hcju/proofwikiqa huggingface dataset:
By Equivalence of Definitions of Normal Subset: 3 iff 4, $S$ being a normal subset of $G$ by Definition 3 and 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 set equality: :$\forall g \in G: g \circ S \circ g^{-1} = S$ By $(2)$ and $(4)$ and definition of set equality: :$\forall g \in G: g^{-1} \circ S \circ g = S$ {{qed}}
==== 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 associative and by the definition of inverse: :$y \circ x \in S$ {{qed|lemma}} ==== 5 implies 3 ==== Suppose that $S$ is a normal subset of $G$ by 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 inverse and the definition of identity: :$\paren {g \circ g^{-1} } \circ x \circ \paren {g \circ g^{-1} } \in S$ By 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 group axiom $G1$: associativity, group axiom $G2$: identity and 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 3. {{qed}}
=== Necessary Condition === Let $N$ be 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 subset product of $g$ with $N$. Thus $N$ is a normal subset of $G$ (by definition 1): :$\forall g \in G: g \circ N = N \circ g$ === Sufficient Condition === Let $N$ be a 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 normal subgroup of $G$ (by definition 1). {{Qed}}
By the definition of a monoid, $\left({G, \circ}\right)$ is a semigroup. By Power Set of Semigroup under Induced Operation is Semigroup, $\left({\mathcal P \left({G}\right), \circ_\mathcal P}\right)$ is a semigroup. By Subset Product by Identity Singleton, $\left\{{e}\right\}$ is an 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 monoid with identity $\left\{{e}\right\}$. {{qed}} Category:Subset Products Category:Monoids Category:Power Set q9eexcxnhi0m3m95ocbqo08278o3zk0
We apply the Method of Truth Tables to the proposition $\neg \left({p \land \neg p}\right)$. As can be seen by inspection, the truth value of the main connective, that is $\neg$, is $T$ for each boolean interpretation for $p$. $\begin{array}{|ccccc|} \hline \neg & (p & \land & \neg & p)\\ \hline T & F & F & T & F \\ T & T & F & F & T \\ \hline \end{array}$ {{qed}}
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 magma. It follows from Subset Product within Semigroup is Associative that $\circ_\mathcal P$ is 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 semigroup. {{qed}} Category:Semigroups Category:Power Set Category:Subset Products du1auei87vrkart49t0xnax2tuwa8o9
By the definition of a group, $\left({G, \circ}\right)$ is a 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
Let $\left({S, \circ}\right)$ be a 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 magma }} {{eqn | o=\implies       | r=A \circ B \subseteq S       | c=Definition of Subset Product }} {{eqn | o=\implies       | r=A \circ_\mathcal P B \subseteq S       | c=Definition of 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 Power Set }} {{end-eqn}} Thus $\left({\mathcal P \left({S}\right), \circ_\mathcal P}\right)$ is a magma. {{qed}} Category:Power Set Category:Magmas Category:Subset Products 6utulg4gigeh20ityr45exebhjlaz1d
{{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 subgroup }} {{end-eqn}} {{Qed}} </onlyinclude>
Let $\struct {G, *}$ be the 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}}
{{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 of Extension }} {{end-eqn}} {{qed}} Category:Group Theory tknfd1imwrrao9s0kojknorjgzf0gmx
{{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 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 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 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 Exclusive Or}} {{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. $\begin{array}{|ccc||ccc|} \hline p & \oplus & q & q & \oplus & p \\ \hline F & F & F & F & F & F \\ F & T & T & T & T & F \\ T & T & F & F & T & T \\ T & F & T & T & F & T \\ \hline \end{array}$ {{qed}}
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. $\begin{array}{|ccccc||ccccc|} \hline p & \oplus & (q & \oplus & r) & (p & \oplus & q) & \oplus & r \\ \hline F & F & F & F & F & F & F & F & F & F \\ F & T & F & T & T & F & F & F & T & T \\ F & T & T & T & F & F & T & T & T & F \\ F & F & T & F & T & F & T & T & F & T \\ T & T & F & F & F & T & T & F & T & F \\ T & F & F & T & T & T & T & F & F & T \\ T & F & T & T & F & T & F & T & F & F \\ T & T & T & F & T & T & F & T & T & T \\ \hline \end{array}$ {{qed}}
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connective match for each boolean interpretations. $\begin{array}{|ccc|} \hline p & \oplus & p \\ \hline F & F & F \\ T & F & T \\ \hline \end{array}$ {{qed}}
=== Forward Implication: Proof === {{:Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Forward Implication/Proof}} === Reverse Implication: Proof === {{:Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Reverse Implication/Proof}}
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. $\begin{array}{|cccc||ccccccccc|} \hline \neg & (p & \iff & q) & (\neg & p & \land & q) & \lor & (p & \land & \neg & q) \\ \hline F & F & T & F & T & F & F & F & F & F & F & T & F \\ T & F & F & T & T & F & T & T & T & F & F & F & T  \\ T & T & F & F & F & T & F & F & T & T & T & T & F  \\ F & T & T & T & F & T & F & T & F & T & F & F & T  \\ \hline \end{array}$ {{qed}}
{{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}}
{{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
{{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}}
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. $\begin{array}{|cccc||ccccccccc|} \hline \neg & (p & \iff & q) & (p & \lor & q) & \land & (\neg & p & \lor & \neg & q) \\ \hline F & F & T & F & F & F & F & F & T & F & T & T & F \\ T & F & F & T & F & T & T & T & T & F & T & F & T \\ T & T & F & F & T & T & F & T & F & T & T & T & F \\ F & T & T & T & T & T & T & F & F & T & F & F & T \\ \hline \end{array}$ {{qed}}
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
By the definition of order isomorphism, $\phi$ is a bijection. Thus to show $\phi$ is a homeomorphism it remains to be shown that: :$\phi$ is continuous and: :$\phi^{-1}$ is continuous. First it is shown that $\phi^{-1}$ is continuous. By Order Isomorphism Preserves Initial Segments and its dual, $\phi$ maps open rays in $\struct {S_1, \le_1}$ to open rays in $\struct {S_2, \le_2}$. By Continuity Test using Sub-Basis, $\phi^{-1}$ is continuous. Next it is shown that $\phi^{-1}$ is continuous. By Inverse of Order Isomorphism is Order Isomorphism, $\phi^{-1}$ is an order isomorphism. Applying the above shows that $\phi = \paren {\phi^{-1} }^{-1}$ is continuous. Both $\phi$ and $\phi^{-1}$ have been shown to be continuous. Thus, by definition, $\phi$ is a homeomorphism. {{qed}} Category:Linearly Ordered Spaces dnyofjqvk2qgg6aiwah84iv1pyw9vl2
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. $\begin{array}{|ccccc||ccccc|} \hline p & \land & (q & \land & r) & (p & \land & q) & \land & r \\ \hline F & F & F & F & F & F & F & F & F & F \\ F & F & F & F & T & F & F & F & F & T \\ F & F & T & F & F & F & F & T & F & F \\ F & F & T & T & T & F & F & T & F & T \\ T & F & F & F & F & T & F & F & F & F \\ T & F & F & F & T & T & F & F & F & T \\ T & F & T & F & F & T & T & T & F & F \\ T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$ {{qed}}
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. $\begin{array}{|ccccc||ccccc|} \hline p & \lor & (q & \lor & r) & (p & \lor & q) & \lor & r \\ \hline F & F & F & F & F & F & F & F & F & F \\ F & T & F & T & T & F & F & F & T & T \\ F & T & T & T & F & F & T & T & T & F \\ F & T & T & T & T & F & T & T & T & T \\ T & T & F & F & F & T & T & F & T & F \\ T & T & F & T & T & T & T & F & T & T \\ T & T & T & T & F & T & T & T & T & F \\ T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$ {{qed}}
{{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: 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: 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}}
=== Proof of Forward Implication === {{:Principle of Commutation/Forward Implication/Formulation 1/Proof}} === Proof of Reverse Implication === {{:Principle of Commutation/Reverse Implication/Formulation 1/Proof}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth values under the main connectives match for all 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}}
=== Proof of Forward Implication === {{:Principle of Commutation/Forward Implication/Formulation 2/Proof 1}} === 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}} {{TheoremIntro|2|\paren {q \implies \paren {p \implies r} } \implies \paren {p \implies \paren {q \implies r} }|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}}
{{BeginTableau|\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {q \implies \paren {p \implies r} } }} {{Assumption|1|p \implies \paren {q \implies r} }} {{SequentIntro|2|1|q \implies \paren {p \implies r}|1|Principle of Commutation: Forward Implication: Formulation 1}} {{Implication|3||\paren {p \implies \paren {q \implies r} } \implies \paren {q \implies \paren {p \implies r} }|1|2}} {{EndTableau}} {{qed}}
{{BeginTableau|\vdash \paren {q \implies \paren {p \implies r} } \implies \paren {p \implies \paren {q \implies r} } }} {{Assumption   |1|q \implies \paren {p \implies r} }} {{SequentIntro |2|1|p \implies \paren {q \implies r}|1|Principle of Commutation: Reverse Implication: Formulation 1}} {{Implication  |3| |\paren {q \implies \paren {p \implies r} } \implies \paren {p \implies \paren {q \implies r} }|1|2}} {{EndTableau}} {{qed}}
The polygon $P$ has $n$ sides, where $n \in \N$. Denote the vertices of $P$ as $A_1, \ldots, A_n$ and its sides as $S_1, \ldots, S_n$, such that each vertex $A_i$ has adjacent sides $S_{i - 1}$ and $S_i$. We use the conventions that $S_0 = S_n$, and $A_{n + 1} = A_1$. As each side $S_i$ is a line segment joining $A_i$ and $A_{i + 1}$, when we define the path $\gamma_i: \closedint 0 1 \to \R^2$ by: :$\map {\gamma_i} t = \paren {1 - t} A_i + t A_{i + 1}$ the image of $\gamma_i$ is equal to the side $S_i$. Define $\gamma: \closedint 0 1 \to \R^2$ as the concatenation $\paren {\cdots \paren {\paren {\gamma_1 * \gamma_2} * \gamma_3} * \ldots * \gamma_{n - 1} } * \gamma_n$. Then each point in the image of $\gamma$ corresponds to a point in a side of $P$. As $\map \gamma 0 = A_1 = \map \gamma 1$ by our definition of $A_{n + 1}$, it follows that $\gamma$ is a closed path. It follows from the definition of polygon that the sides of $P$ do not intersect, except at the vertices. For $i \ne 1$, each vertex $A_i$ is the initial point of $\gamma_i$ and the final point of $\gamma_{i - 1}$, and is equal to exactly one point $\map \gamma {2^{-n - 1 + i} }$ in the image of $\gamma$. Then we have that $\gamma$ restricted to $\hointr 0 1$ is injective. Hence $\gamma$ is a Jordan curve. {{qed}} Category:Topology 7lfmlxoo0byynfdg2uvcwlcosct733d
{{BeginTableau|\vdash \neg \paren {p \implies q} \implies p}} {{Assumption|1|\neg \paren {p \implies q} }} {{SequentIntro|2|1|p \land \neg q|1|Conjunction with Negative Equivalent to Negation of Implication}} {{Simplification|3|1|p|2|1}} {{Implication|4||\neg \paren {p \implies q} \implies p|1|3}} {{EndTableau}} {{Qed}}
{{BeginTableau|\vdash \neg \left({p \implies q}\right) \implies \neg q}} {{Assumption|1|\neg \left({p \implies q}\right)}} {{SequentIntro|2|1|p \land \neg q|1|Conjunction with Negative Equivalent to Negation of Implication}} {{Simplification|3|1|\neg q|2|2}} {{Implication|4||\neg \left({p \implies q}\right) \implies \neg q|1|3}} {{EndTableau}} {{Qed}}
=== Sufficient Condition === Suppose for some $i, j$: :$\left[{r s}\right]_{i j} = T$  Then by the definition of $\lor$ there must exist some $k$ for which: :$\left[{r}\right]_{i k} \land \left[{s}\right]_{k j} = T$ which by our definition implies: :$\left({i, k}\right) \in \mathcal R$ :$\left({k, j}\right) \in \mathcal S$ Then by the definition of a composite relation: $(\left({i, j}\right) \in \mathcal R \circ \mathcal S$ === Necessary Condition === Suppose for some $i, j$: :$\left({i, j}\right) \in \mathcal S \circ \mathcal R$ Then there exists a $k$ for which: :$\left({i, k}\right) \in \mathcal R$ :$\left({k, j}\right) \in \mathcal S$ Hence: :$\left[{r}\right]_{i k} = T$ :$\left[{s}\right]_{k j} = T$ and there exists some $k$ for which: :$\left[{r}\right]_{i k} \land \left[{s}\right]_{k j} = T$ Hence by the definition of $\lor$: :$\left[{r s}\right]_{i j} = T$ {{qed}} Category:Relation Theory 7uk4uhou0t7d40rr3ek2kiswutnihi1
=== $(1)$ implies $(2)$ === This follows immediately from the definition of compactness. {{qed|lemma}} === $(2)$ implies $(1)$ === Suppose that $(2)$ holds. Let $\AA$ be an open cover of $X$. Let $f: \AA \to \powerset B$ map each element of $\AA$ to the set of all elements of $B$ it contains. Since each element of $\AA$ is open, $A = \bigcup f(A)$ for each $A \in \AA$. Let $\AA' = \bigcup \map f \AA$. Then $\AA'$ is an open cover of $X$ consisting solely of elements of $B$. By the premise, $\AA'$ has a finite subset $\FF'$ that covers $X$. Let $g: \FF' \to \AA$ map each element of $\FF'$ to an element of $\AA$ containing it. Note that since $\FF'$ is finite, this does not require the Axiom of Choice. Let $\FF = \map g {\FF'}$. Then $\FF$ is a finite subcover of $\AA$. Since every open cover of $X$ has a finite subcover, $X$ is compact. {{qed}} Category:Compact Spaces a377rwn6vjz6p4kspxbkb0a6naooxx1
{{BeginTableau|\vdash \paren {p \implies q} \land \paren {\neg p \implies q} \implies q}} {{Assumption|1|\paren {p \implies q} \land \paren {\neg p \implies q} }} {{SequentIntro|2|1|q|1|Principle of Dilemma: Formulation 1}} {{Implication|3||\paren {p \implies q} \land \paren {\neg p \implies q} \implies q|1|2}} {{EndTableau|qed}}
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connective is true for all boolean interpretations. $\begin{array}{|ccc|c|ccccc|} \hline (p & \implies & q) & \iff & (\neg & (p & \land & \neg & q)) \\ \hline F & T & F & T & T & F & F & T & F \\ F & T & T & T & T & F & F & F & T \\ T & F & F & T & F & T & T & T & F \\ T & T & T & T & T & T & F & F & T \\ \hline \end{array}$ {{qed}}
Let $S$ be a set. === Necessary Condition === Let $\mathcal C$ be a cover for $S$: :$S \subseteq \bigcup \mathcal C$ Suppose: :$\set {\relcomp S X: X \in \mathcal C}$ is not free. Then there exists an $x \in S$ such that: :$\forall X \in \mathcal C: x \in \relcomp S X$ But by the definition of (relative) complement: :$\forall X \in \mathcal C: x \notin X$ That is: :$x \notin \bigcup \mathcal C$ which contradicts: :$S \subseteq \bigcup \mathcal C$ {{qed|lemma}} === Sufficient Condition === Let: :$\set {\relcomp S X: X \in \mathcal C}$ be free. Suppose: :$\bigcup \mathcal C$ is not a cover for $S$. Then there exists some $x \in S$ such that: :$\forall X \in \mathcal C: x \notin X$ Hence: :$\forall \relcomp S X \in \mathcal C: x \in X$ which contradicts the supposition that:  :$\set {\relcomp S X: X \in \mathcal C}$ is free. {{qed}} Category:Set Theory j3kwwh8bhy821r0podg8mwieey7rh2c
{{BeginTableau|\vdash \left({\neg \left({p \land \neg q}\right)}\right) \implies \left({p \implies q}\right)}} {{Assumption|1|\neg \left({p \land \neg q}\right)}} {{SequentIntro|2|1|p \implies q|1|Implication Equivalent to Negation of Conjunction with Negative: Formulation 1}} {{Implication|3||\left({\neg \left({p \land \neg q}\right)}\right) \implies \left({p \implies q}\right)|1|2}} {{EndTableau}} {{qed}} {{LEM|Implication Equivalent to Negation of Conjunction with Negative/Formulation 1/Reverse Implication}} Category:Implication Equivalent to Negation of Conjunction with Negative estsbm4ogmgfblagpg6qq8budnw25nl
We apply the Method of Truth Tables. $\begin{array}{|ccccccc||ccc|} \hline (p & \implies & q) & \land & (q & \implies & p) & p & \iff & q\\ \hline F & T & F & T & F & T & F & F & T & F \\ F & T & T & F & T & F & F & F & F & T \\ T & F & F & F & F & T & T & T & F & F \\ T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$ As can be seen, only when both $p \implies q$ and $q \implies p$ are true, then so is $p \iff q$. {{qed}}
=== Form 1 === {{:Biconditional Elimination/Sequent Form/Proof 1/Form 1}} === Form 2 === {{:Biconditional Elimination/Sequent Form/Proof 1/Form 2}}
We apply the Method of Truth Tables. $\begin{array}{|ccc||ccc|ccc|} \hline p & \iff & q & p & \implies & q & q & \implies & p \\ \hline F & T & F & F & T & F & F & T & F \\ F & F & T & F & T & T & T & F & F \\ T & F & F & T & F & F & F & T & T \\ T & T & T & T & F & T & T & T & T \\ \hline \end{array}$ As can be seen, when $p \iff q$ is true so are both $p \implies q$ and $q \implies p$. {{qed}}