hcju/proofwikiqa huggingface dataset:
Denote the open ball of $0$ with radius $r \in \R_{>0}$ as $B_r \left({0}\right)$. Let $z \in D$. By the Alternative Differentiability Condition, it follows that there exists $r \in \R_{>0}$ such that for all $h \in B_r \left({0}\right) \setminus \left\{ {0}\right\}$: :$f \left({z + h}\right) = f \left({z}\right) + h \left({f' \left({z}\right) + \epsilon \left({h}\right) }\right)$ where $\epsilon: B_r \left({0}\right) \setminus \left\{ {0}\right\} \to \C$ is a continuous functions with $\displaystyle \lim_{h \mathop \to 0} \epsilon \left({h}\right) = 0$. Then: :$w f \left({z + h}\right) = w f \left({z}\right) + h \left({w f' \left({z}\right) + w \epsilon \left({h}\right) }\right)$ From Multiple Rule for Continuous Functions, it follows that $w \epsilon$ is a continuous function. From Multiple Rule for Limits of Functions, it follows that $\displaystyle \lim_{h \mathop \to 0} w \epsilon \left({h}\right) = \left({ \lim_{h \mathop \to 0} w }\right) \left({ \lim_{h \mathop \to 0} \epsilon \left({h}\right) }\right) = 0$. Then the Alternative Differentiability Condition shows that: :$\left({w f}\right)' \left({z}\right) = w f' \left({z}\right)$ {{qed}}
By Ordering in terms of Join, we have that: :$a \vee \left({a \wedge b}\right) = a$ {{iff}} $a \wedge b \preceq a$ The result thus follows from Meet Precedes Operands. {{qed}}
By Dual Pairs (Order Theory), we observe that the theorem statement is dual to that of Join Absorbs Meet. The result follows by the Global Duality Principle. {{qed}}
=== Proof of Forward Implication === {{:Rule of Exportation/Forward Implication/Formulation 2/Proof 1}} === Proof of Reverse Implication === {{:Rule of Exportation/Reverse Implication/Formulation 2/Proof 1}}
{{BeginTableau|\vdash \left({\left ({p \land q}\right) \implies r}\right) \implies \left({p \implies \left ({q \implies r}\right)}\right)}} {{Assumption|1|\left ({p \land q}\right) \implies r}} {{SequentIntro|2|1|p \implies \left ({q \implies r}\right)|1|Rule of Exportation: Forward Implication: Formulation 1}} {{Implication|3||\left({\left ({p \land q}\right) \implies r}\right) \implies \left({p \implies \left ({q \implies r}\right)}\right)|1|2}} {{EndTableau}} {{qed}}
{{BeginTableau|\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \land q} \implies r} }} {{Assumption|1|p \implies \paren {q \implies r} }} {{SequentIntro|2|1|\paren {p \land q} \implies r|1|Rule of Exportation: Reverse Implication: Formulation 1}} {{Implication|3||\paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \land q} \implies r}|1|2}} {{EndTableau|qed}}
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|ccc||ccc|} \hline p & \land & q & q & \land & p \\ \hline F & F & F & F & F & F \\ F & F & T & T & F & F \\ T & F & F & F & F & T \\ T & T & T & T & T & T \\ \hline \end{array}$ {{qed}}
We apply the Method of Truth Tables. As can be seen by inspection, in both cases, the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|ccc||ccc|} \hline p & \lor & q & q & \lor & p \\ \hline \F & \F & \F & \F & \F & \F \\ \F & \T & \T & \T & \T & \F \\ \T & \T & \F & \F & \T & \T \\ \T & \T & \T & \T & \T & \T \\ \hline \end{array}$ {{qed}}
Denote the open ball of $0$ by radius $r \in \R_{>0}$ as $B_r \left({0}\right)$. Let $z \in D$. By the Alternative Differentiability Condition, it follows that there exists $r \in \R_{>0}$ such that for all $h \in B_r \left({0}\right) \setminus \left\{ {0}\right\}$: :$f \left({z + h}\right) = f \left({z}\right) + h \left({f' \left({z}\right) + \epsilon_f \left({h}\right) }\right)$ :$g \left({z + h}\right) = g \left({z}\right) + h \left({g' \left({z}\right) + \epsilon_g \left({h}\right) }\right)$ where $\epsilon_f, \epsilon_g: B_r \left({0}\right) \setminus \left\{ {0}\right\} \to \C$ are continuous functions that converge to $0$ as $h$ tends to $0$. Then: {{begin-eqn}} {{eqn | l = \left({fg}\right) \left({z + h}\right)        | r = f \left({z}\right) g \left({z}\right) + h \left({f \left({z}\right) g' \left({z}\right) + f \left({z}\right) \epsilon_g \left({h}\right) }\right) + h \left({g \left({z}\right) f' \left({z}\right) + g \left({z}\right) \epsilon_f \left({h}\right) }\right) + h^2 \left({f' \left({z}\right) + \epsilon_f \left({h}\right) }\right) \left({g' \left({z}\right) + \epsilon_g \left({h}\right) }\right) }} {{eqn | r = \left({fg}\right) \left({z}\right) + h \left({f' \left({z}\right) g \left({z}\right) + f \left({z}\right) g' \left({z}\right) + h \left({f' \left({z}\right) + \epsilon_f \left({h}\right) }\right) \left({g' \left({z}\right) + \epsilon_g \left({h}\right) }\right) }\right) }} {{end-eqn}} Define $\epsilon: B_r \left({0}\right) \setminus \left\{ {0}\right\} \to \C $ by $\epsilon \left({h}\right) = h \left({f' \left({z}\right) + \epsilon_f \left({h}\right) }\right) \left({g' \left({z}\right) + \epsilon_g \left({h}\right) }\right)$. From Product Rule for Continuous Functions and Combined Sum Rule for Continuous Functions, it follows that $\epsilon$ is continuous. From Product Rule for Limits of Functions and Combined Sum Rule for Limits of Functions, it follows that: :$\displaystyle \lim_{h \to 0} \epsilon \left({h}\right) = \left({\lim_{h \to 0} h}\right) \left({\lim_{h \to 0} \left({f' \left({z}\right) + \epsilon_f \left({h}\right) }\right) }\right) \left({\lim_{h \to 0} \left({g' \left({z}\right) + \epsilon_g \left({h}\right) }\right) }\right)= 0$ Then the Alternative Differentiability Condition shows that: :$\left({f g}\right)' \left({z}\right) = f' \left({z}\right) g \left({z}\right) + f \left({z}\right) g' \left({z}\right)$ {{qed}}
By assumption, $\struct {X, f}$ is a Dedekind completion of $S$. Also, $g: S \to Y$ is an order embedding and $Y$ is Dedekind complete. Hence by definition of Dedekind completion, there exists a unique $\phi: X \to Y$ such that: :$\phi \circ f = g$ It only remains to show that $\psi$ is an order isomorphism. By reversing the roles of $X$ and $Y$ in the above, we obtain a unique $\psi: Y \to X$ subject to: :$\psi \circ g = f$ Therefore: :$\psi \circ \phi \circ f = f$ :$\phi \circ \psi \circ g = g$ By the uniqueness clause of the definition of Dedekind completion, it follows that: :$\psi \circ \phi = I_X$ :$\phi \circ \psi = I_Y$ where $I_X$ and $I_Y$ are the identity mappings on $X$ and $Y$ respectively. Hence $\psi$ is an order isomorphism. {{qed}}
{{begin-eqn}} {{eqn | l = c \dfrac \d {\d z} f \left({z}\right) + w \dfrac \d {\d z} g \left({z}\right)       | r = \dfrac \d {\d z} \left({c f}\right) \left({z}\right) + \dfrac \d {\d z} \left({w g}\right) \left({z}\right)       | c = Combination Theorem for Complex-Differentiable Functions: Multiple Rule }} {{eqn | r = \dfrac \d {\d z} \left({c f + w g}\right) \left({z}\right)       | c = Combination Theorem for Complex-Differentiable Functions:Sum Rule }} {{end-eqn}} {{qed}}
For all subsets $I \subseteq S$, define: :$\operatorname U \left({I}\right) = \left\{{x \in S: x}\right.$ is an upper bound for $\left.{I}\right\}$ :$\operatorname L \left({I}\right) = \left\{{x \in S: x}\right.$ is a lower bound for $\left.{I}\right\}$ Note that, for all $I, J \subseteq S$: :$I \subseteq \operatorname {LU} \left({I}\right)$ :$I \subseteq J \implies \operatorname {LU} \left({I}\right) \subseteq \operatorname {LU} \left({J}\right)$ :$\operatorname {ULU} \left({I}\right) = \operatorname U \left({I}\right)$ All of the above follow directly from the relevant definitions. Let: :$\tilde S = \left\{{I \subseteq S: I = \operatorname {LU} \left({I}\right)}\right.$, $I$ is non-empty and bounded above$\left.{}\right\}$ We have that $\left({\tilde S, \subseteq}\right)$ is an ordered set. Let $A \subseteq \tilde S$ be non-empty and bounded above. By Union is Smallest Superset: General Result, $\displaystyle \bigcup A$ is non-empty and bounded above. It follows that: :$\displaystyle \operatorname{LU} \left({\bigcup A}\right) \in \tilde S$ From Set is Subset of Union: General Result, it follows that $\displaystyle \operatorname{LU} \left({\bigcup A}\right)$ is an upper bound for $A$. If $I \in \tilde S$ is an upper bound for $A$, then, by Union is Smallest Superset: General Result: :$\displaystyle \bigcup A \subseteq I$ It follows that: :$\displaystyle \operatorname{LU} \left({\bigcup A}\right) \subseteq \operatorname{LU} \left({I}\right) = I$ Hence: :$\displaystyle \operatorname{LU} \left({\bigcup A}\right) = \sup A$ By definition, $\left({\tilde S, \subseteq}\right)$ is Dedekind complete. Let $\phi: S \to \tilde S$ be the order embedding defined as: :$\forall x \in S: \phi \left({x}\right) = x^\preceq$ where $x^\preceq$ denotes the weak lower closure of $x$ (in $S$). Let $\left({X, \preccurlyeq}\right)$ be a Dedekind complete ordered set, and let $f: S \to X$ be an order embedding. Let $\tilde f: \tilde S \to X$ be the increasing mapping defined as: :$\forall I \in \tilde S: \tilde f \left({I}\right) = \sup f \left({I}\right)$ where $f \left({I}\right)$ denotes the image of $I$ under $f$. Then $\tilde f \circ \phi = f$. Let $I, J \in \tilde S$ such that $\tilde f \left({I}\right) \preccurlyeq \tilde f \left({J}\right)$. Then $\forall x \in I: \forall u \in \operatorname U \left({J}\right): f \left({x}\right) \preccurlyeq \tilde f \left({J}\right) \preccurlyeq f \left({u}\right)$. Since $f$ is an order embedding, it follows that $I \subseteq \operatorname {LU} \left({J}\right) = J$. Hence, $\tilde f$ is an order embedding. {{qed}}
To show that $\mathcal R^\ne$ is a strict ordering, it is sufficient to show that $\mathcal R^\ne$ is antireflexive and transitive. === Antireflexive === Follows from Reflexive Reduction is Antireflexive. {{qed|lemma}} === Transitive === Let $a, b, c \in S$. Let $a \mathrel{\mathcal R^\ne} b$ and $b \mathrel{\mathcal R^\ne} c$. By the definition of reflexive reduction: :$a \ne b$ and $a \mathrel{\mathcal R} b$ :$b \ne c$ and $b \mathrel{\mathcal R} c$ Since $\mathrel{\mathcal R}$ is transitive: :$a \mathrel{\mathcal R} c$ Suppose, with a view to deriving a contradiction, that $a = c$. Since $a \mathrel{\mathcal R} b$ it follows that $c \mathrel{\mathcal R} b$. Since $c \mathrel{\mathcal R} b$, $b \mathrel{\mathcal R} c$, and $\mathrel{\mathcal R}$ is antisymmetric, $b = c$. But this contradicts $b \ne c$. The conclusion is that $a \ne c$. Recall that $a \mathrel{\mathcal R} c$. By the definition of reflexive reduction: :$a \mathrel{\mathcal R^\ne} c$ {{qed}} Category:Reflexive Reductions e85znurcjyy8ehftu2tbp6tfjbm71we
Aiming for a contradiction, suppose that $\mathcal R^\ne$ is not asymmetric. That is: :$\exists a, b \in S: a \mathrel{\mathcal R^\ne} b$ and $b \mathrel{\mathcal R^\ne} a$ Then by the definition of reflexive reduction: : $a \mathrel{\mathcal R} b$, $b \mathrel{\mathcal R} a$ and $a \ne b$. But this contradicts the antisymmetry of $\mathcal R$. Thus, by definition, $\mathcal R^\ne$ is an asymmetric relation. {{qed}} Category:Reflexive Reductions b35ioqr8w71pfuxq9jktyw96i2tj0lb
From Supremum of Empty Set is Smallest Element, we have that: :$\bot := \sup \varnothing$ satisfies $\bot \preceq a$ for all $a \in S$. By Ordering in terms of Join, this is equivalent to: :$\forall a \in S: a \vee \bot = a = \bot \vee a$ where the last equality follows as Join is Commutative. Thus $\bot$ is an identity element for $\vee$, and conversely. That $\inf \varnothing$ is an identity element for $\wedge$, and conversely, follows by the Duality Principle. Hence the result.  {{qed}}
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|cccc||c|} \hline \neg & p & \implies & \bot & p \\ \hline T & F & F & F & F \\ F & T & T & F & T \\ \hline \end{array}$ {{qed}}
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|ccccccc||c|} \hline \neg & p & \implies & (q & \land & \neg & q) & p \\ \hline T & F & F & F & F & T & F & F \\ T & F & F & T & F & F & T & F \\ F & T & T & F & F & T & F & T \\ F & T & T & T & F & F & T & T \\ \hline \end{array}$ {{qed}}
{{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|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}}
{{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}}
=== Proof of Forward Implication === {{:Implication is Left Distributive over Conjunction/Forward Implication/Formulation 1/Proof}} === Proof of Reverse Implication === {{:Implication is Left Distributive over Conjunction/Reverse Implication/Formulation 1/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: 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}}
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: 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}}
=== Proof of Forward Implication === {{:Implication is Left Distributive over Conjunction/Forward Implication/Formulation 2/Proof}} === Proof of Reverse Implication === {{:Implication is Left Distributive over Conjunction/Reverse Implication/Formulation 2/Proof}}
As we presume $\circ$ to be continuous, we need only prove that $\struct {R, +, \tau}$ is a topological group. As we presume $+$ to be continuous, we need only show that negation is continuous. As $\struct {R, \circ}$ is a semigroup and $\circ$ is continuous: :$\struct{R, \circ, \tau}$ is a topological semigroup. From Identity Mapping is Homeomorphism, the identity mapping $I_R : \struct {R, \tau} \to \struct {R, \tau}$ is 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 continuous. From Product with Ring Negative, for each $b \in R : -b = \paren {-1_R} \circ b$. Hence negation is continuous. {{qed}} i43n6rn55jasvjiw2e43wgnc6882ov7
By the definition of a relation from $S$ to $T$, each element of $\mathcal F$ is a subset of $S \times T$. By Union of Subsets is Subset: Set of Sets: : $\mathcal R \subseteq S \times T$ Therefore, by the definition of a relation from $S$ to $T$, $\mathcal R$ is a relation from $S$ to $T$. {{qed}} Category:Relation Theory 82xv7p14zwlckbngevugib4r8b1ufv4
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 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 continuous at $z$ by definition. {{qed}} Category:Complex Analysis Category:Continuous Mappings iuzm878f6b7h15nhuq8shtj2nvj1snw
{{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}}
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|ccccccc||ccccc|} \hline (p & \implies & r) & \land & (q & \implies & r) & (p & \lor & q) & \implies & r \\ \hline F & T & F & T & F & T & F & F & F & F & T & F \\ F & T & T & T & F & T & T & F & F & F & T & T \\ F & T & F & F & T & F & F & F & T & T & F & F \\ F & T & T & T & T & T & T & F & T & T & T & T \\ T & F & F & F & F & T & F & T & T & F & F & F \\ T & T & T & T & F & T & T & T & T & F & T & T \\ T & F & F & F & T & F & F & T & T & T & F & F \\ T & T & T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$ {{qed}}
From Closed Ball is Closed in Metric Space, it follows that $\map {B_\epsilon^-} x$ is closed in $\R^n$. For all $a \in \map {B_\epsilon^-} x$ we have $\map d {x, a} \le \epsilon$, where $d$ denotes the Euclidean metric. Then $\map {B_\epsilon^-} x$ is bounded in $\R^n$. From the Heine-Borel Theorem, it follows that $B_\epsilon^- \left({x}\right)$ is compact. {{qed}} Category:Euclidean Space o4quf79inub19hlagn3qnzxwn1iwyuh
Let $\epsilon \in \R_{>0}$. By the definition of 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 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 Cauchy sequence. {{qed}} Category:Isometries Category:Cauchy Sequences fsicej092lpjzhzgi3ez1wi0h3sxij4
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}}
Let $x \in S_1 \cup S_2$. By the definition of union, either $x \in S_1$ or $x \in S_2$. By hypothesis, $S_1 \subseteq T$ and $S_2 \subseteq T$. By definition of 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 subset. {{qed}}
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: : $p \implies r, q \implies r \vdash p \lor q \implies r \lor r$ From the Rule of Idempotence we have: : $r \lor r \vdash r$ and the result follows by Hypothetical Syllogism. {{qed}}
By definition of 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 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 bounded sequence. Then $\sequence {y_n}$ is also a 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 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 subsequence $\sequence {\tuple {x_{n_r}, y_{n_r} } }_{r \mathop \in N}$ that converges to a 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 disjoint, it follows that $x \ne y$. Hence: :$0 < \map d {x, y} = \map d {S, T}$ {{qed}}
Let $a \in S$, and suppose that $b, c \in S$ are complements for $a$. Then: {{begin-eqn}} {{eqn | l = b       | r = \top \wedge b       | c = $\top$ is the identity for $\wedge$ }} {{eqn | r = \left({c \vee a}\right) \wedge b       | c = $c$ is a complement for $a$ }} {{eqn | r = \left({c \wedge b}\right) \vee \left({a \wedge b}\right)       | c = $S$ is a distributive lattice }} {{eqn | r = \left({c \wedge b}\right) \vee \bot       | c = $b$ is a complement for $a$ }} {{eqn | r = c \wedge b       | c = $\bot$ is the 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
By definition, $\top$ is the 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 identity for $\wedge$. The result follows. {{qed}}
By definition, $\bot$ is the 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 identity for $\vee$. The result follows. {{qed}}
By definition, a top for $S$ is a greatest element. The result follows from Greatest Element is Unique. {{qed}} Category:Lattice Theory 683yerk2ng1obdpxb8ac23vneubyuzt
By definition, a bottom for $S$ is a smallest element. The result follows from Smallest Element is Unique. {{qed}} Category:Lattice Theory g46t8oi3wi4enzdn1w1f57fg1p6zcp2