hcju/proofwikiqa huggingface dataset:
An integral domain is a commutative and unitary ring with no proper zero divisors. The result follows from Degree of Product of Polynomials over Ring: Corollary 1. {{qed}}
By Power Set is Complete Lattice, $\struct {\powerset S, \cap, \cup, \subseteq}$ is a complete lattice. Thus the theorem holds by the Knaster-Tarski Lemma. {{qed}} {{Namedfor|Bronisław Knaster|name2 = Alfred Tarski|cat = Knaster|cat2 = Tarski}}
By definition an integral domain is a commutative ring with unity.  From Ring of Polynomial Forms is Commutative Ring with Unity it follows that $\struct {D \sqbrk X, +, \circ}$ is a commutative ring with unity. Suppose $f, g \in D \sqbrk X$ such that neither $f$ nor $g$ are the null polynomial. Let $\map \deg f = n$ and $\map \deg g = m$. From Degree of Product of Polynomials over Integral Domain the degree of $f \odot g$ is $n + m$. Thus by definition $f \odot g$ is not the null polynomial of $D \sqbrk X$. Thus neither $f$ nor $g$ is a proper zero divisor of $D \sqbrk X$. This holds for any two arbitrary non-null polynomials elements of $D \sqbrk X$. Hence $\struct {D \sqbrk X, \oplus, \odot}$ is a commutative ring with unity with no proper zero divisors. That is, $\struct {D \sqbrk X, \oplus, \odot}$ is an integral domain. {{qed}}
Let $\mathcal Q = \mathcal R_1 \cup \mathcal R_2$. Then $Q \subseteq (S_1 \times T_1) \cup (S_2 \times T_2) \subseteq (S_1 \cup S_2) \times (T_1 \cup T_2)$. Thus $Q$ is a relation on $(S_1 \cup S_2) \times (T_1 \cup T_2)$. Let $T'_1$ and $T'_2$ be the images of $\mathcal R_1$ and $\mathcal R_2$, respectively. Let $(x_1, y), (x_2, y) \in Q$. Then $y \in T'_1$ or $y \in T'_2$. If $y \in T'_1$ then $y \notin T'_2$, so neither $(x_1, y)$ nor $(x_2, y)$ is in $\mathcal R_2$, so these pairs are both in $\mathcal R_1$. As $\mathcal R_1$ is one-to-many, $x_1 = x_2$. A similar argument leads to the same result for $y \in T'_2$. As this holds for all such $x_1, x_2, y$: $Q$ is a one-to-many relation. {{qed}} Category:Relation Theory l56ztdhi5bg9c7f5cwi8xib4au1autp
Let $\RR = \RR_1 \cup \RR_2$. Let $\tuple {x, y_1}, \tuple {x, y_2} \in \RR$. By the definition of union, $\tuple {x, y_1}$ and $\tuple {x, y_2}$ are each in $\RR_1$ or $\RR_2$. Suppose that both are in $\RR_1$. Then since $\RR_1$ is a many-to-one relation, $y_1 = y_2$. Suppose that $\tuple {x, y_1} \in \RR_1$ and $\tuple {x, y_2} \in \RR_2$. Then $x$ is in the domain of $\RR_1$ and that of $\RR_2$, contradicting the premise, so this cannot occur. The other two cases are precisely similar. Thus in all cases $y_1 = y_2$. As this holds for all such pairs, $\RR$ is many-to-one. {{qed}} Category:Relation Theory 93vaix66p8b4htn9iiddor48mxwxzmb
We prove the contrapositive: if the $n$th root of $x$ is rational, it must be an integer. By Existence of Canonical Form of Rational Number, there exist an integer $a$ and a natural number $b$ which are coprime such that: {{begin-eqn}} {{eqn | l = x^{1/n}       | r = \frac a b }} {{eqn | ll= \leadsto       | l = x       | r = \frac {a^n} {b^n} }} {{end-eqn}} Since $a$ and $b$ are coprime, $a^n$ and $b^n$ are coprime by Powers of Coprime Numbers are Coprime. Hence $\dfrac {a^n} {b^n}$ is by definition in canonical form. Suppose $b \ne 1$. As the denominator of $\dfrac {a^n} {b^n}$ is not $1$, $x = \dfrac {a^n} {b^n}$ is not an integer. This is a contradiction. Thus $b = 1$, and thus: :$x^{1/n} = a$ which is an integer. {{qed}}
By the definition of bijection, $f$ and $g$ are many-to-one and one-to-many relations. By Union of Many-to-One Relations with Disjoint Domains is Many-to-One and Union of One-to-Many Relations with Disjoint Images is One-to-Many: : $f \cup g$ is many-to-one and one-to-many. Thus to show $f \cup g$ is a bijection requires us only to demonstrate that it is both left-total and right-total. We will show that $f \cup g$ is left-total. Let $x \in A \cup B$. Then $x \in A$ or $x \in B$. If $x \in A$ then since $f$ is left-total there is a $y \in C$ such that $\left({x, y}\right) \in f$. By the definition of union, $\left({x, y}\right) \in f \cup g$. If $x \in B$ then since $g$ is left-total there is a $y \in D$ such that $\left({x, y}\right) \in g$. Then by the definition of union, $\left({x, y}\right) \in f \cup g$. As this holds for all $x$, $f \cup g$ is left-total. The proof that $f \cup g$ is right-total is similar. Thus it has been demonstrated that: : $f \cup g$ is many-to-one : $f \cup g$ is one-to-many : $f \cup g$ is left-total : $f \cup g$ is right-total and therefore, by definition, a bijection. {{qed}} Category:Mapping Theory b0uu79b9e81evrjh8lvkh3j7vtlt6ak
Let $I = \left[{a \,.\,.\, b}\right]$. Let $S \subseteq I$. If $S = \varnothing$, then it has a supremum in $I$ of $a$ and an infimum in $I$ of $b$. Let $S \ne \varnothing$. Since $S \subseteq I$, $a$ is a lower bound of $S$ and $b$ is an upper bound of $S$. Since $L$ is a complete lattice, $S$ has an infimum, $p$, and a supremum, $q$, in $L$. Thus by the definitions of infimum and supremum: : $a \preceq p$ and $q \preceq b$ Let $x \in S$. Since an infimum is a lower bound: : $p \preceq x$ Since a supremum is an upper bound: : $x \preceq q$ Thus $a \preceq p \preceq x \preceq q \preceq b$. Since $\preceq$ is an ordering, it is transitive, so by Transitive Chaining: :$a \preceq p \preceq b$ and $a \preceq q \preceq b$. That is, $p, q \in I$. Thus $p$ and $q$ are the infimum and supremum of $S$ in $I$. As every subset of $I$ has a supremum and infimum in $I$, $I$ is a complete lattice. {{qed}}
Let $S \subseteq L$. If $S = \varnothing$, then $S$ has a supremum of $\bot$ and an infimum of $\top$. Let $S \ne \varnothing$. $S$ is bounded above by $\top$. As $\left({L, \preceq}\right)$ is Dedekind complete, $S$ has a supremum. $S$ is bounded below by $\bot$. By Dedekind Completeness is Self-Dual, $S$ has an infimum. Thus every subset of $L$ has a supremum and an infimum. So, by definition, $\left({L, \preceq}\right)$ is a complete lattice. {{qed}} Category:Complete Lattices k3nlnv25ytz2hg1qp0qvvfaaszcffkv
Let $x \in S \setminus B$. Then by the definition of set difference: : $x \in S$ and $x \notin B$ Suppose for the sake of contradiction that $x \in A$. Then since $A$ is a subset (or subclass) of $B$, $x \in B$, a contradiction. Thus $x \notin A$. Since $x \in S$ and $x \notin A$, we conclude that $x \in S \setminus A$. As this holds for all $x \in S \setminus B$: : $S \setminus B \subseteq S \setminus A$ {{qed}} Category:Set Difference t7nq6c7ruyp6dorjy26rdcerzo37il9
Let $P = \left\{{x \in \mathcal P \left({S}\right): x \subseteq f \left({x}\right) }\right\}$. Let $\bigcup P$ be the union of $P$. Let $x \in P$. Then by Set is Subset of Union: General Result, $x \subseteq \bigcup P$. Since $f$ is increasing, $f \left({x}\right) \subseteq f \left({\bigcup P}\right)$. By the definition of $P$, $x \subseteq f \left({x}\right)$. Thus $x \subseteq f \left({\bigcup P}\right)$ by Subset Relation is Transitive: As this holds for all $x \in P$, $\bigcup P \subseteq f \left({\bigcup P}\right)$ by Union is Smallest Superset: General Result. Now since $f$ is increasing: :$f \left({\bigcup P}\right) \subseteq f \left({ f \left({\bigcup P}\right)}\right)$ Thus $f \left({\bigcup P}\right) \in P$ by the definition of $P$. Therefore by Set is Subset of Union: General Result: : $f \left({\bigcup P}\right) \subseteq \bigcup P$ We have that: : $\bigcup P \subseteq f \left({\bigcup P}\right)$ Thus by definition of set equality: :$f \left({\bigcup P}\right) = \bigcup P$ Thus $\bigcup P$ is a fixed point of $f$. {{qed}}
By the Knaster-Tarski Lemma, $f$ has a least fixed point. Thus it has a fixed point. {{qed}} Category:Complete Lattices 1umbgdu1usc8o7l7c7kujl3z3u5u7a7
By the Knaster-Tarski Lemma: Power Set, $f$ has a least fixed point. Thus it has a fixed point. {{qed}}
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connective match for all boolean interpretations. $\begin{array}{|ccc|c|ccc|} \hline (p & \lor & q) & \iff & (q & \lor & p) \\ \hline F & F & F & T & F & F & F \\ F & T & T & T & T & T & F \\ T & T & F & T & F & T & T \\ T & T & T & T & T & T & T \\ \hline \end{array}$ {{qed}}
{{handwaving}} Let $p$ a prime number such that $p$ divides $x^n$. This is possible because $x > 1$, so $x^n > 1$, hence $x^n$ has prime divisors due to Fundamental Theorem of Arithmetic. To prove the statement, we need to show $p$ divides $x$. We will prove this statement by the Principle of Mathematical Induction on $n$. === Basis of the Induction ===   We have $n = 1$ Clearly, since $p$ divides $x^1 = x$, then $p$ divides $x$. {{qed |lemma}} === Inductive Step === Suppose that for a given $n$, if $p$ divides $x^n$ then $p$ divides $x$.  Then, if $p$ divides $x^{n+1}$, by definition of prime number, either $p$ divides $x^n$ or $p$ divides $x$. If $p$ divides $x^n$, we get from induction hypothesis that $p$ divides $x$. The other case trivially leads to our conclusion. {{qed|lemma}} Hence the result, by Principle of Mathematical Induction. {{qed}} Conversely, let $p$ a prime number such that it divides $x$. Then, $p$ divides $x * x^{n-1} = x^n$, as required. {{qed}} Category:Natural Numbers crq4vrwlccilc19abse60k1dm99lvau
We apply the Method of Truth Tables. $\begin{array}{|ccc||c|} \hline p & \land & q & p \\ \hline F & F & F & F \\ F & F & T & F \\ T & F & F & T \\ T & T & T & T \\ \hline \end{array}$ As can be seen, when $p \land q$ is true so is $p$. {{qed}}
We apply the Method of Truth Tables. $\begin{array}{|ccc||c|} \hline p & \land & q & q \\ \hline F & F & F & F \\ F & F & T & T \\ T & F & F & F \\ T & T & T & T \\ \hline \end{array}$ As can be seen, when $p \land q$ is true so is $q$. {{qed}}
=== Forward Implication === Let $S$ be an ordinal. By Alternative Definition of Ordinal, $S$ is transitive and strictly well-ordered by the epsilon relation. By Strict Well-Ordering is Strict Total Ordering, $S$ is strictly totally ordered by $\in$. Thus: :$\forall x, y \in S: \left({ x \in y \lor x = y \lor y \in x }\right)$ {{qed|lemma}} === Reverse Implication === Let $S$ be a transitive set such that for any $x, y \in S$, $x \in y \lor y \in x \lor x = y$. We first show that $\in$ is a strict ordering of $S$. Asymmetric: Let $x, y \in S$. By Epsilon is Foundational, $\{ x,y \}$ has an $\Epsilon$-minimal element. Thus $x \notin t$ or $y \notin x$. Transitive: Let $x, y, z \in S$ with $x \in y$ and $y \in z$. By assumption, $x = z$, $x \in z$, or $z \in x$. Suppose for the sake of contradiction that $x = z$. Then $x \in y$ and $y \in x$, contradicting the fact that $\Epsilon$ is asymmetric. Suppose that $z \in x$. Then $x \in y$, $y \in z$, and $z \in x$. Thus the set $\left\{ {x, y, z}\right\}$ has no $\Epsilon$-minimal element, contradicting Epsilon is Foundational. Thus $x \in z$. Thus $\in$ is a strict ordering of $S$. Let $T$ be a non-empty subset of $S$. By Epsilon is Foundational, $S$ has an $\Epsilon$-minimal element, $m$. Since a minimal element of a strictly totally ordered set is the smallest element, $\Epsilon$ strictly well-orders $S$. {{LinkWanted|Is there a link to the above statement?}} Thus by Alternative Definition of Ordinal, $S$ is an ordinal. {{qed}}
Since $S$ is non-empty, it has an element $x$. By the definition of ordering, $\preceq$ is a reflexive relation. Thus $x \preceq x$. By Foundational Relation is Antireflexive, $\preceq$ is not a foundational relation. {{qed}} Category:Foundational Relations qrrdxd0b32u7n4vqpuyl10bmczd2tla
Let $T$ be a non-empty subset of $S$. Since $\preceq$ is a well-founded ordering, $T$ has a minimal element with respect to the ordering $\preceq$. That is, there is an element $m \in T$ such that $\forall x \in T: \left({x \npreceq m}\right) \lor \left({x = m}\right)$. Let $x \in T$. Then $x \npreceq m$ or $x = m$. By the definition of reflexive reduction, $\prec$ is a subset of $\preceq$. Thus if $x \npreceq m$, $x \nprec m$. If $x = m$ then by Reflexive Reduction is Antireflexive, $x \nprec m$. As this holds for all $x \in T$, $m$ is $\prec$-minimal in $T$. As each non-empty subset of $S$ has a $\prec$-minimal, $\prec$ is a foundational relation on $S$. {{qed}} Category:Reflexive Reductions Category:Order Theory Category:Foundational Relations 52y9rgrl5pcrhyplotzmqa4ol0yzujt
{{NotZFC}} Let $x \in \mathbb U$. Then by the Axiom of Extension: : $x = \Epsilon^{-1} \left({x}\right)$ where $\Epsilon^{-1} \left({x}\right)$ denotes the preimage of $x$ under $\Epsilon$. Since $x$ is a set, $\prec^{-1} \left({x}\right) = x$ is a set. As this holds for all $x \in \mathbb U$, $\left({\mathbb U, \Epsilon}\right)$ is a proper relational structure. {{qed}} Category:Class Theory 770yldin2t1ypcot2usopgmypzljk5b
We have that: :$x \in a$ and $a$ is transitive. So: :$x \subseteq a$ Thus by the definition of $b$ and of smallest set: :$b \subseteq a$ Since we also have $x \in a$: :$b \cup \set x \subseteq a$ $x \in \set x$, so: :$x \in b \cup \set x$ $b \cup \set x$ is transitive: If $p \in b$ then: :$p \subseteq b \subseteq b \cup \set x$. If $p \in \set x$ then: :$p = x$ So by the definition of $b$: :$p \subseteq b \subseteq b \cup \set x$ Thus by the definition of $a$: :$a \subseteq b \cup \set x$ Thus the theorem holds by definition of set equality. {{qed}} Category:Set Theory l7ziwup2v5r1puow7hgnuvd76gft23u
By Successor Set of Ordinal is Ordinal, the successor of $x$ is an ordinal. That is, $x^+ = x \cup \set x$ is an ordinal. By Set is Element of Successor, $x \in x^+$. Because $x^+$ is an ordinal, it is strictly well-ordered by the epsilon restriction $\Epsilon {\restriction_{x^+} }$. Because a strict ordering is antireflexive and $x \in x^+$, we conclude that $x \notin x$. {{qed}}
By the definition of successor set: : $x^+ = x \cup \{x\}$. By the definition of singleton, $x \in \{x\}$. Thus by the definition of union, $x \in x^+$. {{qed}} Category:Ordinals 7sif0tawwnckd2ghx1u58z7u5b04ugc
By the definition of ordinal, $n$ is transitive. Thus $m \subseteq n$. By Subset of Strictly Well-Ordered Set is Strictly Well-Ordered, it follows that $m$ is strictly well-ordered by the epsilon restriction $\Epsilon {\restriction_m}$. It is now to be shown that $m$ is transitive. If $m = \varnothing$ then the result follows by Empty Set is Transitive. If $m \ne \varnothing$, then let $x \in m$. If $x = \varnothing$, then $x \subseteq m$ by Empty Set is Subset of All Sets. If $x \ne \varnothing$, then let $y \in x$. It suffices to show that $y \in m$. Since $m \subseteq n$, it follows that $x \in n$. Also, $y \in x \land x \in n \implies y \in n$ because $n$ is transitive. And so $x \in n$, $y \in n$, and $m \in n$. A strict well-ordering is transitive by definition. Therefore: :$y \in x \land x \in m \implies y \in m$ Hence the result. {{qed}} Category:Ordinals 17fv3n075gb8vz4h5nkxs3ksn9aqhpm
{{begin-eqn}} {{eqn | l = e^{i x}       | r = \cos x + i \sin x       | c = Euler's Formula }} {{eqn | ll= \leadsto       | l = \cmod {e^{i x} }       | r = \cmod {\cos x + i \sin x} }} {{eqn | r = \sqrt {\paren {\map \Re {\cos x + i \sin x} }^2 + \paren {\map \Im {\cos x + i \sin x} }^2}       | c = {{Defof|Complex Modulus}} }} {{eqn | r = \sqrt {\cos^2 x + \sin^2 x}       | c = as $x$ is wholly real }} {{eqn | r = 1       | c = Sum of Squares of Sine and Cosine }} {{end-eqn}} {{qed}} Category:Complex Modulus Category:Modulus of Exponential of Imaginary Number is One 4e54obxfjkhpr31zu5htm7dromjp3mj
If $x = 0$, the theorem clearly holds, by the definition of powers of zero. Suppose $x \ne 0$. We use the interpretation of real numbers as wholly real complex numbers. Likewise we interpret the absolute value of $x$ as the modulus of $x$. Then $x$ can be expressed in polar form: :$x = r e^{i\theta}$ where $r = \size x$ and $\theta$ is an argument of $x$. Then: {{begin-eqn}} {{eqn | l = x       | r = r e^{i\theta} }} {{eqn | ll= \leadsto       | l = x^y       | r = \left(r{e^{i\theta} }\right)^y }} {{eqn | r = r^y e^{i \theta y} }} {{eqn | ll= \leadsto       | l = \size {x^y}       | r = \size {r^y e^{i \theta y} } }} {{eqn | r = \size {r^y} \size {e^{i \theta y} }       | c = Modulus of Product }} {{eqn | r = \size {r^y}       | c = Modulus of Exponential of Imaginary Number is One }} {{eqn | r = \size {\size x^y}       | c = by definition of $r$ }} {{eqn | r = \size x^y       | c = as $\size x^y \ge 0$ }} {{end-eqn}} {{qed}} {{MissingLinks|exponential properties for $\C$ and that $\size x^y \ge 0$}} Category:Absolute Value Function g0rl07jenlyqentmmvi3q5l1rpraizk
In a truth table, one row is needed for each boolean interpretation of $P$. Let $S$ be the set of different letters used in $P$. The result then follows from applying Number of Boolean Interpretations for Finite Set of Variables to $S$. {{qed}}
Let $x^t$ be the transitive closure of $x$ by Definition 2. Let the mapping $G$ be defined as on that definition page. === $x \in x^t$ === $x \in \set x$ by the definition of singleton. Since $\map G 0 = \set 0$: :$\set x \in \map G \N$ Thus $x \in x^t$ by the definition of union. {{qed|lemma}} === $x^t$ is a Set === By Denumerable Class is Set, the image of $G$ is a set. Thus $x^t$ is a set by the Axiom of Unions. {{qed|lemma}} === $x^t$ is a Transitive Set === Let $y \in x^t$ and let $z \in y$. By the definition of $x^t$: :$\exists n \in \N: y \in \map G n$ Then by definition of union: :$\displaystyle z \in \bigcup \map G n$ But by the definition of $G$: :$z \in \map G {n^+}$ Thus by the definition of $x^t$: :$z \in x^t$ As this holds for all such $y$ and $z$, $x^t$ is transitive. {{qed|lemma}} === $x^t$ is Smallest === Let $m$ be a transitive set such that $x \in m$. We will show by induction that $\map G n \subseteq m$ for each $n \in \N$. By Union is Smallest Superset, that will show that $x^t \subseteq m$. Because  $x \in m$: :$\map G 0 = \set x \subseteq m$ Suppose that $\map G n \subseteq m$. Then by Union is Increasing: :$\displaystyle \bigcup \map G n \subseteq \bigcup m$ {{explain|Transitive set includes its union.}} Thus: :$\displaystyle \bigcup \map G n \subseteq m$ {{qed|lemma}} By Smallest Element is Unique, $x^t$ is the only set satisfying $(2)$. {{qed}} Category:Set Theory iqtibc9o9fsqisxkwxh0qhrkjxcvxyn
By the Axiom of Infinity, $\N$ is a set. Thus by the Axiom of Replacement, $A$ is also a set. {{qed}} Category:Set Theory 5hmyb1p303uf7bgmuq7hv5xmn9nuw12
{{begin-eqn}} {{eqn | l = A       | o = \subseteq       | r = B       | c =  }} {{eqn | ll= \leadstoandfrom       | l = A \cap B       | r = A       | c = Intersection with Subset is Subset }} {{eqn | ll= \leadstoandfrom       | l = \relcomp S {A \cap B}       | r = \relcomp S A       | c = Relative Complement of Relative Complement }} {{eqn | ll= \leadstoandfrom       | l = \relcomp S A \cup \relcomp S B       | r = \relcomp S A       | c = De Morgan's Laws: Complement of Intersection }} {{eqn | ll= \leadstoandfrom       | l = \relcomp S B       | o = \subseteq       | r = \relcomp S A       | c = Union with Superset is Superset }} {{end-eqn}} {{qed}}
Suppose that $\forall i \in I: S_i \subseteq X$. Consider any $\displaystyle x \in \bigcup_{i \mathop \in I} S_i$. By definition of set union: :$\exists i \in I: x \in S_i$ But as $S_i \subseteq X$ it follows that $x \in X$. Thus it follows that: :$\displaystyle \bigcup_{i \mathop \in I} S_i \subseteq X$ So: :$\displaystyle \paren {\forall i \in I: S_i \subseteq X} \implies \bigcup_{i \mathop \in I} S_i \subseteq X$ {{qed}}
Let $\displaystyle s \in \bigcup \left({A \cup B}\right)$. Then by definition of union of set of sets: :$\exists X \in A \cup B: s \in X$ By definition of set union, either: :$X \in A$ or: :$X \in B$ If $X \in A$, then: :$s \in \left\{{x: \exists X \in A: x \in X}\right\}$ If $X \in B$, then: :$s \in \left\{{x: \exists X \in B: x \in X}\right\}$ Thus by definition of union of set of sets, either: :$\displaystyle s \in \bigcup A$ or: :$\displaystyle s \in \bigcup B$ So by definition of set union: :$\displaystyle s \in \left({\bigcup A}\right) \cup \left({\bigcup B}\right)$ So by definition of subset: :$\displaystyle \bigcup \left({A \cup B}\right) \subseteq \left({\bigcup A}\right) \cup \left({\bigcup B}\right)$ Now let $\displaystyle s \in \left({\bigcup A}\right) \cup \left({\bigcup B}\right)$. By definition of set union, either: :$\displaystyle s \in \bigcup A$ or: :$\displaystyle s \in \bigcup B$ That is, by definition of union of set of sets, either: :$s \in \left\{{x: \exists X \in A: x \in X}\right\}$ or: :$s \in \left\{{x: \exists X \in B: x \in X}\right\}$ {{WLOG}}, let $s \in X$ such that $X \in A$. Then by Set is Subset of Union: :$s \in X$ such that $X \in A \cup B$ That is: :$\displaystyle s \in \bigcup \left({A \cup B}\right)$ Similarly if $x \in X$ such that $X \in B$. So by definition of subset: :$\displaystyle \left({\bigcup A}\right) \cup \left({\bigcup B}\right) \subseteq \bigcup \left({A \cup B}\right)$ Hence by definition of equality of sets: :$\displaystyle \bigcup \left({A \cup B}\right) = \left({\bigcup A}\right) \cup \left({\bigcup B}\right)$ {{qed}} Category:Union Distributes over Union dfrc8pf2hupl9cy8eu3n6tg24uz9uvl
{{begin-eqn}} {{eqn | l = x       | o = \in       | r = \map {\bigcup_{\alpha \mathop \in I} } {A_\alpha \cup B_\alpha}       | c =  }} {{eqn | lll=\leadsto       | ll= \exists \beta \in I:       | l = x       | o = \in       | r = A_\beta \cup B_\beta       | c = {{Defof|Union of Family}} }} {{eqn | lll=\leadsto       | l = x       | o = \in       | r = A_\beta       | c = {{Defof|Set Union}} }} {{eqn | lo= \lor       | l = x       | o = \in       | r = B_\beta       | c =  }} {{eqn | lll=\leadsto       | l = x       | o = \in       | r = \bigcup_{\alpha \mathop \in I} A_\alpha       | c = Set is Subset of Union }} {{eqn | lo= \lor       | l = x       | o = \in       | r = \bigcup_{\alpha \mathop \in I} B_\alpha       | c = Set is Subset of Union }} {{eqn | lll=\leadsto       | l = x       | o = \in       | r = \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cup \paren {\bigcup_{\alpha \mathop \in I} B_\alpha}       | c = {{Defof|Set Union}} }} {{end-eqn}} Thus by definition of subset: :$\displaystyle \map {\bigcup_{\alpha \mathop \in I} } {A_\alpha \cup B_\alpha} \subseteq \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cup \paren {\bigcup_{\alpha \mathop \in I} B_\alpha}$ {{qed|lemma}} {{begin-eqn}} {{eqn | l = x       | o = \in       | r = \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cup \paren {\bigcup_{\alpha \mathop \in I} B_\alpha}       | c =  }} {{eqn | lll=\leadsto       | l = x       | o = \in       | r = \bigcup_{\alpha \mathop \in I} A_\alpha       | c = {{Defof|Set Union}} }} {{eqn | lo= \lor       | l = x       | o = \in       | r = \bigcup_{\alpha \mathop \in I} B_\alpha       | c =  }} {{eqn | lll=\leadsto       | ll= \exists \beta \in I:       | l = x       | o = \in       | r = A_\beta       | c = {{Defof|Union of Family}} }} {{eqn | lo= \lor       | ll= \exists \beta \in I:       | l = x       | o = \in       | r = B_\beta       | c =  }} {{eqn | lll=\leadsto       | ll= \exists \beta \in I:       | l = x       | o = \in       | r = A_\beta \cup B_\beta       | c = {{Defof|Union of Family}} }} {{eqn | lll=\leadsto       | l = x       | o = \in       | r = \map {\bigcup_{\alpha \mathop \in I} } {A_\alpha \cup B_\alpha}       | c =  }} {{end-eqn}} Thus by definition of subset: :$\displaystyle \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cup \paren {\bigcup_{\alpha \mathop \in I} B_\alpha} \subseteq \map {\bigcup_{\alpha \mathop \in I} } {A_\alpha \cup B_\alpha}$ {{qed|lemma}} By definition of set equality: :$\displaystyle \map {\bigcup_{\alpha \mathop \in I} } {A_\alpha \cup B_\alpha} = \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cup \paren {\bigcup_{\alpha \mathop \in I} B_\alpha}$ {{qed}}