{ProofByCases|13|1|\left({p \land q}\right) \lor \left({\neg p \land \neg q}\right)|1|5|8|9|12}}
+{{EndTableau}}
+{{qed}}
+{{LEM}}
+[[Category:Biconditional as Disjunction of Conjunctions]]
+egkw4jhwu57bih1jc81ct20djsatuf4
+\end{proof}<|endoftext|>
+\section{Biconditional as Disjunction of Conjunctions/Formulation 1/Reverse Implication}
+Tags: Biconditional as Disjunction of Conjunctions
+
+\begin{theorem}
+: $\left({p \land q}\right) \lor \left({\neg p \land \neg q}\right) \vdash p \iff q$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\left({p \land q}\right) \lor \left({\neg p \land \neg q}\right) \vdash p \iff q}}
+{{Premise|1|\left({p \land q}\right) \lor \left({\neg p \land \neg q}\right)}}
+{{Assumption|2|p \land q}}
+{{Assumption|3|p}}
+{{Simplification|4|2|q|2|2}}
+{{Implication|5|2|p \implies q|3|4}}
+{{Assumption|6|q}}
+{{Simplification|7|2|p|2|1}}
+{{Implication|8|2|q \implies p|6|7}}
+{{BiconditionalIntro|9|2|p \iff q|5|8}}
+{{Assumption|10|\neg p \land \neg q}}
+{{Assumption|11|\neg p}}
+{{Simplification|12|10|\neg q|10|2}}
+{{Implication|13|10|\neg p \implies \neg q|11|12}}
+{{SequentIntro|14|10|q \implies p|13|[[Rule of Transposition]]}}
+{{Assumption|15|\neg q}}
+{{Simplification|16|10|\neg p|10|1}}
+{{Implication|17|10|\neg q \implies \neg p|15|16}}
+{{SequentIntro|18|10|p \implies q|17|[[Rule of Transposition]]}}
+{{BiconditionalIntro|19|10|p \iff q|18|14}}
+{{ProofByCases|20|1|p \iff q|1|2|9|10|19}}
+{{EndTableau}}
+{{qed}}
+[[Category:Biconditional as Disjunction of Conjunctions]]
+kpkzs6086yvdup4ehfnuooz56hz7783
+\end{proof}<|endoftext|>
+\section{Biconditional as Disjunction of Conjunctions/Formulation 1}
+Tags: Biconditional as Disjunction of Conjunctions
+
+\begin{theorem}
+: $p \iff q \dashv \vdash \paren {p \land q} \lor \paren {\neg p \land \neg q}$
+\end{theorem}<|endoftext|>
+\section{Biconditional as Disjunction of Conjunctions/Formulation 2}
+Tags: Biconditional as Disjunction of Conjunctions
+
+\begin{theorem}
+:$\vdash \paren {p \iff q} \iff \paren {\paren {p \land q} \lor \paren {\neg p \land \neg q} }$
+\end{theorem}<|endoftext|>
+\section{Biconditional as Disjunction of Conjunctions}
+Tags: Biconditional as Disjunction of Conjunctions, Biconditional, Disjunction, Conjunction
+
+\begin{theorem}
+==== [[Biconditional as Disjunction of Conjunctions/Formulation 1|Formulation 1]] ====
+{{:Biconditional as Disjunction of Conjunctions/Formulation 1}}
+==== [[Biconditional as Disjunction of Conjunctions/Formulation 2|Formulation 2]] ====
+{{:Biconditional as Disjunction of Conjunctions/Formulation 2}}
+\end{theorem}<|endoftext|>
+\section{Biconditional Equivalent to Biconditional of Negations/Formulation 1/Forward Implication}
+Tags: Biconditional Equivalent to Biconditional of Negations
+
+\begin{theorem}
+: $p \iff q \vdash \neg p \iff \neg q$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|p \iff q \vdash \neg p \iff \neg q}}
+{{Premise|1|p \iff q}}
+{{BiconditionalElimination|2|1|p \implies q|1|1}}
+{{SequentIntro|3|1|\neg q \implies \neg p|2|[[Rule of Transposition]]}}
+{{BiconditionalElimination|4|1|q \implies p|1|2}}
+{{SequentIntro|5|1|\neg p \implies \neg q|4|[[Rule of Transposition]]}}
+{{BiconditionalIntro|6|1|\neg p \iff \neg q|5|3}}
+{{EndTableau}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Biconditional Equivalent to Biconditional of Negations/Formulation 1/Reverse Implication}
+Tags: Biconditional Equivalent to Biconditional of Negations
+
+\begin{theorem}
+:$\neg p \iff \neg q \vdash p \iff q$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\neg p \iff \neg q \vdash p \iff q}}
+{{Premise|1|\neg p \iff \neg q}}
+{{BiconditionalElimination|2|1|\neg p \implies \neg q|1|1}}
+{{SequentIntro|3|1|\neg \neg q \implies \neg \neg p|2|[[Rule of Transposition]]}}
+{{DoubleNegElimination|4|1|q \implies p|3|(twice)}}
+{{BiconditionalElimination|5|1|\neg q \implies \neg p|1|2}}
+{{SequentIntro|6|1|\neg \neg p \implies \neg \neg q|5|[[Rule of Transposition]]}}
+{{DoubleNegElimination|7|1|p \implies q|3|(twice)}}
+{{BiconditionalIntro|8|1|p \iff q|7|4}}
+{{EndTableau|qed}}
+{{LEM|Double Negation Elimination}}
+\end{proof}<|endoftext|>
+\section{Biconditional Equivalent to Biconditional of Negations/Formulation 1}
+Tags: Biconditional Equivalent to Biconditional of Negations, Truth Table Proofs
+
+\begin{theorem}
+: $p \iff q \dashv \vdash \neg p \iff \neg q$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]].
+As can be seen by inspection, in all cases the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
+$\begin{array}{|ccc||ccccc|} \hline
+p & \iff & q & \neg & p & \iff & \neg & q \\
+\hline
+F & T & F & T & F & T & T & F \\
+F & F & T & T & F & F & F & T \\
+T & F & F & F & T & F & T & F \\
+T & T & T & F & T & T & F & T \\
+\hline
+\end{array}$
+{{qed}}
+[[Category:Biconditional Equivalent to Biconditional of Negations]]
+[[Category:Truth Table Proofs]]
+gi5u0dj3jca25l8ylyo15gtegynwcki
+\end{proof}<|endoftext|>
+\section{Biconditional Equivalent to Biconditional of Negations/Formulation 2}
+Tags: Biconditional Equivalent to Biconditional of Negations
+
+\begin{theorem}
+: $\vdash \left({p \iff q}\right) \iff \left({\neg p \iff \neg q}\right)$
+\end{theorem}<|endoftext|>
+\section{Biconditional Equivalent to Biconditional of Negations}
+Tags: Biconditional, Negation
+
+\begin{theorem}
+==== [[Biconditional Equivalent to Biconditional of Negations/Formulation 1|Formulation 1]] ====
+{{:Biconditional Equivalent to Biconditional of Negations/Formulation 1}}
+==== [[Biconditional Equivalent to Biconditional of Negations/Formulation 2|Formulation 2]] ====
+{{:Biconditional Equivalent to Biconditional of Negations/Formulation 2}}
+\end{theorem}<|endoftext|>
+\section{Biconditional iff Disjunction implies Conjunction/Formulation 1/Forward Implication}
+Tags: Biconditional iff Disjunction implies Conjunction
+
+\begin{theorem}
+: $p \iff q \vdash \left({p \lor q}\right) \implies \left({p \land q}\right)$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|p \iff q \vdash \left({p \lor q}\right) \implies \left({p \land q}\right)}}
+{{Premise|1|p \iff q}}
+{{SequentIntro|2|1|\left({p \land q}\right) \lor \left({\neg p \land \neg q}\right)|1|[[Biconditional as Disjunction of Conjunctions]]}}
+{{DeMorgan|3|1|\left({p \land q}\right) \lor \neg \left({p \lor q}\right)|2|Conjunction of Negations}}
+{{Commutation|4|1|\neg \left({p \lor q}\right) \lor \left({p \land q}\right)|3|Disjunction}}
+{{SequentIntro|5|1|\left({p \lor q}\right) \implies \left({p \land q}\right)|4|[[Rule of Material Implication]]}}
+{{EndTableau}}
+{{qed}}
+[[Category:Biconditional iff Disjunction implies Conjunction]]
+bjqipywjtsnh5w8bf5072djd3n8usvn
+\end{proof}<|endoftext|>
+\section{Biconditional iff Disjunction implies Conjunction/Formulation 1/Reverse Implication}
+Tags: Biconditional iff Disjunction implies Conjunction
+
+\begin{theorem}
+: $\left({p \lor q}\right) \implies \left({p \land q}\right) \vdash p \iff q$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\left({p \lor q}\right) \implies \left({p \land q}\right) \vdash p \iff q}}
+{{Premise|1|\left({p \lor q}\right) \implies \left({p \land q}\right)}}
+{{SequentIntro|2|1|\neg \left({p \lor q}\right) \lor \left({p \land q}\right)|1|[[Rule of Material Implication]]}}
+{{Commutation|3|1|\left({p \land q}\right) \lor \neg \left({p \lor q}\right)|2|Disjunction}}
+{{DeMorgan|4|1|\left({p \land q}\right) \lor \left({\neg p \land \neg q}\right)|3|Conjunction of Negations}}
+{{SequentIntro|5|1|\left ({p \implies q}\right) \land \left ({q \implies p}\right)|4
+  |[[Biconditional as Disjunction of Conjunctions]]}}
+{{SequentIntro|6|1|p \iff q|5|[[Rule of Material Equivalence]]}}
+{{EndTableau}}
+{{qed}}
+{{LEM|Rule of Material Implication}}
+[[Category:Biconditional iff Disjunction implies Conjunction]]
+fs5pn4li3zbnvspkihv62321uytc5kk
+\end{proof}<|endoftext|>
+\section{Biconditional iff Disjunction implies Conjunction/Formulation 1}
+Tags: Biconditional iff Disjunction implies Conjunction
+
+\begin{theorem}
+:$p \iff q \dashv \vdash \paren {p \lor q} \implies \paren {p \land q}$
+\end{theorem}<|endoftext|>
+\section{Biconditional iff Disjunction implies Conjunction/Formulation 2}
+Tags: Biconditional iff Disjunction implies Conjunction
+
+\begin{theorem}
+: $\vdash \left({p \iff q}\right) \iff \left({\left({p \lor q}\right) \implies \left({p \land q}\right)}\right)$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\vdash \left({p \iff q}\right) \iff \left({\left({p \lor q}\right) \implies \left({p \land q}\right)}\right)}}
+{{Assumption|1|p \iff q}}
+{{SequentIntro|2|1|\left({p \lor q}\right) \implies \left({p \land q}\right)|1|[[Biconditional iff Disjunction implies Conjunction/Formulation 1|Biconditional iff Disjunction implies Conjunction: Formulation 1]]}}
+{{Implication|3||\left({p \iff q}\right) \implies \left({\left({p \lor q}\right) \implies \left({p \land q}\right)}\right)|1|2}}
+{{Assumption|4|\left({p \lor q}\right) \implies \left({p \land q}\right)}}
+{{SequentIntro|5|4|p \iff q|4|[[Biconditional iff Disjunction implies Conjunction/Formulation 1|Biconditional iff Disjunction implies Conjunction: Formulation 1]]}}
+{{Implication|6||\left({\left({p \lor q}\right) \implies \left({p \land q}\right)}\right) \implies \left({p \iff q}\right)|4|5}}
+{{BiconditionalIntro|7||\left({p \iff q}\right) \iff \left({\left({p \lor q}\right) \implies \left({p \land q}\right)}\right)|3|6}}
+{{EndTableau}}
+{{qed}}
+[[Category:Biconditional iff Disjunction implies Conjunction]]
+nd5xg35tyu3oguip3lvunbq029yqv73
+\end{proof}<|endoftext|>
+\section{Biconditional iff Disjunction implies Conjunction}
+Tags: Biconditional, Disjunction, Conjunction
+
+\begin{theorem}
+==== [[Biconditional iff Disjunction implies Conjunction/Formulation 1|Formulation 1]] ====
+{{:Biconditional iff Disjunction implies Conjunction/Formulation 1}}
+==== [[Biconditional iff Disjunction implies Conjunction/Formulation 2|Formulation 2]] ====
+{{:Biconditional iff Disjunction implies Conjunction/Formulation 2}}
+\end{theorem}<|endoftext|>
+\section{Biconditional is Transitive/Formulation 1/Proof 2}
+Tags: Biconditional is Transitive, Truth Table Proofs
+
+\begin{theorem}
+:$p \iff q, q \iff r \vdash p \iff r$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]].
+As can be seen for all [[Definition:Boolean Interpretation|boolean interpretations]] by inspection, where the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] on the {{LHS}} is $T$, that under the one on the {{RHS}} is also $T$:
+$\begin{array}{|ccccccc||ccc|} \hline
+(p & \iff & q) & \land & (q & \iff & r) & p & \iff & r \\
+\hline
+F & T & F & T & F & T & F & F & T & F \\
+F & T & F & F & F & F & T & F & F & T \\
+F & F & T & F & T & F & F & F & T & F \\
+F & F & T & F & T & T & T & F & F & T \\
+T & F & F & F & F & T & F & T & F & F \\
+T & F & F & F & F & F & T & T & T & T \\
+T & T & T & F & T & F & F & T & F & F \\
+T & T & T & T & T & T & T & T & T & T \\
+\hline
+\end{array}$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Biconditional is Transitive/Formulation 1/Proof 1}
+Tags: Biconditional is Transitive
+
+\begin{theorem}
+: $p \iff q, q \iff r \vdash p \iff r$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|p \iff q, q \iff r \vdash p \iff r}}
+{{Premise|1|p \iff q}}
+{{Premise|2|q \iff r}}
+{{BiconditionalElimination|3|1|p \implies q|1|1}}
+{{BiconditionalElimination|4|2|q \implies r|2|1}}
+{{SequentIntro|5|1, 2|p \implies r|1, 2|[[Hypothetical Syllogism/Formulation 1|Hypothetical Syllogism: Formulation 1]]}}
+{{BiconditionalElimination|6|1|q \implies p|1|2}}
+{{BiconditionalElimination|7|2|r \implies q|2|2}}
+{{SequentIntro|8|1, 2|r \implies p|7, 6|[[Hypothetical Syllogism/Formulation 1|Hypothetical Syllogism: Formulation 1]]}}
+{{BiconditionalIntro|9|1, 2|p \iff r|5|8}}
+{{EndTableau}}
+\end{proof}<|endoftext|>
+\section{Order of Squares in Totally Ordered Ring without Proper Zero Divisors}
+Tags: Totally Ordered Rings
+
+\begin{theorem}
+Let $\struct {R, +, \circ, \le}$ be a [[Definition:Totally Ordered Ring|totally ordered ring]] without [[Definition:Proper Zero Divisor|proper zero divisors]] whose [[Definition:Ring Zero|zero]] is $0_R$.
+Let $x, y \in R$ be [[Definition:Positive|positive]], that is, $0_R \le x, y$.
+Then $x \le y \iff x \circ x \le y \circ y$.
+That is, the [[Definition:Square Mapping|square mapping]] is an [[Definition:Order Embedding|order embedding]] of $\struct {R_{\ge 0}, \le}$ into itself.
+When $R$ is one of the standard [[Definition:Number|sets of numbers]] $\Z, \Q, \R$, then this translates into:
+:If $x, y$ are [[Definition:Positive Number|positive]], then $x \le y \iff x^2 \le y^2$.
+\end{theorem}
+
+\begin{proof}
+From [[Order of Squares in Ordered Ring]], we have:
+:$x \le y \implies x \circ x \le y \circ y$
+To prove the opposite [[Definition:Implication|implication]], we use a [[Proof by Contradiction]].
+{{AimForCont}} $x \circ x \le y \circ y$ but $x \not\le y$.
+Since $\le$ is a [[Definition:Total Ordering|total ordering]], this means $y < x$.
+Since $0_R \le y$, [[Extended Transitivity]] shows that $0_R < x$, so in particular $x ≠ 0_R$.
+As $\le$ is [[Definition:Ordering Compatible with Ring Structure|compatible]] with the [[Definition:Ring (Abstract Algebra)|ring]] structure $\struct {R, +, \circ}$ and $x$ and $y$ are both [[Definition:Positive|positive]], we have:
+:$y \circ y \le y \circ x \le x \circ x$
+Since we assume $x \circ x \le y \circ y$, [[Ordering Cycle implies Equality]] shows that:
+:$y \circ y = x \circ x = y \circ x$
+Therefore:
+{{begin-eqn}}
+{{eqn | l = 0_R
+      | r = x \circ x + \paren {-\paren {y \circ x} }
+      | c = {{Defof|Ring Negative}}
+}}
+{{eqn | r = x \circ x + \paren {-y} \circ x
+      | c = [[Product with Ring Negative]]
+}}
+{{eqn | r = \paren {x + \paren {-y} } \circ x
+      | c = {{Ring-axiom|D}}
+}}
+{{end-eqn}}
+Since $x \ne y$, we have that $x + \paren {-y} \ne 0_R$; otherwise, by {{Ring-axiom|A1}}, we would have $x = \paren {x + \paren {-y} } + y = y$.
+We have already shown that $x \ne 0_R$.
+But then $x$ and $x + \paren {-y}$ are [[Definition:Proper Zero Divisor|proper zero divisors]] of $R$, [[Definition:Contradiction|contradicting]] the [[Definition:Hypothesis|hypothesis]] that $R$ has no [[Definition:Proper Zero Divisor|proper zero divisors]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Order of Squares in Ordered Field}
+Tags: Ordered Fields
+
+\begin{theorem}
+Let $\struct {R, +, \circ, \le}$ be an [[Definition:Ordered Field|ordered field]] whose [[Definition:Field Zero|zero]] is $0_R$ and whose [[Definition:Unity of Field|unity]] is $1_R$.
+Suppose that $\forall a \in R: 0 < a \implies 0 < a^{-1}$.
+Let $x, y \in \struct {R, +, \circ, \le}$ such that $0_R \le x, y$.
+Then $x \le y \iff x \circ x \le y \circ y$.
+That is, the square function is an [[Definition:Order Embedding|order embedding]] of $\struct {R_{\ge 0}, \le}$ into itself.
+When $R$ is one of the [[Definition:Standard Number Field|standard fields of numbers]] $\Q$ and $\R$, then this translates into:
+:If $x, y$ are [[Definition:Positive|positive]] then $x \le y \iff x^2 \le y^2$.
+\end{theorem}
+
+\begin{proof}
+From [[Order of Squares in Ordered Ring]], we have:
+:$x \le y \implies x \circ x \le y \circ y$
+To prove the reverse implication, suppose that $x \circ x \le y \circ y$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = x \circ x
+      | o = \le
+      | r = y \circ y
+      | c = 
+}}
+{{eqn | ll=\leadsto
+      | l = x \circ x + \paren {-\paren {x \circ x} }
+      | o = \le
+      | r = y \circ y + \paren {-\paren {x \circ x} }
+      | c = 
+}}
+{{eqn | ll=\leadsto
+      | l = 0_R
+      | o = \le
+      | r = y \circ y + \paren {-\paren {x \circ x} }
+      | c = 
+}}
+{{eqn | ll=\leadsto
+      | l = 0_R
+      | o = \le
+      | r = \paren {y + \paren {-x} } \circ \paren {y + x}
+      | c = [[Difference of Two Squares]], which applies because a field is a [[Definition:Commutative Ring|commutative ring]].
+}}
+{{end-eqn}}
+As $0_R \le x, y$ we have $0_R \le x + y$.
+Hence by the premise we have $0_R \le \paren {x + y}^{-1}$.
+So as $0_R \le \paren {y + \paren {-x} } \circ \paren {y + x}$ we can multiply both sides by $\paren {x + y}^{-1}$ and get $0_R \le \paren {y + \paren {-x} }$.
+Adding $-x$ to both sides gives us $x \le y$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Law of Identity/Formulation 1/Proof 1}
+Tags: Law of Identity
+
+\begin{theorem}
+: $p \vdash p$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|p \vdash p}}
+{{Premise|1|p}}
+{{EndTableau}}
+{{qed}}
+This is the shortest [[Definition:Tableau Proof (Formal Systems)|tableau proof]] possible.
+\end{proof}<|endoftext|>
+\section{Law of Identity/Formulation 1}
+Tags: Law of Identity
+
+\begin{theorem}
+:$p \vdash p$
+\end{theorem}<|endoftext|>
+\section{Law of Identity/Formulation 1/Proof 2}
+Tags: Law of Identity, Truth Table Proofs
+
+\begin{theorem}
+: $p \vdash p$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]] (trivially) to the proposition.
+$\begin{array}{|c|c|} \hline
+p & p \\
+\hline
+F & F \\
+T & T \\
+\hline
+\end{array}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Law of Identity/Formulation 2/Proof 1}
+Tags: Law of Identity
+
+\begin{theorem}
+: $\vdash p \implies p$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\vdash p \implies p}}
+{{Premise|1|p}}
+{{Implication|2||p \implies p|1|1}}
+{{EndTableau}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Law of Identity/Formulation 2/Proof 2}
+Tags: Law of Identity, Truth Table Proofs
+
+\begin{theorem}
+: $\vdash p \implies p$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]] to the proposition.
+As can be seen by inspection, the [[Definition:Truth Value|truth value]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] is $T$ throughout.
+$\begin{array}{|ccc|} \hline
+p & \implies & p \\
+\hline
+F & T & F \\
+T & T & T \\
+\hline
+\end{array}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Law of Identity/Formulation 2}
+Tags: Law of Identity
+
+\begin{theorem}
+Every [[Definition:Proposition|proposition]] entails itself:
+:$\vdash p \implies p$
+\end{theorem}<|endoftext|>
+\section{Vandermonde Determinant/Alternative Formulations}
+Tags: Vandermonde Determinant
+
+\begin{theorem}
+The '''Vandermonde determinant of order $n$''' is the [[Definition:Determinant of Matrix|determinant]] defined as follows:
+:$V_n = \begin{vmatrix}
+     1 &    x_1 &  x_1^2 & \cdots & x_1^{n - 2} & x_1^{n - 1} \\
+     1 &    x_2 &  x_2^2 & \cdots & x_2^{n - 2} & x_2^{n - 1} \\
+\vdots & \vdots & \vdots & \ddots &      \vdots &      \vdots \\
+     1 &    x_n &  x_n^2 & \cdots & x_n^{n - 2} & x_n^{n - 1}
+\end{vmatrix}$
+Its value is given by:
+:$\displaystyle V_n = \prod_{1 \mathop \le i \mathop < j \mathop \le n} \paren {x_j - x_i}$
+\end{theorem}<|endoftext|>
+\section{Neighborhood Condition for Coarser Topology}
+Tags: Topology
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\tau_1$ and $\tau_2$ be two [[Definition:Topology|topologies]] on $S$.
+Suppose that for all $z \in S$ and for all [[Definition:Open Neighborhood|open]] [[Definition:Neighborhood of Point|neighborhoods]] $N_z$ of $z$ with respect to $\tau_1$, there exists $U \in \tau_2$ such that $U \subseteq N_z$.
+Then $\tau_1$ is [[Definition:Coarser Topology|coarser]] than $\tau_2$.
+\end{theorem}
+
+\begin{proof}
+Let $V \in \tau_1$.
+For all $z \in V$, we have that $V$ is an [[Definition:Open Neighborhood|open]] [[Definition:Neighborhood of Point|neighborhood]] of $z$ with respect to $\tau_1$.
+Then for all $z \in V$, we can find $U_z \in \tau_2$ such that $z \in U_z \subseteq V$.
+Then:
+:$\displaystyle V = \bigcup_{z \mathop \in V } \left\{ {z}\right\} \subseteq \bigcup_{z \mathop \in V } U_z \subseteq V$
+By definition of [[Definition:Set Equality/Definition 2|set equality]], it follows that:
+: $\displaystyle V = \bigcup_{z \mathop \in V} U_z \in \tau_2$
+Hence:
+: $\tau_1 \subseteq \tau_2$
+which by definition means that $\tau_1$ is [[Definition:Coarser Topology|coarser]] than $\tau_2$.
+{{qed}}
+[[Category:Topology]]
+btap2mwfh280od8bb1mk6thrc4uzu1m
+\end{proof}<|endoftext|>
+\section{Absorption Laws (Logic)/Conjunction Absorbs Disjunction/Proof 2}
+Tags: Absorption Laws (Logic)
+
+\begin{theorem}
+:$p \land \left({p \lor q}\right) \dashv \vdash p$
+\end{theorem}
+
+\begin{proof}
+By calculation:
+{{begin-eqn}}
+{{eqn | l=p \land \left({p \lor q}\right)
+      | r=\left({p \lor \bot}\right) \land \left({p \lor q}\right)
+      | c=[[Disjunction with Contradiction]]
+}}
+{{eqn | r=p \lor \left({\bot \land q}\right)
+      | c=[[Disjunction is Left Distributive over Conjunction]]
+}}
+{{eqn | r=p \lor \bot
+      | c=[[Conjunction with Contradiction]]
+}}
+{{eqn | r=p
+      | c=[[Disjunction with Contradiction]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Absorption Laws (Logic)/Conjunction Absorbs Disjunction/Forward Implication}
+Tags: Absorption Laws (Logic)
+
+\begin{theorem}
+:$p \land \left({p \lor q}\right) \vdash p$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|p \land \left({p \lor q}\right) \vdash p}}
+{{Premise|1|p \land \left({p \lor q}\right)}}
+{{Simplification|2|1|p|1|1}}
+{{EndTableau}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Absorption Laws (Logic)/Conjunction Absorbs Disjunction/Reverse Implication}
+Tags: Absorption Laws (Logic)
+
+\begin{theorem}
+:$p \vdash p \land \left({p \lor q}\right)$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|p \vdash p \land \left({p \lor q}\right)}}
+{{Premise|1|p}}
+{{Addition|2|1|p \lor q|1|1}}
+{{Conjunction|3|1|p \land \left({p \lor q}\right)|1|2}}
+{{EndTableau}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Absorption Laws (Logic)/Conjunction Absorbs Disjunction}
+Tags: Absorption Laws (Logic)
+
+\begin{theorem}
+:$p \land \paren {p \lor q} \dashv \vdash p$
+\end{theorem}<|endoftext|>
+\section{Absorption Laws (Logic)/Disjunction Absorbs Conjunction/Proof 1}
+Tags: Truth Table Proofs, Absorption Laws (Logic)
+
+\begin{theorem}
+:$p \lor \left ({p \land q}\right) \dashv \vdash p$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]].
+As can be seen by inspection, the appropriate [[Definition:Truth Value|truth values]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
+$\begin{array}{|ccccc||c|} \hline
+p & \lor & (p & \land & q) & p \\
+\hline
+F & F & F & F & F & F \\
+F & F & F & F & T & F \\
+T & T & T & F & F & T \\
+T & T & T & T & T & T \\
+\hline
+\end{array}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Absorption Laws (Logic)/Disjunction Absorbs Conjunction/Proof 2}
+Tags: Absorption Laws (Logic)
+
+\begin{theorem}
+:$p \lor \left ({p \land q}\right) \dashv \vdash p$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l=p \lor \left({p \land q}\right)
+      | r=\left({p \land \top}\right) \lor \left({p \land q}\right)
+      | c=[[Conjunction with Tautology]]
+}}
+{{eqn | r=p \land \left({\top \lor q}\right)
+      | c=[[Conjunction is Left Distributive over Disjunction]]
+}}
+{{eqn | r=p \land \top
+      | c=[[Disjunction with Tautology]]
+}}
+{{eqn | r=p
+      | c=[[Conjunction with Tautology]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Absorption Laws (Logic)/Disjunction Absorbs Conjunction/Forward Implication}
+Tags: Absorption Laws (Logic)
+
+\begin{theorem}
+:$p \lor \paren {p \land q} \vdash p$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|p \lor \paren {p \land q} \vdash p}}
+{{Premise|1|p \lor \paren {p \land q} }}
+{{Assumption|2|p}}
+{{Assumption|3|p \land q}}
+{{Simplification|4|3|p|3|1}}
+{{ProofByCases|5|1|p|1|2|2|3|4}}
+{{EndTableau}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Absorption Laws (Logic)/Disjunction Absorbs Conjunction/Reverse Implication}
+Tags: Absorption Laws (Logic)
+
+\begin{theorem}
+:$p \vdash p \lor \paren {p \land q}$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|p \vdash p \lor \paren {p \land q} }}
+{{Premise|1|p}}
+{{Addition|2|1|p \lor \paren {p \land q}|1|1}}
+{{EndTableau}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Absorption Laws (Logic)/Disjunction Absorbs Conjunction}
+Tags: Absorption Laws (Logic)
+
+\begin{theorem}
+:$p \lor \left ({p \land q}\right) \dashv \vdash p$
+\end{theorem}<|endoftext|>
+\section{Extended Transitivity}
+Tags: Transitive Relations, Reflexive Closures
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\RR$ be a [[Definition:Transitive Relation|transitive relation]] on $S$.
+Let $\RR^=$ be the [[Definition:Reflexive Closure|reflexive closure]] of $\RR$.
+Let $a, b, c \in S$.
+Then:
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = \paren {a \mathrel \RR b} \land \paren {b \mathrel \RR c}
+      | o = \implies
+      | r = a \mathrel \RR c
+}}
+{{eqn | n = 2
+      | l = \paren {a \mathrel \RR b} \land \paren {b \mathrel {\RR^=} c}
+      | o = \implies
+      | r = a \mathrel \RR c
+}}
+{{eqn | n = 3
+      | l = \paren {a \mathrel {\RR^=} b} \land \paren {b \mathrel \RR c}
+      | o = \implies
+      | r = a \mathrel \RR c
+}}
+{{eqn | n = 4
+      | l = \paren {a \mathrel {\RR^=} b} \land \paren {b \mathrel {\RR^=} c}
+      | o = \implies
+      | r = a \mathrel {\RR^=} c
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+$(1)$ follows from the definition of a [[Definition:Transitive Relation|transitive relation]].
+$(4)$ follows from [[Reflexive Closure of Transitive Relation is Transitive]].
+Suppose that:
+:$\paren {a \mathrel \RR b} \land \paren {b \mathrel {\RR^=} c}$
+By the definition of [[Definition:Reflexive Closure|reflexive closure]]:
+:$b \mathrel \RR c$ or $b = c$
+If $b = c$, then since $a \mathrel \RR b$:
+:$a \mathrel \RR c$
+If $b \mathrel \RR c$ then by [[Definition:Transitive Relation|transitivity]] of $\RR$:
+:$a \mathrel \RR c$
+Thus $(2)$ holds.
+A similar argument proves that $(3)$ holds as well:
+Suppose that:
+:$\paren {a \mathrel {\RR^=} b} \land \paren {b \mathrel \RR c}$
+By the definition of [[Definition:Reflexive Closure|reflexive closure]]:
+:$a \mathrel \RR b$ or $a = b$
+If $a = b$, then since $b \mathrel \RR c$:
+:$a \mathrel \RR c$
+If $a \mathrel \RR b$ then by [[Definition:Transitive Relation|transitivity]] of $\RR$:
+:$a \mathrel \RR c$
+Thus $(3)$ holds.
+{{qed}}
+[[Category:Transitive Relations]]
+[[Category:Reflexive Closures]]
+grgr98m0zamvw910geicyuj15dm47uy
+\end{proof}<|endoftext|>
+\section{Rule of Idempotence/Conjunction}
+Tags: Rule of Idempotence
+
+\begin{theorem}
+{{:Rule of Idempotence/Conjunction/Formulation 1}}
+\end{theorem}<|endoftext|>
+\section{Rule of Idempotence/Disjunction}
+Tags: Rule of Idempotence
+
+\begin{theorem}
+{{:Rule of Idempotence/Disjunction/Formulation 1}}
+\end{theorem}<|endoftext|>
+\section{Rule of Idempotence/Disjunction/Formulation 1/Proof}
+Tags: Truth Table Proofs, Rule of Idempotence
+
+\begin{theorem}
+: $p \dashv \vdash p \lor p$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]].
+As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for each [[Definition:Boolean Interpretation|boolean interpretations]].
+$\begin{array}{|c||ccc|} \hline
+p & p & \lor & p \\
+\hline
+T & T & T & T \\
+F & F & F & F \\
+\hline
+\end{array}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rule of Idempotence/Disjunction/Formulation 1/Forward Implication}
+Tags: Rule of Idempotence
+
+\begin{theorem}
+: $p \vdash p \lor p$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|p \vdash p \lor p}}
+{{Premise|1|p}}
+{{Addition|2|1|p \lor p|1|1}}
+{{EndTableau}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rule of Idempotence/Disjunction/Formulation 1/Reverse Implication}
+Tags: Rule of Idempotence
+
+\begin{theorem}
+: $p \lor p \vdash p$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|p \lor p \vdash p}}
+{{Premise|1|p \lor p}}
+{{Assumption|2|p}}
+{{ProofByCases|3|1|p|1|2|2|2|2}}
+{{EndTableau}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rule of Idempotence/Disjunction/Formulation 1}
+Tags: Rule of Idempotence
+
+\begin{theorem}
+: $p \dashv \vdash p \lor p$
+\end{theorem}<|endoftext|>
+\section{Rule of Idempotence/Disjunction/Formulation 2}
+Tags: Rule of Idempotence
+
+\begin{theorem}
+: $\vdash p \iff \paren {p \lor p}$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau       |p \iff \paren {p \lor p} }}
+{{Assumption         | 1|     p}}
+{{Addition           | 2|   1|p \lor p                    |1|1}}
+{{Implication        | 3|    |p \implies \paren {p \lor p}|1|2}}
+{{Assumption         | 4|     p \lor p}}
+{{Assumption         | 5|     \neg p}}
+{{ModusTollendoPonens| 6|4, 5|p                           |4|5|1}}
+{{NonContradiction   | 7|4, 5|                             6|5}}
+{{Contradiction      | 8|   5|p                           |5|7}}
+{{Implication        | 9|    |\paren {p \lor p} \implies p|4|8}}
+{{BiconditionalIntro |10|    |p \iff \paren {p \lor p}    |3|9}}
+{{EndTableau}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rule of Idempotence/Conjunction/Formulation 1/Proof}
+Tags: Rule of Idempotence, Truth Table Proofs
+
+\begin{theorem}
+: $p \dashv \vdash p \land p$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]].
+As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for each [[Definition:Boolean Interpretation|boolean interpretations]].
+$\begin{array}{|c||ccc|} \hline
+p & p & \land & p \\
+\hline
+T & T & T & T \\
+F & F & F & F \\
+\hline
+\end{array}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rule of Idempotence/Conjunction/Formulation 1/Forward Implication}
+Tags: Rule of Idempotence
+
+\begin{theorem}
+: $p \vdash p \land p$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|p \vdash p \land p}}
+{{Premise|1|p}}
+{{Conjunction|2|1|p \land q|1|1}}
+{{EndTableau}}
+{{qed}}
+[[Category:Rule of Idempotence]]
+blc5sulas1kx4ojnz6z3l99t5mwqqis
+\end{proof}<|endoftext|>
+\section{Rule of Idempotence/Conjunction/Formulation 1/Reverse Implication}
+Tags: Rule of Idempotence
+
+\begin{theorem}
+: $p \land p \vdash p$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|p \land p \vdash p}}
+{{Premise|1|p \land p}}
+{{Simplification|2|1|p|1|1}}
+{{EndTableau}}
+{{qed}}
+[[Category:Rule of Idempotence]]
+fegoi2u2sshc3pgst3ks1onazak1ubx
+\end{proof}<|endoftext|>
+\section{Rule of Idempotence/Conjunction/Formulation 1}
+Tags: Rule of Idempotence
+
+\begin{theorem}
+: $p \dashv \vdash p \land p$
+\end{theorem}<|endoftext|>
+\section{Rule of Idempotence/Conjunction/Formulation 2}
+Tags: Rule of Idempotence
+
+\begin{theorem}
+:$\vdash p \iff \paren {p \land p}$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\vdash p \iff \paren {p \land p} }}
+{{Assumption|1|p}}
+{{Conjunction|2|1|p \land p|1|1}}
+{{Implication|3||p \implies \paren {p \land p}|1|2}}
+{{Assumption|4|p \land p}}
+{{Simplification|5|4|p|4|1}}
+{{Implication|6||\paren {p \land p} \implies p|4|5}}
+{{BiconditionalIntro|7||p \iff \paren {p \land p}|3|6}}
+{{EndTableau}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ordering Cycle implies Equality}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\left({S,\preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Let $x_1$, $x_2$, and $x_3$ be elements of $S$.
+Suppose that
+{{begin-eqn}}
+{{eqn|l = x_1
+     |o = \preceq
+     |r = x_2
+}}
+{{eqn|l = x_2
+     |o = \preceq
+     |r = x_3
+}}
+{{eqn|l = x_3
+     |o = \preceq
+     |r = x_1
+}}
+{{end-eqn}}
+Then $x_1 = x_2 = x_3$.
+\end{theorem}
+
+\begin{proof}
+Since $\preceq$ is an [[Definition:Ordering|ordering]], it is [[Definition:Transitive Relation|transitive]] and [[Definition:Antisymmetric Relation|antisymmetric]].
+By transitivity, $x_1 \preceq x_3$.
+Since $x_1 \preceq x_3$ and $x_3 \preceq x_1$, [[Definition:Antisymmetric Relation|antisymmetry]] allows us to conclude that $x_1 = x_3$.
+Since $x_1 = x_3$ and $x_1 \preceq x_2$, we must have $x_3 \preceq x_2$.
+Since $x_3 \preceq x_2$ and $x_2 \preceq x_3$, antisymmetry allows us to conclude that $x_2 = x_3$.
+Thus $x_1 = x_2 = x_3$.
+{{qed}}
+[[Category:Order Theory]]
+izdvezcs9rfg6c687d0ieugpd04qf8h
+\end{proof}<|endoftext|>
+\section{Euclidean Topology is Tychonoff Topology}
+Tags: Euclidean Space
+
+\begin{theorem}
+Let $T_1 = \left({\R, \tau_1}\right)$ be the [[Definition:Topological Space|topological space]] such that $\tau_1$ is the [[Definition:Euclidean Topology|Euclidean topology]] on $\R$.
+Let $T_n = \left({\R^n, \tau_n}\right)$ be the [[Definition:Topological Space|topological space]] such that $\tau_n$ is the [[Definition:Tychonoff Topology|Tychonoff topology]] on the [[Definition:Cartesian Product|cartesian product]] $\displaystyle \R_n = \prod_{i \mathop = 1}^n \R$.
+Then the [[Definition:Euclidean Topology|Euclidean topology]] on $\R^n$ and the [[Definition:Tychonoff Topology|Tychonoff topology]] on $\R^n$ are the same.
+\end{theorem}
+
+\begin{proof}
+Denote the [[Definition:Euclidean Topology|Euclidean topology]] on $\R^n$ as $\tau$, and denote the [[Definition:Tychonoff Topology|Tychonoff topology]] on $\R^n$ as $\tau'$.
+Let $U \in \tau$, and let $x = \tuple{x_1, \ldots, x_n} \in U$.
+Then there exists $\epsilon \in \R_{>0}$ such that the [[Definition:Open Ball|open ball]] $\map {B_\epsilon} x \subseteq U$.
+We show that:
+:$\displaystyle B' = \prod_{i \mathop = 1}^n \left({x_i - \dfrac \epsilon n \,.\,.\, x_i + \dfrac \epsilon n}\right) \subseteq \map {B_\epsilon} x$
+For if $y = \tuple{y_1, \ldots, y_n} \in B'$, then:
+{{begin-eqn}}
+{{eqn | l = \map d {x, y}
+      | r = \paren{\sum_{i \mathop = 1}^n \size{x_i - y_i}^2}^{1/2}
+      | c = where $d$ denotes the [[Definition:Euclidean Metric/Real Vector Space|Euclidean metric]]
+}}
+{{eqn | o = \le
+      | r = \sum_{i \mathop = 1}^n \size{x_i - y_i}
+      | c = by [[Minkowski's Inequality for Sums]]
+}}
+{{eqn | o = <
+      | r = \sum_{i \mathop = 1}^n \dfrac \epsilon n
+}}
+{{eqn | r = \epsilon
+}}
+{{end-eqn}}
+From [[Natural Basis of Tychonoff Topology of Finite Product]], $B' \in \tau'$.
+As $B' \subseteq \map {B_\epsilon} x \subseteq U$, [[Neighborhood Condition for Coarser Topology]] shows that $\tau' \subseteq \tau$.
+{{qed|lemma}}
+Let $U' \in \tau'$.
+Let $x = \tuple{x_1, \ldots, x_n} \in U'$.
+From [[Natural Basis of Tychonoff Topology of Finite Product]], [[Definition:Set|sets]] of the type $\displaystyle \prod_{i \mathop = 1}^n U'_i$ with $U'_i \in \tau_1$ form an [[Definition:Analytic Basis|analytic basis]] for $\tau'$.
+From [[Equivalence of Definitions of Analytic Basis]], it follows that we can select $U'_1, \ldots, U'_n \in \tau_1$ such that $\displaystyle x \in \prod_{i \mathop = 1}^n U'_i \subseteq U'$.
+By [[Definition:Open Set (Real Analysis)|definition of open set]], it follows that for all $i \in \set {1, \ldots, n}$, we can find $\epsilon_i \in \R_{>0}$ such that $\openint {x_i - \epsilon_i} {x_i + \epsilon_i} \subseteq U'_i$.
+Put $\epsilon = \min \set{\epsilon_i : i = 1, \ldots, n}$.
+We show that the [[Definition:Open Ball|open ball]] $\map {B_\epsilon} x \subseteq U'$.
+For if $y = \tuple{y_1, \ldots, y_n} \in \map {B_\epsilon} x$, then $y_i \in \openint {x_i - \epsilon_i} {x_i + \epsilon_i }$, as:
+:$\size{x_i - y_i} < \epsilon \le \epsilon_i$
+It follows that:
+:$\displaystyle \map {B_\epsilon} x \subseteq \prod_{i \mathop = 1}^n \openint {x_i - \epsilon_i} {x_i + \epsilon_i } \subseteq \prod_{i \mathop = 1}^n U'_i \subseteq U'$
+Then [[Neighborhood Condition for Coarser Topology]] shows that $\tau \subseteq \tau'$.
+{{qed|lemma}}
+It follows that $\tau = \tau'$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Modus Ponendo Tollens/Variant/Formulation 1/Proof}
+Tags: Modus Ponendo Tollens, Truth Table Proofs
+
+\begin{theorem}
+:$\neg \paren {p \land q} \dashv \vdash p \implies \neg q$
+\end{theorem}
+
+\begin{proof}
+As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
+$\begin{array}{|cccc||cccc|} \hline
+\neg & (p & \land & q) & p & \implies & \neg & q \\
+\hline
+T & F & F & F & F & T & T & F \\
+T & F & F & T & F & T & F & T \\
+T & T & F & F & T & T & T & F \\
+F & T & T & T & T & F & F & T \\
+\hline
+\end{array}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Modus Ponendo Tollens/Variant/Formulation 1/Forward Implication}
+Tags: Modus Ponendo Tollens
+
+\begin{theorem}
+: $\neg \left({p \land q}\right) \vdash p \implies \neg q$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\neg \left({p \land q}\right) \vdash p \implies \neg q}}
+{{Premise|1|\neg \left({p \land q}\right)}}
+{{Assumption|2|p}}
+{{Assumption|3|q}}
+{{Conjunction|4|2, 3|p \land q|2|3}}
+{{NonContradiction|5|1, 2, 5|4|1}}
+{{Contradiction|6|1, 2|\neg q|3|5}}
+{{Implication|7|1|p \implies \neg q|2|6}}
+{{EndTableau}}
+{{qed}}
+[[Category:Modus Ponendo Tollens]]
+onntrxn3bw34e75yrimn9eni1iididw
+\end{proof}<|endoftext|>
+\section{Modus Ponendo Tollens/Variant/Formulation 1/Reverse Implication}
+Tags: Modus Ponendo Tollens
+
+\begin{theorem}
+:$p \implies \neg q \vdash \neg \paren {p \land q}$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|p \implies \neg q \vdash \neg \paren {p \land q} }}
+{{Premise|1|p \implies \neg q}}
+{{Assumption|2|p \land q|Assume the opposite of what is to be proved ...}}
+{{Simplification|3|2|p|2|1}}
+{{Simplification|4|2|q|2|2}}
+{{ModusPonens|5|1, 2|\neg q|1|3}}
+{{NonContradiction|6|1, 2|4|5|... and demonstrate a contradiction}}
+{{Contradiction|7|1|\neg \paren {p \land q}|2|6}}
+{{EndTableau|qed}}
+[[Category:Modus Ponendo Tollens]]
+5z8tpfk79jnx16ntyyx18hj42odr2ag
+\end{proof}<|endoftext|>
+\section{Modus Ponendo Tollens/Variant/Formulation 1}
+Tags: Modus Ponendo Tollens
+
+\begin{theorem}
+: $\neg \left({p \land q}\right) \dashv \vdash p \implies \neg q$
+\end{theorem}<|endoftext|>
+\section{Modus Ponendo Tollens/Variant/Formulation 2}
+Tags: Modus Ponendo Tollens
+
+\begin{theorem}
+: $\vdash \left({\neg \left({p \land q}\right)}\right) \iff \left({p \implies \neg q}\right)$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\vdash \left({\neg \left({p \land q}\right)}\right) \iff \left({p \implies \neg q}\right)}}
+{{Assumption|1|\neg \left({p \land q}\right)}}
+{{SequentIntro|2|1|p \implies \neg q|1|[[Modus Ponendo Tollens/Variant/Formulation 1/Forward Implication|Modus Ponendo Tollens: Formulation 1: Forward Implication]]}}
+{{Implication|3||\left({\neg \left({p \land q}\right)}\right) \implies \left({p \implies \neg q}\right)|1|2}}
+{{Assumption|4|p \implies \neg q}}
+{{SequentIntro|5|4|\neg \left({p \land q}\right)|4|[[Modus Ponendo Tollens/Variant/Formulation 1/Reverse Implication|Modus Ponendo Tollens: Formulation 1: Reverse Implication]]}}
+{{Implication|6||\left({p \implies \neg q}\right) \implies \left({\neg \left({p \land q}\right)}\right)|4|5}}
+{{BiconditionalIntro|7||\left({\neg \left({p \land q}\right)}\right) \iff \left({p \implies \neg q}\right)|3|6}}
+{{EndTableau}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Implication Equivalent to Negation of Conjunction with Negative/Formulation 1/Reverse Implication}
+Tags: Implication Equivalent to Negation of Conjunction with Negative
+
+\begin{theorem}
+:$\neg \left({p \land \neg q}\right) \vdash p \implies q$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\neg \left({p \land \neg q}\right) \vdash p \implies q}}
+{{Premise|1|\neg \left({p \land \neg q}\right)}}
+{{DeMorgan|2|1|\neg p \lor \neg \neg q|1|Disjunction of Negations}}
+{{SequentIntro|3|1|p \implies \neg \neg q|2|[[Rule of Material Implication]]}}
+{{Assumption|4|p}}
+{{ModusPonens|5|1, 4|\neg \neg q|3|4}}
+{{DoubleNegElimination|6|1, 4|q|5}}
+{{Implication|7|1|p \implies q|4|6}}
+{{EndTableau}}
+{{qed}}
+{{LEM|Double Negation Elimination}}
+\end{proof}<|endoftext|>
+\section{Implication Equivalent to Negation of Conjunction with Negative/Formulation 1/Forward Implication}
+Tags: Implication Equivalent to Negation of Conjunction with Negative
+
+\begin{theorem}
+:$p \implies q \vdash \neg \left({p \land \neg q}\right)$
+\end{theorem}<|endoftext|>
+\section{Implication Equivalent to Negation of Conjunction with Negative/Formulation 1}
+Tags: Implication Equivalent to Negation of Conjunction with Negative
+
+\begin{theorem}
+:$p \implies q \dashv \vdash \neg \paren {p \land \neg q}$
+\end{theorem}<|endoftext|>
+\section{Implication Equivalent to Negation of Conjunction with Negative}
+Tags: Conjunction, Implication, Negation
+
+\begin{theorem}
+==== [[Implication Equivalent to Negation of Conjunction with Negative/Formulation 1|Formulation 1]] ====
+{{:Implication Equivalent to Negation of Conjunction with Negative/Formulation 1}}
+==== [[Implication Equivalent to Negation of Conjunction with Negative/Formulation 2|Formulation 2]] ====
+{{:Implication Equivalent to Negation of Conjunction with Negative/Formulation 2}}
+\end{theorem}<|endoftext|>
+\section{Zero Derivative implies Constant Complex Function}
+Tags: Complex Analysis
+
+\begin{theorem}
+Let $D \subseteq \C$ be a [[Definition:Connected Domain (Complex Analysis)|connected domain]] of $\C$.
+Let $f: D \to \C$ be a [[Definition:Complex-Differentiable Function|complex-differentiable function]].
+For all $z \in D$, let $\map {f'} z = 0$.
+Then $f$ is [[Definition:Constant Mapping|constant]] on $D$.
+\end{theorem}
+
+\begin{proof}
+Let $u, v: \set {\tuple {x, y} \in \R^2: x + i y = z \in D} \to \R$ be the two [[Definition:Real-Valued Function|real-valued functions]] defined in the [[Cauchy-Riemann Equations]]:
+:$\map u {x, y} = \map \Re {\map f z}$
+:$\map v {x, y} = \map \Im {\map f z}$
+By the [[Cauchy-Riemann Equations]]:
+:$f' = \dfrac {\partial f} {\partial x} = \dfrac {\partial u} {\partial x} + i \dfrac {\partial v} {\partial x}$
+:$f' = -i \dfrac {\partial f} {\partial y} = \dfrac {\partial v} {\partial y} - i \dfrac {\partial u} {\partial y}$
+As $f' = 0$ by assumption, this implies:
+:$0 = \dfrac {\partial u} {\partial x} = \dfrac {\partial u} {\partial y} = \dfrac {\partial v} {\partial x} = \dfrac {\partial v} {\partial y}$
+Then [[Zero Derivative implies Constant Function]] shows that $\map u {x + t, y} = \map u {x, y}$ for all $t \in \R$.
+Similar results apply for the other three [[Definition:Partial Derivative|partial derivatives]].
+Let $z, w \in D$.
+From [[Connected Domain is Connected by Staircase Contours]], it follows that there exists a [[Definition:Staircase Contour|staircase contour]] $C$ in $D$ with [[Definition:Endpoints of Contour (Complex Plane)|endpoints]] $z$ and $w$.
+The contour $C$ is a [[Definition:Concatenation of Contours (Complex Plane)|concatenation]] of [[Definition:Directed Smooth Curve (Complex Plane)|directed smooth curves]] that can be [[Definition:Parameterization of Directed Smooth Curve (Complex Plane)|parameterized]] as [[Definition:Convex Set (Vector Space)/Line Segment|line segments]] on one of these two forms:
+:$(1): \quad \map \gamma t = z_0 + t r$
+:$(2): \quad \map \gamma t = z_0 + i t r$
+for some $z_0 \in D$ and $r \in \R$ for all $t \in \closedint 0 1$.
+If $z_1 \in D$ lies on the same line segment as $z_0$, it follows that for [[Definition:Parameterization of Directed Smooth Curve (Complex Plane)|parameterizations]] of type $(1)$:
+{{begin-eqn}}
+{{eqn | l = \map f {z_1}
+      | r = \map u {z_0 + t r} + \map v {z_0 + t r}
+      | c = for some $t \in \closedint 0 1$
+}}
+{{eqn | r = \map u {z_0} + \map v {z_0}
+      | c = [[Zero Derivative implies Constant Function]]
+}}
+{{eqn | r = \map f {z_0}
+}}
+{{end-eqn}}
+Similarly, for [[Definition:Parameterization of Directed Smooth Curve (Complex Plane)|parameterizations]] of type $(2)$:
+{{begin-eqn}}
+{{eqn | l = \map f {z_1}
+      | r = \map u {z_0 + i t r} + \map v {z_0 + i t r}
+      | c = for some $t \in \closedint 0 1$
+}}
+{{eqn | r = \map f {z_0}
+      | c = [[Zero Derivative implies Constant Function]]
+}}
+{{end-eqn}}
+As the [[Definition:Image of Contour (Complex Plane)|image of $C$]] is connected by these line segments, it follows that for all $z_0$ and $z_1$ in the [[Definition:Image of Contour (Complex Plane)|image]] of $C$:
+:$\map f {z_0} = \map f {z_1}$
+In particular:
+:$\map f z = \map f w$
+so $f$ is [[Definition:Constant Mapping|constant]] on $D$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Relation between P-Product Metric and Chebyshev Distance on Real Vector Space}
+Tags: Chebyshev Distance, P-Product Metrics
+
+\begin{theorem}
+For $n \in \N$, let $\R^n$ be a [[Definition:Euclidean Space|Euclidean space]].
+Let $p \in \R_{\ge 1}$.
+Let $d_p$ be the [[Definition:P-Product Metric/Real Vector Space|$p$-product metric]] on $\R^n$. 
+Let $d_\infty$ be the [[Definition:Chebyshev Distance/Real Vector Space|Chebyshev distance]] on $\R^n$.
+Then
+:$\forall x, y \in \R^n: \map {d_\infty} {x, y} \le \map {d_p} {x, y} \le n^{1/p} \map {d_\infty} {x, y}$
+\end{theorem}
+
+\begin{proof}
+By definition of the [[Definition:Chebyshev Distance/Real Vector Space|Chebyshev distance]] on $\R^n$, we have:
+:$\displaystyle \map {d_\infty} {x, y} = \max_{i \mathop = 1}^n \size {x_i - y_i}$
+where $x = \tuple {x_1, x_2, \ldots, x_n}$ and $y = \tuple {y_1, y_2, \ldots, y_n}$.
+Let $j$ be chosen so that:
+:$\displaystyle \size {x_j - y_j} = \max_{i \mathop = 1}^n \size {x_i - y_i}$
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {d_\infty} {x, y}
+      | r = \paren {\size {x_j - y_j}^p}^{1/p}
+}}
+{{eqn | o = \le
+      | r = \paren {\sum_{i \mathop = 1}^n \size {x_i - y_i}^p}^{1/p}
+}}
+{{eqn | r = \map {d_p} {x, y}
+}}
+{{eqn | o = \le
+      | r = \paren {n \size {x_j - y_j}^p}^{1/p}
+}}
+{{eqn | r = n^{1/p} \size {d_\infty} {x, y}
+}}
+{{end-eqn}}
+\end{proof}<|endoftext|>
+\section{P-Product Metrics on Real Vector Space are Topologically Equivalent/Inequality for General Case}
+Tags: P-Product Metrics
+
+\begin{theorem}
+For $n \in \N$, let $\R^n$ be a [[Definition:Real Vector Space|real vector Space]].
+Let $r, t \in \R_{\ge 1}$.
+Let $d_r$ and $d_t$ be [[Definition:P-Product Metric/Real Vector Space|$p$-product metrics]] on $\R^n$.
+Then $d_r$ and $d_t$ are [[Definition:Topologically Equivalent Metrics|topologically equivalent]].
+\end{theorem}
+
+\begin{proof}
+We show that $\displaystyle d_r \left({x, y}\right) \ge d_t \left({x, y}\right)$, which is equivalent to proving that:
+:$\displaystyle \left({\sum_{i \mathop = 1}^n \left|{x_i - y_i}\right|^r}\right)^{1/r} \ge \left({\sum_{i \mathop = 1}^n \left|{x_i - y_i}\right|^t}\right)^{1/t}$
+Let $\forall i \in \left[{1 \,.\,.\, n}\right]: s_i = \left|{x_i - y_i}\right|$.
+Suppose $s_k = 0$ for some $k \in \left[{1 \,.\,.\, n}\right]$.
+{{improve|Putting off the analysis of s_k being 0 until near the end of the proof should shorten it without losing any clarity.}}
+Then the problem reduces to the equivalent one of showing that:
+: $\displaystyle \left({\sum_{i \mathop = 1}^{n-1} \left|{x_i - y_i}\right|^r}\right)^{1/r} \ge \left({\sum_{i \mathop = 1}^{n-1} \left|{x_i - y_i}\right|^t }\right)^{1/t}$
+that is, of reducing the index by $1$.
+Note that when $n = 1$, from simple algebra $d_r \left({x, y}\right) = d_t \left({x, y}\right)$.
+So, let us start with the assumption that $\forall i \in \left[{1 \,.\,.\, n}\right]: s_i > 0$.
+Let $\displaystyle u = \sum_{i \mathop = 1}^n \left|{x_i - y_i}\right|^r = \sum_{i \mathop = 1}^n s_i^r$, and $v = \dfrac 1 r$.
+From [[Derivative of Function to Power of Function]], $D_r \left({u^v}\right) = v u^{v-1} D_r \left({u}\right) + u^v \ln u D_r  \left({v}\right)$.
+Here:
+:$\displaystyle D_r \left({u}\right) = \sum_{i \mathop = 1}^n s_i^r \ln s_i$ from [[Derivative of Exponential Function]] and [[Sum Rule for Derivatives]]
+:$D_r \left({v}\right) = - \dfrac 1 {r^2}$ from [[Power Rule for Derivatives]]
+In the case where $r=1$, we have:
+:$D_r \left({u^v}\right) = 0$
+When $r > 1$, we have:
+{{begin-eqn}}
+{{eqn | l = D_r \left({\left({\sum_{i \mathop = 1}^n s_i^r}\right)^{1/r} }\right)
+      | r = \dfrac 1 r \left({\sum_{i \mathop = 1}^n s_i^r}\right)^{1/ \left({r - 1}\right) } \left({\sum_{i \mathop = 1}^n s_i^r \ln s_i}\right) - \dfrac {\left({\sum_{i \mathop = 1}^n s_i^r}\right)^{1/r} \ln \left({\sum_{i \mathop = 1}^n s_i^r}\right)} {r^2}
+      | c = 
+}}
+{{eqn | r = \dfrac {\left({\sum_{i \mathop = 1}^n s_i^r}\right)^{1/r} } r \left({\dfrac {\sum_{i \mathop = 1}^n s_i^r \ln s_i} {\sum_{i \mathop = 1}^n s_i^r} - \dfrac {\ln \left({\sum_{i \mathop = 1}^n s_i^r}\right)} r}\right)
+      | c = 
+}}
+{{eqn | r = \dfrac {\left({\sum_{i \mathop = 1}^n s_i^r}\right)^{1/r} } r \left({\dfrac {r \left({\sum_{i \mathop = 1}^n s_i^r \ln s_i}\right) - \left({\sum_{i \mathop = 1}^n s_i^r}\right) \ln \left({\sum_{i \mathop = 1}^n s_i^r}\right)} {r \left({\sum_{i \mathop = 1}^n s_i^r}\right)} }\right)
+      | c = 
+}}
+{{eqn | r = K \left({r \left({\sum_{i \mathop = 1}^n s_i^r \ln s_i}\right) - \left({\sum_{i \mathop = 1}^n s_i^r}\right) \ln \left({\sum_{i \mathop = 1}^n s_i^r}\right)}\right)
+      | c = where $\displaystyle K = \dfrac {\left({\sum_{i \mathop = 1}^n s_i^r}\right)^{1/r} } {r^2 \left({\sum_{i \mathop = 1}^n s_i^r}\right)} > 0$
+}}
+{{eqn | r = K \left({\sum_{i \mathop = 1}^n s_i^r \ln \left({s_i^r}\right) - \left({\sum_{i \mathop = 1}^n s_i^r}\right) \ln \left({\sum_{i \mathop = 1}^n s_i^r}\right)}\right)
+      | c = [[Logarithms of Powers]]
+}}
+{{eqn | r = K \left({\sum_{j \mathop = 1}^n \left({s_j^r \left({\ln \left({s_j^r}\right) - \ln \left({\sum_{i \mathop = 1}^n s_i^r}\right)}\right)}\right)}\right)
+      | c = 
+}}
+{{eqn | r = K \left({\sum_{j \mathop = 1}^n \left({s_j^r \ln \left({\frac {s_j^r} {\sum_{i \mathop = 1}^n s_i^r} }\right)}\right)}\right)
+      | c = 
+}}
+{{end-eqn}}
+where $K > 0$ because all of $s_i, r > 0$.
+For the same reason, $\displaystyle \dfrac{s_j^r} {\sum_{i \mathop = 1}^n s_i^r} < 1$ for all $j \in \left\{ {1, \ldots, n}\right\}$.
+From [[Logarithm of 1 is 0]] and [[Logarithm is Strictly Increasing]], their logarithms are therefore negative.
+So for $r > 1$:
+: $\displaystyle D_r \left({\left({\sum_{i \mathop = 1}^n s_i^r}\right)^{1/r}}\right) < 0$
+So, from [[Derivative of Monotone Function]], it follows that (given the conditions on $r$ and $s_i$) $\displaystyle \left({\sum_{i \mathop = 1}^n s_i^r}\right)^{1/r}$ is [[Definition:Decreasing Real Function|decreasing]].
+As we assumed $r \le t$, we have $d_r \left({x, y}\right) \ge d_t \left({x, y}\right)$.
+{{qed|lemma}}
+\end{proof}<|endoftext|>
+\section{De Morgan's Laws (Logic)/Disjunction of Negations/Formulation 1/Forward Implication}
+Tags: De Morgan's Laws (Logic)
+
+\begin{theorem}
+: $\neg p \lor \neg q \vdash \neg \left({p \land q}\right)$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\neg p \lor \neg q \vdash \neg \left({p \land q}\right)}}
+{{Premise|1|\neg p \lor \neg q}}
+{{Assumption|2|p \land q}}
+{{Simplification|3|2|p|2|1}}
+{{Simplification|4|2|q|2|2}}
+{{Assumption|5|\neg p}}
+{{NonContradiction|6|2, 5|3|5}}
+{{Assumption|7|\neg q}}
+{{NonContradiction|8|2, 7|4|7}}
+{{ProofByCases|9|1, 2|\bot|1|5|6|7|8}}
+{{Contradiction|10|1|\neg \left({p \land q}\right)|2|9}}
+{{EndTableau}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{De Morgan's Laws (Logic)/Disjunction of Negations/Formulation 1/Reverse Implication}
+Tags: De Morgan's Laws (Logic)
+
+\begin{theorem}
+: $\neg \left({p \land q}\right) \vdash \neg p \lor \neg q$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\neg \left({p \land q}\right) \vdash \neg p \lor \neg q}}
+{{Premise|1|\neg \left({p \land q}\right)}}
+{{Assumption|2|\neg \left ({\neg p \lor \neg q}\right)}}
+{{Assumption|3|\neg p}}
+{{Addition|4|3|\neg p \lor \neg q|3|1}}
+{{NonContradiction|5|2, 3|4|2}}
+{{Reductio|6|2|p|3|5}}
+{{Assumption|7|\neg q}}
+{{Addition|8|7|\neg p \lor \neg q|7|2}}
+{{NonContradiction|9|2, 7|8|2}}
+{{Reductio|10|2|q|7|9}}
+{{Conjunction|11|2|p \land q|6|10}}
+{{NonContradiction|12|1, 2|11|1}}
+{{Reductio|13|1|\neg p \lor \neg q|2|12}}
+{{EndTableau}}
+{{qed}}
+{{LEM|Reductio ad Absurdum}}
+\end{proof}<|endoftext|>
+\section{De Morgan's Laws (Logic)/Disjunction of Negations/Formulation 1}
+Tags: De Morgan's Laws (Logic)
+
+\begin{theorem}
+:$\neg p \lor \neg q \dashv \vdash \neg \paren {p \land q}$
+\end{theorem}<|endoftext|>
+\section{De Morgan's Laws (Logic)/Disjunction of Negations/Formulation 2}
+Tags: De Morgan's Laws (Logic)
+
+\begin{theorem}
+:$\vdash \paren {\neg p \lor \neg q} \iff \paren {\neg \paren {p \land q} }$
+\end{theorem}<|endoftext|>
+\section{De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 1/Forward Implication}
+Tags: De Morgan's Laws (Logic)
+
+\begin{theorem}
+:$\neg p \land \neg q \vdash \neg \paren {p \lor q}$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\neg p \land \neg q \vdash \neg \paren {p \lor q} }}
+{{Premise|1|\neg p \land \neg q}}
+{{Simplification|2|1|\neg p|1|1}}
+{{Simplification|3|1|\neg q|1|2}}
+{{Assumption|4|p \lor q}}
+{{Assumption|5|p}}
+{{NonContradiction|6|1, 5|5|2}}
+{{Assumption|7|q}}
+{{NonContradiction|8|1, 7|7|3}}
+{{ProofByCases|9|1, 4|\bot|4|5|6|7|8}}
+{{Contradiction|10|1|\neg \paren {p \lor q}|4|9}}
+{{EndTableau|qed}}
+\end{proof}<|endoftext|>
+\section{De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 1/Reverse Implication}
+Tags: De Morgan's Laws (Logic)
+
+\begin{theorem}
+:$\neg \paren {p \lor q} \vdash \neg p \land \neg q$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\neg \paren {p \lor q} \vdash \neg p \land \neg q}}
+{{Premise|1|\neg \paren {p \lor q} }}
+{{Assumption|5|p}}
+{{Addition|3|2|p \lor q|2|1}}
+{{NonContradiction|4|1, 3|3|1}}
+{{Contradiction|5|1|\neg p|2|4}}
+{{Assumption|6|q}}
+{{Addition|7|6|p \lor q|6|2}}
+{{NonContradiction|8|1, 7|7|1}}
+{{Contradiction|9|1|\neg q|6|8}}
+{{Conjunction|10|1|\neg p \land \neg q|5|9}}
+{{EndTableau|qed}}
+[[Category:De Morgan's Laws (Logic)]]
+7hif3lo8kwg7txhdhfm6bvpm7fym2tn
+\end{proof}<|endoftext|>
+\section{De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 1}
+Tags: De Morgan's Laws (Logic)
+
+\begin{theorem}
+:$\neg p \land \neg q \dashv \vdash \neg \paren {p \lor q}$
+\end{theorem}<|endoftext|>
+\section{De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 2}
+Tags: De Morgan's Laws (Logic)
+
+\begin{theorem}
+:$\vdash \paren {\neg p \land \neg q} \iff \paren {\neg \paren {p \lor q} }$
+\end{theorem}<|endoftext|>
+\section{De Morgan's Laws (Logic)/Conjunction of Negations}
+Tags: De Morgan's Laws (Logic)
+
+\begin{theorem}
+{{:De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 1}}
+\end{theorem}<|endoftext|>
+\section{De Morgan's Laws (Logic)/Conjunction}
+Tags: De Morgan's Laws (Logic)
+
+\begin{theorem}
+{{:De Morgan's Laws (Logic)/Conjunction/Formulation 1}}
+\end{theorem}<|endoftext|>
+\section{De Morgan's Laws (Logic)/Conjunction/Formulation 1/Forward Implication}
+Tags: De Morgan's Laws (Logic)
+
+\begin{theorem}
+: $p \land q \vdash \neg \left({\neg p \lor \neg q}\right)$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|p \land q \vdash \neg \left({\neg p \lor \neg q}\right)}}
+{{Premise|1|p \land q}}
+{{Simplification|2|1|p|1|1}}
+{{Simplification|3|1|q|1|2}}
+{{Assumption|4|\neg p \lor \neg q}}
+{{Assumption|5|\neg p}}
+{{NonContradiction|6|1, 5|2|5}}
+{{Assumption|7|\neg q}}
+{{NonContradiction|8|1, 7|3|7}}
+{{ProofByCases|9|1, 4|\bot|4|5|6|7|8}}
+{{Contradiction|10|1|\neg \left({\neg p \lor \neg q}\right)|4|9}}
+{{EndTableau}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{De Morgan's Laws (Logic)/Conjunction/Formulation 1/Reverse Implication}
+Tags: De Morgan's Laws (Logic)
+
+\begin{theorem}
+: $\neg \left({\neg p \lor \neg q}\right) \vdash p \land q$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\neg \left({\neg p \lor \neg q}\right) \vdash p \land q}}
+{{Premise|1|\neg \left({\neg p \lor \neg q}\right)}}
+{{Assumption|2|\neg \left ({p \land q}\right)}}
+{{DeMorgan|3|2|\neg p \lor \neg q|2|Disjunction of Negations}}
+{{NonContradiction|4|1, 2|3|1}}
+{{Reductio|5|1|p \land q|2|4}}
+|}
+{{qed}}
+{{LEM|Reductio ad Absurdum}}
+\end{proof}<|endoftext|>
+\section{De Morgan's Laws (Logic)/Conjunction/Formulation 1/Proof}
+Tags: De Morgan's Laws (Logic), Truth Table Proofs
+
+\begin{theorem}
+: $p \land q \dashv \vdash \neg \left({\neg p \lor \neg q}\right)$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]].
+As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
+$\begin{array}{|ccc||cccccc|} \hline
+p & \land & q & \neg & (\neg & p & \lor & \neg & q)  \\
+\hline
+F & F & F & F & T & F & T & T & F \\
+F & F & T & F & T & F & T & F & T \\
+T & F & F & F & F & T & T & T & F \\
+T & T & T & T & F & T & F & F & T \\
+\hline
+\end{array}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{De Morgan's Laws (Logic)/Conjunction/Formulation 1}
+Tags: De Morgan's Laws (Logic)
+
+\begin{theorem}
+: $p \land q \dashv \vdash \neg \left({\neg p \lor \neg q}\right)$
+\end{theorem}<|endoftext|>
+\section{De Morgan's Laws (Logic)/Conjunction/Formulation 2}
+Tags: De Morgan's Laws (Logic)
+
+\begin{theorem}
+:$\vdash \paren {p \land q} \iff \paren {\neg \paren {\neg p \lor \neg q} }$
+\end{theorem}<|endoftext|>
+\section{De Morgan's Laws (Logic)/Disjunction}
+Tags: De Morgan's Laws (Logic)
+
+\begin{theorem}
+{{:De Morgan's Laws (Logic)/Disjunction/Formulation 1}}
+\end{theorem}<|endoftext|>
+\section{De Morgan's Laws (Logic)/Disjunction/Formulation 2}
+Tags: De Morgan's Laws (Logic)
+
+\begin{theorem}
+: $\vdash \left({p \lor q}\right) \iff \left({\neg \left({\neg p \land \neg q}\right)}\right)$
+\end{theorem}<|endoftext|>
+\section{De Morgan's Laws (Logic)/Disjunction/Formulation 1/Forward Implication}
+Tags: De Morgan's Laws (Logic)
+
+\begin{theorem}
+: $p \lor q \vdash \neg \left({\neg p \land \neg q}\right)$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|p \lor q \vdash \neg \left({\neg p \land \neg q}\right)}}
+{{Premise|1|p \lor q}}
+{{Assumption|2|\neg p \land \neg q}}
+{{Simplification|3|2|\neg p|2|1}}
+{{Simplification|4|2|\neg q|2|2}}
+{{Assumption|5|p}}
+{{NonContradiction|6|2, 5|5|3}}
+{{Assumption|7|q}}
+{{NonContradiction|8|2, 7|7|4}}
+{{ProofByCases|9|1, 2|\bot|1|5|6|7|8}}
+{{Contradiction|10|1|\neg \left({\neg p \land \neg q}\right)|2|9}}
+{{EndTableau}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{De Morgan's Laws (Logic)/Disjunction/Formulation 1/Reverse Implication}
+Tags: De Morgan's Laws (Logic)
+
+\begin{theorem}
+: $\neg \left({\neg p \land \neg q}\right) \vdash p \lor q$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\neg \left({\neg p \land \neg q}\right) \vdash p \lor q}}
+{{Premise|1|\neg \left({\neg p \land \neg q}\right)}}
+{{Assumption|2|\neg \left ({p \lor q}\right)}}
+{{DeMorgan|3|2|\neg p \land \neg q|2|Conjunction of Negations}}
+{{NonContradiction|4|1, 2|3|1}}
+{{Reductio|5|1|p \lor q|2|4}}
+{{EndTableau}}
+{{qed}}
+{{LEM|Reductio ad Absurdum}}
+\end{proof}<|endoftext|>
+\section{De Morgan's Laws (Logic)/Disjunction/Formulation 1}
+Tags: De Morgan's Laws (Logic)
+
+\begin{theorem}
+:$p \lor q \dashv \vdash \neg \paren {\neg p \land \neg q}$
+\end{theorem}<|endoftext|>
+\section{Homotopy Characterisation of Simply Connected Sets}
+Tags: Homotopy Theory, Connected Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $X$ be a [[Definition:Subset|subset]] of $S$.
+Then $X$ is [[Definition:Simply Connected|simply connected]] {{iff}} the following conditions hold:
+:$(1): \quad $ $X$ is [[Definition:Path-Connected Space|path-connected]].
+:$(2): \quad $ All [[Definition:Path (Topology)|paths]] in $X$ with the same [[Definition:Initial Point of Path|initial points]] and [[Definition:Final Point of Path|final points]] are [[Definition:Free Homotopy|freely homotopic]].
+\end{theorem}
+
+\begin{proof}
+We only have to check condition $(2)$, as a [[Definition:Simply Connected|simply connected set]] is [[Definition:Path-Connected Space|path-connected]] by definition.
+=== Necessary condition ===
+Suppose that $X$ is [[Definition:Simply Connected|simply connected]].
+Let $\gamma_1, \gamma_2: \closedint 0 1 \to X$ be two [[Definition:Path (Topology)|paths]] with $\map {\gamma_1} 0 = \map {\gamma_2} 0$ and $\map {\gamma_1} 1 = \map {\gamma_2} 1$.
+Define the [[Definition:Mapping|mapping]] $-\gamma_2: \closedint 0 1 \to X$ by $-\map {\gamma_2} t = \map {\gamma_2} {1 - t}$.
+From [[Continuity of Composite Mapping]], it follows that $-\gamma_2$ is a path.
+Let $c: \closedint 0 1 \to X$ be the [[Definition:Constant Mapping|constant]] path defined by $\map c t = \map {\gamma_1} 0$.
+When $\equiv$ denotes [[Definition:Equivalence Class|equivalence]] of [[Definition:Homotopy Class|homotopy classes]], we have:
+{{begin-eqn}}
+{{eqn | l = \gamma_1
+      | o = \equiv
+      | r = \gamma_1 * c
+      | c = here, $\gamma_1 * c$ denotes the [[Definition:Concatenation (Topology)|concatenation]] of $\gamma_1$ and $c$
+}}
+{{eqn | o = \equiv
+      | r = \gamma_1 * \paren {-\gamma_2 * \gamma_2 }
+      | c = by assumption, as $\paren {-\gamma_2 * \gamma_2 }$ and $c$ are both [[Definition:Closed Path (Topology)|closed paths]] with [[Definition:Initial Point of Path|initial point]] $\map {\gamma_1} 0$
+}}
+{{eqn | o = \equiv
+      | r = \paren {\gamma_1 * \paren {-\gamma_2} } * \gamma_2 
+}}
+{{eqn | o = \equiv
+      | r = c * \gamma_2
+      | c = by assumption
+}}
+{{eqn | o = \equiv
+      | r = \gamma_2
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $x_0 \in X$.
+We show that the [[Definition:Homotopy Group|fundamental group]] $\map {\pi_1} {X, x_0}$ is [[Definition:Trivial Group|trivial]].
+Let $\gamma_1, \gamma_2: \closedint 0 1 \to X$ be two [[Definition:Closed Path (Topology)|closed paths]] with [[Definition:Initial Point of Path|initial point]] $x_0$.
+By assumption, $\gamma_1$ and $\gamma_2$ are [[Definition:Free Homotopy|freely homotopic]] and belong to the same [[Definition:Homotopy Class|homotopy class]].
+Hence, $\map {\pi_1} {X, x_0}$ has only one [[Definition:Element|element]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rule of Distribution/Disjunction Distributes over Conjunction}
+Tags: Rule of Distribution
+
+\begin{theorem}
+=== [[Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive|Disjunction is Left Distributive over Conjunction]] ===
+{{:Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive}}
+=== [[Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive|Disjunction is Right Distributive over Conjunction]] ===
+{{:Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive}}
+\end{theorem}<|endoftext|>
+\section{Rule of Distribution/Conjunction Distributes over Disjunction}
+Tags: Rule of Distribution
+
+\begin{theorem}
+=== [[Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive|Conjunction is Left Distributive over Disjunction]] ===
+{{:Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive}}
+=== [[Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive|Conjunction is Right Distributive over Disjunction]] ===
+{{:Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive}}
+\end{theorem}<|endoftext|>
+\section{Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1/Proof}
+Tags: Rule of Distribution, Truth Table Proofs
+
+\begin{theorem}
+:$p \land \left({q \lor r}\right) \dashv \vdash \left({p \land q}\right) \lor \left({p \land r}\right)$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]] to the proposition.
+As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
+$\begin{array}{|ccccc||ccccccc|} \hline
+p & \land & (q & \lor & r) & (p & \land & q) & \lor & (p & \land & r) \\
+\hline
+F & F & F & F & F & F & F & F & F & F & F & F \\
+F & F & F & T & T & F & F & F & F & F & F & T \\
+F & F & T & T & F & F & F & T & F & F & F & F \\
+F & F & T & T & T & F & F & T & F & F & F & T \\
+T & F & F & F & F & T & F & F & F & T & F & F \\
+T & T & F & T & T & T & F & F & T & T & T & T \\
+T & T & T & T & F & T & T & T & T & T & F & F \\
+T & T & T & T & T & T & T & T & T & T & T & T \\
+\hline
+\end{array}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2}
+Tags: Rule of Distribution
+
+\begin{theorem}
+:$\vdash \paren {p \land \paren {q \lor r} } \iff \paren {\paren {p \land q} \lor \paren {p \land r} }$
+\end{theorem}<|endoftext|>
+\section{Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive}
+Tags: Rule of Distribution
+
+\begin{theorem}
+==== [[Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1|Formulation 1]] ====
+{{:Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1}}
+==== [[Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2|Formulation 2]] ====
+{{:Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2}}
+\end{theorem}<|endoftext|>
+\section{Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive}
+Tags: Rule of Distribution
+
+\begin{theorem}
+==== [[Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 1|Formulation 1]] ====
+{{:Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 1}}
+==== [[Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 2|Formulation 2]] ====
+{{:Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 2}}
+\end{theorem}<|endoftext|>
+\section{Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 1}
+Tags: Rule of Distribution
+
+\begin{theorem}
+:$\paren {q \lor r} \land p \dashv \vdash \paren {q \land p} \lor \paren {r \land p}$
+\end{theorem}<|endoftext|>
+\section{Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 2}
+Tags: Rule of Distribution
+
+\begin{theorem}
+:$\vdash \paren {\paren {q \lor r} \land p} \iff \paren {\paren {q \land p} \lor \paren {r \land p} }$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\vdash \paren {\paren {q \lor r} \land p} \iff \paren {\paren {q \land p} \lor \paren {r \land p} } }}
+{{Assumption|1|\paren {q \lor r} \land p}}
+{{SequentIntro|2|1|\paren {q \land p} \lor \paren {r \land p}|1|[[Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 1|Conjunction is Right Distributive over Disjunction: Formulation 1]]}}
+{{Implication|3||\paren {\paren {q \lor r} \land p} \implies \paren {\paren {q \land p} \lor \paren {r \land p} }|1|2}}
+{{Assumption|4|\paren {q \land p} \lor \paren {r \land p} }}
+{{SequentIntro|5|4|\paren {q \lor r} \land p|4|[[Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 1|Conjunction is Right Distributive over Disjunction: Formulation 1]]}}
+{{Implication|6||\paren {\paren {q \land p} \lor \paren {r \land p} } \implies \paren {\paren {q \lor r} \land p}|4|5}}
+{{BiconditionalIntro|7||\paren {\paren {q \lor r} \land p} \iff \paren {\paren {q \land p} \lor \paren {r \land p} }|3|6}}
+{{EndTableau}}
+{{qed}
+[[Category:Rule of Distribution]]
+7umzufsmkl5vcnhblyr3fhdma3gi018
+\end{proof}<|endoftext|>
+\section{Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive}
+Tags: Rule of Distribution
+
+\begin{theorem}
+==== [[Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1|Formulation 1]] ====
+{{:Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1}}
+==== [[Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2|Formulation 2]] ====
+{{:Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2}}
+\end{theorem}<|endoftext|>
+\section{Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive}
+Tags: Rule of Distribution
+
+\begin{theorem}
+==== [[Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive/Formulation 1|Formulation 1]] ====
+{{:Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive/Formulation 1}}
+==== [[Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive/Formulation 2|Formulation 2]] ====
+{{:Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive/Formulation 2}}
+\end{theorem}<|endoftext|>
+\section{Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1/Forward Implication}
+Tags: Rule of Distribution
+
+\begin{theorem}
+:$p \lor \paren {q \land r} \vdash \paren {p \lor q} \land \paren {p \lor r}$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|p \lor \paren {q \land r} \vdash \paren {p \lor q} \land \paren {p \lor r} }}
+{{Premise        | 1|p \lor \paren {q \land r} }}
+{{Assumption     | 2|p}}
+{{Addition       | 3|2|p \lor q|2|1}}
+{{Addition       | 4|2|p \lor r|2|1}}
+{{Conjunction    | 5|2|\paren {p \lor q} \land \paren {p \lor r}|3|4}}
+{{Assumption     | 6|q \land r}}
+{{Simplification | 7|6|q|6|1}}
+{{Simplification | 8|6|r|6|2}}
+{{Addition       | 9|6|p \lor q|7|2}}
+{{Addition       |10|6|p \lor r|8|2}}
+{{Conjunction    |11|6|\paren {p \lor q} \land \paren {p \lor r}|7|8}}
+{{ProofByCases   |12|1|\paren {p \lor q} \land \paren {p \lor r}|1|2|5|6|11}}
+{{EndTableau}}
+{{qed}}
+[[Category:Rule of Distribution]]
+i25mmyxvwb862jvl4ipx120n568xzpk
+\end{proof}<|endoftext|>
+\section{Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1/Proof}
+Tags: Rule of Distribution, Truth Table Proofs
+
+\begin{theorem}
+:$p \lor \paren {q \land r} \dashv \vdash \paren {p \lor q} \land \paren {p \lor r}$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]] to the proposition.
+As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
+$\begin{array}{|ccccc||ccccccc|} \hline
+p & \lor & (q & \land & r) & (p & \lor & q) & \land & (p & \lor & r) \\
+\hline
+F & F & F & F & F & F & F & F & F & F & F & F \\
+F & F & F & F & T & F & F & F & F & F & T & T \\
+F & F & T & F & F & F & T & T & F & F & F & F \\
+F & T & T & T & T & F & T & T & T & F & T & T \\
+T & T & F & F & F & T & T & F & T & T & T & F \\
+T & T & F & F & T & T & T & F & T & T & T & T \\
+T & T & T & F & F & T & T & T & T & T & T & F \\
+T & T & T & T & T & T & T & T & T & T & T & T \\
+\hline
+\end{array}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2}
+Tags: Rule of Distribution
+
+\begin{theorem}
+:$\vdash \paren {p \lor \paren {q \land r} } \iff \paren {\paren {p \lor q} \land \paren {p \lor r} }$
+\end{theorem}<|endoftext|>
+\section{Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive/Formulation 1}
+Tags: Rule of Distribution
+
+\begin{theorem}
+:$\paren {q \land r} \lor p \dashv \vdash \paren {q \lor p} \land \paren {r \lor p}$
+\end{theorem}<|endoftext|>
+\section{Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive/Formulation 2}
+Tags: Rule of Distribution
+
+\begin{theorem}
+:$\vdash \paren {\paren {q \land r} \lor p} \iff \paren {\paren {q \lor p} \land \paren {r \lor p} }$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\vdash \paren {\paren {q \land r} \lor p} \iff \paren {\paren {q \lor p} \land \paren {r \lor p} } }}
+{{Assumption|1|\paren {q \land r} \lor p}}
+{{SequentIntro|2|1|\paren {q \lor p} \land \paren {r \lor p}|1|[[Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 1|Conjunction is Right Distributive over Disjunction: Formulation 1]]}}
+{{Implication|3||\paren {\paren {q \land r} \lor p} \implies \paren {\paren {q \lor p} \land \paren {r \lor p} }|1|2}}
+{{Assumption|4|\paren {q \lor p} \land \paren {r \lor p} }}
+{{SequentIntro|5|4|\paren {q \land r} \lor p|4|[[Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 1|Conjunction is Right Distributive over Disjunction: Formulation 1]]}}
+{{Implication|6||\paren {\paren {q \lor p} \land \paren {r \lor p} } \implies \paren {\paren {q \land r} \lor p}|4|5}}
+{{BiconditionalIntro|7||\paren {\paren {q \land r} \lor p} \iff \paren {\paren {q \lor p} \land \paren {r \lor p} }|3|6}}
+{{EndTableau|qed}}
+[[Category:Rule of Distribution]]
+7z7uuzjoo8r26oav447crim4y4t7oxm
+\end{proof}<|endoftext|>
+\section{Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive/Formulation 1/Proof}
+Tags: Rule of Distribution, Truth Table Proofs
+
+\begin{theorem}
+:$\left({q \land r}\right) \lor p \dashv \vdash \left({q \lor p}\right) \land \left({r \lor p}\right)$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]] to the proposition.
+As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
+$\begin{array}{|ccccc||ccccccc|} \hline
+(q & \land & r) & \lor & p & (q & \lor & p) & \land & (r & \lor & p) \\
+\hline
+F & F & F & F & F & F & F & F & F & F & F & F \\
+F & F & F & T & T & F & T & T & T & F & T & T \\
+F & F & T & F & F & F & F & F & F & T & T & F \\
+F & F & T & T & T & F & T & T & T & T & T & T \\
+T & F & F & F & F & T & T & F & F & F & F & F \\
+T & F & F & T & T & T & T & T & T & F & T & T \\
+T & T & T & T & F & T & T & F & T & T & T & F \\
+T & T & T & T & T & T & T & T & T & T & T & T \\
+\hline
+\end{array}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Union of Subsets is Subset/Set of Sets}
+Tags: Union of Subsets is Subset
+
+\begin{theorem}
+Let $T$ be a [[Definition:Set|set]].
+Let $\mathbb S$ be a [[Definition:Set of Sets|set of sets]].
+Suppose that for each $S \in \mathbb S$, $S \subseteq T$.
+Then:
+:$\displaystyle \bigcup \mathbb S \subseteq T$
+\end{theorem}
+
+\begin{proof}
+Let $x \in \displaystyle \bigcup \mathbb S$.
+By the definition of [[Definition:Set Union|union]], there exists an $S \in \mathbb S$ such that $x \in S$.
+By premise, $S \subseteq T$.
+By the definition of [[Definition:Subset|subset]], $x \in T$.
+Since this result holds for each $x \in \displaystyle \bigcup \mathbb S$:
+:$\displaystyle \bigcup \mathbb S \subseteq T$
+{{qed}}
+[[Category:Union of Subsets is Subset]]
+is1e5gkt5f9egiiuv1vwh4v2k7b6qr3
+\end{proof}<|endoftext|>
+\section{Union is Smallest Superset/Set of Sets}
+Tags: Set Union, Subsets
+
+\begin{theorem}
+Let $T$ be a [[Definition:Set|set]].
+Let $\mathbb S$ be a [[Definition:Set of Sets|set of sets]].
+Then:
+:$\displaystyle \paren {\forall X \in \mathbb S: X \subseteq T} \iff \bigcup \mathbb S \subseteq T$
+\end{theorem}
+
+\begin{proof}
+By [[Union of Subsets is Subset/Set of Sets|Union of Subsets is Subset: Set of Sets]]:
+:$\displaystyle \paren {\forall X \in \mathbb S: X \subseteq T} \implies \bigcup \mathbb S \subseteq T$
+{{qed|lemma}}
+For the converse implication, suppose that $\displaystyle \bigcup \mathbb S \subseteq T$.
+Consider any $X \in \mathbb S$ and take any $x \in X$.
+From [[Set is Subset of Union/Set of Sets|Set is Subset of Union: Set of Sets]] we have that $X \subseteq \bigcup \mathbb S$.
+Thus $\displaystyle x \in \bigcup \mathbb S$.
+But $\displaystyle \bigcup \mathbb S \subseteq T$.
+So it follows that $X \subseteq T$.
+So:
+:$\displaystyle \bigcup \mathbb S \subseteq T \implies \paren {\forall X \in \mathbb S: X \subseteq T}$
+{{qed|lemma}}
+Hence:
+:$\displaystyle \paren {\forall X \in \mathbb S: X \subseteq T} \iff \bigcup \mathbb S \subseteq T$
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of Relations is Relation}
+Tags: Relation Theory
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Set|sets]].
+Let $\mathcal F$ be a family of [[Definition:Relation|relations]] from $S$ to $T$.
+Let $\displaystyle \mathcal R = \bigcap \mathcal F$, the [[Definition:Intersection of Set of Sets|intersection]] of all the elements of $\mathcal F$.
+ 
+Then $\mathcal R$ is a [[Definition:Relation|relation]] from $S$ to $T$.
+{{Expand|Binary case}}
+\end{theorem}
+
+\begin{proof}
+By the definition of a [[Definition:Relation|relation]] from $S$ to $T$, each element of $\mathcal F$ is a [[Definition:Subset|subset]] of $S \times T$.
+By [[Intersection of Subsets is Subset/Set of Sets|Intersection of Subsets is Subset: Set of Sets]]:
+: $\mathcal R \subseteq S \times T$
+Therefore, by the definition of a [[Definition:Relation|relation]] from $S$ to $T$, $\mathcal R$ is a relation from $S$ to $T$.
+{{qed}}
+[[Category:Relation Theory]]
+tdt3qiqa479f55i5kwsfnrx83qyh4ag
+\end{proof}<|endoftext|>
+\section{Intersection of Subsets is Subset/Set of Sets}
+Tags: Set Intersection, Subsets
+
+\begin{theorem}
+Let $T$ be a [[Definition:Set|set]].
+Let $\mathbb S$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Set of Sets|set of sets]].
+Suppose that for each $S \in \mathbb S$:
+:$S \subseteq T$
+Then:
+:$\bigcap \mathbb S \subseteq T$
+\end{theorem}
+
+\begin{proof}
+Let $x \in \bigcap \mathbb S$.
+Then by the definition of [[Definition:Set Intersection|intersection]]:
+:$\forall S \in \mathbb S: x \in S$.
+Since $\mathbb S$ is [[Definition:Non-Empty Set|non-empty]] by the premise, it has some [[Definition:Element|element]] $S$.
+Then $x \in S$.
+Since $S \in \mathbb S$, the premise shows that $S \subseteq T$.
+By the definition of [[Definition:Subset|subset]], $x \in T$.
+Since this holds for each $x \in \bigcap \mathbb S$:
+:$\bigcap \mathbb S \subseteq T$
+{{qed}}
+[[Category:Set Intersection]]
+[[Category:Subsets]]
+nvi46jila9jjopv6mlvzhcmdtcs3ngx
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Initial Topology}
+Tags: Initial Topology
+
+\begin{theorem}
+Let $X$ be a [[Definition:Set|set]].
+Let $I$ be an [[Definition:Indexing Set|indexing set]].
+Let $\family {\struct{Y_i, \tau_i}}_{i \mathop \in I}$ be an [[Definition:Indexed Family|indexed family]] of [[Definition:Topological Space|topological spaces]] [[Definition:Indexing Set|indexed]] by $I$.
+Let $\family {f_i: X \to Y_i}_{i \mathop \in I}$ be an [[Definition:Indexed Family|indexed family]] of [[Definition:Mapping|mappings]] [[Definition:Indexing Set|indexed]] by $I$.
+{{TFAE|def = Initial Topology}}
+\end{theorem}
+
+\begin{proof}
+=== [[Equivalence of Definitions of Initial Topology/Definition 1 Implies Definition 2|Definition 1 Implies Definition 2]] ===
+{{:Equivalence of Definitions of Initial Topology/Definition 1 Implies Definition 2}}{{qed|lemma}}
+=== [[Equivalence of Definitions of Initial Topology/Definition 2 Implies Definition 1|Definition 2 Implies Definition 1]] ===
+{{:Equivalence of Definitions of Initial Topology/Definition 2 Implies Definition 1}}{{qed}}
+[[Category:Initial Topology]]
+rmwc5xdissitxao55f9pfktz766vumr
+\end{proof}<|endoftext|>
+\section{Radius of Convergence from Limit of Sequence/Complex Case}
+Tags: Radius of Convergence from Limit of Sequence
+
+\begin{theorem}
+Let $\xi \in \C$ be a [[Definition:Complex Number|complex number]].
+Let $\displaystyle S \paren z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$ be a [[Definition:Complex Power Series|(complex) power series]] about $\xi$.
+Let the [[Definition:Infinite Sequence|sequence]] $\sequence {\cmod {\dfrac {a_{n + 1} } {a_n} } }_{n \mathop \in \N}$ [[Definition:Convergent Sequence (Analysis)|converges]].
+Then $R$ is given by:
+:$\displaystyle \dfrac 1 R = \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1} } {a_n} }$
+If:
+:$\displaystyle \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1} } {a_n} } = 0$
+then the radius of convergence is infinite, and $S \paren z$ is [[Definition:Absolutely Convergent Series|absolutely convergent]] for all $z \in \C$.
+\end{theorem}
+
+\begin{proof}
+Let the [[Definition:Infinite Sequence|sequence]] $\sequence {\cmod {\dfrac {a_{n+1} } {a_n} } }_{n \mathop \in \N}$ [[Definition:Convergent Real Sequence|converge]].
+Let $\epsilon \in \R_{>0}$, and let $z \in \C$.
+Let $\cmod {z - \xi} = R - \epsilon$.
+By definition of [[Definition:Radius of Convergence of Complex Power Series|radius of convergence]], it follows that $S \paren z$ is [[Definition:Absolutely Convergent Series|absolutely convergent]].
+Then from the [[Ratio Test]]:
+:$\lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n+1} \paren {z - \xi}^{n + 1} } {a_n \paren {z - \xi}^n} } \le 1$
+By [[Multiple Rule for Complex Sequences]], this inequality can be rearranged to obtain:
+{{begin-eqn}}
+{{eqn | l = \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1} } {a_n} } 
+      | o = \le
+      | r = \cmod {\dfrac {\paren {z - \xi}^{n + 1} } {\paren {z - \xi}^n } }^{-1}
+}}
+{{eqn | r = \dfrac 1 {\cmod {x - \xi}
+}}
+{{eqn | r = \dfrac 1 {R - \epsilon}
+}}
+{{end-eqn}}
+{{qed|lemma}}
+Let $\cmod {z - \xi} = R + \epsilon$.
+Then $S \paren z$ is [[Definition:Divergent Series|divergent]], so the [[Ratio Test]] shows that:
+:$\displaystyle \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1} \paren {z - \xi}^{n + 1} } {a_n \paren {z - \xi}^n} } \ge 1$
+Similarly, this inequality can be rearranged as:
+:$\displaystyle \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1} } {a_n} } \ge \dfrac 1 {R + \epsilon}$
+{{qed|lemma}}
+As $\epsilon > 0$ was arbitrary, it follows that:
+:$\displaystyle \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1}} {a_n} } = \dfrac 1 R$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Radius of Convergence from Limit of Sequence/Real Case}
+Tags: Radius of Convergence from Limit of Sequence
+
+\begin{theorem}
+Let $\xi \in \R$ be a [[Definition:Real Number|real number]].
+Let $\displaystyle S \paren x = \sum_{n \mathop = 0}^\infty a_n \paren {x - \xi}^n$ be a [[Definition:Power Series|power series]] about $\xi$.
+Then the [[Definition:Radius of Convergence|radius of convergence]] $R$ of $S \paren x$ is given by:
+: $\displaystyle \frac 1 R = \lim_{n \mathop \to \infty} \size {\frac {a_{n + 1} } {a_n} }$
+if this limit exists and is nonzero.
+If
+: $\displaystyle \lim_{n \mathop \to \infty} \size {\frac {a_{n + 1} } {a_n} } = 0$
+then the [[Definition:Radius of Convergence|radius of convergence]] is infinite and therefore the [[Definition:Interval of Convergence|interval of convergence]] is $\R$.
+\end{theorem}
+
+\begin{proof}
+From the [[Ratio Test|ratio test]], $S \paren x$ is [[Definition:Convergent Series|convergent]] if:
+:$\displaystyle \lim_{n \mathop \to \infty} \size {\frac {a_{n + 1} \paren {x - \xi}^{n + 1} } {a_n \paren {x - \xi}^n} } < 1$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \size {\frac {a_{n + 1} \paren {x - \xi}^{n + 1} } {a_n \paren {x - \xi}^n} }
+      | o = <
+      | r = 1
+      | c = 
+}}
+{{eqn | ll= \iff
+      | l = \size {\frac {a_{n + 1} } {a_n} } \size {x - \xi}
+      | o = <
+      | r = 1
+      | c = 
+}}
+{{eqn | ll= \iff
+      | l = \size {\frac {a_{n + 1} } {a_n} }
+      | o = <
+      | r = \frac 1 {\size {x - \xi} }
+      | c = 
+}}
+{{end-eqn}}
+The result follows from the definition of [[Definition:Radius of Convergence|radius of convergence]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Existence of Radius of Convergence of Complex Power Series}
+Tags: Complex Power Series, Convergence
+
+\begin{theorem}
+Let $\xi \in \C$.
+Let $\displaystyle S \paren z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n $ be a [[Definition:Complex Power Series|(complex) power series]] about $\xi$.
+Then there exists a [[Definition:Radius of Convergence of Complex Power Series|radius of convergence]] $R \in \overline \R$ of $S \paren z$.
+\end{theorem}<|endoftext|>
+\section{Conjunction Equivalent to Negation of Implication of Negative/Formulation 1}
+Tags: Conjunction Equivalent to Negation of Implication of Negative
+
+\begin{theorem}
+:$p \land q \dashv \vdash \neg \left({p \implies \neg q}\right)$
+\end{theorem}<|endoftext|>
+\section{Conjunction Equivalent to Negation of Implication of Negative/Formulation 2}
+Tags: Conjunction Equivalent to Negation of Implication of Negative
+
+\begin{theorem}
+:$\vdash \paren {p \land q} \iff \paren {\neg \paren {p \implies \neg q} }$
+\end{theorem}<|endoftext|>
+\section{Conjunction Equivalent to Negation of Implication of Negative/Formulation 1/Proof}
+Tags: Conjunction Equivalent to Negation of Implication of Negative, Truth Table Proofs
+
+\begin{theorem}
+:$p \land q \dashv \vdash \neg \left({p \implies \neg q}\right)$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]].
+As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
+$\begin{array}{|ccc||ccccc|} \hline
+p & \land & q & \neg & (p & \implies & \neg & q) \\
+\hline
+F & F & F & F & F & T & T & F \\
+F & F & T & F & F & T & F & T \\
+T & F & F & F & T & T & T & F \\
+T & T & T & T & T & F & F & T \\
+\hline
+\end{array}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Conjunction Equivalent to Negation of Implication of Negative/Formulation 1/Forward Implication}
+Tags: Conjunction Equivalent to Negation of Implication of Negative
+
+\begin{theorem}
+:$p \land q \vdash \neg \left({p \implies \neg q}\right)$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|p \land q \vdash \neg \left({p \implies \neg q}\right)}}
+{{Premise|1|p \land q}}
+{{Assumption|2|p \implies \neg q|Assume the opposite of what is to be proved ...}}
+{{Simplification|3|1|p|1|1}}
+{{Simplification|4|1|q|1|2}}
+{{ModusPonens|5|1, 2|\neg q|2|3}}
+{{NonContradiction|6|1, 2|4|5| ... and demonstrate a contradiction}}
+{{Contradiction|7|1|\neg \left({p \implies \neg q}\right)|2|6}}
+{{EndTableau}}
+{{qed}}
+[[Category:Conjunction Equivalent to Negation of Implication of Negative]]
+jautv9faa6k5bg2pu70mqy11iszhijj
+\end{proof}<|endoftext|>
+\section{Conjunction Equivalent to Negation of Implication of Negative/Formulation 1/Reverse Implication}
+Tags: Conjunction Equivalent to Negation of Implication of Negative
+
+\begin{theorem}
+:$\neg \left({p \implies \neg q}\right) \vdash p \land q$
+\end{theorem}<|endoftext|>
+\section{Rule of Material Implication/Formulation 1/Forward Implication}
+Tags: Rule of Material Implication
+
+\begin{theorem}
+:$p \implies q \vdash \neg p \lor q$
+\end{theorem}<|endoftext|>
+\section{Rule of Material Implication/Formulation 1}
+Tags: Rule of Material Implication
+
+\begin{theorem}
+:$p \implies q \dashv \vdash \neg p \lor q$
+\end{theorem}<|endoftext|>
+\section{Rule of Material Implication/Formulation 1/Reverse Implication}
+Tags: Rule of Material Implication
+
+\begin{theorem}
+:$\neg p \lor q \vdash p \implies q$
+\end{theorem}<|endoftext|>
+\section{Rule of Material Implication/Formulation 1/Proof}
+Tags: Rule of Material Implication, Truth Table Proofs
+
+\begin{theorem}
+:$p \implies q \dashv \vdash \neg p \lor q$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]].
+As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
+$\begin{array}{|ccc||cccc|} \hline
+p & \implies & q & \neg & p & \lor & q \\
+\hline
+F & T & F & T & F & T & F \\
+F & T & T & T & F & T & T \\
+T & F & F & F & T & F & F \\
+T & T & T & F & T & T & T \\
+\hline
+\end{array}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rule of Material Implication/Formulation 2}
+Tags: Rule of Material Implication
+
+\begin{theorem}
+:$\vdash \paren {p \implies q} \iff \paren {\neg p \lor q}$
+\end{theorem}<|endoftext|>
+\section{Modus Tollendo Ponens/Variant/Formulation 2}
+Tags: Modus Tollendo Ponens
+
+\begin{theorem}
+:$\vdash \paren {p \lor q} \iff \paren {\neg p \implies q}$
+\end{theorem}<|endoftext|>
+\section{Modus Tollendo Ponens/Variant/Formulation 1}
+Tags: Modus Tollendo Ponens
+
+\begin{theorem}
+:$p \lor q \dashv \vdash \neg p \implies q$
+\end{theorem}<|endoftext|>
+\section{Modus Tollendo Ponens/Sequent Form}
+Tags: Modus Tollendo Ponens
+
+\begin{theorem}
+==== [[Modus Tollendo Ponens/Sequent Form/Case 1|Case 1]] ====
+{{:Modus Tollendo Ponens/Sequent Form/Case 1}}
+==== [[Modus Tollendo Ponens/Sequent Form/Case 2|Case 2]] ====
+{{:Modus Tollendo Ponens/Sequent Form/Case 2}}
+\end{theorem}<|endoftext|>
+\section{Modus Tollendo Ponens/Variant/Formulation 1/Proof}
+Tags: Modus Tollendo Ponens, Truth Table Proofs
+
+\begin{theorem}
+:$p \lor q \dashv \vdash \neg p \implies q$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]].
+As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
+$\begin{array}{|ccc||cccc|} \hline
+p & \lor & q & \neg & p & \implies & q \\
+\hline
+F & F & F & T & F & F & F \\
+F & T & T & T & F & T & T \\
+T & T & F & F & T & T & F \\
+T & T & T & F & T & T & T \\
+\hline
+\end{array}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Modus Tollendo Ponens/Variant/Formulation 1/Forward Implication}
+Tags: Modus Tollendo Ponens
+
+\begin{theorem}
+:$p \lor q \vdash \neg p \implies q$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|p \lor q \vdash \neg p \implies q}}
+{{Premise|1|p \lor q}}
+{{Assumption|2|p|Pick the first of the disjuncts ...}}
+{{Assumption|3|\neg p|Assume its negation ...}}
+{{NonContradiction|4|2, 3|2|3|... which introduces a contradiction}}
+{{Explosion|5|2, 3|q|4| ... from a falsehood, ''any'' statement can be derived - pick $q$}}
+{{Implication|6|2|\neg p \implies q|3|5}}
+{{Assumption|7|q|Pick the second of the disjuncts ...}}
+{{Assumption|8|\neg p|... again assume the negation of $p$}}
+{{IdentityLaw|9|7|q|7|$q$ still holds}}
+{{Implication|10|7|\neg p \implies q|8|9}}
+{{ProofByCases|11|1|\neg p \implies q|1|2|6|7|10}}
+{{EndTableau}}
+{{qed}}
+[[Category:Modus Tollendo Ponens]]
+4cs2ci8pnukzednwbx1w4bdyc16iswy
+\end{proof}<|endoftext|>
+\section{Modus Tollendo Ponens/Variant/Formulation 1/Reverse Implication}
+Tags: Modus Tollendo Ponens
+
+\begin{theorem}
+:$\neg p \implies q \vdash p \lor q$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\neg p \implies q \vdash p \lor q}}
+{{Premise|1|\neg p \implies q}}
+{{ExcludedMiddle|2|p \lor \neg p}}
+{{Assumption|3|p}}
+{{Addition|4|3|p \lor q|3|1}}
+{{Assumption|5|\neg p}}
+{{ModusPonens|6|1, 5|q|1|5}}
+{{Addition|7|1, 5|p \lor q|6|2}}
+{{ProofByCases|8|1|p \lor q|2|3|4|5|7}}
+{{EndTableau|qed}}
+{{LEM}}
+\end{proof}<|endoftext|>
+\section{Radius of Convergence of Power Series over Factorial/Complex Case}
+Tags: Radius of Convergence of Power Series over Factorial
+
+\begin{theorem}
+Let $\xi \in \C$ be a [[Definition:Complex Number|complex number]].
+Let $\displaystyle \map f z = \sum_{n \mathop = 0}^\infty \dfrac {\paren {z - \xi}^n} {n!}$.
+Then $\map f z$ [[Definition:Absolutely Convergent Series|converges absolutely]] for all $z \in \C$.
+That is, the [[Definition:Radius of Convergence of Complex Power Series|radius of convergence]] of the [[Definition:Complex Power Series|power series]] $\displaystyle \sum_{n \mathop = 0}^\infty \frac {\paren {z - \xi}^n} {n!}$ is infinite.
+\end{theorem}
+
+\begin{proof}
+This is a [[Definition:Complex Power Series|power series]] in the form $\displaystyle \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$ where $\sequence {a_n} = \sequence {\dfrac 1 {n!} }$.
+Applying [[Radius of Convergence from Limit of Sequence/Complex Case|Radius of Convergence from Limit of Sequence: Complex Case]], we find that:
+{{begin-eqn}}
+{{eqn | l = \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1} } {a_n} }
+      | r = \lim_{n \mathop \to \infty} \cmod {\dfrac {\frac 1 {\paren {n + 1}!} } {\frac 1 {n!} } }
+      | c = 
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \cmod {\dfrac {n!} {\paren {n + 1}!} }
+      | c = 
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \cmod {\dfrac 1 {n + 1} }
+      | c = 
+}}
+{{eqn | r = 0
+      | c = [[Sequence of Powers of Reciprocals is Null Sequence]]
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Radius of Convergence of Power Series over Factorial/Real Case}
+Tags: Radius of Convergence of Power Series over Factorial
+
+\begin{theorem}
+Let $\xi \in \R$ be a [[Definition:Real Number|real number]].
+Let $\displaystyle f \left({x}\right) = \sum_{n \mathop = 0}^\infty \frac {\left({x - \xi}\right)^n} {n!}$.
+Then $f \left({x}\right)$ [[Definition:Convergent Function|converges]] for all $x \in \R$.
+That is, the [[Definition:Interval of Convergence|interval of convergence]] of the [[Definition:Power Series|power series]] $\displaystyle \sum_{n \mathop = 0}^\infty \frac {\left({x - \xi}\right)^n} {n!}$ is $\R$.
+\end{theorem}
+
+\begin{proof}
+This is a [[Definition:Power Series|power series]] in the form $\displaystyle \sum_{n \mathop= 0}^\infty a_n \left({x - \xi}\right)^n$ where $\left \langle {a_n} \right \rangle = \left \langle {\dfrac 1 {n!}} \right \rangle$.
+Applying [[Radius of Convergence from Limit of Sequence]], we find that:
+{{begin-eqn}}
+{{eqn | l = \lim_{n \mathop \to \infty} \left\vert{\dfrac {a_{n+1} } {a_n} }\right\vert
+      | r = \lim_{n \mathop \to \infty} \left\vert{\dfrac {\dfrac 1 {\left({n + 1}\right)!} } {\dfrac 1 {n!} } }\right\vert
+      | c = 
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \left\vert{\dfrac {n!} {\left({n + 1}\right)!} }\right\vert
+      | c = 
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \left\vert{\dfrac 1 {n+1} }\right\vert
+      | c = 
+}}
+{{eqn | r = 0
+      | c = [[Sequence of Powers of Reciprocals is Null Sequence]]
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Logarithms/Natural Logarithm}
+Tags: Sum of Logarithms
+
+\begin{theorem}
+Let $x, y \in \R$ be [[Definition:Strictly Positive Real Number|strictly positive real numbers]].
+Then:
+:$\ln x + \ln y = \map \ln {x y}$
+where $\ln$ denotes the [[Definition:General Logarithm|natural logarithm]].
+\end{theorem}<|endoftext|>
+\section{Sum of Logarithms/General Logarithm}
+Tags: Sum of Logarithms
+
+\begin{theorem}
+Let $x, y, b \in \R$ be [[Definition:Strictly Positive|strictly positive]] [[Definition:Real Number|real numbers]] such that $b > 1$.
+Then:
+:$\log_b x + \log_b y = \map {\log_b} {x y}$
+where $\log_b$ denotes [[Definition:General Logarithm|the logarithm to base $b$]].
+\end{theorem}<|endoftext|>
+\section{Rule of Simplification/Sequent Form/Formulation 1/Proof}
+Tags: Truth Table Proofs, Rule of Simplification
+
+\begin{theorem}
+:$(1): \quad p \land q \vdash p$
+:$(2): \quad p \land q \vdash q$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]].
+$\begin{array}{|ccc||c|c|} \hline
+p & \land & q & p & q \\
+\hline
+F & F & F & F & F \\
+F & F & T & F & T \\
+T & F & F & T & F \\
+T & T & T & T & T \\
+\hline
+\end{array}$
+As can be seen, when $p \land q$ is [[Definition:True|true]] so are both $p$ and $q$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rule of Simplification/Sequent Form}
+Tags: Rule of Simplification
+
+\begin{theorem}
+The [[Rule of Simplification]] can be symbolised by the [[Definition:Sequent|sequents]]:
+\end{theorem}<|endoftext|>
+\section{Rule of Conjunction/Sequent Form}
+Tags: Rule of Conjunction
+
+\begin{theorem}
+The [[Rule of Conjunction]] can be symbolised in [[Definition:Sequent|sequent]] form as follows:
+\end{theorem}<|endoftext|>
+\section{Radius of Convergence of Derivative of Complex Power Series}
+Tags: Power Series, Complex Analysis
+
+\begin{theorem}
+Let $\xi \in \C$.
+For all $z \in \C$, define the [[Definition:Complex Power Series|power series]]:
+: $\displaystyle S \paren z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$
+and:
+: $\displaystyle S' \paren z = \sum_{n \mathop = 1}^\infty n a_n \paren {z - \xi}^{n - 1}$
+Let $R$ be the [[Definition:Radius of Convergence of Complex Power Series|radius of convergence]] of $S \paren z$, and let $R'$ be the radius of convergence of $S' \paren z$.
+Then $R =R'$.
+\end{theorem}
+
+\begin{proof}
+Suppose that $z \in \C$ with $\cmod {z - \xi} < R'$.
+Then $S' \paren z$ [[Definition:Absolutely Convergent Series|converges absolutely]] by [[Existence of Radius of Convergence of Complex Power Series]], so:
+{{begin-eqn}}
+{{eqn | l = 1
+      | o = \ge
+      | r = \limsup_{n \mathop \to \infty} \cmod {n a_n \paren {z - \xi}^{n - 1} }^{1 / n}
+      | c = by [[Nth Root Test|$n$th Root Test]]
+}}
+{{eqn | o = >
+      | r = \limsup_{n \mathop \to \infty} \cmod {a_n \paren {z - \xi}^n }^{1/n}
+      | c = as $\cmod {n  \paren {z - \xi}^{n - 1} } > \cmod { \paren {z - \xi}^n }$ for all $n > \cmod {z - \xi}$
+}}
+{{end-eqn}}
+From the [[Nth Root Test|$n$th Root Test]], it follows that $S \paren z$ [[Definition:Absolutely Convergent Series|converges absolutely]].
+Hence, $R \ge R'$.
+{{qed|lemma}}
+Suppose that $z \in \C$ with $\cmod {z - \xi} < R$.
+Find $z_o \in \C$ such that $\cmod {z - \xi} < \cmod {z_0 - \xi} < R$, so $S \paren {z_0}$ [[Definition:Absolutely Convergent Series|converges absolutely]].
+From the [[Nth Root Test|$n$th Root Test]], it follows that $\cmod {a_n \paren {z_0 - \xi}^n }^{1/n} < 1$ for all $n \ge N$ for some $N \in \N$.
+Then:
+{{begin-eqn}}
+{{eqn | l = 1
+      | r = \limsup_{n \mathop \to \infty} \cmod n^{1/n}
+      | c = by [[Limit of Integer to Reciprocal Power]]
+}}
+{{eqn | o = >
+      | r = \limsup_{n \mathop \to \infty} \cmod n^{1/n} \cmod {\frac {z - \xi} {z_0 - \xi} }
+}}
+{{eqn | r = \limsup_{n \mathop \to \infty} \cmod {n \paren {\frac {z - \xi} {z_0 - \xi} }^{n - 1} }^{1/n} 
+      | c = as $\displaystyle \lim_{n \mathop \to \infty} \paren {\paren {\frac {z - \xi} {z_0 - \xi} }^{n - 1} }^{1/n} = 1$ 
+}}
+{{eqn | o = \ge
+      | r = \limsup_{n \mathop \to \infty} \cmod {n \paren {\frac {z - \xi} {z_0 - \xi} }^{n - 1} a_n \paren {z_0 - \xi}^{n - 1} }^{1/n}
+      | c = from the [[Nth Root Test|$n$th Root Test]], as remarked above
+}}
+{{eqn | r = \limsup_{n \to \mathop \infty} \cmod {n a_n \paren {z - \xi}^{n - 1} }^{1/n}
+}}
+{{end-eqn}}
+From the [[Nth Root Test|$n$th Root Test]], it follows that $S' \paren z$ [[Definition:Absolutely Convergent Series|converges absolutely]].
+Hence, $R' \ge R$, so $R' = R$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rule of Addition/Sequent Form}
+Tags: Disjunction, Rule of Addition
+
+\begin{theorem}
+The [[Rule of Addition]] can be symbolised by the [[Definition:Sequent|sequents]]:
+:$(1): \quad p \vdash p \lor q$
+:$(2): \quad q \vdash p \lor q$
+\end{theorem}<|endoftext|>
+\section{Rule of Addition/Sequent Form/Proof 2}
+Tags: Truth Table Proofs, Rule of Addition
+
+\begin{theorem}
+:$(1): \quad p \vdash p \lor q$
+:$(2): \quad q \vdash p \lor q$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]].
+$\begin{array}{|c|c||ccc|} \hline
+p & q & p & \lor & q\\
+\hline
+F & F & F & F & F \\
+F & T & F & T & T \\
+T & F & T & T & F \\
+T & T & T & T & T \\
+\hline
+\end{array}$
+As can be seen, whenever either $p$ or $q$ (or both) are [[Definition:True|true]], then so is $p \lor q$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rule of Addition/Sequent Form/Proof 1}
+Tags: Rule of Addition
+
+\begin{theorem}
+The [[Rule of Addition]] can be symbolised by the [[Definition:Sequent|sequents]]:
+:$(1): \quad p \vdash p \lor q$
+:$(2): \quad q \vdash p \lor q$
+\end{theorem}
+
+\begin{proof}
+=== [[Rule of Addition/Sequent Form/Proof 1/Form 1|Form 1]] ===
+{{:Rule of Addition/Sequent Form/Proof 1/Form 1}}
+=== [[Rule of Addition/Sequent Form/Proof 1/Form 2|Form 2]] ===
+{{:Rule of Addition/Sequent Form/Proof 1/Form 2}}
+\end{proof}<|endoftext|>
+\section{Rule of Addition/Sequent Form/Form 2/Proof 1}
+Tags: Rule of Addition
+
+\begin{theorem}
+:$q \vdash p \lor q$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|q \vdash p \lor q}}
+{{Premise|1|q}}
+{{Addition|2|1|p \lor q|1|2}}
+{{EndTableau}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Proof by Cases/Sequent Form}
+Tags: Proof by Cases
+
+\begin{theorem}
+[[Proof by Cases]] can be symbolised by the [[Definition:Sequent|sequent]]:
+:$p \lor q, \paren {p \vdash r}, \paren {q \vdash r} \vdash r$
+\end{theorem}<|endoftext|>
+\section{Modus Ponendo Ponens/Sequent Form}
+Tags: Modus Ponendo Ponens
+
+\begin{theorem}
+:$p \implies q, p \vdash q$
+\end{theorem}<|endoftext|>
+\section{Logarithm of Power/Natural Logarithm}
+Tags: Logarithm of Power
+
+\begin{theorem}
+Let $x \in \R$ be a [[Definition:Strictly Positive|strictly positive]] [[Definition:Real Number|real number]].
+Let $a \in \R$ be a [[Definition:Real Number|real number]] such that $a > 1$.
+Let $r \in \R$ be any [[Definition:Real Number|real number]].
+Let $\ln x$ be the [[Definition:Natural Logarithm|natural logarithm]] of $x$.
+Then:
+:$\map \ln {x^r} = r \ln x$
+\end{theorem}<|endoftext|>
+\section{Logarithm of Power/General Logarithm}
+Tags: Logarithm of Power
+
+\begin{theorem}
+Let $x \in \R$ be a [[Definition:Strictly Positive|strictly positive]] [[Definition:Real Number|real number]].
+Let $a \in \R$ be a [[Definition:Real Number|real number]] such that $a > 1$.
+Let $r \in \R$ be any [[Definition:Real Number|real number]].
+Let $\log_a x$ be the [[Definition:General Logarithm|logarithm to the base $a$]] of $x$.
+Then:
+:$\map {\log_a} {x^r} = r \log_a x$
+\end{theorem}
+
+\begin{proof}
+Let $y = r \log_a x$.
+Then:
+{{begin-eqn}}
+{{eqn | l = a^y
+      | r = a^{r \log_a x}
+      | c = 
+}}
+{{eqn | r = \paren {a^{\log_a x} }^r
+      | c = [[Exponent Combination Laws]]
+}}
+{{eqn | r = x^r
+      | c = {{Defof|General Logarithm|Logarithm base $a$}}
+}}
+{{end-eqn}}
+The result follows by [[Definition:General Logarithm|taking logs]] base $a$ of both sides.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Power Set with Union and Intersection forms Boolean Algebra}
+Tags: Power Set, Boolean Algebras
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]], and let $\powerset S$ be its [[Definition:Power Set|power set]].
+Denote with $\cup$, $\cap$ and $\complement$ the operations of [[Definition:Set Union|union]], [[Definition:Set Intersection|intersection]] and [[Definition:Complement|complement]] on $\powerset S$, respectively.
+Then $\struct {\powerset S, \cup, \cap, \complement}$ is a [[Definition:Boolean Algebra|Boolean algebra]].
+\end{theorem}
+
+\begin{proof}
+Taking the criteria for [[Definition:Boolean Algebra/Definition 1|definition 1 of a Boolean algebra]] in turn:
+=== $(\text {BA} 0):$ Closure ===
+$\powerset S$ is [[Definition:Closed Algebraic Structure|closed]] under both $\cup$ and $\cap$:
+:[[Power Set is Closed under Intersection]]
+:[[Power Set is Closed under Union]]
+:[[Power Set is Closed under Complement]]
+{{qed|lemma}}
+=== $(\text {BA} 1):$ Commutativity ===
+Both $\cup$ and $\cap$ are [[Definition:Commutative Operation|commutative]] from [[Intersection is Commutative]] and [[Union is Commutative]].
+{{qed|lemma}}
+=== $(\text {BA} 2):$ Distributivity ===
+Both $\cup$ and $\cap$ [[Definition:Distributive Operation|distribute]] over the other, from [[Union Distributes over Intersection]] and [[Intersection Distributes over Union]].
+{{qed|lemma}}
+=== $(\text {BA} 3):$ Identity Elements ===
+Both $\cup$ and $\cap$ have [[Definition:Identity Element|identities]]:
+From [[Power Set with Intersection is Monoid]], $S$ is the [[Definition:Identity Element|identity]] for $\cap$. 
+From [[Power Set with Union is Monoid]], $\O$ is the [[Definition:Identity Element|identity]] for $\cup$.
+{{qed|lemma}}
+=== $(\text {BA} 4):$ Complements ===
+From [[Union with Complement]]:
+:$\forall A \in S: A \cup \map \complement A = S$
+which is the [[Definition:Identity Element|identity]] for $\cap$.
+From [[Intersection with Complement]]:
+:$\forall A \in S: A \cap \map \complement A = \O$
+which is the [[Definition:Identity Element|identity]] for $\cup$.
+{{qed|lemma}}
+All the criteria for a [[Definition:Boolean Algebra/Definition 1|Boolean algebra]] are therefore fulfilled.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rule of Implication/Sequent Form}
+Tags: Rule of Implication
+
+\begin{theorem}
+The [[Rule of Implication]] can be symbolised by the [[Definition:Sequent|sequent]]:
+:$\left({p \vdash q}\right) \vdash p \implies q$
+\end{theorem}<|endoftext|>
+\section{Rule of Implication/Sequent Form/Proof 2}
+Tags: Truth Table Proofs, Rule of Implication
+
+\begin{theorem}
+:$\left({p \vdash q}\right) \vdash p \implies q$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]].
+$\begin{array}{|c|c||ccc|} \hline
+p & q & p & \implies & q\\
+\hline
+F & F & F & T & F \\
+F & T & F & T & T \\
+T & F & T & F & F \\
+T & T & T & T & T \\
+\hline
+\end{array}$
+As can be seen by inspection, only when $p$ is [[Definition:True|true]] and $q$ is [[Definition:False|false]], then so is $p \implies q$ also [[Definition:False|false]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Absolutely Convergent Series}
+Tags: Absolute Convergence
+
+\begin{theorem}
+Let $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ and $\displaystyle \sum_{n \mathop = 1}^\infty b_n$ be two [[Definition:Real Number|real]] or [[Definition:Complex Number|complex]] [[Definition:Series|series]] that are [[Definition:Absolutely Convergent Series|absolutely convergent]].
+Then the series $\displaystyle \sum_{n \mathop = 1}^\infty \paren {a_n + b_n}$ is [[Definition:Absolutely Convergent Series|absolutely convergent]], and:
+:$\displaystyle \sum_{n \mathop = 1}^\infty \paren {a_n + b_n} = \displaystyle \sum_{n \mathop = 1}^\infty a_n + \sum_{n \mathop = 1}^\infty b_n$
+\end{theorem}
+
+\begin{proof}
+Let $\epsilon \in \R_{>0}$.
+From [[Tail of Convergent Series tends to Zero]], it follows that there exists $M \in \N$ such that:
+:$\displaystyle \sum_{n \mathop = M + 1}^\infty \cmod {a_n} < \dfrac \epsilon 2$
+and:
+:$\displaystyle \sum_{n \mathop = M + 1}^\infty \cmod {b_n} < \dfrac \epsilon 2$
+For all $m \ge M$, it follows that:
+{{begin-eqn}}
+{{eqn | l = \cmod {\sum_{n \mathop = 1}^\infty a_n + \sum_{n \mathop = 1}^\infty b_n - \sum_{n \mathop = 1}^m \paren {a_n + b_n} }
+      | r = \cmod {\sum_{n \mathop = m + 1}^\infty a_n + \sum_{n \mathop = m + 1}^\infty b_n}
+}}
+{{eqn | o = \le
+      | r = \sum_{n \mathop = m + 1}^\infty \cmod {a_n} + \sum_{n \mathop = m + 1}^\infty \cmod {b_n}
+      | c = by [[Triangle Inequality]]
+}}
+{{eqn | o = \le
+      | r = \sum_{n \mathop = M + 1}^\infty \cmod {a_n} + \sum_{n \mathop = M + 1}^\infty \cmod {b_n}
+}}
+{{eqn | o = <
+      | r = \epsilon
+}}
+{{end-eqn}}
+By [[Definition:Convergent Series|definition of convergent series]], it follows that:
+{{begin-eqn}}
+{{eqn | l = \sum_{n \mathop = 1}^\infty a_n + \sum_{n \mathop = 1}^\infty b_n
+      | r = \lim_{m \mathop \to \infty} \sum_{n \mathop = 1}^m \paren {a_n + b_n}
+}}
+{{eqn | r = \sum_{n \mathop = 1}^\infty \paren {a_n + b_n}
+}}
+{{end-eqn}}
+To show that $\displaystyle \sum_{n \mathop = 1}^\infty \paren {a_n + b_n}$ is [[Definition:Absolutely Convergent Series|absolutely convergent]], note that:
+{{begin-eqn}}
+{{eqn | l = \sum_{n \mathop = 1}^\infty \cmod {a_n} + \sum_{n \mathop = 1}^\infty \cmod {b_n}
+      | r = \sum_{n \mathop = 1}^\infty \paren {\cmod {a_n} + \cmod {b_n} }
+      | c = as shown above
+}}
+{{eqn | o = \ge
+      | r = \sum_{n \mathop = 1}^\infty \cmod {a_n + b_n}
+      | c = by [[Triangle Inequality]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rule of Implication/Sequent Form/Proof 1}
+Tags: Rule of Implication
+
+\begin{theorem}
+:$\left({p \vdash q}\right) \vdash p \implies q$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\left({p \vdash q}\right) \vdash p \implies q}}
+{{Premise|1|p}}
+|-
+| align="right" | 2 ||
+| align="right" | 1
+| $q$
+| [[Definition:By Hypothesis|By hypothesis]]
+| 1
+| as $p \vdash q$
+{{Implication|3|1|p \implies q|1|2}}
+{{EndTableau}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Rule of Simplification/Sequent Form/Formulation 1}
+Tags: Rule of Simplification
+
+\begin{theorem}
+:$(1): \quad p \land q \vdash p$
+:$(2): \quad p \land q \vdash q$
+\end{theorem}<|endoftext|>
+\section{Rule of Simplification/Sequent Form/Formulation 2}
+Tags: Rule of Simplification
+
+\begin{theorem}
+:$(1): \quad \vdash p \land q \implies p$
+:$(2): \quad \vdash p \land q \implies q$
+\end{theorem}<|endoftext|>
+\section{Rule of Simplification/Sequent Form/Formulation 2/Proof 1/Form 1}
+Tags: Rule of Simplification
+
+\begin{theorem}
+:$\vdash p \land q \implies p$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\vdash p \land q \implies p}}
+{{Assumption|1|p \land q}}
+{{Simplification|2|1|p|1|1}}
+{{Implication|3||p \land q \implies p|1|2}}
+{{EndTableau}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Rule of Simplification/Sequent Form/Formulation 2/Proof 1/Form 2}
+Tags: Rule of Simplification
+
+\begin{theorem}
+:$\vdash p \land q \implies q$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\vdash p \land q \implies q}}
+{{Assumption|1|p \land q}}
+{{Simplification|2|1|q|1|2}}
+{{Implication|3||p \land q \implies q|1|2}}
+{{EndTableau}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Complement of Complement in Uniquely Complemented Lattice}
+Tags: Bounded Lattices
+
+\begin{theorem}
+Let $\left({S, \wedge, \vee, \preceq}\right)$ be a [[Definition:Uniquely Complemented Lattice|uniquely complemented lattice]].
+For each $x \in S$, let $\neg x$ be the [[Definition:Complement (Lattice Theory)|complement]] of $x$.
+Then for each $x \in S$:
+:$\neg \neg x = x$
+\end{theorem}
+
+\begin{proof}
+By the definition of a [[Definition:Complement (Lattice Theory)|complement]] of $x$:
+:$\neg x \vee x = \top$
+:$\neg x \wedge x = \bot$
+Since $\vee$ and $\wedge$ are [[Definition:Commutative Operation|commutative]]:
+:$x \vee \neg x = \top$
+:$x \wedge \neg x = \bot$
+Thus by the definition of [[Definition:Complement (Lattice Theory)|complement]], $x$ is a complement of $\neg x$.
+By the definition of a [[Definition:Uniquely Complemented Lattice|uniquely complemented lattice]], $x = \neg \neg x$.
+{{qed}}
+[[Category:Bounded Lattices]]
+762w9prslhhnoeouofbbnd3v1cv3473
+\end{proof}<|endoftext|>
+\section{De Morgan's Laws imply Uniquely Complemented Lattice is Boolean Lattice}
+Tags: Bounded Lattices
+
+\begin{theorem}
+Let $\left({S, \wedge, \vee, \preceq}\right)$ be a [[Definition:Uniquely Complemented Lattice|uniquely complemented lattice]].
+Then the following are equivalent:
+$(1):\quad \forall p, q \in S: \neg p \vee \neg q = \neg \left({p \wedge q}\right)$
+$(2):\quad \forall p, q \in S: \neg p \wedge \neg q = \neg \left({p \vee q}\right)$
+$(3):\quad \forall p, q \in S: p \preceq q \iff \neg q \preceq \neg p$
+$(4):\quad \left({S, \wedge, \vee, \preceq}\right)$ is a [[Definition:Distributive Lattice|distributive lattice]].
+\end{theorem}
+
+\begin{proof}
+=== $(1)$ implies $(2)$ ===
+Suppose:
+:$\forall p, q \in S: \neg p \vee \neg q = \neg \left({p \wedge q}\right)$
+Then applying this to $\neg p$ and $\neg q$:
+:$\neg \neg p \vee \neg \neg q = \neg \left({\neg p \wedge \neg q}\right)$
+By [[Complement of Complement in Uniquely Complemented Lattice]], $\neg \neg p = p$ and $\neg \neg q = q$.
+Thus:
+:$p \vee q = \neg \left({\neg p \wedge \neg q}\right)$.
+Taking [[Definition:Complement (Lattice Theory)|complements]] of both sides:
+:$\neg \left({p \vee q}\right) = \neg \neg \left({\neg p \wedge \neg q}\right)$
+Again applying [[Complement of Complement in Uniquely Complemented Lattice]]:
+:$\neg \left({p \vee q}\right) = \neg p \wedge \neg q$
+{{qed|lemma}}
+=== $(2)$ implies $(1)$ ===
+By [[Dual Pairs (Order Theory)]], $\wedge$ and $\vee$ are [[Definition:Dual Statement (Order Theory)|dual]].
+Thus this implication follows from the above by [[Duality Principle (Order Theory)|Duality]].
+{{qed|lemma}}
+=== $(1)$ implies $(3)$ ===
+By the definition of a [[Definition:Lattice/Definition 3|lattice]]:
+:$p \preceq q \iff p \vee q = q$
+Applying this to $\neg q$ and $\neg p$:
+:$\neg q \preceq \neg p \iff \neg q \vee \neg p = \neg p$
+By $(1)$:
+:$\neg q \vee \neg p = \neg \left({q \wedge p}\right)$
+So:
+:$\neg q \preceq \neg p \iff \neg \left({q \wedge p}\right) = \neg p$
+Taking the [[Definition:Complement (Lattice Theory)|complements]] of both sides of the equation on the right, and applying [[Complement of Complement in Uniquely Complemented Lattice]]:
+:$\neg q \preceq \neg p \iff q \wedge p = p$
+But the right side is equivalent to $p \preceq q$ {{explain|We define the ordering on a lattice (definition 3) based on joins. We need an equivalent one based on meets, or maybe we have it somewhere already.}}
+Therefore:
+:$\neg q \preceq \neg p \iff p \preceq q$
+{{qed|lemma}}
+=== $(3)$ implies $(1)$ ===
+{{improve|This is ugly}}
+{{MissingLinks}}
+Suppose that $p \preceq q \iff \neg q \preceq \neg p$
+By the definition of join:
+:$\neg p, \neg q \preceq \neg p \vee \neg q$
+Thus $\neg \left({\neg p \vee \neg q}\right) \preceq p, q$.
+By the definition of meet:
+:$\neg \left({\neg p \vee \neg q}\right) \preceq p \wedge q$
+Thus:
+:$\neg\left({p \wedge q}\right) \preceq \neg\neg \left({\neg p \vee \neg q}\right)$
+By [[Complement of Complement in Uniquely Complemented Lattice]]:
+$*\quad \neg\left({p \wedge q}\right) \preceq \neg p \vee \neg q$
+Dually:
+:$\neg x \wedge \neg y \preceq \neg \left({x \vee y}\right)$
+Letting $x = \neg p$ and $y = \neg q$:
+:$\neg \neg p \wedge \neg \neg q \preceq \neg \left({\neg p \vee \neg q}\right)$
+By [[Complement of Complement in Uniquely Complemented Lattice]]:
+:$p \wedge q \preceq \neg \left({\neg p \vee \neg q}\right)$
+By the premise and [[Complement of Complement in Uniquely Complemented Lattice]], then:
+$**\quad \neg p \vee \neg q \preceq \neg \left({p \wedge q}\right)$
+By $*$ and $**$:
+$\quad \neg\left({p \wedge q}\right) = \neg p \vee \neg q$
+{{qed|lemma}}
+=== $(1)$, $(2)$, and $(3)$ together imply $(4)$ ===
+$b, c \preceq b \vee c$, so
+:$a \wedge b \preceq a \wedge \left({b \vee c}\right)$
+:$a \wedge c \preceq a \wedge \left({b \vee c}\right)$
+By the definition of join:
+:$\left({a \wedge b}\right) \vee \left({a \wedge c}\right) \preceq a \wedge \left({b \vee c}\right)$
+{{finish}}
+\end{proof}<|endoftext|>
+\section{Complement of Bottom/Boolean Algebra}
+Tags: Boolean Algebras
+
+\begin{theorem}
+Let $\left({S, \vee, \wedge, \neg}\right)$ be a [[Definition:Boolean Algebra|Boolean algebra]].
+Then $\neg \bot = \top$.
+\end{theorem}
+
+\begin{proof}
+Since $\bot$ is the [[Definition:Identity Element|identity]] for $\vee$, the first condition for $\neg \bot$:
+:$\bot \vee \neg \bot = \top$
+implies that $\neg \bot = \top$ is the only possibility.
+Since $\top$ is the [[Definition:Identity Element|identity]] for $\wedge$, it follows that:
+:$\bot \wedge \top = \bot$
+and we conclude that:
+:$\neg \bot = \top$
+as desired.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complement of Bottom/Bounded Lattice}
+Tags: Bounded Lattices
+
+\begin{theorem}
+Let $\left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Bounded Lattice|bounded lattice]].
+Then the [[Definition:Bottom (Lattice Theory)|bottom]] $\bot$ has a [[Definition:Unique|unique]] [[Definition:Complement (Lattice Theory)|complement]], namely $\top$, [[Definition:Top (Lattice Theory)|top]].
+\end{theorem}
+
+\begin{proof}
+We know that $\bot$ is the [[Definition:Identity Element|identity]] for $\vee$.
+Therefore, from the condition that:
+:$\bot \vee a = \top$
+for a [[Definition:Complement (Lattice Theory)|complement]] $a$ of $\bot$, it follows that $a = \top$ is the only possibility.
+Since also:
+:$\bot \wedge \top = \bot$
+as $\top$ is the [[Definition:Identity Element|identity]] for $\wedge$, the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complement of Top/Bounded Lattice}
+Tags: Bounded Lattices
+
+\begin{theorem}
+Let $\left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Bounded Lattice|bounded lattice]].
+Then the [[Definition:Top (Lattice Theory)|top]] $\top$ has a [[Definition:Unique|unique]] [[Definition:Complement (Lattice Theory)|complement]], namely $\bot$, [[Definition:Top (Lattice Theory)|bottom]].
+\end{theorem}
+
+\begin{proof}
+By [[Dual Pairs (Order Theory)]], $\top$ is [[Definition:Dual Statement (Order Theory)|dual]] to $\bot$.
+The result follows from the [[Duality Principle (Order Theory)|Duality Principle]] and [[Complement of Bottom (Bounded Lattice)|Complement of Bottom]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complement of Top/Boolean Algebra}
+Tags: Boolean Algebras
+
+\begin{theorem}
+Let $\struct {S, \vee, \wedge, \neg}$ be a [[Definition:Boolean Algebra|Boolean algebra]].
+Then $\neg \top = \bot$.
+\end{theorem}
+
+\begin{proof}
+Since $\top$ is the [[Definition:Identity Element|identity]] for $\wedge$, the second condition for $\neg \top$:
+:$\top \wedge \neg \top = \bot$
+implies that $\neg \top = \bot$ is the only possibility.
+Since $\bot$ is the [[Definition:Identity Element|identity]] for $\vee$, it follows that:
+:$\top \vee \bot = \top$
+and we conclude that:
+:$\neg \top = \bot$
+as desired.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Duality Principle (Boolean Algebras)}
+Tags: Boolean Algebras, Named Theorems
+
+\begin{theorem}
+Let $\struct {S, \vee, \wedge}$ be a [[Definition:Boolean Algebra|Boolean algebra]].
+Then any theorem in $\struct {S, \vee, \wedge}$ remains valid if both $\vee$ and $\wedge$ are interchanged, and also $\bot$ and $\top$ are interchanged throughout the whole theorem.
+\end{theorem}
+
+\begin{proof}
+Let us take the axioms of a [[Definition:Boolean Algebra|Boolean algebra]] $\struct {S, \wedge, \vee}$:
+{{begin-axiom}}
+{{axiom | n = \text {BA} 0
+        | lc=
+        | t = $S$ is [[Definition:Closed Algebraic Structure|closed]] under both $\vee$ and $\wedge$
+}}
+{{axiom | n = \text {BA} 1
+        | lc=
+        | t = Both $\vee$ and $\wedge$ are [[Definition:Commutative Operation|commutative]]
+}}
+{{axiom | n = \text {BA} 2
+        | lc=
+        | t = Both $\vee$ and $\wedge$ [[Definition:Distributive Operation|distribute]] over the other
+}}
+{{axiom | n = \text {BA} 3
+        | lc=
+        | t = Both $\vee$ and $\wedge$ have [[Definition:Identity Element|identities]] $\bot$ and $\top$ respectively
+}}
+{{axiom | n = \text {BA} 4
+        | lc=
+        | t = $\forall a \in S: \exists \neg a \in S: a \vee \neg a = \top, a \wedge \neg a = \bot$
+}}
+{{end-axiom}}
+It can be seen by inspection, that exchanging $\wedge$ and $\vee$, and $\bot$ and $\top$ throughout does not change the axioms.
+Thus, what you get is a [[Definition:Boolean Algebra|Boolean algebra]] again.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Identities of Boolean Algebra also Zeroes}
+Tags: Boolean Algebras
+
+\begin{theorem}
+Let $\struct {S, \vee, \wedge, \neg}$ be a [[Definition:Boolean Algebra/Definition 1|Boolean algebra, defined as in Definition 1]].
+Let the [[Definition:Identity Element|identity]] for $\vee$ be $\bot$ and the [[Definition:Identity Element|identity]] for $\wedge$ be $\top$.
+Then:
+:$(1): \quad \forall x \in S: x \vee \top = \top$
+:$(2): \quad \forall x \in S: x \wedge \bot = \bot$
+That is, $\bot$ is a [[Definition:Zero Element|zero element]] for $\wedge$, and $\top$ is a [[Definition:Zero Element|zero element]] for $\vee$.
+\end{theorem}
+
+\begin{proof}
+Let $x \in S$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x \vee \top
+      | r = \paren {x \vee \top} \wedge \top
+      | c = [[Definition:Boolean Algebra/Axioms/Definition 1|Boolean Algebra: Axiom $(\text {BA}_1 3)$]]: $\top$ is the [[Definition:Identity Element|identity]] of $\wedge$
+}}
+{{eqn | r = \paren {x \vee \top} \wedge \paren {x \vee \neg x}
+      | c = [[Definition:Boolean Algebra/Axioms/Definition 1|Boolean Algebra: Axiom $(\text {BA}_1 4)$]]: $x \vee x' = \top$
+}}
+{{eqn | r = x \vee \paren {\top \wedge x'}
+      | c = [[Definition:Boolean Algebra/Axioms/Definition 1|Boolean Algebra: Axiom $(\text {BA}_1 2)$]]: both $\vee$ and $\wedge$ [[Definition:Distributive Operation|distribute]] over the other
+}}
+{{eqn | r = x \vee x'
+      | c = [[Definition:Boolean Algebra/Axioms/Definition 1|Boolean Algebra: Axiom $(\text {BA}_1 3)$]]: $\top$ is the [[Definition:Identity Element|identity]] of $\wedge$
+}}
+{{eqn | r = \top
+      | c = [[Definition:Boolean Algebra/Axioms/Definition 1|Boolean Algebra: Axiom $(\text {BA}_1 4)$]] $x \vee x' = \top$
+}}
+{{end-eqn}}
+So $x \vee \top = \top$.
+{{qed|lemma}}
+The result $x \wedge \bot = \bot$ follows from the [[Duality Principle (Boolean Algebras)|Duality Principle]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complement of Complement (Boolean Algebras)}
+Tags: Boolean Algebras
+
+\begin{theorem}
+Let $\left({S, \vee, \wedge, \neg}\right)$ be a [[Definition:Boolean Algebra|Boolean algebra]].
+Then for all $a \in S$:
+:$\neg (\neg a) = a$
+\end{theorem}
+
+\begin{proof}
+Follows directly from [[Complement in Boolean Algebra is Unique]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{De Morgan's Laws (Boolean Algebras)}
+Tags: Boolean Algebras, De Morgan's Laws
+
+\begin{theorem}
+:$\neg \paren {a \vee b} = \neg a \wedge \neg b$
+:$\neg \paren {a \wedge b} = \neg a \vee \neg b$
+\end{theorem}
+
+\begin{proof}
+By virtue of [[Complement in Boolean Algebra is Unique]], it will suffice to verify:
+:$\paren {a \vee b} \wedge \paren {\neg a \wedge \neg b} = \bot$
+:$\paren {a \vee b} \vee \paren {\neg a \wedge \neg b} = \top$
+For the first of these, compute:
+{{begin-eqn}}
+{{eqn | l = \paren {a \vee b} \wedge \paren {\neg a \wedge \neg b}
+      | r = \paren {\paren {a \vee b} \wedge \neg a} \wedge \neg b
+      | c = $\wedge$ is [[Definition:Associative|associative]]
+}}
+{{eqn | r = \paren {\paren {a \wedge \neg a} \vee \paren {b \wedge \neg a} } \wedge \neg b
+      | c = $\wedge$ [[Definition:Distributive Operation|distributes]] over $\vee$
+}}
+{{eqn | r = \paren {\bot \vee \paren {b \wedge \neg a} } \wedge \neg b
+      | c = Axiom $(\text {BA} \ 4)$ for [[Definition:Boolean Algebra|Boolean algebras]]
+}}
+{{eqn | r = \paren {b \wedge \neg a} \wedge \neg b
+      | c = $\bot$ is the [[Definition:Identity Element|identity]] for $\vee$
+}}
+{{eqn | r = \neg a \wedge \paren {b \wedge \neg b}
+      | c = $\wedge$ is [[Definition:Commutative Operation|commutative]] and [[Definition:Associative|associative]]
+}}
+{{eqn | r = \neg a \wedge \bot
+      | c = Axiom $(\text {BA} \ 4)$
+}}
+{{eqn | r = \bot
+      | c = [[Identities of Boolean Algebra also Zeroes]]
+}}
+{{end-eqn}}
+By the [[Duality Principle (Boolean Algebras)|Duality Principle]], we also conclude:
+:$\paren {a \wedge b} \vee \paren {\neg a \vee \neg b} = \top$
+Substituting $\neg a$ and $\neg b$ for $a$ and $b$, respectively, this becomes:
+:$\paren {\neg a \wedge \neg b} \vee \paren {a \vee b} = \top$
+where [[Complement of Complement (Boolean Algebras)]] was applied.
+Since $\vee$ is [[Definition:Commutative Operation|commutative]], we have also derived the second of the two equations above.
+Hence:
+:$\neg \paren {a \vee b} = \neg a \wedge \neg b$
+{{qed|lemma}}
+Now by again the [[Duality Principle (Boolean Algebras)|Duality Principle]]:
+:$\neg \paren {a \wedge b} = \neg a \vee \neg b$
+follows immediately.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complement in Boolean Algebra is Unique}
+Tags: Boolean Algebras
+
+\begin{theorem}
+Let $\left({S, \vee, \wedge}\right)$ be a [[Definition:Boolean Algebra|Boolean algebra]].
+Then for all $a \in S$, there is a [[Definition:Unique|unique]] $b \in S$ such that:
+:$a \wedge b = \bot, a \vee b = \top$
+i.e., a valid choice for $\neg a$ as in axiom $(BA \ 4)$ for [[Definition:Boolean Algebra|Boolean algebras]].
+\end{theorem}
+
+\begin{proof}
+Suppose $b, c \in S$ both satisfy the identities.
+Then:
+{{begin-eqn}}
+{{eqn | l = b
+      | r = b \wedge \top
+      | c = $\top$ is the [[Definition:Identity Element|identity]] for $\wedge$
+}}
+{{eqn | r = b \wedge \left({a \vee c}\right)
+      | c = by hypothesis
+}}
+{{eqn | r = \left({b \wedge a}\right) \vee \left({b \wedge c}\right)
+      | c = $\wedge$ [[Definition:Distributive Operation|distributes]] over $\vee$
+}}
+{{eqn | r = \bot \vee \left({b \wedge c}\right)
+      | c = by hypothesis
+}}
+{{eqn | r = \left({a \wedge c}\right) \vee \left({b \wedge c}\right)
+      | c = Axiom $(BA \ 4)$
+}}
+{{eqn | r = \left({a \vee b}\right) \wedge c
+      | c = $\wedge$ [[Definition:Distributive Operation|distributes]] over $\vee$
+}}
+{{eqn | r = \top \wedge c
+      | c = by hypothesis
+}}
+{{eqn | r = c
+      | c = $\top$ is the [[Definition:Identity Element|identity]] for $\wedge$
+}}
+{{end-eqn}}
+That is, $b = c$.
+The result follows.
+{{qed}}
+[[Category:Boolean Algebras]]
+4vs574sm211kodsk7csg2129n0tmgmh
+\end{proof}<|endoftext|>
+\section{Cancellation of Join in Boolean Algebra}
+Tags: Boolean Algebras
+
+\begin{theorem}
+Let $\left({S, \vee, \wedge, \neg}\right)$ be a [[Definition:Boolean Algebra|Boolean algebra]].
+Let $a, b, c \in S$, and suppose that:
+{{begin-eqn}}
+{{eqn | l = a \vee c
+      | r = b \vee c
+}}
+{{eqn | l = a \vee \neg c
+      | r = b \vee \neg c
+}}
+{{end-eqn}}
+Then $a = b$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = a
+      | r = a \vee \bot
+      | c = $\bot$ is the [[Definition:Identity Element|identity]] for $\vee$
+}}
+{{eqn | r = a \vee \left({c \wedge \neg c}\right)
+      | c = [[Definition:Boolean Algebra/Axioms/Definition 1|Boolean Algebra: Axiom $(BA_1 \ 4)$]]
+}}
+{{eqn | r = \left({a \vee c}\right) \wedge \left({a \vee \neg c}\right)
+      | c = $\vee$ [[Definition:Distributive Operation|distributes]] over $\wedge$
+}}
+{{eqn | r = \left({b \vee c}\right) \wedge \left({b \vee \neg c}\right)
+      | c = by hypothesis
+}}
+{{eqn | r = b \vee \left({c \wedge \neg c}\right)
+      | c = $\vee$ [[Definition:Distributive Operation|distributes]] over $\wedge$
+}}
+{{eqn | r = b \vee \bot
+      | c = [[Definition:Boolean Algebra/Axioms/Definition 1|Boolean Algebra: Axiom $(BA_1 \ 4)$]]
+}}
+{{eqn | r = b
+      | c = $\bot$ is the [[Definition:Identity Element|identity]] for $\vee$
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cancellation of Meet in Boolean Algebra}
+Tags: Boolean Algebras
+
+\begin{theorem}
+Let $\struct {S, \vee, \wedge, \neg}$ be a [[Definition:Boolean Algebra|Boolean algebra]].
+Let $a, b, c \in S$.
+Let:
+{{begin-eqn}}
+{{eqn | l = a \wedge c
+      | r = b \wedge c
+}}
+{{eqn | l = a \wedge \neg c
+      | r = b \wedge \neg c
+}}
+{{end-eqn}}
+Then:
+: $a = b$
+\end{theorem}
+
+\begin{proof}
+Follows from [[Cancellation of Join in Boolean Algebra]] through the [[Duality Principle (Boolean Algebras)|Duality Principle]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Operations of Boolean Algebra are Associative}
+Tags: Boolean Algebras
+
+\begin{theorem}
+Let $\struct {S, \vee, \wedge, \neg}$ be a [[Definition:Boolean Algebra/Definition 1|Boolean algebra, defined as in Definition 1]].
+Then:
+:$\forall a, b, c \in S: \paren {a \wedge b} \wedge c = a \wedge \paren {b \wedge c}$
+:$\forall a, b, c \in S: \paren {a \vee b} \vee c = a \vee \paren {b \vee c}$
+That is, both $\vee$ and $\wedge$ are [[Definition:Associative Operation|associative operations]].
+\end{theorem}
+
+\begin{proof}
+Let $a, b, c \in S$.
+Let:
+:$x = a \wedge \paren {b \wedge c}$
+:$y = \paren {a \wedge b} \wedge c$
+Then:
+{{begin-eqn}}
+{{eqn | l = a \vee x
+      | r = a \vee \paren {a \wedge \paren {b \wedge c} }
+      | c = 
+}}
+{{eqn |r = \paren {a \vee a} \wedge \paren {a \vee \paren {b \wedge c} }
+      | c = [[Definition:Boolean Algebra/Axioms/Definition 1|Boolean Algebra: Axiom $(\text {BA}_1 \ 2)$]]: both $\vee$ and $\wedge$ [[Definition:Distributive Operation|distribute]] over the other
+}}
+{{eqn | r = a \wedge \paren {a \vee \paren {b \wedge c} }
+      | c = [[Operations of Boolean Algebra are Idempotent]]
+}}
+{{eqn | r = a
+      | c = [[Absorption Laws (Boolean Algebras)]]
+}}
+{{end-eqn}}
+Similarly:
+{{begin-eqn}}
+{{eqn | l = a \vee y
+      | r = a \vee \paren {\paren {a \wedge b} \wedge c}
+      | c = 
+}}
+{{eqn | r = \paren {a \vee \paren {a \wedge b} } \wedge \paren {a \vee c}
+      | c = [[Definition:Boolean Algebra/Axioms/Definition 1|Boolean Algebra: Axiom $(\text {BA}_1 \ 2)$]]: both $\vee$ and $\wedge$ [[Definition:Distributive Operation|distribute]] over the other
+}}
+{{eqn | r = a \wedge \paren {a \vee c}
+      | c = [[Absorption Laws (Boolean Algebras)]]
+}}
+{{eqn | r = a
+      | c = [[Absorption Laws (Boolean Algebras)]]
+}}
+{{end-eqn}}
+Thus we see we have $a \vee x = a \vee y$.
+Next:
+{{begin-eqn}}
+{{eqn | l = \neg a \vee x
+      | r = \neg a \vee \paren {a \wedge \paren {b \wedge c} }
+      | c = 
+}}
+{{eqn | r = \paren {\neg a \vee a} \wedge \paren {\neg a \vee \paren {b \wedge c} }
+      | c = [[Definition:Boolean Algebra/Axioms/Definition 1|Boolean Algebra: Axiom $(\text {BA}_1 \ 2)$]]: both $\vee$ and $\wedge$ [[Definition:Distributive Operation|distribute]] over the other
+}}
+{{eqn | r = \top \wedge \paren {\neg a \vee \paren {b \wedge c} }
+      | c = [[Definition:Boolean Algebra/Axioms/Definition 1|Boolean Algebra: Axiom $(\text {BA}_1 \ 4)$]]: $\neg a \vee a = \top$
+}}
+{{eqn | r = \neg a \vee \paren {b \wedge c}
+      | c = [[Definition:Boolean Algebra/Axioms/Definition 1|Boolean Algebra: Axiom $(\text {BA}_1 \ 3)$]]: $\top$ is the [[Definition:Identity Element|identity]] of $\wedge$
+}}
+{{end-eqn}}
+Similarly:
+{{begin-eqn}}
+{{eqn | l = \neg a \vee y
+      | r = \neg a \vee \paren {\paren {a \wedge b} \wedge c}
+      | c = 
+}}
+{{eqn | r = \paren {\neg a \vee \paren {a \wedge b} } \wedge \paren {\neg a \vee c}
+      | c = [[Definition:Boolean Algebra/Axioms/Definition 1|Boolean Algebra: Axiom $(\text {BA}_1 \ 2)$]]: both $\vee$ and $\wedge$ [[Definition:Distributive Operation|distribute]] over the other
+}}
+{{eqn | r = \paren {\paren {\neg a \vee a} \wedge \paren {\neg a \vee b} } \wedge \paren {\neg a \vee c}
+      | c = [[Definition:Boolean Algebra/Axioms/Definition 1|Boolean Algebra: Axiom $(\text {BA}_1 \ 2)$]]: both $\vee$ and $\wedge$ [[Definition:Distributive Operation|distribute]] over the other
+}}
+{{eqn | r = \paren {\top \wedge \paren {\neg a \vee b} } \wedge \paren {\neg a \vee c}
+      | c = [[Definition:Boolean Algebra/Axioms/Definition 1|Boolean Algebra: Axiom $(\text {BA}_1 \ 4)$]]: $\neg a \vee a = \top$
+}}
+{{eqn | r = \paren {\neg a \vee b} \wedge \paren {\neg a \vee c}
+      | c = [[Definition:Boolean Algebra/Axioms/Definition 1|Boolean Algebra: Axiom $(\text {BA}_1 \ 3)$]]: $\top$ is the [[Definition:Identity Element|identity]] of $\wedge$
+}}
+{{eqn | r = \neg a \vee \paren {b \wedge c}
+      | c = [[Definition:Boolean Algebra/Axioms/Definition 1|Boolean Algebra: Axiom $(\text {BA}_1 \ 2)$]]: both $\vee$ and $\wedge$ [[Definition:Distributive Operation|distribute]] over the other
+}}
+{{end-eqn}}
+Thus we see we have $\neg a \vee x = \neg a \vee y$.
+In conclusion, we have:
+:$a \vee x = a \vee y$
+:$\neg a \vee x = \neg a \vee y$
+Hence $x = y$ by [[Cancellation of Join in Boolean Algebra]], that is:
+:$\paren {a \wedge b} \wedge c = a \wedge \paren {b \wedge c}$
+{{qed|lemma}}
+The result:
+:$\paren {a \vee b} \vee c = a \vee \paren {b \vee c}$
+follows from the [[Duality Principle (Boolean Algebras)|Duality Principle]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Absorption Laws (Boolean Algebras)}
+Tags: Boolean Algebras
+
+\begin{theorem}
+Let $\struct {S, \vee, \wedge}$ be a [[Definition:Boolean Algebra/Definition 1|Boolean algebra, defined as in Definition 1]].
+Then for all $a, b \in S$:
+:$a = a \vee \paren {a \wedge b}$
+:$a = a \wedge \paren {a \vee b}$
+That is, $\vee$ [[Definition:Absorb|absorbs]] $\wedge$, and $\wedge$ [[Definition:Absorb|absorbs]] $\vee$.
+\end{theorem}
+
+\begin{proof}
+Let $a, b \in S$.
+Then:
+{{begin-eqn}}
+{{eqn | l = a \vee \paren {a \wedge b}
+      | r = \paren {a \wedge \top} \vee \paren {a \wedge b}
+      | c = [[Definition:Boolean Algebra/Axioms/Definition 1|Boolean Algebra: Axiom $(BA_1 \ 3)$]]: $\top$ is the [[Definition:Identity Element|identity]] for $\wedge$
+}}
+{{eqn | r = a \wedge \paren {\top \vee b}
+      | c = [[Definition:Boolean Algebra/Axioms/Definition 2|Boolean Algebra: Axiom $(BA_1 \ 2)$]]: $\wedge$ [[Definition:Distributive Operation|distributes]] over $\vee$
+}}
+{{eqn | r = a \wedge \top
+      | c = [[Identities of Boolean Algebra also Zeroes]]
+}}
+{{eqn | r = a
+      | c = [[Definition:Boolean Algebra/Axioms/Definition 1|Boolean Algebra: Axiom $(BA_1 \ 3)$]] $\top$ is the [[Definition:Identity Element|identity]] for $\wedge$
+}}
+{{end-eqn}}
+as desired.
+{{qed|lemma}}
+The result:
+:$a = a \wedge \paren {a \vee b}$
+follows from the [[Duality Principle (Boolean Algebras)|Duality Principle]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Difference of Absolutely Convergent Series}
+Tags: Absolute Convergence
+
+\begin{theorem}
+Let $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ and $\displaystyle \sum_{n \mathop = 1}^\infty b_n$ be two [[Definition:Real Number|real]] or [[Definition:Complex Number|complex]] [[Definition:Series|series]] that are [[Definition:Absolutely Convergent Series|absolutely convergent]].
+Then the series $\displaystyle \sum_{n \mathop = 1}^\infty \paren {a_n - b_n}$ is [[Definition:Absolutely Convergent Series|absolutely convergent]], and:
+:$\displaystyle \sum_{n \mathop = 1}^\infty \paren {a_n - b_n} = \sum_{n \mathop = 1}^\infty a_n - \sum_{n \mathop = 1}^\infty b_n$
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Series|series]] $\displaystyle \sum_{n \mathop = 1}^\infty \paren {-b_n}$ is [[Definition:Absolutely Convergent Series|absolutely convergent]], as $\cmod {-b_n} = \cmod {b_n}$ for all $n \in \N$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \sum_{n \mathop = 1}^\infty a_n - \sum_{n \mathop = 1}^\infty b_n
+      | r = \sum_{n \mathop = 1}^\infty a_n + \paren {-1} \sum_{n \mathop = 1}^\infty b_n
+}}
+{{eqn | r = \sum_{n \mathop = 1}^\infty a_n + \sum_{n \mathop = 1}^\infty \paren {-1} b_n
+      | c = [[Manipulation of Absolutely Convergent Series/Scale Factor|Manipulation of Absolutely Convergent Series: Scale Factor]]
+}}
+{{eqn | r = \sum_{n \mathop = 1}^\infty \paren {a_n - b_n}
+      | c = [[Sum of Absolutely Convergent Series]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{True Statement is implied by Every Statement/Formulation 1}
+Tags: True Statement is implied by Every Statement
+
+\begin{theorem}
+:$p \vdash q \implies p$
+\end{theorem}<|endoftext|>
+\section{True Statement is implied by Every Statement/Formulation 2}
+Tags: True Statement is implied by Every Statement
+
+\begin{theorem}
+:$\vdash q \implies \paren {p \implies q}$
+\end{theorem}<|endoftext|>
+\section{True Statement is implied by Every Statement}
+Tags: Paradoxes of Material Implication, Implication, True Statement is implied by Every Statement
+
+\begin{theorem}
+==== [[True Statement is implied by Every Statement/Formulation 1|Formulation 1]] ====
+{{:True Statement is implied by Every Statement/Formulation 1}}
+==== [[True Statement is implied by Every Statement/Formulation 2|Formulation 2]] ====
+{{:True Statement is implied by Every Statement/Formulation 2}}
+\end{theorem}<|endoftext|>
+\section{False Statement implies Every Statement}
+Tags: Implication, Paradoxes of Material Implication, False Statement implies Every Statement
+
+\begin{theorem}
+==== [[False Statement implies Every Statement/Formulation 1|Formulation 1]] ====
+{{:False Statement implies Every Statement/Formulation 1}}
+==== [[False Statement implies Every Statement/Formulation 2|Formulation 2]] ====
+{{:False Statement implies Every Statement/Formulation 2}}
+\end{theorem}<|endoftext|>
+\section{False Statement implies Every Statement/Formulation 1}
+Tags: False Statement implies Every Statement
+
+\begin{theorem}
+: $\neg p \vdash p \implies q$
+\end{theorem}<|endoftext|>
+\section{False Statement implies Every Statement/Formulation 1/Proof 1}
+Tags: False Statement implies Every Statement
+
+\begin{theorem}
+:$\neg p \vdash p \implies q$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\neg p \vdash p \implies q}}
+{{Premise|1|\neg p}}
+{{Addition|2|1|\neg p \lor q|1|1}}
+{{SequentIntro|3|1|p \implies q|2|[[Rule of Material Implication/Formulation 1/Reverse Implication|Rule of Material Implication]]}}
+{{EndTableau}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{False Statement implies Every Statement/Formulation 1/Proof 2}
+Tags: False Statement implies Every Statement, Truth Table Proofs
+
+\begin{theorem}
+:$\neg p \vdash p \implies q$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]].
+As can be seen by inspection, where the [[Definition:Truth Value|truth value]] in the relevant column on the {{LHS}} is $T$, that under the one on the {{RHS}} is also $T$:
+$\begin{array}{|cc||ccc|} \hline
+\neg & p & p & \implies & q \\
+\hline
+T & F & F & T & F \\
+T & F & F & T & T \\
+F & F & T & F & F \\
+F & F & T & T & T \\
+\hline
+\end{array}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{False Statement implies Every Statement/Formulation 2}
+Tags: False Statement implies Every Statement
+
+\begin{theorem}
+:$\vdash \neg p \implies \paren {p \implies q}$
+\end{theorem}<|endoftext|>
+\section{Smullyan's Drinking Principle}
+Tags: Paradoxes, Logic
+
+\begin{theorem}
+Suppose that at least one person is in the pub.
+Then there is a person $x$ in the pub with the property that if $x$ is drinking, then everyone in the pub is drinking.
+\end{theorem}
+
+\begin{proof}
+There are two cases:
+:$(1): \quad$ Everyone in the pub is drinking.
+:$(2): \quad$ Someone in the pub is not drinking.
+Suppose first that everyone in the pub is drinking.
+Then $x$ can be chosen to be any person in the pub.
+Suppose instead that someone in the pub is not drinking.
+Then $x$ can be chosen to be any person in the pub who is not drinking.
+{{qed}}
+{{Namedfor|Raymond Merrill Smullyan|cat = Smullyan}}
+[[Category:Paradoxes]]
+[[Category:Logic]]
+hv93wwf21gkw1kb9kszxod78i3sag3o
+\end{proof}<|endoftext|>
+\section{Modus Tollendo Ponens/Sequent Form/Case 1}
+Tags: Modus Tollendo Ponens
+
+\begin{theorem}
+:$p \lor q, \neg p \vdash q$
+\end{theorem}<|endoftext|>
+\section{Modus Tollendo Ponens/Sequent Form/Case 2}
+Tags: Modus Tollendo Ponens
+
+\begin{theorem}
+:$p \lor q, \neg q \vdash p$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|p \lor q, \neg q \vdash p}}
+{{Premise|1|p \lor q}}
+{{Premise|2|\neg q}}
+{{Assumption|3|q}}
+{{SequentIntro|4|2|q \implies p|2|[[False Statement implies Every Statement/Formulation 1|False Statement implies Every Statement]]}}
+{{ModusPonens|5|2, 3|p|4|3}}
+{{Assumption|6|p}}
+{{ProofByCases|7|1, 2|p|1|3|5|6|6}}
+{{EndTableau}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Two is Boolean Algebra}
+Tags: Boolean Algebras
+
+\begin{theorem}
+Let $\mathbf 2$ denote [[Definition:Two (Boolean Algebra)|two]].
+Then $\mathbf 2$ is a [[Definition:Boolean Algebra|Boolean algebra]].
+\end{theorem}
+
+\begin{proof}
+It is useful to first state the [[Definition:Cayley Table|Cayley tables]] for the three logical operations $\lor$, $\land$ and $\neg$:
+:$\begin{array}{c|cc}
+ \lor & \bot & \top \\ \hline
+ \bot & \bot & \top \\
+ \top & \top & \top
+\end{array} \qquad \begin{array}{c|cc}
+ \land & \bot & \top \\ \hline
+ \bot  & \bot & \bot \\
+ \top  & \bot & \top
+\end{array} \qquad \begin{array}{c|cc}
+      & \bot & \top \\ \hline
+ \neg & \top & \bot
+\end{array}$
+Let us now verify the axioms for a [[Definition:Boolean Algebra|Boolean algebra]] in turn.
+=== $(BA \ 0)$: Closure ===
+It is immediate from the [[Definition:Cayley Table|Cayley tables]] that $S$ is [[Definition:Closed Algebraic System|closed]] under $\lor$, $\land$ and $\neg$.
+{{qed|lemma}}
+=== $(BA \ 1)$: Commutativity ===
+Follows from the [[Rule of Commutation]].
+{{qed|lemma}}
+=== $(BA \ 2)$: Distributivity ===
+Follows from the [[Rule of Distribution]]
+{{qed|lemma}}
+=== $(BA \ 3)$: Identities ===
+Follows from [[Conjunction with Tautology]] and [[Disjunction with Contradiction]].
+{{qed|lemma}}
+=== $(BA \ 4)$: Complements ===
+Follows from [[Contradiction is Negation of Tautology]] and [[Tautology is Negation of Contradiction]].
+{{qed|lemma}}
+Having verified all axioms, we conclude $\mathbf 2$ is a [[Definition:Boolean Algebra|Boolean algebra]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Two-Valued Functions form Boolean Algebra}
+Tags: Boolean Algebras
+
+\begin{theorem}
+Let $\mathbf 2$ be the [[Definition:Boolean Algebra|Boolean algebra]] [[Definition:Two (Boolean Algebra)|two]], and let $X$ be a [[Definition:Set|set]].
+Let $\mathbf 2^X$ be the [[Definition:Set of All Mappings|set of all mappings]] $p: X \to \mathbf 2$.
+Define the operations $\vee$, $\wedge$ and $\neg$ on $\mathbf 2^X$ in [[Definition:Pointwise Operation|pointwise]] fashion thus:
+:$\vee: \mathbf 2^X \times \mathbf 2^X \to \mathbf 2^X, \left({p \vee q}\right) (x) := p (x) \vee q (x)$
+:$\wedge: \mathbf 2^X \times \mathbf 2^X \to \mathbf 2^X, \left({p \wedge q}\right) (x) := p (x) \wedge q (x)$
+:$\neg: \mathbf 2^X \to \mathbf 2^X, \left({\neg p}\right) (x) := \neg p (x)$
+Furthermore, write $\bot$ and $\top$ for the [[Definition:Constant Mapping|constant mappings]] with these values, viz:
+:$\bot: X \to \mathbf 2, \bot (x) := \bot$
+:$\top: X \to \mathbf 2, \top (x) := \top$
+Then $\left({\mathbf 2^X, \vee, \wedge, \neg}\right)$ is a [[Definition:Boolean Algebra|Boolean algebra]], with $\bot$ and $\top$ as [[Definition:Identity Element|identities]] for $\vee$ and $\wedge$, respectively.
+\end{theorem}
+
+\begin{proof}
+Let us verify the axioms for a [[Definition:Boolean Algebra|Boolean algebra]] in turn.
+=== $(BA \ 0)$: Closure ===
+Follows from [[Induced Operations Preserve Closure]].
+{{qed|lemma}}
+=== $(BA \ 1)$: Commutativity ===
+Follows from [[Induced Operations Preserve Commutativity]].
+{{qed|lemma}}
+=== $(BA \ 2)$: Distributivity ===
+Follows from [[Induced Operations Preserve Distributivity]].
+{{qed|lemma}}
+=== $(BA \ 3)$: Identities ===
+Follows from [[Identity for Induced Operation]].
+{{qed|lemma}}
+=== $(BA \ 4)$: Complements ===
+Follows from [[Induced Operations Preserve Identities]].
+{{qed|lemma}}
+Having verified all the axioms, we conclude $\mathbf 2^X$ is a [[Definition:Boolean Algebra|Boolean algebra]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Power Set is Boolean Ring}
+Tags: Power Set, Boolean Rings
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]], and let $\mathcal P \left({S}\right)$ be its [[Definition:Power Set|power set]].
+Denote with $*$ and $\cap$ [[Definition:Symmetric Difference|symmetric difference]] and [[Definition:Set Intersection|intersection]], respectively.
+Then $\left({S, *, \cap}\right)$ is a [[Definition:Boolean Ring|Boolean ring]].
+\end{theorem}
+
+\begin{proof}
+From [[Symmetric Difference with Intersection forms Ring]], $\left({S, *, \cap}\right)$ is a [[Definition:Ring (Abstract Algebra)|ring]] [[Definition:Ring with Unity|with unity]].
+By [[Intersection is Idempotent]], $\cap$ is [[Definition:Idempotent Operation|idempotent]].
+It follows that $\left({S, *, \cap}\right)$ is a [[Definition:Boolean Ring|Boolean ring]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Power Set and Two-Valued Functions are Isomorphic Boolean Rings}
+Tags: Power Set, Boolean Rings
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\mathbf 2$ be the [[Definition:Boolean Ring|Boolean ring]] [[Definition:Two Ring|two]].
+Let $\mathcal P \left({S}\right)$ be the [[Definition:Power Set|power set]] of $S$; by [[Power Set is Boolean Ring]], it is a Boolean ring.
+Let $\mathbf 2^S$ be the [[Definition:Set of All Mappings|set of all mappings]] $f: S \to \mathbf 2$; by [[Two-Valued Functions form Boolean Ring]], it is also a Boolean ring.
+Let $\chi_{\left({\cdot}\right)}: \mathcal P \left({S}\right) \to \mathbf 2^S$ be the [[Definition:Characteristic Function Operation|characteristic function operation]].
+Then $\chi_{\left({\cdot}\right)}$ is a [[Definition:Ring Isomorphism|ring isomorphism]].
+\end{theorem}
+
+\begin{proof}
+From [[Support Operation Inverse to Characteristic Function Operation]], $\chi_{\left({\cdot}\right)}$ is a [[Definition:Bijection|bijection]].
+It therefore suffices to establish it is a [[Definition:Ring Homomorphism|ring homomorphism]].
+By [[Characteristic Function of Symmetric Difference]]:
+:$\chi_{A * B} = \chi_A + \chi_B - 2 \chi_A \chi_B$
+Since $\mathbf 2^S$ is a [[Definition:Boolean Ring|Boolean ring]], by [[Idempotent Ring has Characteristic Two]], the right-hand side reduces to:
+:$\chi_{A * B} = \chi_A + \chi_B$
+showing that $\chi_{\left({\cdot}\right)}$ preserves [[Definition:Ring Addition|ring addition]].
+By [[Characteristic Function of Intersection/Variant 1|Characteristic Function of Intersection: Variant 1]]:
+:$\chi_{A \cap B} = \chi_A \cdot \chi_B$
+showing that $\chi_{\left({\cdot}\right)}$ preserves the [[Definition:Ring Product|ring product]].
+Hence $\chi_{\left({\cdot}\right)}$ is a [[Definition:Ring Homomorphism|ring homomorphism]].
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Peirce's Law/Formulation 1}
+Tags: Peirce's Law
+
+\begin{theorem}
+: $\left({p \implies q}\right) \implies p \vdash p$
+\end{theorem}<|endoftext|>
+\section{Peirce's Law/Formulation 2/Proof 2}
+Tags: Peirce's Law, Truth Table Proofs
+
+\begin{theorem}
+: $\vdash \left({\left({p \implies q}\right) \implies p}\right) \implies p$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]] to the proposition.
+As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] are $T$ for all [[Definition:Boolean Interpretation|boolean interpretations]].
+$\begin{array}{|ccccc|c|c|}\hline
+((p & \implies & q) & \implies & p) & \implies & p \\
+\hline
+F & T & F & F & F & T & F \\
+F & T & T & F & F & T & F \\
+T & F & F & T & T & T & T \\
+T & T & T & T & T & T & T \\
+\hline
+\end{array}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Peirce's Law/Formulation 2}
+Tags: Peirce's Law
+
+\begin{theorem}
+: $\vdash \left({\left({p \implies q}\right) \implies p}\right) \implies p$
+\end{theorem}<|endoftext|>
+\section{Peirce's Law/Strong Form/Formulation 1}
+Tags: Peirce's Law, Truth Table Proofs
+
+\begin{theorem}
+:$\paren {\paren {p \implies q} \implies p} \dashv \vdash p$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]].
+As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
+$\begin{array}{|ccccc||c|}\hline
+((p & \implies & q) & \implies & p) & p \\
+\hline
+F & T & F & F & F & F \\
+F & T & T & F & F & F \\
+T & F & F & T & T & T \\
+T & T & T & T & T & T \\
+\hline
+\end{array}$
+{{qed}}
+{{Namedfor|Charles Sanders Peirce}}
+[[Category:Peirce's Law]]
+[[Category:Truth Table Proofs]]
+h176q3qbxl3f7z8ak724c46xngv1otc
+\end{proof}<|endoftext|>
+\section{Peirce's Law/Strong Form/Formulation 2}
+Tags: Peirce's Law
+
+\begin{theorem}
+:$\vdash \paren {\paren {p \implies q} \implies p} \iff p$
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+{{Namedfor|Charles Sanders Peirce}}
+[[Category:Peirce's Law]]
+epgggbe7bnanpn0c0yyejgf5h701w5q
+\end{proof}<|endoftext|>
+\section{Peirce's Law/Strong Form}
+Tags: Peirce's Law
+
+\begin{theorem}
+{{:Peirce's Law/Strong Form/Formulation 1}}
+\end{theorem}<|endoftext|>
+\section{Conjunction with Negative Equivalent to Negation of Implication}
+Tags: Conjunction, Implication
+
+\begin{theorem}
+==== [[Conjunction with Negative Equivalent to Negation of Implication/Formulation 1|Formulation 1]] ====
+{{:Conjunction with Negative Equivalent to Negation of Implication/Formulation 1}}
+==== [[Conjunction with Negative Equivalent to Negation of Implication/Formulation 2|Formulation 2]] ====
+{{:Conjunction with Negative Equivalent to Negation of Implication/Formulation 2}}
+\end{theorem}<|endoftext|>
+\section{Conjunction with Negative Equivalent to Negation of Implication/Formulation 1}
+Tags: Conjunction with Negative Equivalent to Negation of Implication
+
+\begin{theorem}
+:$p \land \neg q \dashv \vdash \neg \left({p \implies q}\right)$
+\end{theorem}<|endoftext|>
+\section{Conjunction with Negative Equivalent to Negation of Implication/Formulation 1/Forward Implication}
+Tags: Conjunction with Negative Equivalent to Negation of Implication
+
+\begin{theorem}
+:$p \land \neg q \vdash \neg \paren {p \implies q}$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|p \land \neg q \vdash \neg \paren {p \implies q} }}
+{{Premise|1|p \land \neg q}}
+{{Assumption|2|p \implies q| Assume the opposite of what is to be proved ...}}
+{{Simplification|3|1|p|1|1}}
+{{Simplification|4|1|\neg q|1|2}}
+{{ModusPonens|5|1, 2|q|2|3}}
+{{NonContradiction|6|1, 2|5|4|... and demonstrate a contradiction}}
+{{Contradiction|7|1|\neg \paren {p \implies q}|2|6}}
+{{EndTableau}}
+{{qed}}
+[[Category:Conjunction with Negative Equivalent to Negation of Implication]]
+3llyeaaoyezyiibkb14p0clorrmu0zz
+\end{proof}<|endoftext|>
+\section{Conjunction with Negative Equivalent to Negation of Implication/Formulation 1/Reverse Implication}
+Tags: Conjunction with Negative Equivalent to Negation of Implication
+
+\begin{theorem}
+:$\neg \left({p \implies q}\right) \vdash p \land \neg q$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\neg \left({p \implies q}\right) \vdash p \land \neg q}}
+{{Premise|1|\neg \left({p \implies q}\right)}}
+{{Assumption|2|\neg \left({p \land \neg q}\right)}}
+{{SequentIntro|3|2|p \implies q|2|[[Implication Equivalent to Negation of Conjunction with Negative]]}}
+{{NonContradiction|4|1, 2|3|1}}
+{{Reductio|5|1|p \land \neg q|2|4}}
+{{EndTableau}}
+{{qed}}
+{{LEM|Reductio ad Absurdum}}
+[[Category:Conjunction with Negative Equivalent to Negation of Implication]]
+74vn9lygaw57ac3s1upxd6d0jrjstff
+\end{proof}<|endoftext|>
+\section{Conjunction with Negative Equivalent to Negation of Implication/Formulation 1/Proof}
+Tags: Conjunction with Negative Equivalent to Negation of Implication
+
+\begin{theorem}
+:$p \land \neg q \dashv \vdash \neg \left({p \implies q}\right)$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]].
+As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
+$\begin{array}{|cccc||cccc|} \hline
+p & \land & \neg & q & \neg & (p & \implies & q) \\
+\hline
+F & F & T & F & F & F & T & F \\
+F & F & F & T & F & F & T & T \\
+T & T & T & F & T & T & F & F \\
+T & F & F & T & F & T & T & T \\
+\hline
+\end{array}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Conjunction with Negative Equivalent to Negation of Implication/Formulation 2}
+Tags: Conjunction with Negative Equivalent to Negation of Implication
+
+\begin{theorem}
+:$\vdash \paren {p \land \neg q} \iff \paren {\neg \paren {p \implies q} }$
+\end{theorem}<|endoftext|>
+\section{Reflexive Circular Relation is Equivalence}
+Tags: Equivalence Relations, Circular Relations, Reflexive Relations
+
+\begin{theorem}
+Let $\mathcal R \subseteq S \times S$ be a [[Definition:Reflexive Relation|reflexive]] and [[Definition:Circular Relation|circular relation]] in $S$.
+Then $\mathcal R$ is an [[Definition:Equivalence Relation|equivalence relation]].
+\end{theorem}
+
+\begin{proof}
+To prove a [[Definition:Relation|relation]] is an [[Definition:Equivalence Relation|equivalence]], we need to prove it is [[Definition:Reflexive Relation|reflexive]], [[Definition:Symmetric Relation|symmetric]] and [[Definition:Transitive Relation|transitive]].
+So, checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]:
+=== Reflexive ===
+[[Definition:By Hypothesis|By hypothesis]] $\mathcal R$ is [[Definition:Reflexive Relation|reflexive]].
+{{Qed|lemma}}
+=== Symmetric ===
+By [[Definition:Reflexive Relation|reflexivity]]:
+:$\tuple {x, x} \in \mathcal R$
+If $\tuple {x, y} \in \mathcal R$ then by the definition of [[Definition:Circular Relation|circular relation]] $\tuple {y, x} \in \mathcal R$.
+Hence $\mathcal R$ is [[Definition:Symmetric Relation|symmetric]].
+{{Qed|lemma}}
+=== Transitive ===
+Let $\tuple {x, y}, \tuple {y, z} \in \mathcal R$.
+By definition of [[Definition:Circular Relation|circular relation]]:
+:$\tuple {z, x} \in \mathcal R$
+By $\mathcal R$ being [[Definition:Symmetric Relation|symmetric]]:
+:$\tuple {x, z} \in \mathcal R$
+Hence $\mathcal R$ is [[Definition:Transitive Relation|transitive]].
+{{Qed|lemma}}
+Thus is $\mathcal R$ is [[Definition:Reflexive Relation|reflexive]], [[Definition:Symmetric Relation|symmetric]] and [[Definition:Transitive Relation|transitive]], and therefore by definition an [[Definition:Equivalence Relation|equivalence]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Inverse of Algebraic Structure Isomorphism is Isomorphism/General Result}
+Tags: Isomorphisms
+
+\begin{theorem}
+Let $\phi: \left({S, \circ_1, \circ_2, \ldots, \circ_n}\right)$ and $\left({T, *_1, *_2, \ldots, *_n}\right)$ be [[Definition:Algebraic Structure|algebraic structures]].
+Let $\phi: \left({S, \circ_1, \circ_2, \ldots, \circ_n}\right) \to \left({T, *_1, *_2, \ldots, *_n}\right)$ be a [[Definition:Mapping|mapping]].
+Then:
+: $\phi: \left({S, \circ_1, \circ_2, \ldots, \circ_n}\right) \to \left({T, *_1, *_2, \ldots, *_n}\right)$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]]
+{{iff}}:
+: $\phi^{-1}: \left({T, *_1, *_2, \ldots, *_n}\right) \to \left({S, \circ_1, \circ_2, \ldots, \circ_n}\right)$ is also an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
+\end{theorem}
+
+\begin{proof}
+{{proof wanted}}
+[[Category:Isomorphisms]]
+jvirzcgscz2cg3gja3i6o20wa5d6fu2
+\end{proof}<|endoftext|>
+\section{Fundamental Theorem of Calculus for Complex Riemann Integrals}
+Tags: Complex Analysis, Fundamental Theorem of Calculus
+
+\begin{theorem}
+Let $\closedint a b$ be a [[Definition:Closed Real Interval|closed real interval]].
+Let $F, f: \closedint a b \to \C$ be [[Definition:Complex Function|complex functions]].
+Suppose that $F$ is a [[Definition:Complex Primitive|primitive]] of $f$.
+Then the [[Definition:Complex Riemann Integral|complex Riemann integral]] of $f$ satisfies:
+:$\displaystyle \int_a^b \map f t \rd t = \map F b - \map F a$
+\end{theorem}
+
+\begin{proof}
+Let $u, v: \closedint a b \times \set 0 \to \R$ be defined as in the [[Cauchy-Riemann Equations]]:
+:$\map u {t, y} = \map \Re {\map F z}$
+:$\map v {t, y} = \map \Im {\map F z}$
+where:
+:$\map \Re {\map F z}$ denotes the [[Definition:Real Part|real part]] of $\map F z$
+:$\map \Im {\map F z}$ denotes the [[Definition:Imaginary Part|imaginary part]] of $\map F z$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \int_a^b \map f t \rd t
+      | r = \int_a^b \map {F'} {t + 0 i} \rd t
+      | c = by assumption
+}}
+{{eqn | r = \int_a^b \paren {\map {\dfrac {\partial u} {\partial t} } {t, 0} + i \map {\dfrac {\partial v} {\partial t} } {t, 0} } \rd t
+      | c = [[Cauchy-Riemann Equations]]
+}}
+{{eqn | r = \int_a^b \map {\dfrac {\partial u} {\partial t} } {t, 0} \rd t + i \int_a^b \map {\dfrac {\partial v} {\partial t} } {t, 0} \rd t
+      | c = {{Defof|Complex Riemann Integral}}
+}}
+{{eqn | r = \map u {b, 0} - \map u {a, 0} + i \paren {\map v {b, 0} - \map v {a, 0} }
+      | c = [[Fundamental Theorem of Calculus]] 
+}}
+{{eqn | r = \map F b - \map F a
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fundamental Theorem of Calculus for Contour Integrals}
+Tags: Contour Integration
+
+\begin{theorem}
+Let $F, f: D \to \C$ be [[Definition:Complex Function|complex functions]], where $D$ is a [[Definition:Connected Domain (Complex Analysis)|connected domain]].
+Let $C$ be a [[Definition:Contour (Complex Plane)|contour]] that is a [[Definition:Concatenation of Contours (Complex Plane)|concatenation]] of the [[Definition:Directed Smooth Curve (Complex Plane)|directed smooth curves]] $C_1, \ldots, C_n$.
+Let $C_k$ be [[Definition:Parameterization of Directed Smooth Curve (Complex Plane)|parameterized]] by the [[Definition:Smooth Path (Complex Analysis)|smooth path]] $\gamma_k: \closedint {a_k} {b_k} \to D$ for all $k \in \set {1, \ldots, n}$.
+Suppose that $F$ is a [[Definition:Complex Primitive|primitive]] of $f$.
+If $C$ has [[Definition:Start Point of Contour (Complex Plane)|start point]] $z$ and [[Definition:End Point of Contour (Complex Plane)|end point]] $w$, then:
+:$\displaystyle \int_C \map f z \rd z = \map F w - \map F z$
+If $C$ is a [[Definition:Closed Contour (Complex Plane)|closed contour]], then:
+:$\displaystyle \oint_C \map f z \rd z = 0$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int_C \map f z
+      | r = \sum_{k \mathop = 1}^n \int_{a_k}^{b_k} \map f {\map {\gamma_k} t} \map {\gamma_k'} t \rd t
+      | c = {{Defof|Complex Contour Integral}}
+}}
+{{eqn | r = \sum_{k \mathop = 1}^n \int_{a_k}^{b_k} \paren {\dfrac \rd {\rd t} \map F {\map {\gamma_k} t} } \rd t
+      | c = [[Derivative of Complex Composite Function]]
+}}
+{{eqn | r = \sum_{k \mathop = 1}^n \paren {\map F {\map {\gamma_k} {b_k} } - \map F {\map {\gamma_k} {a_k} } }
+      | c = [[Fundamental Theorem of Calculus for Complex Riemann Integrals]]
+}}
+{{eqn | r = \map F {\map {\gamma_n} {b_n} } - \map F {\map {\gamma_1} {a_1} }
+      | c = the sum is [[Definition:Telescoping Series|telescoping]]
+}}
+{{eqn | r = \map F w - \map F z
+      | c = {{Defof|Endpoints of Contour (Complex Plane)}}
+}}
+{{end-eqn}}
+If $C$ is a [[Definition:Closed Contour (Complex Plane)|closed contour]], then $z = w$.
+It follows that:
+:$\map F w - \map F z = 0$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Power Series is Taylor Series}
+Tags: Power Series, Taylor Series
+
+\begin{theorem}
+Let $\displaystyle f \left({z}\right) = \sum_{n \mathop = 0}^\infty a_n \left({z - \xi}\right)^n$ be a [[Definition:Complex Power Series|complex power series]] about $\xi \in \C$.
+Let $R$ be the [[Definition:Radius of Convergence of Complex Power Series|radius of convergence]] of $f$.
+Then, $f$ is of [[Definition:Differentiability Class|differentiability class $C^\infty$]].
+For all $n \in \N$:
+:$a_n = \dfrac{f^{\left({n}\right) } \left({\xi}\right) }{ n! }$
+Hence, $f$ is equal to its [[Definition:Taylor Series|Taylor series expansion]] about $\xi$:
+:$\displaystyle \forall z \in \C, \left\vert{z - \xi}\right\vert < R: \quad f \left({z}\right) = \sum_{n \mathop = 0}^\infty \dfrac{\left({z - \xi}\right)^n}{n!} f^{\left({n}\right) } \left({\xi}\right)$
+\end{theorem}
+
+\begin{proof}
+First, we prove by [[Principle of Mathematical Induction|induction]] over $k \in \N_{\ge 1}$ that:
+:$\displaystyle f^{\left({k}\right) } \left({z}\right) = \sum_{n \mathop = k}^\infty a_n \left({z - \xi}\right)^{n - k} n^{\underline k}$
+where $n^{\underline k}$ denotes the [[Definition:Falling Factorial|falling factorial]].
+=== Basis for the Induction ===
+For $k = 1$, it follows from [[Derivative of Complex Power Series]] that 
+:$\displaystyle f^{\left({k}\right) } \left({z}\right) = \sum_{n \mathop = 1}^\infty n a_n \left({z - \xi}\right)^{n - 1}$
+As $n = n^{\underline 1}$, this proves the hypothesis.
+From [[Radius of Convergence of Derivative of Complex Power Series]], it follows that the equation holds for all $z \in \C$ with $\left\vert{z - \xi}\right\vert < R$.
+=== Induction Hypothesis ===
+For fixed $k \in \N_{\ge 1}$, the hypothesis is that:
+:$\displaystyle f^{\left({k}\right) } \left({z}\right) = \sum_{n \mathop = k}^\infty a_n \left({z - \xi}\right)^{n-k} n^{\underline k}$
+and $f^{\left({k}\right) }$ has [[Definition:Radius of Convergence of Complex Power Series|radius of convergence]] $R$.
+=== Induction Step ===
+Note that $f^{\left({k+1}\right) }$ is the [[Definition:Derivative of Complex Function|derivative]] of $f^{\left({k}\right) }$.
+From [[Radius of Convergence of Derivative of Complex Power Series]], it follows that $f^{\left({k+1}\right) }$ has [[Definition:Radius of Convergence of Complex Power Series|radius of convergence]] $R$.
+For $z \in \C$ with $\left\vert{z - \xi}\right\vert < R$, we have:
+{{begin-eqn}}
+{{eqn | l = f^{\left({k+1}\right) }
+      | r = \dfrac{\mathrm d}{\mathrm dz} \left({\sum_{n \mathop = k}^\infty a_n \left({z - \xi}\right)^{n-k} n^{\underline k} }\right)
+      | c = [[Power Series is Taylor Series#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = \dfrac{\mathrm d}{\mathrm dz} \left({\sum_{n \mathop = 0}^\infty a_{n+k} \left({z - \xi}\right)^{n} \left({n+k}\right)^{\underline k} }\right)
+      | c = subtracting $k$ from $n$
+}}
+{{eqn | r = \sum_{n \mathop = 1}^\infty n a_{n+k} \left({z - \xi}\right)^{n-1} \left({n+k}\right)^{\underline k}
+      | c = [[Derivative of Complex Power Series]]
+}}
+{{eqn | r = \sum_{n \mathop = 1}^\infty a_{n+k} \left({z - \xi}\right)^{n-1} \left({n+k}\right)^{\underline{k+1} }
+      | c = {{Defof|Falling Factorial}}
+}}
+{{eqn | r = \sum_{n \mathop = k+1}^\infty a_{n} \left({z - \xi}\right)^{n - \left({k+1}\right) } n^{\underline{k+1} }
+      | c = adding $k$ to $n$
+}}
+{{end-eqn}}
+{{qed|lemma}}
+With $k \in \N$, we have:
+{{begin-eqn}}
+{{eqn | l = f^{\left({k}\right)} \left({\xi}\right)
+      | r = \sum_{n \mathop = k}^\infty a_n \left({\xi - \xi}\right)^{n - k} n^{\underline k}
+}}
+{{eqn | r = a_k 0^0 k^{\underline k}
+}}
+{{eqn | r = a_k k!
+}}
+{{end-eqn}}
+Hence, it follows that $a_n = \dfrac{f^{\left({n}\right)} \left({\xi}\right)} {n!}$.
+By definition of [[Definition:Taylor Series|Taylor series]], it follows that $f$ is equal to its [[Definition:Taylor Series|Taylor series expansion]] about $\xi$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Image of Complex Exponential Function}
+Tags: Exponential Function
+
+\begin{theorem}
+The [[Definition:Image of Mapping|image]] of the [[Definition:Complex Exponential Function|complex exponential function]] is $\C \setminus \left\{ {0}\right\}$.
+\end{theorem}
+
+\begin{proof}
+Let $z \in \C \setminus \left\{ {0}\right\}$.
+Let $r = \cmod z$ be the [[Definition:Complex Modulus|modulus]] of $z$, and let $\theta = \arg \left({z}\right)$ be the [[Definition:Argument of Complex Number|argument]] of $z$.
+Then $r>0$.
+Let $\ln$ denote the [[Definition:Natural Logarithm|real natural logarithm]], and let $e$ denote the [[Definition:Real Exponential Function|real exponential function]].
+Then:
+{{begin-eqn}}
+{{eqn | l = \exp \left({\ln r + i \theta}\right)
+      | r = e^{ \ln r } \left({\cos \theta + i \sin \theta }\right)
+      | c = {{Defof|Exponential Function/Complex|subdef = Real Functions|Complex Exponential Function}}
+}}
+{{eqn | r = r \left({\cos \theta + i \sin \theta }\right)
+      | c = [[Exponential of Natural Logarithm]]
+}}
+{{eqn | r = z
+      | c = {{Defof|Polar Form of Complex Number}}
+}}
+{{end-eqn}}
+Hence, $z \in \operatorname{Im} \left({\exp}\right)$.
+Suppose instead that $z=0$.
+Let $z_0 = r_0 \left({\cos \theta_0 + i \sin \theta_0 }\right) \in \C$.
+From [[Exponential Tends to Zero and Infinity]], it follows that $e^{ r_0 } \ne 0$.
+As $\cmod {\cos \theta_0 + i \sin \theta_0} = 1$, it follows that $\cos \theta_0 + i \sin \theta_0 \ne 0$.
+Then this equation has no solutions:
+:$0 = \exp z_0 = e^{ r_0 } \left({\cos \theta_0 + i \sin \theta_0 }\right)$
+Hence, $\operatorname{Im} \left({\exp}\right) = \C \setminus \left\{ {0}\right\}$-
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Reciprocal of Complex Exponential}
+Tags: Exponential Function, Reciprocals
+
+\begin{theorem}
+:$\dfrac 1 {\map \exp z} = \map \exp {-z}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \exp {-z}
+      | r = \dfrac {\map \exp {-z} } {\map \exp 0}
+      | c = as $\map \exp 0 = 1$ by [[Exponential of Zero]]
+}}
+{{eqn | r = \dfrac {\map \exp {-z} } {\map \exp {z - z} }
+}}
+{{eqn | r = \dfrac {\map \exp {-z} } {\map \exp z \, \map \exp {-z} }
+      | c = [[Exponential of Sum/Complex Numbers|Exponential of Sum: Complex Numbers]]
+}}
+{{eqn | r = \dfrac 1 {\map \exp z}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Power Function Strictly Preserves Ordering in Ordered Group}
+Tags: Ordered Semigroups
+
+\begin{theorem}
+Let $n \in \N_{>0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
+\end{theorem}
+
+\begin{proof}
+By the definition of an [[Definition:Ordered Group|ordered group]], $\preceq$ is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
+By the definition of an [[Definition:Ordering|ordering]], $\preceq$ is [[Definition:Transitive Relation|transitive]].
+By [[Reflexive Reduction of Relation Compatible with Group Operation is Compatible]], $\prec$ is also compatible with $\circ$.
+By [[Reflexive Reduction of Transitive Antisymmetric Relation is Strict Ordering]], $\prec$ is also transitive.
+By the definition of an ordered group, $\left({S,\circ}\right)$ is a group, and therefore a [[Definition:Semigroup|semigroup]].
+Thus the theorem holds by [[Transitive Relation Compatible with Semigroup Operation Relates Powers of Related Elements]].
+{{qed}}
+[[Category:Ordered Semigroups]]
+hce7aaxz91tgmtbcgbfykv7efadcsbd
+\end{proof}<|endoftext|>
+\section{Period of Complex Exponential Function}
+Tags: Exponential Function
+
+\begin{theorem}
+:$\map \exp {z + 2 k \pi i} = \map \exp z$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \exp {z + 2 k \pi i}
+      | r = \map \exp z \, \map \exp {2 k \pi i}
+      | c = [[Exponential of Sum/Complex Numbers|Exponential of Sum: Complex Numbers]]
+}}
+{{eqn | r = \map \exp z \times 1
+      | c = [[Euler's Formula/Examples/e^2 k i pi|Euler's Formula Example: $e^{2 k i \pi}$]]
+}}
+{{eqn | r = \map \exp z
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Zero Staircase Integral Condition for Primitive}
+Tags: Complex Analysis
+
+\begin{theorem}
+Let $f: D \to \C$ be a [[Definition:Continuous Complex Function|continuous complex function]], where $D$ is a [[Definition:Connected Domain (Complex Analysis)|connected domain]].
+Let $z_0 \in D$.
+Suppose that $\displaystyle \oint_C \map f z \rd z = 0$ for all [[Definition:Closed Contour (Complex Plane)|closed]] [[Definition:Staircase Contour|staircase contours]] $C$ in $D$.
+Then $f$ has a [[Definition:Complex Primitive|primitive]] $F: D \to \C$ defined by:
+:$\displaystyle \map F w = \int_{C_w} \map f z \rd z$
+where $C_w$ is any [[Definition:Staircase Contour|staircase contour]] in $D$ with [[Definition:Start Point of Contour (Complex Plane)|start point]] $z_0$ and [[Definition:End Point of Contour (Complex Plane)|end point]] $w$.
+\end{theorem}
+
+\begin{proof}
+From [[Connected Domain is Connected by Staircase Contours]], it follows that there exists a [[Definition:Staircase Contour|staircase contour]] $C_w$ in $D$ with [[Definition:Start Point of Contour (Complex Plane)|start point]] $z_0$ and [[Definition:End Point of Contour (Complex Plane)|end point]] $w$.
+If $C_w'$ is another staircase contour with the same [[Definition:Endpoints of Contour (Complex Plane)|endpoints]] as $C_w$, then $C_w' \cup \paren {-C_w}$ is a [[Definition:Closed Contour (Complex Plane)|closed]] [[Definition:Staircase Contour|staircase contour]].
+Then the definition of $F$ is independent of the choice of [[Definition:Contour (Complex Plane)|contour]], as:
+{{begin-eqn}}
+{{eqn | l = \int_{C_w} \map f z \rd z
+      | r = \int_{C_w} \map f z \rd z + \int_{C_w' \cup \paren {-C_w} } \map f z \rd z
+      | c = by assumption
+}}
+{{eqn | r = \int_{C_w} \map f z \rd z + \int_{C_w'} \map f z \rd z + \int_{-C_w} \map f z \rd z
+      | c = [[Contour Integral of Concatenation of Contours]]
+}}
+{{eqn | r = \int_{C_w} \map f z \rd z + \int_{C_w'} \map f z \rd z - \int_{C_w} \map f z \rd z
+      | c = [[Contour Integral along Reversed Contour]]
+}}
+{{eqn | r = \int_{C_w'} \map f z \rd z
+}}
+{{end-eqn}}
+We now show that $F$ is the [[Definition:Complex Primitive|primitive]] of $f$.
+Let $\epsilon \in \R_{>0}$.
+By [[Definition:Continuous Complex Function|definition of continuity]], there exists $r \in \R_{>0}$ such that the [[Definition:Open Ball|open ball]] $\map {B_r} w \subseteq D$, and for all $z \in \map {B_r} w$:
+:$\size {\map f z - \map f w} < \dfrac \epsilon 2$
+Let $h = x+iy \in \C \setminus \set 0$ with $x, y \in \R$ such that $\size h < r$.
+Let $\LL$ be the staircase contour that goes in a horizontal [[Definition:Convex Set (Vector Space)/Line Segment|line]] from $w$ to $w + x$, and continues in a vertical line from $w + x$ to $w + h$.
+As $w + x, w + h \in \map {B_r} w$, it follows from [[Open Ball is Convex Set]] that $\LL$ is a contour in $\map {B_r} w$.
+Then $C_w \cup \LL$ is a staircase contour from $z_0$ to $w + h$, so:
+{{begin-eqn}}
+{{eqn | l = \map F {w + h} - \map F w
+      | r = \int_{C_w \cup \LL} \map f z \rd z - \int_{C_w} \map f z \rd z
+}}
+{{eqn | r = \int_\LL \map f z \rd z
+      | c = [[Contour Integral of Concatenation of Contours]]
+}}
+{{end-eqn}}
+From [[Derivative of Complex Polynomial]], it follows that $\dfrac \rd {\rd z} \map f w z = \map f w$, so:
+{{begin-eqn}}
+{{eqn | l = \int_\LL \map f w \rd z
+      | r = \map f w \paren {w + h} - \map f w w
+      | c = [[Fundamental Theorem of Calculus for Contour Integrals]]
+}}
+{{eqn | r = h \map f w
+}}
+{{end-eqn}}
+We can now show that $\map {F'} w = \map f w$, as:
+{{begin-eqn}}
+{{eqn | l = \size {\dfrac {\map F {w + h} - \map F w} h - \map f w}
+      | r = \size {\dfrac 1 h \int_\LL \map f z \rd z - \dfrac 1 h h \map f w}
+      | c = by the above calculations
+}}
+{{eqn | r = \size {\dfrac 1 h} \size {\int_\LL \paren {\map f z - \map f w} \rd z}
+      | c = [[Linear Combination of Contour Integrals]]
+}}
+{{eqn | o = < 
+      | r = \size {\dfrac 1 h} \dfrac \epsilon 2 \map L \LL
+      | c = [[Estimation Lemma]], as $z \in \map {B_r} w$
+}}
+{{eqn | r = \dfrac {\size x + \size y} {\size h} \dfrac \epsilon 2 
+      | c = the lengths of the line segments are $\size x$ and $\size y$
+}}
+{{eqn | o = \le
+      | r = \epsilon
+      | c = [[Modulus Larger than Real Part and Imaginary Part]]
+}}
+{{end-eqn}}
+When $h$ tends to $0$, we have $\map {F'} w = \map f w$ by [[Definition:Complex-Differentiable at Point|definition of differentiability]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Complex Polynomial}
+Tags: Complex Differential Calculus
+
+\begin{theorem}
+Let $a_n \in \C$ for $n \in \left\{ {0, 1, \ldots, N}\right\}$, where $N \in \N$.
+Let $f: \C \to \C$ be a [[Definition:Complex Number|complex]] [[Definition:Polynomial|polynomial]] defined by $\displaystyle f \left({z}\right) = \sum_{n \mathop = 0}^N a_n z^n$.
+Then $f$ is [[Definition:Complex-Differentiable Function|complex differentiable]] and its [[Definition:Derivative of Complex Function|derivative]] is:
+:$\displaystyle f' \left({z}\right) = \sum_{n \mathop = 1}^N n a_n z^{n-1}$
+\end{theorem}
+
+\begin{proof}
+For all $n > N$, put $a_n = 0$.
+Then:
+:$\displaystyle f \left({z}\right) = \sum_{n \mathop = 0}^\infty a_n z^n$
+The result now follows from [[Derivative of Complex Power Series]].
+{{qed}}
+[[Category:Complex Differential Calculus]]
+t58tbo04g3a75np4x3mq3x7asipi2is
+\end{proof}<|endoftext|>
+\section{Nested Sphere Theorem}
+Tags: Complete Metric Spaces, Named Theorems
+
+\begin{theorem}
+Let $M = \left({A, d}\right)$ be a [[Definition:Complete Metric Space|complete metric space]].
+Let $\left\langle{S_n}\right\rangle$ be a [[Definition:Sequence|sequence]] of [[Definition:Closed Ball|closed balls]] in $M$ defined by:
+: $S_n = B^-_{\rho_n} \left({x_n}\right)$
+where $\rho_n \to 0$ as $n \to \infty$ and:
+: $S_1 \supseteq S_2 \supseteq \cdots \supseteq S_n \supseteq \cdots$
+Then there exists $x \in A$ such that:
+:$\displaystyle \bigcap_{n \mathop = 1}^\infty S_n = \left\{{x}\right\}$
+\end{theorem}
+
+\begin{proof}
+Let $S_n = B^-_{\rho_n} \left({x_n}\right)$ be the [[Definition:Closed Ball|closed ball]] of [[Definition:Radius of Closed Ball|radius]] $\rho_n$ about the point $x_n$.
+That is, let $S_n = \left\{{x \in A: d \left({x_n, x}\right) \le \rho_n}\right\}$.
+Then the sequence $\left\langle{x_n}\right\rangle$ forms a [[Definition:Cauchy Sequence (Metric Space)|Cauchy sequence]]:
+:$d \left({x_n, x_{n+p}}\right) < \rho_n$
+for any $p \ge 0$ since $S_{n+p} \subseteq S_n$.
+However, $\rho_n \to 0$ as $n \to \infty$ and therefore $d \left({x_n, x_{n+p}}\right) \to 0$ as $n \to \infty$ for any $p \ge 0$.
+Since the [[Definition:Metric Space|space]] $M$ is [[Definition:Complete Metric Space|complete]], there exists $x \in X$ such that $x_n \to x$ as $n \to \infty$.
+Then since the [[Definition:Subsequence|subsequence]] $\left\langle{x_k}\right\rangle_{k = n}^\infty$ is contained entirely in $S_n$ and [[Definition:Convergent Sequence (Metric Space)|converges]] to $x$, $x$ is a member of the [[Definition:Closure (Topology)|closure]] of $S_n$ by [[Closure of Subset of Metric Space by Convergent Sequence]].
+Since $S_n$ is [[Definition:Closed Set (Topology)|closed]], by [[Closed Set Equals its Closure]], we have:
+:$\forall n \in \N: x \in S_n$
+Hence:
+:$\displaystyle x \in \bigcap_{n \mathop = 1}^\infty S_n$
+Suppose now that $y \ne x$.
+Then it follows that $d \left({x, y}\right) > 0$, and hence that for some $n$:
+:$d \left({x, y}\right) > 2 \rho_n$
+Since $x, x_n \in S_n$, it then follows that:
+:$d \left({x, x_n}\right) \le \rho_n$
+Now:
+{{begin-eqn}}
+{{eqn|l = 2 \rho_n
+     |o = <
+     |r = d \left({x, y}\right)
+}}
+{{eqn|o = \le
+     |r = d \left({x, x_n}\right) + d \left({x_n, y}\right) 
+     |c = [[Definition:Metric Space Axioms|Axiom $(M2)$ for metrics]]
+}}
+{{eqn|o = \le
+     |r = \rho_n + d \left({x_n, y}\right)
+}}
+{{end-eqn}}
+From the above, it follows that:
+:$d \left({x_n, y}\right) > \rho_n$
+so that $y \notin S_n$, and consequently:
+:$\displaystyle y \notin \bigcap_{i \mathop = 1}^\infty S_n$
+Hence:
+:$\displaystyle \bigcap_{n \mathop = 1}^\infty S_n = \left\{{x}\right\}$
+{{qed}}
+[[Category:Complete Metric Spaces]]
+[[Category:Named Theorems]]
+pzhppcz0hslzecdycn1u21bshgjp17h
+\end{proof}<|endoftext|>
+\section{Principle of Dilemma/Formulation 1}
+Tags: Principle of Dilemma
+
+\begin{theorem}
+:$\paren {p \implies q} \land \paren {\neg p \implies q} \dashv \vdash q$
+\end{theorem}<|endoftext|>
+\section{Principle of Dilemma/Formulation 1/Forward Implication}
+Tags: Principle of Dilemma
+
+\begin{theorem}
+: $\left({p \implies q}\right) \land \left({\neg p \implies q}\right) \vdash q$
+\end{theorem}<|endoftext|>
+\section{Principle of Dilemma/Formulation 1/Forward Implication/Proof 3}
+Tags: Principle of Dilemma
+
+\begin{theorem}
+: $\left({p \implies q}\right) \land \left({\neg p \implies q}\right) \vdash q$
+\end{theorem}
+
+\begin{proof}
+From the [[Constructive Dilemma]] we have:
+: $p \implies q, r \implies s \vdash p \lor r \implies q \lor s$
+from which, changing the names of letters strategically:
+: $p \implies q, \neg p \implies q \vdash p \lor \neg p \implies q \lor q$
+From [[Law of Excluded Middle]], we have:
+: $\vdash p \lor \neg p$
+From the [[Rule of Idempotence]] we have:
+: $q \lor q \vdash q$
+and the result follows by [[Hypothetical Syllogism]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Principle of Dilemma/Formulation 1/Reverse Implication}
+Tags: Principle of Dilemma
+
+\begin{theorem}
+: $q \vdash \left({p \implies q}\right) \land \left({\neg p \implies q}\right)$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|q \vdash \left({p \implies q}\right) \land \left({\neg p \implies q}\right)}}
+{{Premise|1|q}}
+{{SequentIntro|2|1|p \implies q|1|[[True Statement is implied by Every Statement]]}}
+{{SequentIntro|3|1|\neg p \implies q|1|[[True Statement is implied by Every Statement]]}}
+{{Conjunction|4|1|\left({p \implies q}\right) \land \left({\neg p \implies q}\right)|2|3}}
+{{EndTableau}}
+{{qed}}
+[[Category:Principle of Dilemma]]
+9yblx1tnx4w60wc2mrbf3rgkgc15rnc
+\end{proof}<|endoftext|>
+\section{Principle of Dilemma/Formulation 2}
+Tags: Principle of Dilemma
+
+\begin{theorem}
+:$\vdash \paren {p \implies q} \land \paren {\neg p \implies q} \iff q$
+\end{theorem}<|endoftext|>
+\section{Principle of Dilemma/Formulation 2/Forward Implication}
+Tags: Principle of Dilemma
+
+\begin{theorem}
+:$\vdash \paren {p \implies q} \land \paren {\neg p \implies q} \implies q$
+\end{theorem}<|endoftext|>
+\section{Principle of Dilemma/Formulation 2/Reverse Implication}
+Tags: Principle of Dilemma
+
+\begin{theorem}
+:$\vdash q \implies \paren {\paren {p \implies q} \land \paren {\neg p \implies q} }$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\vdash q \implies \paren {\paren {p \implies q} \land \paren {\neg p \implies q} } }}
+{{Assumption|1|q|}}
+{{SequentIntro|2|1|\paren {p \implies q} \land \paren {\neg p \implies q}|1|[[Principle of Dilemma/Formulation 1/Reverse Implication|Principle of Dilemma: Formulation 1: Reverse Implication]]}}
+{{Implication|3||q \implies \paren {\paren {p \implies q} \land \paren {\neg p \implies q} }|1|2}}
+{{EndTableau}}
+{{qed}}
+[[Category:Principle of Dilemma]]
+d9q4uvsexr57d799yi2aupfipflfpf2
+\end{proof}<|endoftext|>
+\section{Properties of Complex Exponential Function}
+Tags: Exponential Function
+
+\begin{theorem}
+Let $z \in \C$ be a [[Definition:Complex Number|complex number]].
+Let $\exp z$ be the [[Definition:Complex Exponential Function|exponential of $z$]].
+Then:
+\end{theorem}<|endoftext|>
+\section{Open Domain is Connected iff it is Path-Connected}
+Tags: Complex Analysis
+
+\begin{theorem}
+Let $D \subseteq \C$ be a [[Definition:Open Set (Complex Analysis)|open]] [[Definition:Subset|subset]] of the [[Definition:Complex Number|set of complex numbers]].
+Then $D$ is [[Definition:Connected Set (Topology)|connected]] {{iff}} $D$ is [[Definition:Path-Connected|path-connected]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+[[Complex Plane is Metric Space]] shows that $\C$ is [[Definition:Topologically Equivalent Metric Spaces|topologically equivalent]] to the [[Definition:Euclidean Space|Euclidean space]] $\R^2$.
+The result follows from [[Connected Open Subset of Euclidean Space is Path-Connected]].
+{{explain|It still needs to be established that connectedness is preserved by a [[Definition:Topologically Equivalent Metric Spaces|homeomorphism]].}}
+{{qed|lemma}}
+=== Sufficient Condition ===
+The result follows from [[Path-Connected Space is Connected]].
+{{qed}}
+[[Category:Complex Analysis]]
+t5htvab26mdou1b95gqf3n80posp7i4
+\end{proof}<|endoftext|>
+\section{Proof by Contradiction/Variant 2/Formulation 1}
+Tags: Proof by Contradiction
+
+\begin{theorem}
+:$p \implies q, p \implies \neg q \vdash \neg p$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|p \implies q, p \implies \neg q \vdash \neg p}}
+{{Premise|1|p \implies q}}
+{{Premise|2|p \implies \neg q}}
+{{Assumption|3|p}}
+{{ModusPonens|4|1, 3|q|1|3}}
+{{ModusPonens|5|2, 3|\neg q|2|3}}
+{{NonContradiction|6|1, 2, 3|4|5}}
+{{Contradiction|7|1, 2|\neg p|3|6}}
+{{EndTableau}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Proof by Contradiction/Variant 2/Formulation 2}
+Tags: Proof by Contradiction
+
+\begin{theorem}
+:$\vdash \paren {\paren {p \implies q} \land \paren {p \implies \neg q} } \implies \neg p$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau  |\vdash \paren {\paren {p \implies q} \land \paren {p \implies \neg q} } \implies \neg p}}
+{{Assumption    |1|\paren {p \implies q} \land \paren {p \implies \neg q} }}
+{{Simplification|2|1|p \implies q|1|1}}
+{{Simplification|3|1|p \implies \neg q|1|2}}
+{{SequentIntro  |4|1|\neg p|2, 3|[[Proof by Contradiction/Variant 2/Formulation 1|Proof by Contradiction: Formulation 1]]}}
+{{Implication   |5| |\paren {\paren {p \implies q} \land \paren {p \implies \neg q} } \implies \neg p|1|4}}
+{{EndTableau}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Proof by Contradiction/Variant 3/Formulation 1}
+Tags: Proof by Contradiction
+
+\begin{theorem}
+:$p \implies \neg p \vdash \neg p$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|p \implies \neg p \vdash \neg p}}
+{{Premise|1|p \implies \neg p}}
+{{Assumption|2|p}}
+{{ModusPonens|3|1, 2|\neg p|1|2}}
+{{NonContradiction|4|1, 2|2|3}}
+{{Contradiction|5|1, 2|\neg p|2|4}}
+{{EndTableau|qed}}
+\end{proof}<|endoftext|>
+\section{Proof by Contradiction/Variant 3/Formulation 2}
+Tags: Proof by Contradiction
+
+\begin{theorem}
+:$\vdash \paren {p \implies \neg p} \implies \neg p$
+\end{theorem}<|endoftext|>
+\section{Primitive of Function on Connected Domain}
+Tags: Complex Analysis
+
+\begin{theorem}
+Let $f: D \to \C$ be a [[Definition:Continuous Complex Function|continuous complex function]], where $D$ is a [[Definition:Connected Domain (Complex Analysis)|connected domain]].
+Then the following three conditions are equivalent:
+:$(1): \quad$ $f$ has a [[Definition:Complex Primitive|primitive]].
+:$(2): \quad$ For any two [[Definition:Contour (Complex Plane)|contours]] $C_1, C_2$ in $D$ with identical [[Definition:Start Point of Contour (Complex Plane)|start points]] $z_1 \in D$ and [[Definition:End Point of Contour (Complex Plane)|end points]] $z_2 \in D$, we have:
+::$\displaystyle \int_{C_1} \map f z \rd z = \int_{C_2} \map f z \rd z$
+:$(3): \quad$ For all [[Definition:Closed Contour (Complex Plane)|closed contours]] $C$ in $D$, we have:
+::$\displaystyle \oint_C \map f z \rd z = 0$
+If the conditions hold, we can choose any $z_0 \in D$ and define a primitive $F: D \to \C$ of $f$ by:
+:$\displaystyle \map F w = \int_{C_w} \map f z \rd z$
+where $C_w$ is any [[Definition:Contour (Complex Plane)|contour]] in $D$ with [[Definition:Start Point of Contour (Complex Plane)|start point]] $z_0$ and [[Definition:End Point of Contour (Complex Plane)|end point]] $w$.
+\end{theorem}
+
+\begin{proof}
+=== $(1)$ implies $(2)$ ===
+If $F$ is a [[Definition:Complex Primitive|primitive]] of $f$, we have:
+{{begin-eqn}}
+{{eqn | l = \int_{C_1} \map f z \rd z 
+      | r = \map F {z_2} - \map F {z_1}
+      | c = [[Fundamental Theorem of Calculus for Contour Integrals]]
+}}
+{{eqn | r = \int_{C_2} \map f z \rd z 
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== $(2)$ implies $(3)$ ===
+Let $C$ be a [[Definition:Closed Contour (Complex Plane)|closed contour]] in $D$ with [[Definition:Endpoints of Contour (Complex Plane)|endpoints]] $z_0$.
+Let the [[Definition:Constant Mapping|constant function]] $\gamma: \closedint 0 1 \to D$ with $\map \gamma t = z_0$ be the [[Definition:Parameterization of Contour (Complex Plane)|parameterization of a contour]] $C_0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \oint_C \map f z \rd z
+      | r = \oint_{C_0} \map f z \rd z
+      | c = by assumption
+}}
+{{eqn | r = \int_0^1 \map f {\map \gamma t} \map {\gamma'} t \rd t
+      | c = {{Defof|Complex Contour Integral}}
+}}
+{{eqn | r = 0
+      | c = by [[Derivative of Complex Polynomial]], as $\gamma$ is constant
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== $(3)$ implies $(1)$ ===
+This follows from [[Zero Staircase Integral Condition for Primitive]].
+{{qed|lemma}}
+=== Construction of a Primitive ===
+If the conditions hold, we choose $z_0 \in D$ and define a [[Definition:Complex Function|function]] $F: D \to \C$ of $f$ by:
+:$\displaystyle \map F w = \int_{C_w} \map f z \rd z$
+where $C_w$ is any [[Definition:Contour (Complex Plane)|contour]] in $D$ with [[Definition:Start Point of Contour (Complex Plane)|start point]] $z_0$ and [[Definition:End Point of Contour (Complex Plane)|end point]] $w$.
+From [[Connected Domain is Connected by Staircase Contours]], it follows that we can choose $C_w$ to be a [[Definition:Staircase Contour|staircase contour]].
+If $C_w'$ is another [[Definition:Contour (Complex Plane)|contour]] in $D$ with the same [[Definition:Endpoint of Contour (Complex Plane)|endpoints]] as $C_w$, then:
+:$\displaystyle \int_{C_w'} \map f z \rd z = \int_{C_w} \map f z \rd z$
+by condition $(2)$, so $F$ is well-defined.
+From [[Zero Staircase Integral Condition for Primitive]], it follows that $F$ is a [[Definition:Complex Primitive|primitive]] of $f$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Mind the Gap}
+Tags: Total Orderings
+
+\begin{theorem}
+Let $\left({S, \preccurlyeq}\right)$ be a [[Definition:Totally Ordered Set|totally ordered set]].
+Let $a, b \in S$ with $a \prec b$.
+Suppose that:
+:$\left\{ {x \in S: a \prec x \prec b}\right\} = \varnothing$
+Then:
+: $a^\succ = b^\succcurlyeq$
+and
+: $b^\prec = a^\preccurlyeq$
+where:
+: $a^\succ$ is the [[Definition:Strict Upper Closure of Element|strict upper closure]] of $a$
+: $b^\succcurlyeq$ is the [[Definition:Weak Upper Closure of Element|weak upper closure]] of $b$
+: $b^\prec$ is the [[Definition:Strict Lower Closure of Element|strict lower closure]] of $b$
+: $a^\preccurlyeq$ is the [[Definition:Weak Lower Closure of Element|weak lower closure]] of $a$.
+\end{theorem}
+
+\begin{proof}
+Let:
+:$p \in a^\succ$
+By the definition of strict upper closure:
+: $a \prec p$
+By the definition of total ordering, $p \prec b$ or $p \succcurlyeq b$.
+But if $p \prec b$ then $a \prec p \prec b$, contradicting the premise.
+Thus $p \succcurlyeq b$, so $p \in b^\succcurlyeq$.
+By definition of [[Definition:Subset|subset]]:
+:$a^\succ \subseteq b^\succcurlyeq$
+Let:
+: $p \in b^\succcurlyeq$
+By the definition of weak upper closure:
+:$b \preccurlyeq p$
+Since $a \prec b$, [[Extended Transitivity]] shows that $a \prec p$.
+Thus:
+:$p \in a^\succ$
+By definition of [[Definition:Subset|subset]]:
+:$b^\succcurlyeq \subseteq a^\succ$
+Therefore by definition of [[Definition:Set Equality|set equality]]:
+:$a^\succ = b^\succcurlyeq$
+{{qed|lemma}}
+A similar argument shows that $b^\prec = a^\preccurlyeq$.
+{{qed}}
+[[Category:Total Orderings]]
+icypkz4hxwvkb4jujqksn97vn3531ez
+\end{proof}<|endoftext|>
+\section{Proof by Contradiction/Sequent Form}
+Tags: Proof by Contradiction
+
+\begin{theorem}
+:$\paren {p \vdash \bot} \vdash \neg p$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\paren {p \vdash \bot} \vdash \neg p}}
+{{Premise      |1|p \vdash \bot}}
+{{Assumption   |2|p}}
+{{Contradiction|3|1|\neg p|2|2}}
+{{EndTableau}}
+{{Qed}}
+[[Category:Proof by Contradiction]]
+s213zzvegzbnzr2r7p55j1sgvbbcsk8
+\end{proof}<|endoftext|>
+\section{Left or Right Inverse of Matrix is Inverse}
+Tags: Inverse Matrices
+
+\begin{theorem}
+Let $\mathbf A, \mathbf B$ be [[Definition:Square Matrix|square matrices of order $n$]] over a [[Definition:Commutative and Unitary Ring|commutative ring with unity]] $\left({R, +, \circ}\right)$.
+Suppose that:
+: $\mathbf A \mathbf B = \mathbf I_n$
+where $\mathbf I_n$ is the [[Definition:Unit Matrix|unit matrix of order $n$]].
+Then $\mathbf A$ and $\mathbf B$ are [[Definition:Invertible Matrix|invertible matrices]], and furthermore:
+:$\mathbf B = \mathbf A^{-1}$
+where $\mathbf A^{-1}$ is the [[Definition:Inverse Matrix|inverse]] of $\mathbf A$.
+\end{theorem}
+
+\begin{proof}
+When $1_R$ denotes the [[Definition:Unity of Ring|unity]] of $R$, we have:
+{{begin-eqn}}
+{{eqn | l = 1_R
+      | r = \map \det {\mathbf I_n}
+      | c = [[Determinant of Unit Matrix]]
+}}
+{{eqn | r = \map \det {\mathbf A \mathbf B}
+      | c = by assumption
+}}
+{{eqn | r = \map \det {\mathbf A} \map \det {\mathbf B}
+      | c = [[Determinant of Matrix Product]]
+}}
+{{end-eqn}} 
+From [[Matrix is Invertible iff Determinant has Multiplicative Inverse]], it follows that $\mathbf A$ and $\mathbf B$ are [[Definition:Invertible Matrix|invertible]].
+Then:
+{{begin-eqn}}
+{{eqn | l = \mathbf B
+      | r = \mathbf I_n \mathbf B
+      | c = [[Unit Matrix is Unity of Ring of Square Matrices]]
+}}
+{{eqn | r = \paren {\mathbf A^{-1} \mathbf A} \mathbf B
+      | c = {{Defof|Inverse Matrix}}
+}}
+{{eqn | r = \mathbf A^{-1} \paren {\mathbf A \mathbf B}
+      | c = [[Matrix Multiplication is Associative]]
+}}
+{{eqn | r = \mathbf A^{-1} \mathbf I_n
+      | c = by assumption
+}}
+{{eqn | r = \mathbf A^{-1}
+      | c = [[Unit Matrix is Unity of Ring of Square Matrices]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Banach-Alaoglu Theorem}
+Tags: Functional Analysis
+
+\begin{theorem}
+Let $X$ be a [[Definition:Separable Space|separable]] [[Definition:Normed Linear Space|normed space]].
+Then the closed unit sphere in its [[Definition:Dual Space|dual]] $X^*$ is [[Definition:Initial Topology|weak*]] [[Definition:Sequential Compactness|sequentially compact]].
+\end{theorem}
+
+\begin{proof}
+We have to show the following: given a bounded sequence in $X^*$, there is a weakly convergent subsequence.
+Let $\sequence {l_n}_{n \mathop \in \N}$ be a bounded sequence in $X^*$.
+Let $\sequence {x_n}_{n \mathop \in \N}$ be a countable dense subset of $X$.
+Choose subsequences $\N \supset \Lambda_1 \supset \Lambda_2 \supset \ldots$ such that:
+:$\forall j \in \N: \map {l_k} {x_j} \to a_j =: \map l {x_j}$
+as $k \to \infty$, $k \in \Lambda_j$.
+Let $\Lambda$ be the diagonal sequence.
+=== Claim 1 ===
+$l$ can be extended to an element of $X^*$.
+=== Proof ===
+$l$ can be extended in the obvious way to a linear function on $M = \operatorname {span} \set {x_j}_{j \mathop \in \N}$.
+We extend it to a functional in $X^*$ by pointwise limit (notice that $M$ is dense in $X$).
+We have:
+:$\displaystyle \size {\map l x} = \lim_{k \mathop \to \infty} \size {\map l {x_k} } \le \limsup_{k \mathop \to \infty} \norm {l_k}_{X^*} \norm x_X$
+where $x_k \to x$ as $k \to \infty$.
+Since $\set {l_k}_k \mathop \in \N$ was bounded, $l$ is bounded and thus continuous.
+{{qed|lemma}}
+=== Claim 2 ===
+$l_k \stackrel {\omega^*} {\to} l$ as $k \to \infty$, $k \in \Lambda$.
+=== Proof ===
+Let:
+:$\displaystyle X \ni x = \lim_{\substack {j \mathop \to \infty \\ j \mathop \in J} } x_j$
+where $J$ is some subset of $\N$.
+We have then, for every $j \in J$:
+{{begin-eqn}}
+{{eqn | l = \size {\map {l_k} x - \map l x}
+      | o = \le
+      | r = \size {\map {l_k} {x - x_j} } + \size {\map l {x - x_j} } + \size {\map {l_k} {x_j} - \map l {x_j} }
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \paren {\sup_{i \mathop \in \Lambda} \norm {l_i}_{X^*} + \norm l_{X^*} } \norm {x - x_j}_X + \size {\map {l_k} {x_j} - \map l {x_j} } 
+      | c = 
+}}
+{{end-eqn}}
+Now given $\epsilon >0$, we find a $j \in J$ such that the first term is less than $\epsilon / 2$. 
+For fixed $j$, we have by construction of $l$ that $\map {l_k} {x_j}$ converges to $\map l {x_j}$.
+Therefore, we can find a $k_\epsilon$ such that for $k \ge k_\epsilon$ we have:
+:$\size {\map {l_k} x - \map l x} < \epsilon$ 
+as requested.
+{{qed}}
+{{Namedfor|Stefan Banach|name2 = Leonidas Alaoglu|cat = Banach|cat2 = Alaoglu}}
+\end{proof}<|endoftext|>
+\section{Empty Product is Terminal Object}
+Tags: Limits and Colimits, Category Theory
+
+\begin{theorem}
+Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
+Suppose $\mathbf C$ admits a [[Definition:Product (Category Theory)|product]] $\displaystyle \prod \O$ for the [[Definition:Empty Set|empty set]].
+Then $\displaystyle \prod \O$ is a [[Definition:Terminal Object|terminal object]] of $\mathbf C$.
+\end{theorem}
+
+\begin{proof}
+By definition, $\displaystyle \prod \O$ is the [[Definition:Limit (Category Theory)|limit]] of the empty [[Definition:Subcategory|subcategory]] $\mathbf 0$ of $\mathbf C$.
+The result follows from [[Terminal Object as Limit]].
+{{qed}}
+[[Category:Limits and Colimits]]
+[[Category:Category Theory]]
+lfph52fpugtjevxom96agpp4cyf6qnn
+\end{proof}<|endoftext|>
+\section{Unary Product for Object is Itself}
+Tags: Category Theory
+
+\begin{theorem}
+Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
+Let $C$ be an [[Definition:Object (Category Theory)|object]] of $\mathbf C$.
+Then $\displaystyle \prod \set C = C$, where $\displaystyle \prod$ denotes [[Definition:Product (Category Theory)/General Definition|product]].
+\end{theorem}
+
+\begin{proof}
+Follows directly from [[Limit of Singleton]].
+{{qed}}
+[[Category:Category Theory]]
+2335z4jfrkhosg2mgzcixrrjl2slywj
+\end{proof}<|endoftext|>
+\section{Alaoglu's Theorem}
+Tags: Functional Analysis
+
+\begin{theorem}
+The [[Definition:Closed Unit Ball|closed unit ball]] of the [[Definition:Dual Space|dual]] of a [[Definition:Normed Linear Space|normed space]] is [[Definition:Compact Topological Space|compact]] with respect to the [[Definition:Initial Topology|weak* topology]].
+{{rewrite|Rephrase to match usual presentation}}
+\end{theorem}
+
+\begin{proof}
+Let $X$ be a [[Definition:Normed Vector Space|normed vector space]].
+Denote by $B$ the [[Definition:Closed Unit Ball|closed unit ball]] in $X$.
+Let $X^*$ be the dual of $X$.
+Denote by $B^*$ the [[Definition:Closed Unit Ball|closed unit ball]] in $X^*$.
+Let $\map {\mathcal F} B = \closedint {-1} 1^B$ be the topological space of functions from $B$ to $\closedint {-1} 1$.
+By [[Tychonoff's Theorem]], $\map {\mathcal F} B$ is compact with respect to the product topology.
+We define the restriction map:
+:$R: B^*\to \map {\mathcal F} B$
+by $\map R \psi = \psi \restriction_B$.
+=== Lemma 1 ===
+$\map R {B^*}$ is a closed subset of $\map {\mathcal F} B$.
+=== Proof ===
+{{ProofWanted}}
+{{qed|lemma}}
+=== Lemma 2 ===
+$R$ is a homeomorphism from $B^*$ with the weak* topology to its image $\map R {B^*}$ seen as a subset of $\map {\mathcal F} B$ with the product topology.
+=== Proof ===
+{{ProofWanted}}
+{{qed|lemma}}
+Thus by lemma 2, $B^*$ in the weak* topology is homeomorphic with $\map R {B^*}$.
+This is a closed subset of $\map {\mathcal F} B$ (by lemma 1) and thus compact.
+{{qed}}
+{{Namedfor|Leonidas Alaoglu|cat = Alaoglu}}
+\end{proof}<|endoftext|>
+\section{Principle of Non-Contradiction/Sequent Form/Formulation 1}
+Tags: Principle of Non-Contradiction
+
+\begin{theorem}
+:$p, \neg p \vdash \bot$
+\end{theorem}<|endoftext|>
+\section{Vector Cross Product is Orthogonal to Factors}
+Tags: Vector Cross Product
+
+\begin{theorem}
+Let $\mathbf a$ and $\mathbf b$ be [[Definition:Vector (Euclidean Space)|vectors]] in the [[Definition:Real Euclidean Space|Euclidean space]] $\R^3$.
+Let $\mathbf a \times \mathbf b$ denote the [[Definition:Vector Cross Product|vector cross product]].
+Then:
+:$(1): \quad$ $\mathbf a$ and $\mathbf a \times \mathbf b$ are [[Definition:Orthogonal (Linear Algebra)|orthogonal]].
+:$(2): \quad$ $\mathbf b$ and $\mathbf a \times \mathbf b$ are [[Definition:Orthogonal (Linear Algebra)|orthogonal]].
+\end{theorem}
+
+\begin{proof}
+Let $\mathbf a = \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix}$, and $\mathbf b = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}$.
+Then the [[Definition:Dot Product|dot product]] of $\mathbf a$ and $\mathbf a \times \mathbf b$ is:
+{{begin-eqn}}
+{{eqn | l = \mathbf a \cdot \left({\mathbf a \times \mathbf b}\right)
+      | r = a_1 \left({a_2 b_3 - a_3 b_2}\right) + a_2 \left({a_3 b_1 - a_1 b_3}\right) + a_3 \left({a_1 b_2 - a_2 b_1}\right)
+      | c = {{Defof|Dot Product}} and {{Defof|Vector Cross Product}}
+}}
+{{eqn | r = a_1 a_2 b_3 - a_1 a_3 b_2 + a_2 a_3 b_1 - a_1 a_2 b_3 + a_1 a_3 b_2 - a_2 a_3 b_1
+}}
+{{eqn | r = 0
+}}
+{{end-eqn}}
+Since the [[Definition:Dot Product|dot product]] is equal to zero, the vectors are [[Definition:Orthogonal (Linear Algebra)|orthogonal]] by definition.
+Similarly, $\mathbf b$ and $\mathbf a \times \mathbf b$ are [[Definition:Orthogonal (Linear Algebra)|orthogonal]]:
+{{begin-eqn}}
+{{eqn | l = \mathbf b \cdot \left({\mathbf a \times \mathbf b}\right)
+      | r = b_1 \left({a_2 b_3 - a_3 b_2}\right) + b_2 \left({a_3 b_1 - a_1 b_3}\right) + b_3 \left({a_1 b_2 - a_2 b_1}\right)
+}}
+{{eqn | r = a_2 b_1 b_3 - a_3 b_1 b_2 + a_3 b_1 b_2 - a_1 b_2 b_3 + a_1 b_2 b_3 - a_2 b_1 b_3
+}}
+{{eqn | r = 0
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Squeeze Theorem/Sequences/Metric Spaces}
+Tags: Metric Spaces, Limits of Sequences
+
+\begin{theorem}
+Let $M = \left({S, d}\right)$ be a [[Definition:Metric Space|metric space]] or [[Definition:Pseudometric Space|pseudometric space]].
+Let $\left\langle{x_n}\right\rangle$ be a [[Definition:Infinite Sequence|sequence]] in $S$.
+Let $p \in S$. 
+Let $\left\langle{r_n}\right\rangle$ be a sequence in $\R_{\ge 0}$.
+Let $\left\langle{r_n}\right\rangle$ [[Definition:Limit of Sequence (Metric Space)|converge]] to $0$.
+For each $n$, let $d \left({p, x_n}\right) \le r_n$.
+Then $\left\langle{x_n}\right\rangle$ converges to $p$.
+\end{theorem}
+
+\begin{proof}
+Let $\epsilon > 0$.
+Since $r$ [[Definition:Limit of Sequence (Metric Space)|converges]] to $0$, there exists an $N$ such that $n > N \implies r_n < \epsilon$.
+But then by [[Extended Transitivity]], $n > N \implies d \left({p, x_n}\right) < \epsilon$.
+Thus $\left\langle{x_n}\right\rangle$ converges to $p$.
+{{qed}}
+[[Category:Metric Spaces]]
+[[Category:Limits of Sequences]]
+ttjzk3l7syew0kp9sci703bxd57re3f
+\end{proof}<|endoftext|>
+\section{Squeeze Theorem/Sequences/Linearly Ordered Space}
+Tags: Order Topology, Limits of Sequences
+
+\begin{theorem}
+Let $\left({S, \le, \tau}\right)$ be a [[Definition:Linearly Ordered Space|linearly ordered space]].
+Let $\left\langle{x_n}\right\rangle$, $\left\langle{y_n}\right\rangle$, and $\left\langle{z_n}\right\rangle$ be sequences in $S$.
+Let $p \in S$.
+Let $\left\langle{x_n}\right\rangle$ and $\left\langle{z_n}\right\rangle$ both converge to $p$
+For each $n$, let $x_n \le y_n \le z_n$.
+Then $\left\langle{y_n}\right\rangle$ converges to $p$.
+\end{theorem}
+
+\begin{proof}
+Let $m \in S$ and $m < p$.
+Then $\left\langle{x_n}\right\rangle$ eventually succeeds $m$.
+Thus by [[Extended Transitivity]], $\left\langle{y_n}\right\rangle$ eventually succeeds $m$.
+A similar argument using $\left\langle{z_n}\right\rangle$ proves the dual statement.
+Thus $\left\langle{y_n}\right\rangle$ is eventually in each ray containing $p$, so it converges to $p$.
+{{qed}}
+[[Category:Order Topology]]
+[[Category:Limits of Sequences]]
+skzmfnjs6c5wl2fah2fqawkhp58tt2k
+\end{proof}<|endoftext|>
+\section{Squeeze Theorem for Filter Bases}
+Tags: Filter Theory, Order Topology, Convergence
+
+\begin{theorem}
+Let $\struct {S, \le, \tau}$ be a [[Definition:Linearly Ordered Space|linearly ordered space]].
+Let $F_1$, $F_2$, and $F_3$ be [[Definition:Filter Basis|filter bases]] in $S$.
+Let:
+:$\forall T \in F_1: \exists M \in F_2: \forall x \in M: \exists y \in T: y \le x$
+That is:
+:for each $T \in F_1$, $F_2$ has an element $M$ such that all elements of $M$ succeed some element of $T$.
+Similarly, let:
+:$\forall U \in F_3: \exists N \in F_2: \forall x \in N: \exists y \in U: x \le y$
+That is:
+:for each $U \in F_3$, $F_2$ has an element $N$ such that all elements of $N$ precede some element of $U$.
+Let $F_1$ and $F_3$ each converge to a point $p \in S$.
+Then $F_2$ converges to $p$.
+\end{theorem}
+
+\begin{proof}
+Let $q \in S$ such that $q < p$.
+We will show that $F_2$ has an element which is a subset of $q^\ge$.
+{{explain|$q^\ge$ is the [[Definition:Upper Closure of Element|upper closure]] in what set?}}
+Since $F_1$ converges to $p$, it has an element:
+:$A \subseteq q^\ge$.
+Thus there is an element $k$ in $A$ and an element $M$ in $F_2$ such that all elements of $M$ succeed $k$.
+Then by [[Extended Transitivity]]:
+:$M \subseteq q^\ge$
+A similar argument using $F_3$ proves the dual statement.
+Thus $F_2$ converges to $p$.
+{{explain}}
+{{qed}}
+[[Category:Filter Theory]]
+[[Category:Order Topology]]
+[[Category:Convergence]]
+ia5fwo0pwzub1pc2obn184j9ojlmykf
+\end{proof}<|endoftext|>
+\section{Non-Zero Vectors Orthogonal iff Perpendicular}
+Tags: Linear Algebra, Analytic Geometry
+
+\begin{theorem}
+Let $\mathbf u$, $\mathbf v$ be non-[[Definition:Zero Vector|zero]] [[Definition:Vector (Euclidean Space)|vectors]] in the [[Definition:Real Euclidean Space|Euclidean space]] $\R^n$.
+Then $\mathbf u$ and $\mathbf v$ are [[Definition:Orthogonal (Linear Algebra)|orthogonal]] {{iff}} they are [[Definition:Perpendicular (Linear Algebra)|perpendicular]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+When $\theta$ denotes the [[Definition:Angle Between Vectors|angle]] between $\mathbf u$ and $\mathbf v$ measured in [[Definition:Radian|radians]], we have:
+{{begin-eqn}}
+{{eqn | l = 0
+      | r = \mathbf u \cdot \mathbf v
+      | c = {{Defof|Orthogonal Vectors}}
+}}
+{{eqn | r = \left\Vert{\mathbf u}\right\Vert \left\Vert{\mathbf v}\right\Vert \cos \theta
+      | c = [[Cosine Formula for Dot Product]]
+}}
+{{end-eqn}}
+where $\left\Vert{\cdot}\right\Vert$ denotes the [[Definition:Euclidean Norm|Euclidean norm]].
+As $\mathbf u, \mathbf v$ are non-[[Definition:Zero Vector|zero vectors]], it follows from the [[Definition:Norm on Vector Space|norm axiom of positive definiteness]] that $\left\Vert{\mathbf u}\right\Vert \ne 0$, and $\left\Vert{\mathbf v}\right\Vert \ne 0$.
+Then $\cos \theta = 0$.
+It follows from [[Zeroes of Sine and Cosine]] that $\theta = \dfrac \pi 2$, given that $\theta \in \left[{0\,.\,.\,\pi}\right]$ by the convention of the [[Definition:Angle Between Vectors|definition of angle between vectors]].
+Then $\mathbf u$ and $\mathbf v$ are [[Definition:Perpendicular (Linear Algebra)|perpendicular]] by the [[Definition:Right Angle|measurement of a right angle]].
+{{qed|lemma}}
+=== Sufficient Condition ===
+If $\mathbf u$ and $\mathbf v$ are [[Definition:Perpendicular (Linear Algebra)|perpendicular]], then the [[Definition:Angle Between Vectors|angle]] between them measures $\dfrac \pi 2$ in [[Definition:Radian|radians]].
+Then $\cos \theta = 0$, and the calculations above show that $\mathbf u \cdot \mathbf v = 0$.
+Hence, $\mathbf u$ and $\mathbf v$ are [[Definition:Orthogonal (Linear Algebra)|orthogonal]].
+{{qed}}
+[[Category:Linear Algebra]]
+[[Category:Analytic Geometry]]
+ojyl8da3qmaktopioezmvwcsytw6juf
+\end{proof}<|endoftext|>
+\section{Exponential Function is Continuous/Complex}
+Tags: Exponential Function is Continuous
+
+\begin{theorem}
+:$\forall z_0 \in \C: \displaystyle \lim_{z \mathop \to z_0} \exp z = \exp z_0$
+\end{theorem}
+
+\begin{proof}
+This proof depends on the [[Definition:Exponential Function/Complex/Differential Equation|differential equation definition of the exponential function]].
+The result follows from [[Complex-Differentiable Function is Continuous]].
+{{qed}}
+[[Category:Exponential Function is Continuous]]
+445lwmth2kcf596w3ms1rmazvjrik3d
+\end{proof}<|endoftext|>
+\section{Relation Compatible with Group Operation is Reflexive or Antireflexive}
+Tags: Compatible Relations, Group Theory
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
+Let $\RR$ be a [[Definition:Endorelation|relation]] on $G$ that is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
+Then $\RR$ is [[Definition:Reflexive Relation|reflexive]] or [[Definition:Antireflexive Relation|antireflexive]].
+\end{theorem}
+
+\begin{proof}
+Suppose that $\RR$ is not [[Definition:Antireflexive Relation|antireflexive]].
+Then there is some $x \in G$ such that $x \mathrel \RR x$.
+Let $y \in G$.
+Then by the definition of [[Definition:Relation Compatible with Operation|compatibility]]:
+:$\paren {x \circ \paren {x^{-1} \circ y} } \mathrel \RR \paren {x \circ \paren {x^{-1} \circ y} }$
+By {{GroupAxiom|1}} and {{GroupAxiom|3}}:
+:$y \mathrel \RR y$
+Since this holds for arbitrary $y$, $\RR$ is [[Definition:Reflexive Relation|reflexive]].
+Thus $\mathcal R$ is either [[Definition:Reflexive Relation|reflexive]] or [[Definition:Antireflexive Relation|antireflexive]].
+{{qed}}
+{{explain|While the above demonstrates that if such a relation is not antireflexive it must be reflexive, it does not approach the task of demonstrating that such a relation can actually *be* antireflexive.}}
+[[Category:Compatible Relations]]
+[[Category:Group Theory]]
+ii85i0461niyz1id47zcu28phlmvv0h
+\end{proof}<|endoftext|>
+\section{Order Topology on Convex Subset is Subspace Topology}
+Tags: Order Topology, Topological Subspaces
+
+\begin{theorem}
+Let $\left({S, \preceq,\tau}\right)$ be a [[Definition:Linearly Ordered Space|linearly ordered space]].
+Let $A \subseteq S$ be a [[Definition:Convex Set (Order Theory)|convex set]] in $S$.
+Let $\upsilon$ be the [[Definition:Order Topology|order topology]] on $A$.
+Let $\tau'$ be the $\tau$-relative [[Definition:Subspace Topology|subspace topology]] on $A$.
+Then $\upsilon = \tau'$.
+\end{theorem}
+
+\begin{proof}
+{{improve|Reword so as to be less wordy. Add pictures so as to be easier to visualize.}}
+By the definition of the [[Definition:Order Topology|order topology]], the sets of [[Definition:Open Ray|open rays]] in $\left({S, \preceq}\right)$ and $\left({A, \preceq}\right)$ form [[Definition:Sub-Basis|sub-bases]] for $\tau$ and $\upsilon$, respectively.
+By [[Sub-Basis for Topological Subspace]], we need only show that the set of [[Definition:Open Ray|open rays]] in $\left({S, \preceq}\right)$ induces a subbase for $\upsilon$.
+Specifically, we will show that each [[Definition:Open Ray|open ray]] in $A$ is the intersection of $A$ with an [[Definition:Open Ray|open ray]] in $S$, and we will show that the intersection of any [[Definition:Open Ray|open ray]] in $S$ with $A$ is either an [[Definition:Open Ray|open ray]] in $A$, the empty set, or $A$.
+By [[Dual Pairs (Order Theory)]], strict upper closure is dual to strict lower closure, so by the [[Duality Principle (Order Theory)/Local Duality|duality principle]] we need show this only for the [[Definition:Upward-Pointing Ray|upward-pointing rays]].
+For each $p \in S$, let $p^{\succ S}$ be the [[Definition:Strict Upper Closure of Element|strict upper closure]] of $p$ in $S$.
+For each $p \in A$, let $p^{\succ A}$ be the [[Definition:Strict Upper Closure of Element|strict upper closure]] of $p$ in $A$.
+Let $p \in A$.
+Then by [[Strict Upper Closure in Restricted Ordering]]:
+:$p^{\succ A} = A \cap p^{\succ S}$
+Let $q \in S$.
+If $q \in A$, then $A \cap q^{\succ S} = q^{\succ A}$, so $A \cap q^{\succ S}$ is an [[Definition:Open Ray|open ray] in $A$.
+Suppose instead that $q \notin A$.
+If $A \cap q^{\succ S} = \varnothing$, the proof is complete.
+Otherwise, $A \cap q^{\succ S}$ must have some element $x$.
+Suppose [[Proof by Contradiction|for the sake of contradiction]] that there is a $y \in A \setminus \left({ A \cap q^{\succ S} }\right) = A \cap q^{\preceq S}$.
+Then since $q \notin A$, it follows that $y \prec q$.
+Thus $y \prec q \prec x$, $y, x \in A$, and $q \notin A$, contradicting the fact that $A$ is [[Definition:Convex Set (Order Theory)|convex]].
+Thus $A \cap q^{\succ S} = A$.
+{{qed}}
+[[Category:Order Topology]]
+[[Category:Topological Subspaces]]
+27m938ooaqvwwhgxxqyy2t2qkuhvld9
+\end{proof}<|endoftext|>
+\section{Exponential Function is Continuous/Real Numbers}
+Tags: Exponential Function is Continuous, Continuous Real Functions
+
+\begin{theorem}
+The [[Definition:Real Exponential Function|real exponential function]] is [[Definition:Continuous Function|continuous]].
+That is:
+:$\forall x_0 \in \R: \displaystyle \lim_{x \mathop \to x_0} \ \exp x = \exp x_0$
+\end{theorem}<|endoftext|>
+\section{Edge of Tree is Bridge}
+Tags: Tree Theory
+
+\begin{theorem}
+Let $T$ be a [[Definition:Tree (Graph Theory)|tree]].
+Let $e$ be an [[Definition:Edge of Graph|edge]] of $T$.
+Then $e$ is a [[Definition:Bridge|bridge]] of $T$.
+\end{theorem}
+
+\begin{proof}
+From [[Condition for Edge to be Bridge]], $e$ is a [[Definition:Bridge|bridge]] [[Definition:Iff|iff]] $e$ does not lie on any [[Definition:Circuit|circuit]].
+Since $T$ is a [[Definition:Tree (Graph Theory)|tree]], there are no [[Definition:Circuit|circuits]] in $T$.
+The result follows.
+{{qed}}
+[[Category:Tree Theory]]
+d6vql0l2gsxkaymoedoq8ja7dk2y05u
+\end{proof}<|endoftext|>
+\section{Sub-Basis for Topological Subspace}
+Tags: Topological Bases, Topological Subspaces
+
+\begin{theorem}
+Let $\left({X,\tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Let $K$ be a [[Definition:Sub-Basis|sub-basis]] for $\tau$.
+Let $\left({S, \tau'}\right)$ be a [[Definition:Topological Subspace|subspace]] of $\left({X, \tau}\right)$.
+Let $K' = \left\{{ U \cap S: U \in K }\right\}$.
+That is, $K'$ consists of the [[Definition:Open Set (Topology)|$\tau'$-open sets]] in $S$ corresponding to [[Definition:Element|elements]] of $K$.
+Then $K'$ is a [[Definition:Sub-Basis|sub-basis]] for $\tau'$.
+\end{theorem}
+
+\begin{proof}
+Let $B$ be the [[Definition:Basis (Topology)|basis]] for $\tau$ generated by $K$.
+By [[Basis for Topological Subspace]], $B$ generates a basis, $B\,'$, for $\tau'$.
+We will show that $K'$ generates $B\,'$.
+Let $V' \in B\,'$.
+Then for some $V \in B$:
+:$V' = S \cap V$
+By the definition of a [[Definition:Sub-Basis|sub-basis]], there is a [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] $K_V$ of $K$ such that:
+:$\displaystyle V = \bigcap K_V$
+Let $K_V' = \left\{{S \cap U: U \in K_V }\right\}$.
+Then:
+:$\displaystyle V' = S \cap V = S \cap \bigcap K_V$
+so:
+:$\displaystyle V' = \bigcap_{P \mathop \in K_V} \left({S \cap P}\right) = \bigcap K_V'$
+Thus $B\,'$ is generated by $K'$.
+{{qed}}
+[[Category:Topological Bases]]
+[[Category:Topological Subspaces]]
+cwtvjt3c1l9iu0xs6z0bd8w714lclmo
+\end{proof}<|endoftext|>
+\section{Modus Ponendo Ponens/Variant 1}
+Tags: Modus Ponendo Ponens
+
+\begin{theorem}
+:  $p \vdash \left({p \implies q}\right) \implies q$
+\end{theorem}<|endoftext|>
+\section{Modus Ponendo Ponens/Variant 1/Proof 2}
+Tags: Truth Table Proofs, Modus Ponendo Ponens
+
+\begin{theorem}
+:$p \vdash \left({p \implies q}\right) \implies q$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]].
+$\begin{array}{|c|ccccc|} \hline
+p & (p & \implies & q) & \implies & q\\
+\hline
+F & F & T & F & F & F \\
+F & F & T & T & T & T \\
+T & T & F & F & T & F \\
+T & T & T & T & T & T \\
+\hline
+\end{array}$
+As can be seen by inspection, when $p$ is [[Definition:True|true]], the [[Definition:Truth Value|value]] of the [[Definition:Main Connective|main connective]] is also [[Definition:True|true]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Modus Ponendo Ponens/Variant 2}
+Tags: Modus Ponendo Ponens
+
+\begin{theorem}
+:  $\vdash p \implies \left({\left({p \implies q}\right) \implies q}\right)$
+\end{theorem}<|endoftext|>
+\section{Modus Ponendo Ponens/Variant 2/Proof 1}
+Tags: Modus Ponendo Ponens
+
+\begin{theorem}
+:$\vdash p \implies \left({\left({p \implies q}\right) \implies q}\right)$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\vdash p \implies \left({\left({p \implies q}\right) \implies q}\right)}}
+{{Assumption|1|p}}
+{{Assumption|2|p \implies q}}
+{{ModusPonens|3|1, 2|q|2|1}}
+{{Implication|4|1|\left({p \implies q}\right) \implies q|2|3}}
+{{Implication|5||p \implies \left({\left({p \implies q}\right) \implies q}\right)|1|4}}
+{{EndTableau}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Modus Ponendo Ponens/Variant 2/Proof 2}
+Tags: Truth Table Proofs, Modus Ponendo Ponens
+
+\begin{theorem}
+:$\vdash p \implies \paren {\paren {p \implies q} \implies q}$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]].
+$\begin{array}{|c|c|ccccc|} \hline
+p & \implies & ((p & \implies & q) & \implies & q)\\
+\hline
+F & T & F & T & F & F & F \\
+F & T & F & T & T & T & T \\
+T & T & T & F & F & T & F \\
+T & T & T & T & T & T & T \\
+\hline
+\end{array}$
+As can be seen by inspection, the [[Definition:Main Connective|main connective]] is [[Definition:True|true]] throughout.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rule of Commutation/Conjunction}
+Tags: Conjunction, Rule of Commutation
+
+\begin{theorem}
+[[Definition:Conjunction|Conjunction]] is [[Definition:Commutative Operation|commutative]]:
+=== [[Rule of Commutation/Conjunction/Formulation 1|Formulation 1]] ===
+{{:Rule of Commutation/Conjunction/Formulation 1}}
+=== [[Rule of Commutation/Conjunction/Formulation 2|Formulation 2]] ===
+{{:Rule of Commutation/Conjunction/Formulation 2}}
+\end{theorem}<|endoftext|>
+\section{Rule of Commutation/Disjunction}
+Tags: Disjunction, Rule of Commutation
+
+\begin{theorem}
+[[Definition:Disjunction|Disjunction]] is [[Definition:Commutative Operation|commutative]]:
+=== [[Rule of Commutation/Disjunction/Formulation 1|Formulation 1]] ===
+{{:Rule of Commutation/Disjunction/Formulation 1}}
+=== [[Rule of Commutation/Disjunction/Formulation 2|Formulation 2]] ===
+{{:Rule of Commutation/Disjunction/Formulation 2}}
+\end{theorem}<|endoftext|>
+\section{Factor Principles/Conjunction on Right/Formulation 1}
+Tags: Factor Principles
+
+\begin{theorem}
+: $p \implies q \vdash \left({p \land r}\right) \implies \left ({q \land r}\right)$
+\end{theorem}<|endoftext|>
+\section{Factor Principles/Conjunction on Right/Formulation 2}
+Tags: Factor Principles
+
+\begin{theorem}
+: $\vdash \left({p \implies q}\right) \implies \left({\left({p \land r}\right) \implies \left ({q \land r}\right)}\right)$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\vdash \left({p \implies q}\right) \implies \left({\left({p \land r}\right) \implies \left ({q \land r}\right)}\right)}}
+{{Assumption|1|p \implies q}}
+{{SequentIntro|2|1|\left({p \land r}\right) \implies \left ({q \land r}\right)|1
+  |[[Factor Principles/Conjunction on Right/Formulation 1|Factor Principles: Conjunction on Right: Formulation 1]]}}
+{{Implication|3|1|\left({p \implies q}\right) \implies \left({\left({p \land r}\right) \implies \left ({q \land r}\right)}\right)|1|2}}
+{{EndTableau}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Interior of Convex Angle is Convex Set}
+Tags: Vector Spaces
+
+\begin{theorem}
+Let $\mathbf v, \mathbf w$ be two non-[[Definition:Zero Vector|zero]] [[Definition:Vector (Linear Algebra)|vectors]] in $\R^2$, and let $p$ be a [[Definition:Point|point]] in $\R^2$.
+Suppose that the [[Definition:Angle Between Vectors|angle]] between $\mathbf v$ and $\mathbf w$ is a [[Definition:Convex Angle|convex angle]].
+Then the [[Definition:Set|set]] 
+:$U = \left\{ {p + st \mathbf v + \left({1-s}\right) t \mathbf w : s \in \left({0\,.\,.\,1}\right) , t \in \R_{>0} }\right\}$
+is a [[Definition:Convex Set (Vector Space)|convex set]].
+{{expand|It'd be really nice to have a picture of $U$ to support intuition and connect with the page title}}
+\end{theorem}
+
+\begin{proof}
+Let $p_1 ,p_2 \in U$.
+Then for $i \in \left\{ {1, 2}\right\}$, $p_i = p + s_i t_i \mathbf v + \left({1 - s_i}\right) t_i \mathbf w$ for some $s_i \in \left({0\,.\,.\,1}\right) , t_i \in \R_{>0}$.
+[[Definition:WLOG|WLOG]] assume that $t_1 \le t_2$.
+Suppose that $q \in \R^2$ lies on the [[Definition:Convex Set (Vector Space)/Line Segment|line segment]] joining $p_1$ and $p_2$, so:
+{{begin-eqn}}
+{{eqn |l= q
+      |r= p + s_1 t_1 \mathbf v + \left({1 - s_1}\right) t_1 \mathbf w + s \left({ p + s_2 t_2 \mathbf v + \left({1 - s_2}\right) t_2 \mathbf w - p - s_1 t_1 \mathbf v - \left({1 - s_1}\right) t_1 \mathbf w }\right)
+      |c= for some $s \in \left({0\,.\,.\,1}\right)$
+}}
+{{eqn |r= p + \left({ \left({1 - s}\right) s_1 t_1 + s s_2 t_2}\right) \mathbf v +  \left({ \left({1 - s}\right) \left({1 - s_1}\right) t_1 + s \left({1 - s_2}\right) t_2}\right) \mathbf w
+}}
+{{eqn |r= p + \dfrac{\left({1 - s}\right) s_1 t_1 + s s_2 t_2}{r}r \mathbf v + \dfrac{t_1 + st_2 - st_1 - \left({1 - s}\right) s_1 t_1 - s s_2 t_2}{r} r \mathbf w
+      |c= where $r = t_1 + s \left({t_2 - t_1}\right)$
+}}
+{{eqn |r= p + \dfrac{\left({1 - s}\right) s_1 t_1 + s s_2 t_2}{r}r \mathbf v + \left({ 1 - \dfrac{\left({1 - s}\right) s_1 t_1 + s s_2 t_2}{r} }\right) r \mathbf w
+}}
+{{end-eqn}}
+As $t_1 \le t_2$, it follows that $r \in \R_{>0}$.
+We have $\dfrac{ \left({1 - s}\right) s_1 t_1 + s s_2 t_2}{r}> 0$, and:
+{{begin-eqn}}
+{{eqn |l= 1 - \dfrac{\left({1 - s}\right) s_1 t_1 + s s_2 t_2}{r}
+      |r= \dfrac{ \left({1 - s}\right) \left({1 - s_1}\right) t_1 + s \left({1 - s_2}\right) t_2}{r}
+}}
+{{eqn |o= >
+      |r= 0
+}}
+{{end-eqn}}
+It follows that $\dfrac{ \left({1 - s}\right) s_1 t_1 + s s_2 t_2}{r} \in \left({0\,.\,.\,1}\right)$.
+Then $q \in U$.
+By [[Definition:Convex Set (Vector Space)|definition of convex set]], it follows that $U$ is convex.
+{{qed}}
+[[Category:Vector Spaces]]
+9bmr4r3gkhrw6nlhcxeeovr0al1t8ls
+\end{proof}<|endoftext|>
+\section{Factor Principles/Conjunction on Right}
+Tags: Factor Principles
+
+\begin{theorem}
+==== [[Factor Principles/Conjunction on Right/Formulation 1|Formulation 1]] ====
+{{:Factor Principles/Conjunction on Right/Formulation 1}}
+==== [[Factor Principles/Conjunction on Right/Formulation 2|Formulation 2]] ====
+{{:Factor Principles/Conjunction on Right/Formulation 2}}
+\end{theorem}<|endoftext|>
+\section{Factor Principles/Conjunction on Left}
+Tags: Factor Principles
+
+\begin{theorem}
+==== [[Factor Principles/Conjunction on Left/Formulation 1|Formulation 1]] ====
+{{:Factor Principles/Conjunction on Left/Formulation 1}}
+==== [[Factor Principles/Conjunction on Left/Formulation 2|Formulation 2]] ====
+{{:Factor Principles/Conjunction on Left/Formulation 2}}
+\end{theorem}<|endoftext|>
+\section{Factor Principles/Conjunction on Left/Formulation 1}
+Tags: Factor Principles
+
+\begin{theorem}
+: $p \implies q \vdash \left({r \land p}\right) \implies \left ({r \land q}\right)$
+\end{theorem}<|endoftext|>
+\section{Factor Principles/Conjunction on Left/Formulation 2}
+Tags: Factor Principles
+
+\begin{theorem}
+:$\vdash \paren {p \implies q} \implies \paren {\paren {r \land p} \implies \paren {r \land q} }$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\vdash \paren {p \implies q} \implies \paren {\paren {r \land p} \implies \paren {r \land q} } }}
+{{Assumption|1|p \implies q}}
+{{SequentIntro|2|1|\paren {\paren {r \land p} \implies \paren {r \land q} }|1
+  |[[Factor Principles/Conjunction on Left/Formulation 1|Factor Principles: Conjunction on Left: Formulation 1]]}}
+{{Implication|3|1|\paren {p \implies q} \implies \paren {\paren {r \land p} \implies \paren {r \land q} }|1|2}}
+{{EndTableau}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Triangle is Convex Set}
+Tags: Vector Spaces
+
+\begin{theorem}
+The [[Definition:Interior of Jordan Curve|interior]] of a [[Definition:Triangle (Geometry)|triangle]] embedded in $\R^2$ is a [[Definition:Convex Set (Vector Space)|convex set]].
+\end{theorem}
+
+\begin{proof}
+Denote the [[Definition:Triangle (Geometry)|triangle]] as $\triangle$, and the [[Definition:Interior of Jordan Curve|interior]] of the [[Definition:Boundary (Geometry)|boundary]] of $\triangle$ as $\Int \triangle$.
+From [[Boundary of Polygon is Jordan Curve]], it follows that the [[Definition:Boundary (Geometry)|boundary]] of $\triangle$ is equal to the [[Definition:Image of Mapping|image]] of a [[Definition:Jordan Curve|Jordan curve]], so $\Int \triangle$ is well-defined.
+Denote the [[Definition:Vertex (Geometry)|vertices]] of $\triangle$ as $A_1, A_2, A_3$.
+For $i \in \set {1, 2, 3}$, put $j = i \bmod 3 + 1$, $k = \paren {i + 1} \bmod 3 + 1$, and:
+:$U_i = \set {A_i + s t \paren {A_j - A_i} + \paren {1 - s} t \paren {A_k - A_i} : s \in \openint 0 1, t \in \R_{>0} }$
+Suppose that the [[Definition:Angle Between Vectors|angle]] $\angle A_i$ between is $A_j - A_i$ and $A_k - A_i$ is [[Definition:Convex Angle|non-convex]].
+As $\angle A_i$ is an [[Definition:Internal Angle|internal angle]] in $\triangle$, it follows from [[Definition:Polygon|definition of polygon]] that $\angle A_i$ cannot be [[Definition:Zero Angle|zero]] or [[Definition:Straight Angle|straight]].
+Then $\angle A_i$ is larger than a [[Definition:Straight Angle|straight angle]], which is impossible by [[Sum of Angles of Triangle Equals Two Right Angles]].
+It follows that $\angle A_i$ is [[Definition:Convex Angle|convex]].
+From [[Characterization of Interior of Triangle]], it follows that:
+:$\displaystyle \Int \triangle = \bigcap_{i \mathop = 1}^3 U_i$
+From [[Interior of Convex Angle is Convex Set]], it follows for $i \in \set {1, 2, 3}$ that $U_i$ is a [[Definition:Convex Set (Vector Space)|convex set]].
+The result now follows from [[Intersection of Convex Sets is Convex Set (Vector Spaces)]].
+{{qed}}
+[[Category:Vector Spaces]]
+62yculhuqp577pbsv8offiogbpmwcf4
+\end{proof}<|endoftext|>
+\section{Norm of Vector Cross Product}
+Tags: Vector Cross Product
+
+\begin{theorem}
+Let $\mathbf a$ and $\mathbf b$ be [[Definition:Vector (Euclidean Space)|vectors]] in the [[Definition:Real Euclidean Space|Euclidean space]] $\R^3$.
+Let $\times$ denote the [[Definition:Vector Cross Product|vector cross product]].
+Then:
+:$(1): \quad$ $\left\Vert{ \mathbf a \times \mathbf b }\right\Vert^2 = \left\Vert{\mathbf a}\right\Vert^2 \left\Vert{\mathbf b}\right\Vert^2 - \left({\mathbf a \cdot \mathbf b}\right)^2$
+:$(2): \quad$ $\left\Vert{ \mathbf a \times \mathbf b }\right\Vert = \left\Vert{\mathbf a}\right\Vert \left\Vert{\mathbf b}\right\Vert \left\vert{\sin \theta}\right\vert$
+where $\theta$ is the [[Definition:Angle Between Vectors|angle between $\mathbf a$ and $\mathbf b$]], or an arbitrary [[Definition:Real Number|number]] if $\mathbf a$ or $\mathbf b$ is the [[Definition:Zero Vector|zero vector]].
+\end{theorem}
+
+\begin{proof}
+Let $\mathbf a = \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix}$, and $\mathbf b = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \left\Vert{ \mathbf a \times \mathbf b }\right\Vert^2
+      | r = \left({ \mathbf a \times \mathbf b }\right) \cdot \left({ \mathbf a \times \mathbf b }\right)
+      | c = {{Defof|Euclidean Norm}}
+}}
+{{eqn | r = \begin{bmatrix} a_2 b_3 - a_3 b_2 \\ a_3 b_1 - a_1 b_3 \\ a_1 b_2 - a_2 b_1 \end{bmatrix} \cdot \begin{bmatrix} a_2 b_3 - a_3 b_2 \\ a_3 b_1 - a_1 b_3 \\ a_1 b_2 - a_2 b_1 \end{bmatrix}
+      | c = {{Defof|Vector Cross Product}}
+}}
+{{eqn | r = a_2^2 b_3^2 + a_3^2 b_2^2 - 2a_2 a_3 b_2 b_3 + a_3^2 b_1^2 + a_1^2 b_3^2 - 2a_1 a_3 b_1 b_3 + a_1^2 b_2^2 + a_2^2 b_1^2 - 2a_1 a_2 b_1 b_2
+      | c = {{Defof|Dot Product}}
+}}
+{{eqn | r = \left({a_1^2 + a_2^2 + a_3^2}\right) \left({b_1^2 + b_2^2 + b_3^2}\right) - \left({a_1 b_1 + a_2 b_2 + a_3 b_3}\right)^2
+      | c = by algebraic manipulations
+}}
+{{eqn | r = \left\Vert{\mathbf a}\right\Vert^2 \left\Vert{\mathbf b}\right\Vert^2 - \left({\mathbf a \cdot \mathbf b}\right)^2
+}}
+{{end-eqn}}
+This proves $(1)$.
+{{qed|lemma}}
+If $\mathbf a$ or $\mathbf b$ is the [[Definition:Zero Vector|zero vector]], then $\left\Vert{\mathbf a}\right\Vert = 0$, or $\left\Vert{\mathbf b}\right\Vert = 0$ by the [[Definition:Norm on Vector Space|positive definiteness norm axiom]].
+By calculation, it follows that $\mathbf a \times \mathbf b$ is also the zero vector, so $\left\Vert{\mathbf a \times \mathbf b}\right\Vert = 0$.
+Hence, equality $(2)$ holds.
+If both $\mathbf a$ or $\mathbf b$ are non-zero [[Definition:Vector|vectors]], we continue the calculations from the first section:
+{{begin-eqn}}
+{{eqn | l = \left\Vert{ \mathbf a \times \mathbf b }\right\Vert^2
+      | r = \left\Vert{\mathbf a}\right\Vert^2 \left\Vert{\mathbf b}\right\Vert^2 - \left({\mathbf a \cdot \mathbf b}\right)^2
+}}
+{{eqn | r = \left\Vert{\mathbf a}\right\Vert^2 \left\Vert{\mathbf b}\right\Vert^2 - \left\Vert{\mathbf a}\right\Vert^2 \left\Vert{\mathbf b}\right\Vert^2 \cos^2 \theta 
+      | c = [[Cosine Formula for Dot Product]]
+}}
+{{eqn | r = \left\Vert{\mathbf a}\right\Vert^2 \left\Vert{\mathbf b}\right\Vert^2 \left({1 - \cos^2 \theta}\right)
+}}
+{{eqn | r = \left\Vert{\mathbf a}\right\Vert^2 \left\Vert{\mathbf b}\right\Vert^2 \sin^2 \theta
+      | c = [[Sum of Squares of Sine and Cosine]]
+}}
+{{end-eqn}}
+Equality $(2)$ now follows after taking the [[Definition:Principal Square Root|square root]] of both sides of the equality.
+This is possible as [[Square of Real Number is Non-Negative]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subset Relation is Compatible with Subset Product}
+Tags: Compatible Relations, Subset Products
+
+\begin{theorem}
+Let $\left({S,\circ}\right)$ be a [[Definition:magma|magma]].
+Let $\mathcal P \left({S}\right)$ be the [[Definition:Power Set|power set]] of $S$.
+Let $\circ_\mathcal P$ be the [[Definition:Subset Product|operation induced on $\mathcal P \left({S}\right)$ by $\circ$]].
+Then the [[Definition:subset|subset]] relation $\subseteq$ is [[Definition:Relation Compatible with Operation|compatible]] with $\circ_\mathcal P$.
+\end{theorem}
+
+\begin{proof}
+Let $A, B, C \in \mathcal P \left({S}\right)$.
+Let $A \subseteq B$.
+Let $x \in A \circ_\mathcal P C$.
+Then for some $a \in A$ and some $c \in C$:
+:$x = a \circ c$
+Since $A \subseteq B$, $a \in B$.
+Thus $x \in B \circ_\mathcal P C$.
+Since this holds for all $x \in A \circ_\mathcal P C$:
+:$A \circ_\mathcal P C \subseteq B \circ_\mathcal P C$
+The same argument shows that:
+:$C \circ_\mathcal P A \subseteq C \circ_\mathcal P B$
+{{qed}}
+[[Category:Compatible Relations]]
+[[Category:Subset Products]]
+n8a6efuivcr6ue4s3gb8oluuwggujjr
+\end{proof}<|endoftext|>
+\section{Subset Relation is Compatible with Subset Product/Corollary 1}
+Tags: Subset Products, Compatible Relations
+
+\begin{theorem}
+Let $A, B, C, D \in \powerset S$.
+Let $A \subseteq B$ and $C \subseteq D$.
+Then:
+:$A \circ_\PP C \subseteq B \circ_\PP D$
+\end{theorem}
+
+\begin{proof}
+By [[Subset Relation is Compatible with Subset Product]], $\subseteq$ is [[Definition:Relation Compatible with Operation|compatible]] with $\circ_\PP$.
+By [[Subset Relation is Transitive]], $\subseteq$ is [[Definition:Transitive Relation|transitive]].
+Thus the theorem holds by [[Operating on Transitive Relationships Compatible with Operation]].
+{{qed}}
+[[Category:Subset Products]]
+[[Category:Compatible Relations]]
+n5123gorw863wboo6opdwomth1tut18
+\end{proof}<|endoftext|>
+\section{Subset Relation is Compatible with Subset Product/Corollary 2}
+Tags: Subset Products
+
+\begin{theorem}
+Let $A,B \in \mathcal P \left({S}\right)$, the [[Definition:Power Set|power set]] of $S$.
+{{improve|No need to use power set. $A \subseteq B \subseteq S$ sufficient.}}
+Let $A \subseteq B$.
+Let $x \in S$.
+Then:
+:$x \circ A \subseteq x \circ B$
+:$A \circ x \subseteq B \circ x$
+\end{theorem}
+
+\begin{proof}
+This follows from [[Subset Relation is Compatible with Subset Product]] and the definition of the [[Definition:Subset Product with Singleton|subset product with a singleton]].
+{{qed}}
+[[Category:Subset Products]]
+hwov202p0rmiee3qlizogofinxbtbx2
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Normal Subset/1 iff 2}
+Tags: Normal Subsets
+
+\begin{theorem}
+Let $\left({G, \circ}\right)$ be a [[Definition:Group|group]].
+Let $S$ be a [[Definition:Subset|subset]] of $G$.
+Then [[Definition:Normal Subset/Definition 1|Normal Subset/Definition 1]] is equivalent to [[Definition:Normal Subset/Definition 2|Normal Subset/Definition 2]].
+That is, the following three statements are equivalent:
+:$(1)\quad \forall g \in G: g \circ S = S \circ g$
+:$(2)\quad \forall g \in G: g \circ S \circ g^{-1} = S$
+:$(3)\quad \forall g \in G: g^{-1} \circ S \circ g = S$
+\end{theorem}
+
+\begin{proof}
+Let $e$ be the [[Definition:Identity Element|identity]] of $G$.
+First note that:
+:$(4): \quad \left({\forall g \in G: g \circ S \circ g^{-1} = S}\right) \iff \left({\forall g \in G: g^{-1} \circ S \circ g = S}\right)$
+which is shown by, for example, setting $h := g^{-1}$ and substituting.
+=== Necessary Condition ===
+Suppose that $S$ satisfies $(1)$.
+Then:
+{{begin-eqn}}
+{{eqn | lo= \forall g \in G:
+      | l = g \circ S
+      | r = S \circ g
+      | c = $(1)$
+}}
+{{eqn | ll= \implies
+      | l = \left({g \circ S}\right) \circ g^{-1}
+      | r = \left({S \circ g}\right) \circ g^{-1}
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = g \circ S \circ g^{-1}
+      | r = S \circ \left({g \circ g^{-1} }\right)
+      | c = [[Subset Product within Semigroup is Associative/Corollary|Subset Product within Semigroup is Associative: Corollary]]
+}}
+{{eqn | ll= \implies
+      | l = g \circ S \circ g^{-1}
+      | r = S \circ e
+      | c = {{Defof|Inverse Element}}
+}}
+{{eqn | ll= \implies
+      | l = g \circ S \circ g^{-1}
+      | r = S
+      | c = [[Subset Product by Identity Singleton]]
+}}
+{{eqn | lo= \forall g \in G:
+      | l = g^{-1} \circ S \circ g
+      | r = S
+      | c = $(4)$
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $S$ be a [[Definition:Subset|subset of $G$]] such that:
+: $\forall g \in G: g \circ S \circ g^{-1} = S$
+or:
+: $\forall g \in G: g^{-1} \circ S \circ g = S$
+By $(4)$, as long as one of these statements holds, the other one holds as well.
+Then:
+{{begin-eqn}}
+{{eqn | lo= \forall g \in G:
+      | l = g \circ S \circ g^{-1}
+      | r = S
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \left({g \circ S \circ g^{-1} }\right) \circ g
+      | r = S \circ g
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \left({g \circ S}\right) \circ \left({g^{-1} \circ g}\right)
+      | r = S \circ g
+      | c = [[Subset Product within Semigroup is Associative/Corollary|Subset Product within Semigroup is Associative: Corollary]]
+}}
+{{eqn | ll= \implies
+      | l = \left({g \circ S}\right) \circ e
+      | r = S \circ g
+      | c = {{Defof|Inverse Element}}
+}}
+{{eqn | ll= \implies
+      | l = g \circ S
+      | r = S \circ g
+      | c = [[Subset Product by Identity Singleton]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subset Product with Identity}
+Tags: Abstract Algebra
+
+\begin{theorem}
+Let $\left({S, \circ}\right)$ be a [[Definition:Magma|magma]].
+Let $\left({S, \circ}\right)$ have an [[Definition:Identity Element|identity element]] $e$.
+Then $e \circ S = S \circ e = S$, where $\circ$ is understood to be the [[Definition:Subset Product with Singleton|subset product with singleton]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = e \circ S
+      | r = \left\{ {e} \right\} \circ S
+      | c = {{Defof|Subset Product with Singleton}}
+}}
+{{eqn | r = \left\{ {x \circ y: x \in \left\{ {e} \right\}, \, y \in S} \right\}
+      | c = {{Defof|Subset Product}}
+}}
+{{eqn | r = \left\{ {e \circ y: y \in S} \right\}
+      | c = 
+}}
+{{eqn | r = \left\{ {y: y \in S} \right\}
+      | c = {{Defof|Identity Element}}
+}}
+{{eqn | r = S
+      | c =
+}}
+{{end-eqn}}
+Thus:
+:$e \circ S = S$
+A similar argument shows that:
+:$S \circ e = S$
+{{qed}}
+[[Category:Abstract Algebra]]
+gwfbjq2zd305mamxotwx0r838cut2sj
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Normal Subset/2 implies 3}
+Tags: Normal Subsets
+
+\begin{theorem}
+Let $\left({G, \circ}\right)$ be a [[Definition:group|group]].
+Let $S \subseteq G$.
+Let $S$ be a [[Definition:Normal Subset/Definition 2|normal subset of $G$ by Definition 2]].
+Then $S$ is a [[Definition:Normal Subset/Definition 3|normal subset of $G$ by Definition 3]].
+That is, if:
+:$\forall g \in G: g \circ S \circ g^{-1} = S$
+or:
+:$\forall g \in G: g^{-1} \circ S \circ g = S$
+then:
+:$\forall g \in G: g \circ S \circ g^{-1} \subseteq S$
+and:
+:$\forall g \in G: g^{-1} \circ S \circ g \subseteq S$
+\end{theorem}
+
+\begin{proof}
+We have that:
+:$\left({\forall g \in G: g \circ S \circ g^{-1} = S}\right) \iff \left({\forall g \in G: g^{-1} \circ S \circ g = S}\right)$
+The result follows by definition of [[Definition:Set Equality/Definition 2|set equality]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Superset Relation is Compatible with Subset Product}
+Tags: Compatible Relations
+
+\begin{theorem}
+Let $\left({S,\circ}\right)$ be a [[Definition:Magma|magma]].
+Let $\circ_{\mathcal P}$ be the [[Definition:Subset Product|subset product]] on $\mathcal P \left({S}\right)$, the [[Definition:Power Set|power set]] of $S$.
+Then the [[Definition:Superset|superset]] relation $\supseteq$ is [[Definition:Relation Compatible with Operation|compatible]] with $\circ_{\mathcal P}$.
+\end{theorem}
+
+\begin{proof}
+By [[Subset Relation is Compatible with Subset Product]], the [[Definition:Subset|subset]] relation $\subseteq$ is [[Definition:Relation Compatible with Operation|compatible]] with $\circ_{\mathcal P}$.
+From [[Inverse of Subset Relation is Superset]], the [[Definition:Inverse Relation|inverse]] of $\subseteq$ is $\supseteq$.
+The result follows from [[Inverse of Relation Compatible with Operation is Compatible]].
+{{qed}}
+[[Category:Compatible Relations]]
+0ld0w22afe9z6m2t6oiestk41kajtld
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Normal Subset/3 iff 4}
+Tags: Normal Subsets
+
+\begin{theorem}
+Let $\left({G,\circ}\right)$ be a [[Definition:group|group]].
+Let $S \subseteq G$.
+Then:
+: $S$ is a [[Definition:Normal Subset/Definition 3|normal subset of $G$ by Definition 3]]
+[[Definition:Iff|iff]]:
+: $S$ is a [[Definition:Normal Subset/Definition 4|normal subset of $G$ by Definition 4]].
+That is, the following conditions are equivalent:
+: $(1)\quad \forall g \in G: g \circ S \circ g^{-1} \subseteq S$
+: $(2)\quad \forall g \in G: g^{-1} \circ S \circ g \subseteq S$
+: $(3)\quad \forall g \in G: S \subseteq g \circ S \circ g^{-1}$
+: $(4)\quad \forall g \in G: S \subseteq g^{-1} \circ S \circ g$
+\end{theorem}
+
+\begin{proof}
+First note that:
+:$(5): \quad \left({\forall g \in G: g \circ S \circ g^{-1} \subseteq S}\right) \iff \left({\forall g \in G: g^{-1} \circ S \circ g \subseteq S}\right)$
+:$(6): \quad \left({\forall g \in G: S \subseteq g \circ S \circ g^{-1}}\right) \iff \left({\forall g \in G: S \subseteq g^{-1} \circ S \circ g}\right)$
+which is shown by, for example, setting $h := g^{-1}$ and substituting.
+Therefore:
+: conditions $(1)$ and $(2)$ are equivalent
+and:
+: conditions $(3)$ and $(4)$ are equivalent.
+It remains to be shown that condition $(1)$ is equivalent to condition $(3)$.
+Suppose that $(1)$ holds.
+Then:
+{{begin-eqn}}
+{{eqn | lo= \forall g \in G:
+      | l = g \circ S \circ g^{-1}
+      | o = \subseteq
+      | r = S
+}}
+{{eqn | l = g^{-1} \circ \left({ g \circ S \circ g^{-1} }\right)
+      | o = \subseteq
+      | r = g^{-1} \circ S
+      | c = [[Subset Relation is Compatible with Subset Product/Corollary 2]]
+}}
+{{eqn | l = S \circ g^{-1}
+      | o = \subseteq
+      | r = g^{-1} \circ S
+      | c = [[Subset Product within Semigroup is Associative/Corollary]] and the definition of [[Definition:Inverse Element|inverse]]
+}}
+{{eqn | l = \left({ S \circ g^{-1} }\right) \circ g
+      | o = \subseteq
+      | r = \left({ g^{-1} \circ S }\right) \circ g
+      | c = [[Subset Relation is Compatible with Subset Product/Corollary 2]]
+}}
+{{eqn | l = S
+      | o = \subseteq
+      | r = g^{-1} \circ S \circ g
+      | c = [[Subset Product within Semigroup is Associative/Corollary]] and the definition of [[Definition:Inverse Element|inverse]]
+}}
+{{end-eqn}}
+Thus condition $(1)$ implies condition $(3)$.
+The exact same argument, substituting $\supseteq$ for $\subseteq$ and using [[Superset Relation is Compatible with Subset Product]] instead of [[Subset Relation is Compatible with Subset Product]] proves that $(3)$ implies $(1)$.
+{{explain|the necessary corollary is currently only available in subset form, but these arguments are all the same.  Alternatively, the exact same method can be used to move the gs from the right to the left as was used to move them from the left to the right.}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Normal Subset/3 and 4 imply 2}
+Tags: Normal Subsets
+
+\begin{theorem}
+Let $\left({G,\circ}\right)$ be a [[Definition:group|group]].
+Let $S \subseteq G$.
+Let $S$ be a [[Definition:Normal Subset/Definition 3|normal subset of $G$ by Definition 3]] and [[Definition:Normal Subset/Definition 4|Definition 4]].
+Then $S$ is a [[Definition:Normal Subset/Definition 2|normal subset of $G$ by Definition 2]].
+\end{theorem}
+
+\begin{proof}
+By [[Equivalence of Definitions of Normal Subset/3 iff 4|Equivalence of Definitions of Normal Subset: 3 iff 4]], $S$ being a normal subset of $G$ by [[Definition:Normal Subset/Definition 3|Definition 3]] and [[Definition:Normal Subset/Definition 4|Definition 4]] implies that the following hold:
+: $(1)\quad \forall g \in G: g \circ S \circ g^{-1} \subseteq S$
+: $(2)\quad \forall g \in G: g^{-1} \circ S \circ g \subseteq S$
+: $(3)\quad \forall g \in G: S \subseteq g \circ S \circ g^{-1}$
+: $(4)\quad \forall g \in G: S \subseteq g^{-1} \circ S \circ g$
+By $(1)$ and $(3)$ and definition of [[Definition:Set Equality/Definition 2|set equality]]:
+:$\forall g \in G: g \circ S \circ g^{-1} = S$
+By $(2)$ and $(4)$ and definition of [[Definition:Set Equality/Definition 2|set equality]]:
+:$\forall g \in G: g^{-1} \circ S \circ g = S$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Normal Subset/3 iff 5}
+Tags: Normal Subsets
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:group|group]].
+Let $S \subseteq G$.
+Then:
+: $S$ is a [[Definition:Normal Subset/Definition 3|normal subset of $G$ by Definition 3]]
+{{iff}}:
+: $S$ is a [[Definition:Normal Subset/Definition 5|normal subset of $G$ by Definition 5]].
+\end{theorem}
+
+\begin{proof}
+==== 3 implies 5 ====
+Suppose that for each $g \in G$:
+: $g^{-1} \circ S \circ g \subseteq S$
+Let $x, y \in G$ such that $x \circ y \in S$.
+Then:
+: $x^{-1} \circ \paren {x \circ y} \circ x \in S$
+Since $\circ$ is [[Definition:associative|associative]] and by the definition of [[Definition:Inverse Element|inverse]]:
+:$y \circ x \in S$
+{{qed|lemma}}
+==== 5 implies 3 ====
+Suppose that $S$ is a normal subset of $G$ by [[Definition:Normal Subset/Definition 5|Definition 5]].
+Then for each $x, y \in G$:
+:$x \circ y \in S \implies y \circ x \in S$
+Let $g \in G$ and let $n \in S$.
+Then by the definition of [[Definition:Inverse Element|inverse]] and the definition of [[Definition:Identity Element|identity]]:
+:$\paren {g \circ g^{-1} } \circ x \circ \paren {g \circ g^{-1} } \in S$
+By [[Definition:Group Axioms|group axiom $G1$: associativity]]:
+:$g \circ \paren {g^{-1} \circ x \circ g \circ g^{-1} } \in S$
+By the premise:
+:$\paren {g^{-1} \circ x \circ g \circ g^{-1} } \circ g \in S$
+By [[Definition:Group Axioms|group axiom $G1$: associativity]], [[Definition:Group Axioms|group axiom $G2$: identity]] and [[Definition:Group Axioms|group axiom $G3$: inverses]]:
+:$g^{-1} \circ x \circ g \in S$
+Since this holds for all $x$, $S$ is a normal subset by [[Definition:Normal Subset/Definition 3|Definition 3]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Normal Subgroup}
+Tags: Normal Subgroups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $N$ be a [[Definition:Subgroup|subgroup]] of $G$.
+{{TFAE|def = Normal Subgroup}}
+\end{theorem}
+
+\begin{proof}
+
+\end{proof}<|endoftext|>
+\section{Subgroup is Normal iff Normal Subset}
+Tags: Conjugacy, Normal Subgroups
+
+\begin{theorem}
+Let $\left({G, \circ}\right)$ be a [[Definition:Group|group]].
+Let $N$ be a [[Definition:Subgroup|subgroup]] of $G$.
+Then $N$ is [[Definition:Normal Subgroup/Definition 1|normal in $G$ (by definition 1)]] {{iff}} it is a [[Definition:Normal Subset|normal subset]] of $G$.
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $N$ be [[Definition:Normal Subgroup/Definition 1|normal in $G$ (by definition 1)]]:
+Thus for each $g \in G$:
+:$\forall g \in G: g \circ N = N \circ g$
+where $g \circ N$ denotes the [[Definition:Subset Product with Singleton|subset product of $g$ with $N$]].
+Thus $N$ is a [[Definition:Normal Subset/Definition 1|normal subset of $G$ (by definition 1)]]:
+:$\forall g \in G: g \circ N = N \circ g$
+=== Sufficient Condition ===
+Let $N$ be a [[Definition:Normal Subset/Definition 1|normal subset of $G$ (by definition 1)]]:
+:$\forall g \in G: g \circ N = N \circ g$
+Since $N$ is a subgroup, $N$ is a [[Definition:Normal Subgroup/Definition 1|normal subgroup of $G$ (by definition 1)]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Rule of Explosion/Sequent Form}
+Tags: Rule of Explosion
+
+\begin{theorem}
+:$\bot \vdash \phi$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\bot \vdash \phi}}
+{{Premise|1|\bot}}
+{{Explosion|2|1|\phi|1}}
+{{EndTableau}}
+{{Qed}}
+[[Category:Rule of Explosion]]
+qvl3j0govct9cz4i06dono4n38594tf
+\end{proof}<|endoftext|>
+\section{Law of Excluded Middle/Sequent Form}
+Tags: Law of Excluded Middle
+
+\begin{theorem}
+The [[Law of Excluded Middle]] can be symbolised by the [[Definition:Sequent|sequent]]:
+:$\vdash p \lor \neg p$
+\end{theorem}<|endoftext|>
+\section{Power Set of Monoid under Induced Operation is Monoid}
+Tags: Subset Products, Monoids, Power Set
+
+\begin{theorem}
+Let $\left({G, \circ}\right)$ be a [[Definition:Monoid|monoid]] with [[Definition:Identity Element|identity]] $e$.
+Let $\left({\mathcal P \left({G}\right), \circ_\mathcal P}\right)$ be the [[Definition:Algebraic Structure|algebraic structure]] consisting of the [[Definition:Power Set|power set]] of $G$ and the [[Definition:Subset Product|operation induced on $\mathcal P \left({S}\right)$ by $\circ$]].
+Then $\left({\mathcal P \left({G}\right), \circ_\mathcal P}\right)$ is a [[Definition:Monoid|monoid]] with [[Definition:Identity Element|identity]] $\left\{{e}\right\}$.
+\end{theorem}
+
+\begin{proof}
+By the definition of a [[Definition:Monoid|monoid]], $\left({G, \circ}\right)$ is a [[Definition:Semigroup|semigroup]].
+By [[Power Set of Semigroup under Induced Operation is Semigroup]], $\left({\mathcal P \left({G}\right), \circ_\mathcal P}\right)$ is a [[Definition:Semigroup|semigroup]].
+By [[Subset Product by Identity Singleton]], $\left\{{e}\right\}$ is an [[Definition:Identity Element|identity]] for $\left({\mathcal P \left({G}\right), \circ_\mathcal P}\right)$.
+Thus $\left({\mathcal P \left({G}\right), \circ_\mathcal P}\right)$ is a [[Definition:Monoid|monoid]] with identity $\left\{{e}\right\}$.
+{{qed}}
+[[Category:Subset Products]]
+[[Category:Monoids]]
+[[Category:Power Set]]
+q9eexcxnhi0m3m95ocbqo08278o3zk0
+\end{proof}<|endoftext|>
+\section{Principle of Non-Contradiction/Sequent Form/Formulation 2}
+Tags: Principle of Non-Contradiction
+
+\begin{theorem}
+:$\vdash \neg \paren {p \land \neg p}$
+\end{theorem}<|endoftext|>
+\section{Power Set of Semigroup under Induced Operation is Semigroup}
+Tags: Semigroups, Power Set, Subset Products
+
+\begin{theorem}
+Let $\left({S, \circ}\right)$ be a [[Definition:Semigroup|semigroup]].
+Let $\left({\mathcal P \left({S}\right), \circ_\mathcal P}\right)$ be the [[Definition:Algebraic Structure|algebraic structure]] consisting of the [[Definition:Power Set|power set]] of $S$ and the [[Definition:Subset Product|operation induced on $\mathcal P \left({S}\right)$ by $\circ$]].
+Then $\left({\mathcal P \left({S}\right), \circ_\mathcal P}\right)$ is a [[Definition:Semigroup|semigroup]].
+\end{theorem}
+
+\begin{proof}
+From [[Power Set of Magma under Induced Operation is Magma]] we conclude that $\left({\mathcal P \left({S}\right), \circ_\mathcal P}\right)$ is a [[Definition:Magma|magma]].
+It follows from [[Subset Product within Semigroup is Associative]] that $\circ_\mathcal P$ is [[Definition:Associative Operation|associative]] in $\left({\mathcal P \left({S}\right), \circ_\mathcal P}\right)$.
+Thus $\left({\mathcal P \left({S}\right), \circ_\mathcal P}\right)$ is a [[Definition:Semigroup|semigroup]].
+{{qed}}
+[[Category:Semigroups]]
+[[Category:Power Set]]
+[[Category:Subset Products]]
+du1auei87vrkart49t0xnax2tuwa8o9
+\end{proof}<|endoftext|>
+\section{Power Set of Group under Induced Operation is Monoid}
+Tags: Subset Products, Group Theory, Power Set
+
+\begin{theorem}
+Let $\left({G, \circ}\right)$ be a [[Definition:Group|group]] with [[Definition:Identity Element|identity]] $e$.
+Let $\left({\mathcal P \left({G}\right), \circ_\mathcal P}\right)$ be the [[Definition:Algebraic Structure|algebraic structure]] consisting of the [[Definition:Power Set|power set]] of $G$ and the [[Definition:Subset Product|operation induced on $\mathcal P \left({G}\right)$ by $\circ$]].
+Then $\left({\mathcal P \left({G}\right), \circ_\mathcal P}\right)$ is a [[Definition:Monoid|monoid]] with [[Definition:Identity Element|identity]] $\left\{{e}\right\}$.
+\end{theorem}
+
+\begin{proof}
+By the definition of a [[Definition:Group|group]], $\left({G, \circ}\right)$ is a [[Definition:Monoid|monoid]].
+The result follows from [[Power Set of Monoid under Induced Operation is Monoid]].
+{{qed}}
+[[Category:Subset Products]]
+[[Category:Group Theory]]
+[[Category:Power Set]]
+tlnyxsktuounjzecxgxl0yadtiqo45a
+\end{proof}<|endoftext|>
+\section{Power Set of Magma under Induced Operation is Magma}
+Tags: Power Set, Magmas, Subset Products
+
+\begin{theorem}
+Let $\left({S, \circ}\right)$ be a [[Definition:Magma|magma]].
+Let $\left({\mathcal P \left({S}\right), \circ_\mathcal P}\right)$ be the [[Definition:Algebraic Structure|algebraic structure]] consisting of the [[Definition:Power Set|power set]] of $S$ and the [[Definition:Subset Product|operation induced on $\mathcal P \left({S}\right)$ by $\circ$]].
+Then $\left({\mathcal P \left({S}\right), \circ_\mathcal P}\right)$ is a [[Definition:Magma|magma]].
+\end{theorem}
+
+\begin{proof}
+Let $\left({S, \circ}\right)$ be a [[Definition:Magma|magma]].
+Let $A, B \subseteq S$.
+{{begin-eqn}}
+{{eqn | o=
+      | r=\forall a \in A, b \in B: a \circ b \in S
+      | c=Definition of [[Definition:Magma|magma]]
+}}
+{{eqn | o=\implies
+      | r=A \circ B \subseteq S
+      | c=Definition of [[Definition:Subset Product|Subset Product]]
+}}
+{{eqn | o=\implies
+      | r=A \circ_\mathcal P B \subseteq S
+      | c=Definition of [[Definition:Subset Product|operation induced on $\mathcal P \left({S}\right)$ by $\circ$]]
+}}
+{{eqn | o=\implies
+      | r=A \circ_\mathcal P B \in \mathcal P \left({S}\right)
+      | c=Definition of [[Definition:Power Set|Power Set]]
+}}
+{{end-eqn}}
+Thus $\left({\mathcal P \left({S}\right), \circ_\mathcal P}\right)$ is a [[Definition:Magma|magma]].
+{{qed}}
+[[Category:Power Set]]
+[[Category:Magmas]]
+[[Category:Subset Products]]
+6utulg4gigeh20ityr45exebhjlaz1d
+\end{proof}<|endoftext|>
+\section{Element in Left Coset iff Product with Inverse in Subgroup}
+Tags: Cosets
+
+\begin{theorem}
+Let $y H$ denote the [[Definition:Left Coset|left coset]] of $H$ by $y$.
+Then:
+:$x \in y H \iff x^{-1} y \in H$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \in
+      | r = y H
+      | c = 
+}}
+{{eqn | lll=\iff
+      | ll= \exists h \in H:
+      | l = x
+      | r = y h
+      | c = {{Defof|Left Coset}}
+}}
+{{eqn | lll=\iff
+      | ll= \exists h \in H:
+      | l = x^{-1}
+      | r = h^{-1} y^{-1}
+      | c = [[Inverse of Group Product]]
+}}
+{{eqn | lll=\iff
+      | ll= \exists h \in H:
+      | l = x^{-1} y
+      | r = h^{-1}
+      | c = Product with $y$ on the right
+}}
+{{eqn | lll=\iff
+      | l = x^{-1} y
+      | o = \in
+      | r = H
+      | c = $H$ is a [[Definition:Subgroup|subgroup]]
+}}
+{{end-eqn}}
+{{Qed}}
+</onlyinclude>
+\end{proof}<|endoftext|>
+\section{Element in Right Coset iff Product with Inverse in Subgroup}
+Tags: Cosets
+
+\begin{theorem}
+Let $H \circ y$ denote the [[Definition:Right Coset|right coset]] of $H$ by $y$.
+Then:
+:$x \in H y \iff x y^{-1} \in H$
+\end{theorem}
+
+\begin{proof}
+Let $\struct {G, *}$ be the [[Definition:Opposite Group|opposite group]] of $G$.
+Then:
+:$x \in H y \iff x \in y * H$
+:$x y^{-1} \in H \iff y^{-1} * x \in H$ 
+Since $H$ is closed under inverses:
+:$x y^{-1} \in H \iff x^{-1} * y \in H$ 
+By [[Element in Left Coset iff Product with Inverse in Subgroup]]:
+:$x \in y * H \iff x^{-1} * y \in H$
+Hence the result.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Inverse of Inverse of Subset of Group}
+Tags: Group Theory
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
+Let $X \subseteq G$.
+Then:
+: $\paren {X^{-1} }^{-1} = X$.
+where $X^{-1}$ denotes the [[Definition:Inverse of Subset of Group|inverse of $X$]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \paren {X^{-1} }^{-1}
+      | r = \set {x^{-1}: x \in X^{-1} }
+      | c = {{Defof|Inverse of Subset of Group}}
+}}
+{{eqn | r = \set {\paren {x^{-1} }^{-1}: x \in X}
+      | c = {{Defof|Inverse of Subset of Group}}
+}}
+{{eqn | r = \set {x: x \in X}
+      | c = [[Inverse of Group Inverse]]
+}}
+{{eqn | r = X
+      | c = [[Axiom:Axiom of Extension|Axiom of Extension]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Group Theory]]
+tknfd1imwrrao9s0kojknorjgzf0gmx
+\end{proof}<|endoftext|>
+\section{Exclusive Or is Commutative}
+Tags: Exclusive Or, Exclusive Or is Commutative
+
+\begin{theorem}
+: $p \oplus q \dashv \vdash q \oplus p$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|p \oplus q \vdash q \oplus p}}
+{{Premise|1|p \oplus q}}
+{{SequentIntro|2|1|\left({p \lor q} \right) \land \neg \left({p \land q}\right)|1|Definition of [[Non-Equivalence|Exclusive Or]]}}
+{{Commutation|3|1|\left({q \lor p} \right) \land \neg \left({p \land q}\right)|2|Disjunction}}
+{{Commutation|4|1|\left({q \lor p} \right) \land \neg \left({q \land p}\right)|3|Conjunction}}
+{{SequentIntro|5|1|q \oplus p|4|Definition of [[Non-Equivalence|Exclusive Or]]}}
+{{EndTableau}}
+{{BeginTableau|q \oplus p \vdash p \oplus q}}
+{{Premise|1|q \oplus p}}
+{{SequentIntro|2|1|\left({q \lor p} \right) \land \neg \left({q \land p}\right)|1|Definition of [[Non-Equivalence|Exclusive Or]]}}
+{{Commutation|3|1|\left({q \lor p} \right) \land \neg \left({p \land q}\right)|2|Conjunction}}
+{{Commutation|4|1|\left({p \lor q} \right) \land \neg \left({p \land q}\right)|3|Disjunction}}
+{{SequentIntro|5|1|p \oplus q|4|Definition of [[Non-Equivalence|Exclusive Or]]}}
+{{EndTableau}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]].
+As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
+$\begin{array}{|ccc||ccc|} \hline
+p & \oplus & q & q & \oplus & p \\
+\hline
+F & F & F & F & F & F \\
+F & T & T & T & T & F \\
+T & T & F & F & T & T \\
+T & F & T & T & F & T \\
+\hline
+\end{array}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Exclusive Or is Associative}
+Tags: Exclusive Or, Truth Table Proofs
+
+\begin{theorem}
+[[Definition:Exclusive Or|Exclusive or]] is [[Definition:Associative|associative]]:
+: $p \oplus \left({q \oplus r}\right) \dashv \vdash \left({p \oplus q}\right) \oplus r$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]].
+As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
+$\begin{array}{|ccccc||ccccc|} \hline
+p & \oplus & (q & \oplus & r) & (p & \oplus & q) & \oplus & r \\
+\hline
+F & F & F & F & F & F & F & F & F & F \\
+F & T & F & T & T & F & F & F & T & T \\
+F & T & T & T & F & F & T & T & T & F \\
+F & F & T & F & T & F & T & T & F & T \\
+T & T & F & F & F & T & T & F & T & F \\
+T & F & F & T & T & T & T & F & F & T \\
+T & F & T & T & F & T & F & T & F & F \\
+T & T & T & F & T & T & F & T & T & T \\
+\hline
+\end{array}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Exclusive Or with Itself}
+Tags: Exclusive Or, Truth Table Proofs
+
+\begin{theorem}
+[[Definition:Exclusive Or|Exclusive or]] destroys copies of itself:
+: $p \oplus p \dashv \vdash \bot$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]].
+As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] match for each [[Definition:Boolean Interpretation|boolean interpretations]].
+$\begin{array}{|ccc|} \hline
+p & \oplus & p \\
+\hline
+F & F & F \\
+T & F & T \\
+\hline
+\end{array}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Proof 1}
+Tags: Non-Equivalence as Disjunction of Conjunctions
+
+\begin{theorem}
+: $\neg \left ({p \iff q}\right) \dashv \vdash \left({\neg p \land q}\right) \lor \left({p \land \neg q}\right)$
+\end{theorem}
+
+\begin{proof}
+=== [[Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Forward Implication/Proof|Forward Implication: Proof]] ===
+{{:Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Forward Implication/Proof}}
+=== [[Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Reverse Implication/Proof|Reverse Implication: Proof]] ===
+{{:Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Reverse Implication/Proof}}
+\end{proof}<|endoftext|>
+\section{Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Proof 2}
+Tags: Non-Equivalence as Disjunction of Conjunctions, Truth Table Proofs
+
+\begin{theorem}
+: $\neg \left ({p \iff q}\right) \dashv \vdash \left({\neg p \land q}\right) \lor \left({p \land \neg q}\right)$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]].
+As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
+$\begin{array}{|cccc||ccccccccc|} \hline
+\neg & (p & \iff & q) & (\neg & p & \land & q) & \lor & (p & \land & \neg & q) \\
+\hline
+F & F & T & F & T & F & F & F & F & F & F & T & F \\
+T & F & F & T & T & F & T & T & T & F & F & F & T  \\
+T & T & F & F & F & T & F & F & T & T & T & T & F  \\
+F & T & T & T & F & T & F & T & F & T & F & F & T  \\
+\hline
+\end{array}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Non-Equivalence as Disjunction of Conjunctions/Formulation 1}
+Tags: Non-Equivalence as Disjunction of Conjunctions
+
+\begin{theorem}
+: $\neg \left ({p \iff q}\right) \dashv \vdash \left({\neg p \land q}\right) \lor \left({p \land \neg q}\right)$
+\end{theorem}<|endoftext|>
+\section{Exclusive Or is Negation of Biconditional}
+Tags: Exclusive Or, Biconditional
+
+\begin{theorem}
+[[Definition:Exclusive Or|Exclusive or]] is [[Definition:Logically Equivalent|equivalent]] to the [[Definition:Negation|negation]] of the [[Definition:Biconditional|biconditional]]:
+:$p \oplus q \dashv \vdash \neg \paren {p \iff q}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = p \oplus q
+      | o = \dashv \vdash
+      | r = \paren {p \lor q} \land \neg \paren {p \land q}
+      | c = {{Defof|Exclusive Or}}
+}}
+{{eqn | o = \dashv \vdash
+      | r = \neg \paren {p \iff q}
+      | c = [[Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Exclusive Or as Disjunction of Conjunctions}
+Tags: Exclusive Or, Disjunction, Conjunction, Exclusive Or as Disjunction of Conjunctions
+
+\begin{theorem}
+:$p \oplus q \dashv \vdash \paren {\neg p \land q} \lor \paren {p \land \neg q}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = p \oplus q
+      | o = \dashv \vdash
+      | r = \neg \left ({p \iff q}\right)
+      | c = [[Exclusive Or is Negation of Biconditional]]
+}}
+{{eqn | o = \dashv \vdash
+      | r = \left({\neg p \land q}\right) \lor \left({p \land \neg q}\right)
+      | c = [[Non-Equivalence as Disjunction of Conjunctions/Formulation 1|Non-Equivalence as Disjunction of Conjunctions]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]].
+As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
+$\begin{array}{|ccc||ccccccccc|} \hline
+p & \oplus & q & (\neg & p & \land & q) & \lor & (p & \land & \neg & q) \\
+\hline
+F & F & F & T & F & F & F & F & F & F & T & F \\
+F & T & T & T & F & T & T & T & F & F & F & T  \\
+T & T & F & F & T & F & F & T & T & T & T & F  \\
+T & F & T & F & T & F & T & F & T & F & F & T  \\
+\hline
+\end{array}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Non-Equivalence as Disjunction of Negated Implications/Proof 1}
+Tags: Non-Equivalence as Disjunction of Negated Implications
+
+\begin{theorem}
+: $\neg \left ({p \iff q}\right) \dashv \vdash \neg \left({p \implies q}\right) \lor \neg \left({q \implies p}\right)$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\neg \left ({p \iff q}\right) \vdash \neg \left({p \implies q}\right) \lor \neg \left({q \implies p}\right)}}
+{{Premise|1|\neg \left ({p \iff q}\right)}}
+{{SequentIntro|2|1|\neg \left({\left ({p \implies q}\right) \land \left ({q \implies p}\right)}\right)|1
+  |[[Rule of Material Equivalence]]}}
+{{DeMorgan|3|1|\neg \left({p \implies q}\right) \lor \neg \left({q \implies p}\right)|2|Disjunction of Negations}}
+{{EndTableau}}
+{{qed|lemma}}
+{{BeginTableau|\neg \left({p \implies q}\right) \lor \neg \left({q \implies p}\right) \vdash \neg \left ({p \iff q}\right)}}
+{{Premise|1|\neg \left({p \implies q}\right) \lor \neg \left({q \implies p}\right)}}
+{{DeMorgan|2|1|\neg \left({\left ({p \implies q}\right) \land \left ({q \implies p}\right)}\right)|1|Disjunction of Negations}}
+{{SequentIntro|3|1|\neg \left ({p \iff q}\right)|2|[[Rule of Material Equivalence]]}}
+{{EndTableau}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Non-Equivalence as Disjunction of Negated Implications/Proof 2}
+Tags: Non-Equivalence as Disjunction of Negated Implications, Truth Table Proofs
+
+\begin{theorem}
+: $\neg \left ({p \iff q}\right) \dashv \vdash \neg \left({p \implies q}\right) \lor \neg \left({q \implies p}\right)$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]].
+As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
+$\begin{array}{|cccc||ccccccccc|} \hline
+\neg & (p & \iff & q) & \neg & (p & \implies & q) & \lor & \neg & (q & \implies & p) \\
+\hline
+F & F & T & F & F & F & T & F & F & F & F & T & F \\
+T & F & F & T & F & F & T & T & T & T & T & F & F  \\
+T & T & F & F & T & T & F & F & T & F & F & T & T  \\
+F & T & T & T & F & T & T & T & F & F & T & T & T  \\
+\hline
+\end{array}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Non-Equivalence as Disjunction of Negated Implications}
+Tags: Biconditional, Disjunction, Negation, Implication
+
+\begin{theorem}
+: $\neg \left ({p \iff q}\right) \dashv \vdash \neg \left({p \implies q}\right) \lor \neg \left({q \implies p}\right)$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\neg \left ({p \iff q}\right) \vdash \neg \left({p \implies q}\right) \lor \neg \left({q \implies p}\right)}}
+{{Premise|1|\neg \left ({p \iff q}\right)}}
+{{SequentIntro|2|1|\neg \left({\left ({p \implies q}\right) \land \left ({q \implies p}\right)}\right)|1
+  |[[Rule of Material Equivalence]]}}
+{{DeMorgan|3|1|\neg \left({p \implies q}\right) \lor \neg \left({q \implies p}\right)|2|Disjunction of Negations}}
+{{EndTableau}}
+{{qed|lemma}}
+{{BeginTableau|\neg \left({p \implies q}\right) \lor \neg \left({q \implies p}\right) \vdash \neg \left ({p \iff q}\right)}}
+{{Premise|1|\neg \left({p \implies q}\right) \lor \neg \left({q \implies p}\right)}}
+{{DeMorgan|2|1|\neg \left({\left ({p \implies q}\right) \land \left ({q \implies p}\right)}\right)|1|Disjunction of Negations}}
+{{SequentIntro|3|1|\neg \left ({p \iff q}\right)|2|[[Rule of Material Equivalence]]}}
+{{EndTableau}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]].
+As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
+$\begin{array}{|cccc||ccccccccc|} \hline
+\neg & (p & \iff & q) & \neg & (p & \implies & q) & \lor & \neg & (q & \implies & p) \\
+\hline
+F & F & T & F & F & F & T & F & F & F & F & T & F \\
+T & F & F & T & F & F & T & T & T & T & T & F & F  \\
+T & T & F & F & T & T & F & F & T & F & F & T & T  \\
+F & T & T & T & F & T & T & T & F & F & T & T & T  \\
+\hline
+\end{array}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Conjunction of Disjunction with Negation is Conjunction with Negation}
+Tags: Conjunction, Disjunction, Negation
+
+\begin{theorem}
+:$\left({p \lor q}\right) \land \neg q \dashv \vdash p \land \neg q$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|p \land \neg q \vdash \left({p \lor q}\right) \land \neg q}}
+{{Premise|1|p \land \neg q}}
+{{Simplification|2|1|p|1|1}}
+{{Addition|3|1|p \lor q|2|1}}
+{{Simplification|4|1|\neg q|1|2}}
+{{Conjunction|5|1|\left({p \lor q}\right) \land \neg q|3|4}}
+{{EndTableau}}
+{{qed}}
+... and its converse:
+{{BeginTableau|\left({p \lor q}\right) \land \neg q \vdash p \land \neg q}}
+{{Premise|1|\left({p \lor q}\right) \land \neg q}}
+{{SequentIntro|2|1|\left({p \land \neg q}\right) \lor \left({q \land \neg q}\right)|1
+  |[[Conjunction Distributes over Disjunction]]
+}}
+{{Assumption|3|q \land \neg q}}
+{{NonContradiction|4|3|3|3}}
+{{Contradiction|5||\neg \left ({q \land \neg q}\right)|3|4}}
+{{SequentIntro|6|1|p \land \neg q|2, 5|[[Disjunctive Syllogism]]}}
+{{EndTableau}}
+{{qed}}
+[[Category:Conjunction]]
+[[Category:Disjunction]]
+[[Category:Negation]]
+tfhj416ip3giibivxe26q54ehzsfkq3
+\end{proof}<|endoftext|>
+\section{Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction/Proof 1}
+Tags: Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction
+
+\begin{theorem}
+: $\neg \left ({p \iff q}\right) \dashv \vdash \left({p \lor q} \right) \land \neg \left({p \land q}\right)$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\neg \left ({p \iff q}\right) \vdash \left({p \lor q} \right) \land \neg \left({p \land q}\right)}}
+{{Premise|1|\neg \left ({p \iff q}\right)}}
+{{SequentIntro|2|1|\left({\neg p \land q}\right) \lor \left({p \land \neg q}\right)|1
+  |[[Non-Equivalence as Disjunction of Conjunctions/Formulation 1|Non-Equivalence as Disjunction of Conjunctions: Formulation 1]]
+}}
+{{Commutation|3|1|\left({p \land \neg q}\right) \lor \left({\neg p \land q}\right)|2|Disjunction}}
+{{Commutation|4|1|\left({p \land \neg q}\right) \lor \left({q \land \neg p}\right)|3|Conjunction}}
+{{SequentIntro|5|1
+  |\left ({\left({p \lor q}\right) \land \neg q}\right) \lor \left({\left({q \lor p}\right) \land \neg p}\right)
+  |4
+  |[[Conjunction of Disjunction with Negation is Conjunction with Negation]]
+}}
+{{Commutation|6|1
+  |\left ({\left({p \lor q}\right) \land \neg q}\right) \lor \left({\left({p \lor q}\right) \land \neg p}\right)
+  |5|Disjunction}}
+{{SequentIntro|7|1|\left({p \lor q}\right) \land \left({\neg q \lor \neg p}\right)|6
+  |[[Conjunction Distributes over Disjunction]]
+}}
+{{Commutation|8|1|\left({p \lor q}\right) \land \left({\neg p \lor \neg q}\right)|7|Disjunction}}
+{{DeMorgan|9|1|\left({p \lor q}\right) \land \neg \left({\neg \neg p \land \neg \neg q}\right)|8|Disjunction}}
+{{DoubleNegElimination|10|1|\left({p \lor q}\right) \land \neg \left({p \land q}\right)|9}}
+{{EndTableau}}
+{{BeginTableau|\left({p \lor q} \right) \land \neg \left({p \land q}\right) \vdash \neg \left ({p \iff q}\right)}}
+{{Premise|1|\left({p \lor q}\right) \land \neg \left({p \land q}\right)}}
+{{DeMorgan|2|1|\left({p \lor q}\right) \land \left({\neg p \lor \neg q}\right)|1|Disjunction of Negations}}
+{{Commutation|3|1|\left({p \lor q}\right) \land \left({\neg p \lor \neg q}\right)|2|Disjunction}}
+{{SequentIntro|4|1
+  |\left ({\left({p \lor q}\right) \land \neg q}\right) \lor \left({\left({p \lor q}\right) \land \neg p}\right)
+  |3
+  |[[Conjunction Distributes over Disjunction]]
+}}
+{{SequentIntro|5|1|\left({p \land \neg q}\right) \lor \left({q \land \neg p}\right)|4|[[Conjunction of Disjunction with Negation is Conjunction with Negation]]}}
+{{Commutation|6|1|\left({q \land \neg p}\right) \lor \left({p \land \neg q}\right)|5|Disjunction}}
+{{Commutation|7|1|\left({\neg p \land q}\right) \lor \left({p \land \neg q}\right)|6|Conjunction}}
+{{SequentIntro|8|1|\neg \left ({p \iff q}\right)|6|[[Non-Equivalence as Disjunction of Conjunctions/Formulation 1|Non-Equivalence as Disjunction of Conjunctions: Formulation 1]]}}
+{{EndTableau}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction/Proof 2}
+Tags: Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction, Truth Table Proofs
+
+\begin{theorem}
+: $\neg \left ({p \iff q}\right) \dashv \vdash \left({p \lor q} \right) \land \neg \left({p \land q}\right)$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]].
+As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
+$\begin{array}{|cccc||cccccccc|} \hline
+\neg & (p & \iff & q) & (p & \lor & q) & \land & \neg & (p & \land & q) \\
+\hline
+F & F & T & F & F & F & F & F & T & F & F & F \\
+T & F & F & T & F & T & T & T & T & F & F & T  \\
+T & T & F & F & T & T & F & T & T & T & F & F  \\
+F & T & T & T & T & T & T & F & F & T & T & T  \\
+\hline
+\end{array}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction}
+Tags: Conjunction, Disjunction, Negation, Biconditional
+
+\begin{theorem}
+: $\neg \left ({p \iff q}\right) \dashv \vdash \left({p \lor q} \right) \land \neg \left({p \land q}\right)$
+That is, [[Definition:Logical Not|negation]] of the [[Definition:Biconditional|biconditional]] means the same thing as '''either-or but not both''', that is, [[Definition:Exclusive Or|exclusive or]].
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\neg \left ({p \iff q}\right) \vdash \left({p \lor q} \right) \land \neg \left({p \land q}\right)}}
+{{Premise|1|\neg \left ({p \iff q}\right)}}
+{{SequentIntro|2|1|\left({\neg p \land q}\right) \lor \left({p \land \neg q}\right)|1
+  |[[Non-Equivalence as Disjunction of Conjunctions/Formulation 1|Non-Equivalence as Disjunction of Conjunctions: Formulation 1]]
+}}
+{{Commutation|3|1|\left({p \land \neg q}\right) \lor \left({\neg p \land q}\right)|2|Disjunction}}
+{{Commutation|4|1|\left({p \land \neg q}\right) \lor \left({q \land \neg p}\right)|3|Conjunction}}
+{{SequentIntro|5|1
+  |\left ({\left({p \lor q}\right) \land \neg q}\right) \lor \left({\left({q \lor p}\right) \land \neg p}\right)
+  |4
+  |[[Conjunction of Disjunction with Negation is Conjunction with Negation]]
+}}
+{{Commutation|6|1
+  |\left ({\left({p \lor q}\right) \land \neg q}\right) \lor \left({\left({p \lor q}\right) \land \neg p}\right)
+  |5|Disjunction}}
+{{SequentIntro|7|1|\left({p \lor q}\right) \land \left({\neg q \lor \neg p}\right)|6
+  |[[Conjunction Distributes over Disjunction]]
+}}
+{{Commutation|8|1|\left({p \lor q}\right) \land \left({\neg p \lor \neg q}\right)|7|Disjunction}}
+{{DeMorgan|9|1|\left({p \lor q}\right) \land \neg \left({\neg \neg p \land \neg \neg q}\right)|8|Disjunction}}
+{{DoubleNegElimination|10|1|\left({p \lor q}\right) \land \neg \left({p \land q}\right)|9}}
+{{EndTableau}}
+{{BeginTableau|\left({p \lor q} \right) \land \neg \left({p \land q}\right) \vdash \neg \left ({p \iff q}\right)}}
+{{Premise|1|\left({p \lor q}\right) \land \neg \left({p \land q}\right)}}
+{{DeMorgan|2|1|\left({p \lor q}\right) \land \left({\neg p \lor \neg q}\right)|1|Disjunction of Negations}}
+{{Commutation|3|1|\left({p \lor q}\right) \land \left({\neg p \lor \neg q}\right)|2|Disjunction}}
+{{SequentIntro|4|1
+  |\left ({\left({p \lor q}\right) \land \neg q}\right) \lor \left({\left({p \lor q}\right) \land \neg p}\right)
+  |3
+  |[[Conjunction Distributes over Disjunction]]
+}}
+{{SequentIntro|5|1|\left({p \land \neg q}\right) \lor \left({q \land \neg p}\right)|4|[[Conjunction of Disjunction with Negation is Conjunction with Negation]]}}
+{{Commutation|6|1|\left({q \land \neg p}\right) \lor \left({p \land \neg q}\right)|5|Disjunction}}
+{{Commutation|7|1|\left({\neg p \land q}\right) \lor \left({p \land \neg q}\right)|6|Conjunction}}
+{{SequentIntro|8|1|\neg \left ({p \iff q}\right)|6|[[Non-Equivalence as Disjunction of Conjunctions/Formulation 1|Non-Equivalence as Disjunction of Conjunctions: Formulation 1]]}}
+{{EndTableau}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]].
+As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
+$\begin{array}{|cccc||cccccccc|} \hline
+\neg & (p & \iff & q) & (p & \lor & q) & \land & \neg & (p & \land & q) \\
+\hline
+F & F & T & F & F & F & F & F & T & F & F & F \\
+T & F & F & T & F & T & T & T & T & F & F & T  \\
+T & T & F & F & T & T & F & T & T & T & F & F  \\
+F & T & T & T & T & T & T & F & F & T & T & T  \\
+\hline
+\end{array}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Non-Equivalence as Conjunction of Disjunction with Disjunction of Negations}
+Tags: Conjunction, Disjunction, Negation, Biconditional
+
+\begin{theorem}
+:$\neg \paren {p \iff q} \dashv \vdash \paren {p \lor q} \land \paren {\neg p \lor \neg q}$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\neg \paren {p \iff q} \vdash \paren {p \lor q} \land \paren {\neg p \lor \neg q} }}
+{{Premise     |1|\neg \paren {p \iff q} }}
+{{SequentIntro|2|1|\paren {p \lor q} \land \neg \paren {p \land q}|1|[[Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction]]}}
+{{DeMorgan    |3|1|\paren {p \lor q} \land \paren {\neg p \lor \neg q}|2|Disjunction of Negations}}
+{{EndTableau}}
+{{BeginTableau|\paren {p \lor q} \land \paren {\neg p \lor \neg q} \vdash \neg \paren {p \iff q} }}
+{{Premise     |1|\paren {p \lor q} \land \paren {\neg p \lor \neg q} }}
+{{DeMorgan    |2|1|\paren {p \lor q} \land \neg \paren {p \land q}|1|Disjunction of Negations}}
+{{SequentIntro|3|1|\neg \paren {p \iff q}|2|[[Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction]]}}
+{{EndTableau|qed}}
+\end{proof}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]].
+As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
+$\begin{array}{|cccc||ccccccccc|} \hline
+\neg & (p & \iff & q) & (p & \lor & q) & \land & (\neg & p & \lor & \neg & q) \\
+\hline
+F & F & T & F & F & F & F & F & T & F & T & T & F \\
+T & F & F & T & F & T & T & T & T & F & T & F & T \\
+T & T & F & F & T & T & F & T & F & T & T & T & F \\
+F & T & T & T & T & T & T & F & F & T & F & F & T \\
+\hline
+\end{array}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inversion Mapping is Involution}
+Tags: Inversion Mappings
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]], and let $\iota: G \to G$ be the [[Definition:Inversion Mapping|inversion mapping]].
+Then $\iota$ is an [[Definition:Involution (Mapping)|involution]].
+That is:
+:$\forall g \in G: \map \iota {\map \iota g} = g$
+\end{theorem}
+
+\begin{proof}
+Let $g \in G$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \iota {\map \iota g}
+      | r = \paren {g^{-1} }^{-1}
+      | c = {{Defof|Inversion Mapping}}
+}}
+{{eqn | r = g
+      | c = [[Inverse of Group Inverse]]
+}}
+{{end-eqn}}
+which establishes the result.
+{{qed}}
+[[Category:Inversion Mappings]]
+scu8viz33uzskvkzuh04emh1ycpbec3
+\end{proof}<|endoftext|>
+\section{Order Isomorphism between Linearly Ordered Spaces is Homeomorphism}
+Tags: Linearly Ordered Spaces
+
+\begin{theorem}
+Let $\struct {S_1, \le_1, \tau_1}$ and $\struct {S_2, \le_2, \tau_2}$ be [[Definition:Linearly Ordered Space|linearly ordered spaces]].
+Let $\phi: S_1 \to S_2$ be an [[Definition:Order Isomorphism|order isomorphism]] from $\struct {S_1, \le_1}$ to $\struct {S_2, \le_2}$.
+Then $\phi$ is a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]] from $\struct {S_1, \tau_1}$ to $\struct {S_2, \tau_2}$.
+\end{theorem}
+
+\begin{proof}
+By the definition of [[Definition:Order Isomorphism|order isomorphism]], $\phi$ is a [[Definition:Bijection|bijection]].
+Thus to show $\phi$ is a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]] it remains to be shown that:
+:$\phi$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]]
+and:
+:$\phi^{-1}$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+First it is shown that $\phi^{-1}$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+By [[Order Isomorphism Preserves Initial Segments]] and its dual, $\phi$ maps [[Definition:Open Ray|open rays]] in $\struct {S_1, \le_1}$ to [[Definition:Open Ray|open rays]] in $\struct {S_2, \le_2}$.
+By [[Continuity Test using Sub-Basis]], $\phi^{-1}$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+Next it is shown that $\phi^{-1}$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+By [[Inverse of Order Isomorphism is Order Isomorphism]], $\phi^{-1}$ is an [[Definition:Order Isomorphism|order isomorphism]].
+Applying the above shows that $\phi = \paren {\phi^{-1} }^{-1}$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+Both $\phi$ and $\phi^{-1}$ have been shown to be [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+Thus, by definition, $\phi$ is a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]].
+{{qed}}
+[[Category:Linearly Ordered Spaces]]
+dnyofjqvk2qgg6aiwah84iv1pyw9vl2
+\end{proof}<|endoftext|>
+\section{Rule of Association/Conjunction}
+Tags: Rule of Association, Conjunction
+
+\begin{theorem}
+[[Definition:Conjunction|Conjunction]] is [[Definition:Associative Operation|associative]]:
+=== [[Rule of Association/Conjunction/Formulation 1|Formulation 1]] ===
+{{:Rule of Association/Conjunction/Formulation 1}}
+=== [[Rule of Association/Conjunction/Formulation 2|Formulation 2]] ===
+{{:Rule of Association/Conjunction/Formulation 2}}
+\end{theorem}<|endoftext|>
+\section{Rule of Association/Disjunction}
+Tags: Rule of Association, Disjunction
+
+\begin{theorem}
+[[Definition:Disjunction|Disjunction]] is [[Definition:Associative Operation|associative]]:
+=== [[Rule of Association/Disjunction/Formulation 1|Formulation 1]] ===
+{{:Rule of Association/Disjunction/Formulation 1}}
+=== [[Rule of Association/Disjunction/Formulation 2|Formulation 2]] ===
+{{:Rule of Association/Disjunction/Formulation 2}}
+\end{theorem}<|endoftext|>
+\section{Rule of Association/Conjunction/Formulation 1}
+Tags: Rule of Association
+
+\begin{theorem}
+:$p \land \left({q \land r}\right) \dashv \vdash \left({p \land q}\right) \land r$
+\end{theorem}<|endoftext|>
+\section{Rule of Association/Disjunction/Formulation 1}
+Tags: Rule of Association
+
+\begin{theorem}
+:$p \lor \paren {q \lor r} \dashv \vdash \paren {p \lor q} \lor r$
+\end{theorem}<|endoftext|>
+\section{Rule of Association/Conjunction/Formulation 1/Proof 1}
+Tags: Rule of Association
+
+\begin{theorem}
+:$p \land \left({q \land r}\right) \dashv \vdash \left({p \land q}\right) \land r$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|p \land \left({q \land r}\right) \vdash \left({p \land q}\right) \land r}}
+{{Premise|1|p \land \left({q \land r}\right)}}
+{{Simplification|2|1|p|1|1}}
+{{Simplification|3|1|q \land r|1|2}}
+{{Simplification|4|1|q|3|1}}
+{{Simplification|5|1|r|3|2}}
+{{Conjunction|6|1|p \land q|2|4}}
+{{Conjunction|7|1|\left({p \land q}\right) \land r|6|5}}
+{{EndTableau}}
+{{BeginTableau|\left({p \land q}\right) \land r \vdash p \land \left({q \land r}\right)}}
+{{Premise|1|\left({p \land q}\right) \land r}}
+{{Simplification|2|1|p \land q|1|1}}
+{{Simplification|3|1|r|1|2}}
+{{Simplification|4|1|p|2|1}}
+{{Simplification|5|1|q|2|2}}
+{{Conjunction|6|1|q \land r|5|3}}
+{{Conjunction|7|1|p \land \left({q \land r}\right)|4|6}}
+{{EndTableau}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rule of Association/Disjunction/Formulation 1/Proof 2}
+Tags: Rule of Association, Truth Table Proofs
+
+\begin{theorem}
+:$p \lor \left({q \lor r}\right) \dashv \vdash \left({p \lor q}\right) \lor r$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]].
+As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
+$\begin{array}{|ccccc||ccccc|} \hline
+p & \lor & (q & \lor & r) & (p & \lor & q) & \lor & r \\
+\hline
+F & F & F & F & F & F & F & F & F & F \\
+F & T & F & T & T & F & F & F & T & T \\
+F & T & T & T & F & F & T & T & T & F \\
+F & T & T & T & T & F & T & T & T & T \\
+T & T & F & F & F & T & T & F & T & F \\
+T & T & F & T & T & T & T & F & T & T \\
+T & T & T & T & F & T & T & T & T & F \\
+T & T & T & T & T & T & T & T & T & T \\
+\hline
+\end{array}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rule of Association/Conjunction/Formulation 2}
+Tags: Rule of Association
+
+\begin{theorem}
+:$\vdash \paren {p \land \paren {q \land r} } \iff \paren {\paren {p \land q} \land r}$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\vdash \paren {p \land \paren {q \land r} } \iff \paren {\paren {p \land q} \land r} }}
+{{Assumption  |1|p \land \paren {q \land r} }}
+{{SequentIntro|2|1|\paren {p \land q} \land r|1|[[Rule of Association/Conjunction/Formulation 1|Rule of Association: Formulation 1]]}}
+{{Implication |3||\paren {p \land \paren {q \land r} } \implies \paren {\paren {p \land q} \land r}|1|2}}
+{{Assumption  |4|\paren {p \land q} \land r}}
+{{SequentIntro|5|4|p \land \paren {q \land r}|4|[[Rule of Association/Conjunction/Formulation 1|Rule of Association: Formulation 1]]}}
+{{Implication |6||\paren {\paren {p \land q} \land r} \implies \paren {p \land \paren {q \land r} }|4|5}}
+{{BiconditionalIntro|7||\paren {p \land \paren {q \land r} } \iff \paren {\paren {p \land q} \land r}|3|6}}
+{{EndTableau}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rule of Association/Disjunction/Formulation 2}
+Tags: Rule of Association
+
+\begin{theorem}
+:$\vdash \paren {p \lor \paren {q \lor r} } \iff \paren {\paren {p \lor q} \lor r}$
+\end{theorem}<|endoftext|>
+\section{Principle of Commutation/Forward Implication/Formulation 1}
+Tags: Principle of Commutation
+
+\begin{theorem}
+: $p \implies \left({q \implies r}\right) \vdash q \implies \left({p \implies r}\right)$
+\end{theorem}<|endoftext|>
+\section{Principle of Commutation/Formulation 1}
+Tags: Principle of Commutation
+
+\begin{theorem}
+:$p \implies \paren {q \implies r} \dashv \vdash q \implies \paren {p \implies r}$
+\end{theorem}<|endoftext|>
+\section{Principle of Commutation/Reverse Implication/Formulation 1}
+Tags: Principle of Commutation
+
+\begin{theorem}
+:$q \implies \paren {p \implies r} \vdash p \implies \paren {q \implies r}$
+\end{theorem}<|endoftext|>
+\section{Principle of Commutation/Reverse Implication/Formulation 1/Proof}
+Tags: Principle of Commutation
+
+\begin{theorem}
+: $q \implies \left({p \implies r}\right) \vdash p \implies \left({q \implies r}\right)$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|q \implies \left({p \implies r}\right) \vdash p \implies \left({q \implies r}\right)}}
+{{Premise|1|q \implies \left({p \implies r}\right)}}
+{{Assumption|2|p}}
+{{Assumption|3|q}}
+{{ModusPonens|4|1, 3|p \implies r|1|3}}
+{{ModusPonens|5|1, 2, 3|r|2|4}}
+{{Implication|6|1, 2|q \implies r|3|5}}
+{{Implication|7|1|p \implies \left({q \implies r}\right)|2|6}}
+{{EndTableau}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Principle of Commutation/Formulation 1/Proof 1}
+Tags: Principle of Commutation
+
+\begin{theorem}
+: $p \implies \left({q \implies r}\right) \dashv \vdash q \implies \left({p \implies r}\right)$
+\end{theorem}
+
+\begin{proof}
+=== [[Principle of Commutation/Forward Implication/Formulation 1/Proof|Proof of Forward Implication]] ===
+{{:Principle of Commutation/Forward Implication/Formulation 1/Proof}}
+=== [[Principle of Commutation/Reverse Implication/Formulation 1/Proof|Proof of Reverse Implication]] ===
+{{:Principle of Commutation/Reverse Implication/Formulation 1/Proof}}
+\end{proof}<|endoftext|>
+\section{Principle of Commutation/Formulation 1/Proof 2}
+Tags: Principle of Commutation, Truth Table Proofs
+
+\begin{theorem}
+: $p \implies \left({q \implies r}\right) \dashv \vdash q \implies \left({p \implies r}\right)$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]] to the proposition.
+As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
+$\begin{array}{|ccccc||ccccc|} \hline
+p & \implies & (q & \implies & r) & q & \implies & (p & \implies & r) \\
+\hline
+F & T & F & T & F & F & T & F & T & F \\
+F & T & F & T & T & F & T & F & T & T \\
+F & T & T & F & F & T & T & F & T & F \\
+F & T & T & T & T & T & T & F & T & T \\
+T & T & F & T & F & F & T & T & F & F \\
+T & T & F & T & T & F & T & T & T & T \\
+T & F & T & F & F & T & F & T & F & F \\
+T & T & T & T & T & T & T & T & T & T \\
+\hline
+\end{array}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Principle of Commutation/Formulation 2}
+Tags: Principle of Commutation
+
+\begin{theorem}
+:$\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {q \implies \paren {p \implies r} }$
+\end{theorem}
+
+\begin{proof}
+=== [[Principle of Commutation/Forward Implication/Formulation 2/Proof 1|Proof of Forward Implication]] ===
+{{:Principle of Commutation/Forward Implication/Formulation 2/Proof 1}}
+=== [[Principle of Commutation/Reverse Implication/Formulation 2/Proof|Proof of Reverse Implication]] ===
+{{:Principle of Commutation/Reverse Implication/Formulation 2/Proof}}
+{{BeginTableau|\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {q \implies \paren {p \implies r} } }}
+{{TheoremIntro|1|\paren {p \implies \paren {q \implies r} } \implies \paren {q \implies \paren {p \implies r} }|[[Principle of Commutation/Forward Implication/Formulation 2/Proof 1|Principle of Commutation: Forward Implication: Formulation 2]]}}
+{{TheoremIntro|2|\paren {q \implies \paren {p \implies r} } \implies \paren {p \implies \paren {q \implies r} }|[[Principle of Commutation/Reverse Implication/Formulation 2/Proof|Principle of Commutation: Reverse Implication: Formulation 2]]}}
+{{BiconditionalIntro|3||\paren {p \implies \paren {q \implies r} } \iff \paren {q \implies \paren {p \implies r} }|1|2}}
+{{EndTableau}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Principle of Commutation/Forward Implication/Formulation 2}
+Tags: Principle of Commutation
+
+\begin{theorem}
+: $\vdash \left({p \implies \left({q \implies r}\right)}\right) \implies \left({q \implies \left({p \implies r}\right)}\right)$
+\end{theorem}<|endoftext|>
+\section{Principle of Commutation/Reverse Implication/Formulation 2}
+Tags: Principle of Commutation
+
+\begin{theorem}
+: $\vdash \left({q \implies \left({p \implies r}\right)}\right) \implies \left({p \implies \left({q \implies r}\right)}\right)$
+\end{theorem}<|endoftext|>
+\section{Linearly Ordered Space is Connected iff Linear Continuum}
+Tags: Linearly Ordered Spaces, Connected Spaces, Continua
+
+\begin{theorem}
+Let $T = \struct {S, \preceq, \tau}$ be a [[Definition:Linearly Ordered Space|linearly ordered space]].
+Then $S$ is a [[Definition:Connected Topological Space|connected space]] {{iff}} it is a [[Definition:Linear Continuum|linear continuum]].
+\end{theorem}
+
+\begin{proof}
+{{explain|It is not clear how the premises of the subproofs (e.g. "$T$ is [[Definition:Disconnected Space|disconnected]] and [[Definition:Close Packed|close packed]]" directly relate to the premises of the main theorem, e.g.: "$S$ is a [[Definition:Connected (Topology)|connected space]]..."}}
+=== Necessary Condition ===
+Suppose that $T$ is [[Definition:Disconnected Space|disconnected]] and a [[Definition:Linear Continuum|linear continuum]].
+Then there are [[Definition:Non-Empty Set|non-empty]] [[Definition:Open Set (Topology)|open sets]] $U, V$ that [[Definition:Separated Sets|separate]] $T$.
+Let $a \in U$ and $b \in V$.
+{{WLOG}}, suppose that $a \prec b$.
+Let $B = \set {p \in S: \closedint a p \subseteq U}$
+$B$ contains $a$, so it is [[Definition:Non-Empty Set|non-empty]], and it is [[Definition:Bounded Above Set|bounded above]] by $b$.
+In fact, if $y$ is any point not in $U$, it must be an [[Definition:Upper Bound of Set|upper bound]] for $B$:
+Because, if it were not [[Definition:Bounded Above Set|bounded above]] by $y$, there would be some $p \in B$, such that $y < p$. 
+But $p \in B$ means by definition that $\closedint a p \subseteq U$. 
+Now, $a < y < p$, so $y \in \closedint a p$ and so $y \in U$, which was not the case by assumption.
+So any $y \notin U$ is an [[Definition:Upper Bound of Set|upper bound]] for $B$, and so in particular $b$ is.
+So $B$ has a [[Definition:Supremum of Set|supremum]] $m$, as a [[Definition:Bounded Above Set|bounded above]], [[Definition:Non-Empty Set|non-empty set]] has a [[Definition:Supremum of Set|supremum]].
+This is part of the definition of [[Definition:Linear Continuum|linear continuum]], namely being [[Definition:Dedekind Complete|Dedekind complete]].
+First suppose that $m \in B$.
+Then $m \in U$.
+Since $U$ is [[Definition:Open Set (Topology)|open]], there exists a $p \in S$ such that $\hointr m p \subseteq U$.
+Since $S$ is [[Definition:Close Packed|close packed]], it has an [[Definition:Element|element]] $q$ [[Definition:Strictly Between|strictly between]] $m$ and $p$.
+Then $m \in B$ implies that:
+:$\closedint a m \subseteq U$
+We also have that:
+:$\closedint m q \subseteq \hointr m p \subseteq U$
+so:
+:$\closedint a q \subseteq U$
+Then $q \in B$, contradicting the fact that $m$ is an [[Definition:Upper Bound of Set|upper bound]] for $B$.
+Thus $m \notin B$.
+So $\closedint a m \nsubseteq U$ (or else $m \in B$ by definition).
+So there is some [[Definition:Element|element]] $w \in S$ that is in $\closedint a m$ and that is not in $U$.
+Thus $w$ lies in $V$, as these two sets [[Definition:Cover of Set|cover]] $S$. 
+As already proved above, $w \notin U$ implies it is an [[Definition:Upper Bound of Set|upper bound]] for $B$.
+But then $m$ is the minimal one [[Definition:Upper Bound of Set|upper bound]] by the definition of [[Definition:Supremum|supremum]], so $w < m$ is impossible.
+We can only have $w = m$.
+So $m \notin U$, so by necessity, $m \in V$.
+Then as $V$ is [[Definition:Open Set (Topology)|open]], we have some $w' < m$ such that $\hointl {w'} m \subseteq V$.
+Again, by $S$ being [[Definition:Close Packed|close packed]], we have some $w''$ with $w' < w'' < m$. 
+This $w''$ is in $V$, so not in $U$
+So $w''$ is an [[Definition:Upper Bound of Set|upper bound]] for $B$ again, and strictly smaller than $m$.
+This is a [[Definition:Contradiction|contradiction]].
+This shows that the [[Definition:Separated Sets|separation]] cannot exist.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Suppose that $T$ is not a [[Definition:Linear Continuum|linear continuum]].
+Then either $T$ is not [[Definition:Close Packed|close packed]] or $T$ is not [[Definition:Dedekind Complete|Dedekind complete]].
+Suppose first that $T$ is not [[Definition:Close Packed|close packed]].
+Then there are points $a, b \in S$ such that $a \prec b$ and no point lies [[Definition:Strictly Between|strictly between]] $a$ and $b$.
+Thus:
+:$X = a^\preceq \cup b^\succeq$
+and the [[Definition:Component (Topology)|components]] of this [[Definition:Set Union|union]] are [[Definition:Disjoint Sets|disjoint]].
+By [[Mind the Gap]]:
+:$a^\preceq = b^\prec$
+:$b^\succeq = a^\succ$
+where:
+:$a^\preceq$ denotes the [[Definition:Lower Closure of Element|lower closure]] of $a$
+:$b^\prec$ denotes the [[Definition:Strict Lower Closure of Element|strict lower closure]] of $b$
+:$b^\succeq$ denotes the [[Definition:Upper Closure of Element|upper closure]] of $b$
+:$a^\succ$ denotes the [[Definition:Strict Upper Closure of Element|strict upper closure]] of $a$.
+Thus these two [[Definition:Set|sets]] are [[Definition:Open Set (Topology)|open sets]] that [[Definition:Separated Sets|separate]] $T$.
+Therefore $T$ is [[Definition:Disconnected Space|disconnected]].
+Suppose next that $T$ is not [[Definition:Dedekind Complete|Dedekind complete]].
+Then there is a [[Definition:Non-Empty Set|non-empty set]] $B \subset S$ which is [[Definition:Bounded Above Set|bounded above]] in $S$ but has no [[Definition:Supremum of Set|supremum]] in $S$.
+Let $U$ be the set of [[Definition:Upper Bound of Set|upper bounds]] of $B$ ([[Definition:Non-Empty Set|non-empty]] by assumption).
+{{explain|Minor details}}
+Since $B$ has no [[Definition:Supremum of Set|supremum]], $U$ is [[Definition:Open Set (Topology)|open]].
+Let $A = S \setminus U$ be the set of points that are ''not'' [[Definition:Upper Bound of Set|upper bounds]] for $B$.
+Let $p \in S \setminus U$.
+Then there is an [[Definition:Element|element]] of $s \in B$ such that $p \prec s$.
+Then:
+:$p \in s^\prec \subset S \setminus U$
+Thus $S \setminus U$ is also [[Definition:Open Set (Topology)|open]].
+It is [[Definition:Non-Empty Set|non-empty]] because it contains all [[Definition:Element|elements]] of $B$.
+So $U$ and $S \setminus U$ are [[Definition:Open Set (Topology)|open sets]] [[Definition:Separated Sets|separating]] $S$.
+So $T$ is [[Definition:Disconnected Space|disconnected]].
+{{qed}}
+\end{proof}
\ No newline at end of file