hcju/proofwikiqa huggingface dataset:
By Cardinal Class is Subset of Ordinal Class, the set: :$M = \set {\card \BB: \BB \in \mathbb B}$ is a subset of the class of ordinals. By Ordinal Class is Strongly Well-Ordered by Subset: :$M$ is well ordered by the $\subseteq$ relation. By Ordinal Class is Strongly Well-Ordered by Subset there exists a smallest element $m_0 \in M$: :$\forall m \in M: m_0 \subseteq m$ Hence by Smallest Element is Minimal there exists a basis $\mathcal B_0$ of $T$ which has minimal cardinality: :$m_0 = \map {w_2} T$. Let: :$\displaystyle \map {w_1} T = \bigcap_{\BB \mathop \in \mathbb B} \card \BB$ By Intersection is Subset: :$\displaystyle \map {w_1} T = \bigcap M \subseteq m_0$ But by Intersection is Largest Subset: :$\displaystyle \mathfrak m_0 \subseteq \bigcap M$ By definition of set equality: :$\map {w_1} T = \map {w_2} T$ and hence the result. {{qed}}
=== Sufficient Condition === Let $T$ be first-countable. By definition of first-countable: :$\forall x \in T: \exists \mathcal B \in \mathbb B \left({x}\right): \left\vert{\mathcal B}\right\vert \leq \aleph_0$ where $\mathbb B \left({x}\right)$ denotes the set of all local bases at $x$. Then by definition of character of a point: :$\forall x \in T: \chi \left({x, T}\right) \leq \aleph_0$ Hence by definition of character of topogical space: :$\chi \left({T}\right) \leq \aleph_0$ {{qed|lemma}} === Necessary Condition === Let $\chi \left({T}\right) \leq \aleph_0$. By definition of character of topogical space: :$\forall x \in T: \chi \left({x, T}\right) \leq \aleph_0$ Then by definition of character of a point: :$\forall x \in T: \exists \mathcal B \in \mathbb B \left({x}\right): \left\vert{\mathcal B}\right\vert \leq \aleph_0$ Thus by definition of first-countable: :$T$ is first-countable. {{qed}}
From Difference of Two Powers: :$\displaystyle a^n - b^n = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$ The result follows directly by setting $n = 3$. {{qed}}
By definition of weight of $T$ there exists a basis $\mathcal B$ of $T$ such that: :$(1): \quad \left\vert{\mathcal B}\right\vert = w \left({T}\right)$ Let: :$\mathcal B_1 = \left\{{W \in \mathcal B: \exists U \in \mathcal F: W \subseteq U}\right\}$ By definition of subset: :$\mathcal B_1 \subseteq \mathcal B$ Then by Subset implies Cardinal Inequality: :$(2): \quad \left\vert{\mathcal B_1}\right\vert \leq \left\vert{\mathcal B}\right\vert$ By definition of set $\mathcal B_1$: :$\forall W \in \mathcal B_1: \exists U \in \mathcal F: W \subseteq U$ Then by the Axiom of Choice there exists a mapping $f$ from $\mathcal B_1$ into $\mathcal F$ such that :$(3): \quad \forall U \in \mathcal B_1: U \subseteq f \left({U}\right)$ Let: :$\mathcal G = \operatorname{Im} \left({f}\right)$ where $\operatorname{Im} \left({f}\right)$ denotes the image of $f$. By hypothesis: :$\mathcal G \subseteq \mathcal F$ Then by Union of Subset of Family is Subset of Union of Family: :$\displaystyle \bigcup \mathcal G \subseteq \bigcup \mathcal F$ By definition of set equality, to prove $\displaystyle \bigcup \mathcal G = \bigcup \mathcal F$ it is sufficient to show: :$\displaystyle \bigcup \mathcal F \subseteq \bigcup \mathcal G$ Let $\displaystyle x \in \bigcup \mathcal F$. By definition of union there exists a set $A$ such that: :$A \in \mathcal F$ and $x \in A$ Because $A$ is open, then by definition of basis there exists $U \in \mathcal B$ such that: :$x \in U \subseteq A$ By definition of the set $\mathcal B_1$: :$U \in \mathcal B_1$ Because $\mathcal G = \operatorname{Im} \left({f}\right)$: :$f \left({U}\right) \in \mathcal G$ By $(3)$ and Set is Subset of Union: :$\displaystyle U \subseteq f \left({U}\right) \subseteq \bigcup \mathcal G$ Thus by definition of subset: :$\displaystyle x \in \bigcup \mathcal G$ This ends the proof of inclusion. By Cardinality of Image of Mapping not greater than Cardinality of Domain: :$\left\vert{\operatorname{Im} \left({f}\right)}\right\vert \leq \left\vert{\mathcal B_1}\right\vert$ Thus by $(1)$ and $(2)$: :$\left\vert{\mathcal G}\right\vert \leq w \left({T}\right)$ {{qed}}
By Basis for Discrete Topology $\mathcal B$ is a basis of $T$. It remains to be shown that $\mathcal B$ is the smallest basis of $T$. Let $\mathcal C$ be a basis of $T$. Let $A \in \mathcal B$. By definition of the set $\mathcal B$: :$\exists x \in S: A = \left\{{x}\right\}$ By definition of basis: :$\exists B \in \mathcal C: x \in B \subseteq A$ Then by Singleton of Element is Subset: :$\left\{{x}\right\} \subseteq B$ Hence $B = A$ by definition of set equality. Thus $A \in \mathcal C$. {{qed}}
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connective are $T$ for all boolean interpretations. $\begin{array}{|ccc|c|c||c|c|} \hline p & \land & q & p & q & p \land q \implies p & p \land q \implies q \\ \hline F & F & F & F & F & T & T \\ F & F & T & F & T & T & T \\ T & F & F & T & F & T & T \\ T & T & T & T & T & T & T \\ \hline \end{array}$ {{qed}}
We apply the Method of Truth Tables to the proposition: :$\left({p \implies q}\right) \iff \left({\lnot p \implies \lnot q}\right)$ $\begin{array}{|ccc|c|ccc|} \hline p & \implies & q) & \iff & (\lnot & p & \implies & \lnot & q) \\ \hline F & T & F & T & T & F & T & T & F \\ F & T & T & F & T & F & F & F & T \\ T & F & F & F & F & T & T & T & F \\ T & T & T & T & F & T & T & F & T \\ \hline \end{array}$ As can be seen by inspection, the truth values under the main connectives do not match for all boolean interpretations. {{qed}}
By Basis for Discrete Topology the set $\mathcal B = \left\{{\left\{{x}\right\}: x \in S}\right\}$ is a basis of $T$. By Set of Singletons is Smallest Basis of Discrete Space $\mathcal B$ is smallest basis of $T$: :for every basis $\mathcal C$ of $T$, $\mathcal B \subseteq \mathcal C$. Then by Subset implies Cardinal Inequality: :for every basis $\mathcal C$ of $T$, $\left\vert{\mathcal B}\right\vert \leq \left\vert{\mathcal C}\right\vert$. Hence $\left\vert{\mathcal B}\right\vert$ is minimal cardinalty of basis of $T$: :$w \left({T}\right) = \left\vert{\mathcal B}\right\vert$ by definition of weight. Thus by Cardinality of Set of Singletons: :$w \left({T}\right) = \left\vert{S}\right\vert$. {{qed}}
Define a mapping $f: S \to T$: :$\forall x \in S: f \left({x}\right) = \left\{ {x}\right\}$ By Singleton Equality: :$\forall x, y \in S: f \left({x}\right) = f \left({y}\right) \implies x = y$ Then, by definition, $f$ is an injection. By the definition of set $T$: :$\forall y \in T: \exists x \in S: y = f \left({x}\right)$ Then, by definition, $f$ is a surjection. Hence, by definition, $f: S \to T$ is a bijection. Thus, by definition, $S$ and $T$ are equivalent: :$S \sim T$ Thus by definition of cardinality: :$\left\vert T \right\vert = \left\vert S \right\vert$ {{qed}}
{{BeginTableau|\vdash \paren {\neg p \implies p} \implies  p}} {{Premise|1|\neg p \implies p}} {{SequentIntro|2|1|p|1|Clavius's Law: Formulation 1}} {{Implication|3||\paren {\neg p \implies p} \implies  p|1|2}} {{EndTableau|qed}}
We apply the Method of Truth Tables. As can be seen by inspection, the truth value under the main connective is true for all boolean interpretations. :$\begin{array}{|cccc|c|c|} \hline (\neg & p & \implies & p) & \implies & p \\ \hline T & F & F & F & T & F \\ F & T & T & T & T & T \\ \hline \end{array}$ {{qed}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth values under the main connective are true for all boolean interpretations. $\begin{array}{|ccccc|c|ccccc|} \hline (p & \implies & (q & \implies & r)) & \implies & ((p & \land & q) & \implies & r) \\ \hline F & T & F & T & F & T & F & F & F & T & F \\ F & T & F & T & T & T & F & F & F & T & T \\ F & T & T & F & F & T & F & F & T & T & F \\ F & T & T & T & T & T & F & F & T & T & T \\ T & T & F & T & F & T & T & F & F & T & F \\ T & T & F & T & T & T & T & F & F & T & T \\ T & F & T & F & F & T & T & T & T & F & F \\ T & T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$ {{qed}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth values under the main connective are true for all boolean interpretations. $\begin{array}{|ccccc|c|ccccc|} \hline ((p & \land & q) & \implies & r) & \implies & (p & \implies & (q & \implies & r)) \\ \hline F & F & F & T & F & T & F & T & F & T & F \\ F & F & F & T & T & T & F & T & F & T & T \\ F & F & T & T & F & T & F & T & T & F & F \\ F & F & T & T & T & T & F & T & T & T & T \\ T & F & F & T & F & T & T & T & F & T & F \\ T & F & F & T & T & T & T & T & F & T & T \\ T & T & T & F & F & T & T & F & T & F & F \\ T & T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$ {{qed}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth values under the main connective are true for all boolean interpretations. $\begin{array}{|ccccccc|c|ccccc|} \hline ((p & \implies & r) & \land & (q & \implies & r)) & \implies & ((p & \lor & q) & \implies & r) \\ \hline F & T & F & T & F & T & F & T & F & F & F & T & F \\ F & T & T & T & F & T & T & T & F & F & F & T & T \\ F & T & F & F & T & F & F & T & F & T & T & F & F \\ F & T & T & T & T & T & T & T & F & T & T & T & T \\ T & F & F & F & F & T & F & T & T & T & F & F & F \\ T & T & T & T & F & T & T & T & T & T & F & T & T \\ T & F & F & F & T & F & F & T & T & T & T & F & F \\ T & T & T & T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$ {{qed}}
{{BeginTableau|\vdash p \iff \neg \neg p}} {{TheoremIntro|1|p \implies \neg \neg p|Double Negation Introduction: Formulation 2}} {{TheoremIntro|2|\neg \neg p \implies p|Double Negation Elimination: Formulation 2}} {{BiconditionalIntro|3||p \iff \neg \neg p|1|2}} {{EndTableau}} {{qed}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth values under the main connective are true for all boolean interpretations. $\begin{array}{|c|c|ccc|} \hline p & \iff & \neg & \neg & p \\ \hline F & T & F & T & F \\ T & T & T & F & T \\ \hline \end{array}$ {{qed}}
{{BeginTableau|\vdash \left({\neg p \land \neg q}\right) \iff \left({\neg \left({p \lor q}\right)}\right)}} {{Assumption|1|\neg p \land \neg q}} {{SequentIntro|2|1|\neg \left({p \lor q}\right)|1|De Morgan's Laws (Logic): Disjunction of Negations: Formulation 1}} {{Implication|3||\left({\neg p \land \neg q}\right) \implies \left({\neg \left({p \lor q}\right)}\right)|1|2}} {{Assumption|4|\neg \left({p \lor q}\right)}} {{SequentIntro|5|4|\neg p \land \neg q|4|De Morgan's Laws (Logic): Disjunction of Negations: Formulation 1}} {{Implication|6||\left({\neg \left({p \lor q}\right)}\right) \implies \left({\neg p \land \neg q}\right)|4|5}} {{BiconditionalIntro|7||\left({\neg p \land \neg q}\right) \iff \left({\neg \left({p \lor q}\right)}\right)|3|6}} {{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. $\begin{array}{|ccccc|c|cccc|} \hline \neg & p & \land & \neg & q & \iff & \neg & (p & \lor & q) \\ \hline T & F & T & T & F & T & T & F & F & F \\ T & F & F & F & T & T & F & F & T & T \\ F & T & F & T & F & T & F & T & T & F \\ F & T & F & F & T & T & F & T & T & T \\ \hline \end{array}$ {{qed}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth values under the main connective is true for all boolean interpretations. $\begin{array}{|ccccc|c|ccccccc|} \hline p & \land & (q & \lor & r) & \iff & (p & \land & q) & \lor & (p & \land & r) \\ \hline F & F & F & F & F & T & F & F & F & F & F & F & F \\ F & F & F & T & T & T & F & F & F & F & F & F & T \\ F & F & T & T & F & T & F & F & T & F & F & F & F \\ F & F & T & T & T & T & F & F & T & F & F & F & T \\ T & F & F & F & F & T & T & F & F & F & T & F & F \\ T & T & F & T & T & T & T & F & F & T & T & T & T \\ T & T & T & T & F & T & T & T & T & T & T & F & F \\ T & T & T & T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$ {{qed}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth values under the main connective is true for all boolean interpretations. $\begin{array}{|ccccc|c|ccccccc|} \hline p & \lor & (q & \land & r) & \iff & (p & \lor & q) & \land & (p & \lor & r) \\ \hline F & F & F & F & F & T & F & F & F & F & F & F & F \\ F & F & F & F & T & T & F & F & F & F & F & T & T \\ F & F & T & F & F & T & F & T & T & F & F & F & F \\ F & T & T & T & T & T & F & T & T & T & F & T & T \\ T & T & F & F & F & T & T & T & F & T & T & T & F \\ T & T & F & F & T & T & T & T & F & T & T & T & T \\ T & T & T & F & F & T & T & T & T & T & T & T & F \\ T & T & T & T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$ {{qed}}
There are two cases:  :infinite weight :finite weight. === Case when Weight is Infinite === Let $T$ has infinite weight. By definition of weight, there exists a basis $\BB_1$ of $T$ such that: :$(1): \quad \card {\BB_1} = \map w T$ We will prove: :$(2): \quad \displaystyle \forall U \in \BB_1: \exists \AA \subseteq \BB: U = \bigcup \AA \land \card \AA \le \map w T$ Let $U \in \BB_1$. Let $S = \set {W \in \BB: W \subseteq U}$. By definition of subset: :$S \subseteq \BB$ By definition of basis: $\displaystyle \bigcup S = U$ By definition of set $S$, $S$ is set of open subset of $T$. Then by Existence of Subfamily of Cardinality not greater than Weight of Space and Unions Equal there exists a subset $\AA \subseteq S$ such that: :$\displaystyle \bigcup \AA = \bigcup S$ and $\card \AA \le \map w T$ Thus by Subset Relation is Transitive: :$\AA \subseteq \BB$. Thus: :$\displaystyle U = \bigcup \AA \land \card \AA \le \map w T$ This ends proof of $(2)$. By $(2)$ and Axiom of Choice there exists a mapping $f: \BB_1 \to \powerset \BB$ such that: :$(3): \quad \displaystyle \forall U \in \BB_1: U = \bigcup \map f U \land \card {\map f U} \le \map w T$ By Union is Smallest Superset, because $\forall U \in \BB_1: \map f U \subseteq \BB$: :$\displaystyle \bigcup_{U \mathop \in \BB_1} \map f U \subseteq \BB$ Set $\BB_0 := \displaystyle \bigcup_{U \mathop \in \BB_1} \map f U = \bigcup \Img f$ Now we will show that $\BB_0$ is basis of $T$. By definition of basis: :$\BB \subseteq \tau$ Thus by Subset Relation is Transitive: :$\BB_0 \subseteq \tau$ Let $A$ be an open subset of $X$. Let $p$ be a point of $X$ such that $p \in A$. Then by definition of basis there exists $U \in \BB_1$ such that: :$p \in U \subseteq A$. By $(3)$, $U = \displaystyle \bigcup \map f U$. Then by definition of union there exists a set $D$ such that: :$p \in D \in \map f U$ By Set is Subset of Union: :$D \subseteq U$ By definition of union: :$D \in \BB_0$ Thus by Subset Relation is Transitive: :$\exists D \in \BB_0: p \in D \subseteq A$ This by definition of basis ends a proof of basis. By $(1)$ and Cardinality of Image of Mapping not greater than Cardinality of Domain: :$\card {\Img f} \le \map w T$ For every $U \in \BB_1$: :$\card {\map f U} \le \map w T$ Then by Cardinality of Union not greater than Product: :$\card {\BB_0} \le \card {\map w T \times \map w T}$ Thus by Cardinal Product Equal to Maximum, because $\map w T$ is infinite : :$\card {\BB_0} \le \map w T$. Thus by definition of weight: :$\card {\BB_0} = \map w T$. {{qed|lemma}} === Case when Weight is Finite === Let $T$ has finite weight. By Finite Weight Space has Basis equal Image of Mapping of Intersections, there exist a basis $\BB_0$ of $T$ and a mapping $f: X \to \tau$ such that: :$\BB_0 = \Img f)$ and: :$\forall x \in X: \paren {x \in \map f x \land \forall U \in \tau: x \in U \implies \map f x \subseteq U}$ Thus by Image of Mapping of Intersections is Smallest Basis: :$\BB_0 \subseteq \BB$ Thus by Cardinality of Image of Mapping of Intersections is not greater than Weight of Space: :$\card {\BB_0} \le \map w T$ Thus by definition of weight: :$\card {\BB_0} = \map w T$ {{qed}}
$\FF = \O$ or $\FF = \set \O$ or $\O \ne \FF \ne \set \O$. In case when $\FF = \O$ or $\FF = \set \O$: {{begin-eqn}} {{eqn | l = \size {\bigcup \FF}       | r = \size \O       | c = Union of Empty Set, Union of Singleton }} {{eqn | o = \le       | r = \size {\mathbf m \times \mathbf n}       | c = Subset implies Cardinal Inequality because Empty Set is Subset of All Sets }} {{end-eqn}} In case when $\O \ne \FF \ne \{\O\}$: :by Surjection iff Cardinal Inequality there exists a surjection $f: \mathbf m \to \FF$ as $\size {\mathbf m} = \mathbf m$ by Cardinal of Cardinal Equal to Cardinal. $\FF$ contains non empty set $A_0$. :$\size {A_0} > \mathbf 0$. By assumption: :$\size {A_0} \le \mathbf n$. Then: :$\mathbf 0 < \mathbf n$. Hence: :$\size {\set 0} = \mathbf 1 \le \mathbf n$. Define a family $\family {B_A}_{A \mathop \in \FF}$: :$B_A = \begin{cases} A & A \ne \O \\ \set 0 & A = \O \end {cases}$ By Surjection iff Cardinal Inequality define a family $\family {g_A}_{A \mathop \in \FF}$ of surjections: :$\forall A \in \FF: g_A: \mathbf n \to B_A$ is a surjection. Define a mapping $h:\mathbf m \times \mathbf n \to \ds \bigcup_{A \mathop \in \FF} B_A$: :$\forall \alpha \in \mathbf m: \forall \beta \in \mathbf n: \map h {\alpha, \beta} = \map {g_{\map f \alpha} } \beta$ We will show by definition that $h$ is a surjection. Let $x \in \ds \bigcup_{A \mathop \in \FF} B_A$. Then by definition of union: :$\exists A \in \FF: x \in B_A$. By definition of surjection: :$\exists \alpha \in \mathbf m: \map f \alpha = A$. By definition of surjection: :$\exists \beta \in \mathbf n: \map {g_A} \beta = x$ So: :$\map h {\alpha, \beta} = \map {g_{\map f \alpha} } \beta = x$ This ends the proof of surjection. Hence by Surjection iff Cardinal Inequality: :$\ds \size {\bigcup_{A \mathop \in \FF} B_A} \le \size {\mathbf m \times \mathbf n}$ By definition of subset: :$\forall A \in \FF: A \subseteq B_A$. Then by Set Union Preserves Subsets: :$\ds \bigcup \FF \subseteq \bigcup_{A \mathop \in \FF} B_A$. Hence by Subset implies Cardinal Inequality: :$\ds \size {\bigcup \FF} \le \ds \size {\bigcup_{A \mathop \in \FF} B_A}$ Thus the result: :$\size {\bigcup \FF} \le \size {\mathbf m \times \mathbf n}$ Thus by definition of product of cardinals: :$\size {\mathbf m \times \mathbf n} = \mathbf m \mathbf n$ {{qed}}
Let $\mathcal B$ be a basis. Let $V \in \operatorname{Im} \left({f}\right)$. Then by definition of image there exists a point $b \in X$ such that : $V = f \left({b}\right)$. Then $V$ is open because $\operatorname{Im} \left({f}\right) \subseteq \tau$. By assumption of mapping $f$: : $b \in V$. Then by definition of basis there exists a subset $U \in \mathcal B$ such that : $b \in U \subseteq V$. By definition of basis: : $\mathcal B \subseteq \tau$. Then by definition of subset: : $U \in \tau$. Then by assumption of mapping $f$: :$f \left({b}\right) \subseteq U$. Thus by definition of set equality: : $V = U \in \mathcal B$. {{qed}}
By definition of weight there exists a basis $\BB$ of $T$ such that. :$\card \BB = \map w T$ By Image of Mapping of Intersections is Smallest Basis: :$\Img f \subseteq \BB$ Thus by Subset implies Cardinal Inequality: :$\card {\Img f} \le \card \BB = \map w T$ {{qed}}
By definition of weight there exists a basis $\BB$ such that: :$\card \BB = \map w T$ where: :$\map w T$ denotes the weight of $T$ :$\card \BB$ denotes the cardinality of $\BB$. By assumption that weight is finite: :$\card \BB$ is finite Then by Cardinality of Set is Finite iff Set is Finite: :$\BB$ is finite Define a mapping $f: X \to \powerset X$: :$(1): \quad \forall x \in X: \map f x = \bigcap \set {U \in \BB: x \in U}$ By definition of subset: :$\forall x \in X: \set {U \in \BB: x \in U} \subseteq \BB$ By Subset of Finite Set is Finite: :$\forall x \in X: \set {U \in \BB: x \in U}$ is finite Then by General Intersection Property of Topological Space: :$\forall x \in X: \bigcap \set {U \in \BB: x \in U} \in \tau$ So: :$f: X \to \tau$ We will prove that: :$(2): \quad \forall x \in X: \paren {x \in \map f x \land \forall U \in \tau: x \in U \implies \map f x \subseteq U}$ Let $x \in X$. By $(1)$: :$\map f x = \bigcap \set {U \in \BB: x \in U}$ Thus by definition of intersection: :$x \in \map f x$ Let $U$ be an open set of $T$. Let $x \in U$. By definition of basis: :$\exists V \in \BB: x \in V \subseteq U$ Then: :$V \in \set {U \in \BB: x \in U}$ Hence by Intersection is Subset: :$\map f x \subseteq V$ Thus by Subset Relation is Transitive: :$\map f x \subseteq U$ This ends the proof of $(2)$. We will prove that $\Img f$ is a basis of $T$. By $f: X \to \tau$ and definition of image: :$\Img f \subseteq \tau$ Let $U$ be an open set of $T$. Let $x$ be a point $x \in X$ such that: :$x \in U$ By $(2)$: :$\map f x \in \Img f \land x \in \map f x \subseteq U$ By definition of basis this ends the proof of basis. Thus the result. {{qed}}
A cube, by definition, has $6$ faces. Each face is subdivided into $9$ facets. Hence there are $6 \times 9 = 54$ facets in total. {{qed}}
{{begin-eqn}} {{eqn | l = x \in S * T       | o = \iff       | r = x \in S \oplus x \in T       | c = Symmetric Difference: Definition 5 }} {{eqn | r = \left({\neg \left({x \in S}\right) \land \left({x \in T}\right)}\right) \lor \left({\left({x \in S}\right) \land \neg \left({x \in T}\right)}\right)       | o = \iff       | c = Non-Equivalence as Disjunction of Conjunctions }} {{eqn | r = \left({x \in \overline S \land x \in T}\right) \lor \left({x \in S \land x \in \overline T}\right)       | o = \iff       | c = {{Defof|Set Complement}} }} {{eqn | r = \left({x \in \overline S \cup T}\right) \lor \left({x \in S \cup \overline T}\right)       | o = \iff       | c = {{Defof|Set Intersection}} }} {{eqn | r = x \in \left({\overline S \cup T}\right) \cup \left({S \cup \overline T}\right)       | o = \iff       | c = {{Defof|Set Union}} }} {{eqn | r = x \in \left({S \cup \overline T}\right) \cup \left({\overline S \cup T}\right)       | o = \iff       | c = Union is Commutative }} {{end-eqn}} The result follows by definition of set equality.
{{begin-eqn}} {{eqn | l = x \in S * T       | o = \iff       | r = x \in S \oplus x \in T       | c = Symmetric Difference: Definition 5 }} {{eqn | r = \left({x \in S \lor x \in T} \right) \land \neg \left({x \in S \land x \in T}\right)       | o = \iff       | c = Definition of Exclusive Or }} {{eqn | r = \left({x \in S \cup T}\right) \land \left({x \notin S \cap T}\right)       | o = \iff       | c = Definition of Set Intersection and Set Union }} {{eqn | r = x \in \left({S \cup T}\right) \setminus \left({S \cap T}\right)       | o = \iff       | c = Definition of Set Difference }} {{end-eqn}} The result follows by definition of set equality.
:400px 400px In the left hand diagram, $R$ is depicted in yellow and $S \cup T$ is depicted in blue. Their intersection $R \cap \paren {S \cup T}$ where they overlap is depicted in green. In the right hand diagram, $\paren {R \cap S}$ is depicted in yellow and $\paren {R \cap T}$ is depicted in blue. Their intersection, where they overlap,  is depicted in green. Their union is the total shaded area: yellow, blue and green. As can be seen by inspection, the areas are the same.
:400px 400px In the left hand diagram, $R$ is depicted in blue and $S \cap T$ is depicted in yellow. Their intersection, where they overlap, is depicted in green. Their union $R \cup \paren {S \cap T}$ is the total shaded area: yellow, blue and green. In the right hand diagram, $\paren {R \cup S}$ is depicted in yellow and $\paren {R \cup T}$ is depicted in blue. Their intersection, where they overlap, is depicted in green. As can be seen by inspection, the areas are the same.
:400px $\overline T_1$ is depicted in yellow and $\overline T_2$ is depicted in red. Their intersection, where they overlap, is depicted in orange. Their union $\overline T_1 \cup \overline T_2$  is the total shaded area: yellow, red and orange. As can be seen by inspection, this also equals the complement of the intersection of $T_1$ and $T_2$.
:400px $\overline T_1$ is depicted in yellow and $\overline T_2$ is depicted in red. Their intersection, $\overline T_1 \cap \overline T_2$, is depicted in orange. As can be seen by inspection, this also equals the complement of the union of $T_1$ and $T_2$.