hcju/proofwikiqa huggingface dataset:
From Set Difference with Superset is Empty Set we have: : $S \subseteq T \iff S \setminus T = \varnothing$ : $T \subseteq S \iff T \setminus S = \varnothing$ The result follows from definition of set equality: :$S = T \iff \left({S \subseteq T}\right) \land \left({T \subseteq S}\right)$ {{qed}}
Suppose $\map {b_n} x \ge \map {b_{n + 1} } x$ for each $x \in D$. All we need to show is that: :$\displaystyle \sum_{n \mathop = 1}^\infty \size {\map {b_n} x - \map {b_{n + 1} } x}$ converges uniformly on $D$. {{explain|Why is the above sufficient? On the surface of it, no reference has been made to $\map {a_n} x$. Its connection to the above statement needs to be clarified.}} To do this we show that the Cauchy criterion holds. Assign $\epsilon < 0$. Then by definition of uniform convergence: :$\displaystyle \exists N \in \N: \forall x \in D: \forall n \ge N: \size {\map {b_n} x} < \frac \epsilon 2$ Let $x \in D$ and $n > m \ge N$. Then: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = m + 1}^n \size {\map {b_k} x - \map {b_{k + 1} } x}       | r = \sum_{k \mathop = m + 1}^n \paren {\map {b_k} x - \map {b_{k + 1} } x} }} {{eqn | r = \map {b_{m + 1} } x - \map {b_{n + 1} } x }} {{eqn | r = \size {\map {b_{m + 1} } x - \map {b_{n + 1} } x} }} {{eqn | o = \le       | r = \size {\map {b_{m + 1} } x + \map {b_{n + 1} } x} }} {{eqn | o = <       | r = \frac \epsilon 2 + \frac \epsilon 2 }} {{eqn | r = \epsilon }} {{end-eqn}} {{qed}} {{Namedfor|Johann Peter Gustav Lejeune Dirichlet|cat = Dirichlet}}
A mapping is a relation. Thus it follows that Inverse of Inverse Relation can be applied directly. {{qed}}
Let $f: S \to T$ be a bijection. From Composite of Bijection with Inverse is Identity Mapping we have: :$f^{-1} \circ f = I_S$ :$f \circ f^{-1} = I_T$ where $I_S$ and $I_T$ are the identity mappings on $S$ and $T$ respectively. The result follows from Left and Right Inverses of Mapping are Inverse Mapping. {{qed}}
{{begin-eqn}} {{eqn | l = \relcomp S \O       | r = S \setminus \O       | c = {{Defof|Relative Complement}} }} {{eqn | r = S       | c = Set Difference with Empty Set is Self }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \relcomp S S       | r = S \setminus S       | c = {{Defof|Relative Complement}} }} {{eqn | r = \O       | c = Set Difference with Self is Empty Set }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \relcomp S {\relcomp S T}       | r = S \setminus \paren {S \setminus T}       | c = {{Defof|Relative Complement}} }} {{eqn | r = S \cap T       | c = Set Difference with Set Difference }} {{end-eqn}} The definition of the relative complement requires that: :$T \subseteq S$ But from Intersection with Subset is Subset: :$T \subseteq S \iff T \cap S = T$ Thus: :$\relcomp S {\relcomp S T} = T$ follows directly. {{qed}}
By the definition of relative complement: :$\complement_S \left({\complement_S \left({T}\right)}\right) = S \setminus \left({S \setminus T}\right)$ Let $t \in T$. Then by the definition of set difference: :$t \notin S \setminus T$ Since $T \in T$ and $T \subseteq S$, by the definition of subset: :$t \in S$ Thus: :$t \in \left({S \setminus \left({S \setminus T}\right)}\right)$ Suppose instead that: :$t \in \left({S \setminus \left({S \setminus T}\right)}\right)$ Then: :$t \in S$ and: :$\neg \left({t \in \left({S \setminus T}\right)}\right)$ Thus: :$\neg \left({\left({t \in S}\right) \land \neg \left({t \in T}\right)}\right)$ By Implication Equivalent to Negation of Conjunction with Negative: :$t \in S \implies t \in T$ By Modus Ponendo Ponens: :$t \in T$ {{qed}}
{{begin-eqn}} {{eqn | l = T \cap \relcomp S T       | r = \paren {S \setminus T} \cap T       | c = {{Defof|Relative Complement}} }} {{eqn | r = \O       | c = Set Difference Intersection with Second Set is Empty Set }} {{end-eqn}} {{qed}}
=== Step 1 === By the definition of relative complement, we have that: :$\relcomp S T \subseteq S$ and: :$T \subseteq S$ Hence by Union is Smallest Superset: :$\relcomp S T \cup T\subseteq S$ {{qed|lemma}} === Step 2 === Let $x \in S$. By Law of Excluded Middle, one of the following two applies: :$(1): \quad x \in T$ :$(2): \quad x \notin T$ If $(2)$, then by definition of relative complement: :$x \in S \setminus T = \relcomp S T$ So: :$x \in T \lor x \in \relcomp S T$ By definition of set union: :$x \in \relcomp S T \cup T$ Thus: :$x \in S \implies x \in \relcomp S T \cup T$ By definition of subset it follows that: :$S \subseteq \relcomp S T \cup T$ {{qed|lemma}} === Step 3 === From: :$\relcomp S T \cup T\subseteq S$ and: :$S \subseteq \relcomp S T \cup T$ it follows from definition of set equality that: :$S = \relcomp S T \cup T$ {{qed}} {{LEM}}
{{begin-eqn}} {{eqn | l = A \setminus B       | r = \set {x: x \in A \land x \notin B}       | c = {{Defof|Set Difference}} }} {{eqn | r = \set {x: \paren {x \in A \land x \in X} \land x \notin B}       | c = {{Defof|Subset}}, Modus Ponens and Rule of Conjunction }} {{eqn | r = \set {x: x \in A \land \paren {x \in X \land x \notin B} }       | c = Conjunction is Associative }} {{eqn | r = \set {x: x \in A \land x \in \relcomp S B}       | c = {{Defof|Relative Complement}} }} {{eqn | r = A \cap \relcomp S B       | c = {{Defof|Set Intersection}} }} {{end-eqn}} {{qed}}
Suppose we define a group $G$ in the usual way, but make the first pair of axiom replacements listed above: : the existence of a left identity : every element has a left inverse. Let $e \in G$ be a left identity and $g \in G$. Then, from Left Inverse for All is Right Inverse, each left inverse is also a right inverse with respect to the left identity. Also from Left Identity while exists Left Inverse for All is Identity we have that the left identity is also a right identity. Also we have that such an Identity is Unique, so this element can rightly be called '''the''' identity. So we have that: : $G$ has an identity; : each element of $G$ has an element that is both a left inverse and a right inverse with respect to this identity. Therefore, the validity of the two axiom replacements is proved. {{qed}} The proof of the alternate pair of replacements (existence of a right identity and closure under taking right inverses) is similar.
We shall prove that this is true for the first equation: {{begin-eqn}} {{eqn | l = a \circ g       | r = b }} {{eqn | ll= \leadstoandfrom       | l = a^{-1} \circ \paren {a \circ g}       | r = a^{-1} \circ b       | c = $\circ$ is a Cancellable Binary Operation }} {{eqn | ll= \leadstoandfrom       | l = \paren {a^{-1} \circ a} \circ g       | r = a^{-1} \circ b       | c = {{GroupAxiom|1}} }} {{eqn | ll= \leadstoandfrom       | l = e \circ g       | r = a^{-1} \circ b       | c = {{GroupAxiom|3}} }} {{eqn | ll= \leadstoandfrom       | l = g       | r = a^{-1} \circ b       | c = {{GroupAxiom|2}} }} {{end-eqn}} Because the statements: :$a \circ g = b$ and :$g = a^{-1} \circ b$ are equivalent, we may conclude that $g$ is indeed the only solution of the equation. The proof that the unique solution of $h$ is $b \circ a^{-1}$ in the second equation follows similarly. {{qed}}
Suppose that $\exists x, y \in G: a \circ x = b = a \circ y$. {{begin-eqn}} {{eqn | l = a \circ x       | r = a \circ y       | c =  }} {{eqn | ll= \leadsto       | l = a^{-1} \circ \paren {a \circ x}       | r = a^{-1} \circ \paren {a \circ y}       | c = {{GroupAxiom|3}} }} {{eqn | ll= \leadsto       | l = \paren {a^{-1} \circ a} \circ x       | r = \paren {a^{-1} \circ a} \circ y       | c = {{GroupAxiom|1}} }} {{eqn | ll= \leadsto       | l = e \circ x       | r = e \circ y       | c = {{GroupAxiom|3}} }} {{eqn | ll= \leadsto       | l = x       | r = y       | c = {{GroupAxiom|2}} }} {{end-eqn}} So such an element, if it exists, is unique. Now it is demonstrated that $g = a^{-1} b$ satisfies the requirement for $a \circ g = b$ Since $a \in G$, it follows by group axiom $G3$: existence of inverses that $a^{-1} \in G$. {{begin-eqn}} {{eqn | l = a       | o = \in       | r = G }} {{eqn | ll= \leadsto       | l = a^{-1}       | o = \in       | r = G       | c = {{GroupAxiom|3}} }} {{eqn | ll= \leadsto       | l = a^{-1} \circ b       | o = \in       | r = G       | c = {{GroupAxiom|0}} }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = a \circ g       | r = a \circ \paren {a^{-1} \circ b} }} {{eqn | r = \paren {a \circ a^{-1} } \circ b       | c = {{GroupAxiom|1}} }} {{eqn | r = e \circ b       | c = {{GroupAxiom|3}} }} {{eqn | r = b       | c = {{GroupAxiom|2}} }} {{end-eqn}} Thus, such a $g$ exists. The properties of $h$ are proved similarly. {{qed}}
{{begin-eqn}} {{eqn | l = S * T       | r = \paren {S \setminus T} \cup \paren {T \setminus S}       | c = {{Defof|Symmetric Difference}} }} {{eqn | r = \paren {T \setminus S} \cup \paren {S \setminus T}       | c = Union is Commutative }} {{eqn | r = T * S       | c = {{Defof|Symmetric Difference}} }} {{end-eqn}} {{Qed}}
=== $(1)$ iff $(2)$ === {{:Equivalence of Definitions of Symmetric Difference/(1) iff (2)}} {{qed|lemma}} === $(1)$ iff $(3)$ === {{:Equivalence of Definitions of Symmetric Difference/(1) iff (3)}} {{qed|lemma}} === $(2)$ iff $(4)$ === {{:Equivalence of Definitions of Symmetric Difference/(2) iff (4)}} {{qed|lemma}} === $(2)$ iff $(5)$ === {{:Equivalence of Definitions of Symmetric Difference/(2) iff (5)}} {{qed|lemma}} === $(3)$ iff $(5)$ === {{:Equivalence of Definitions of Symmetric Difference/(3) iff (5)}} {{qed}} Category:Symmetric Difference Category:Set Difference Category:Set Intersection Category:Set Union Category:Equivalence of Definitions of Symmetric Difference o4wdikguwfpiw83dtlgtwpjxn0au5v2
{{begin-eqn}} {{eqn | l = S = T       | o = \leadstoandfrom       | r = S \subseteq T \land T \subseteq S       | c = {{Defof|Set Equality}} }} {{eqn | o = \leadstoandfrom       | r = \paren {S \setminus T = \O} \land \paren {T \setminus S = \O}       | c = Set Difference with Superset is Empty Set }} {{eqn | o = \leadstoandfrom       | r = \paren {S \setminus T} \cup \paren {T \setminus S} = \O       | c = Union is Empty iff Sets are Empty }} {{eqn | o = \leadstoandfrom       | r = S * T = \O       | c = {{Defof|Symmetric Difference}} }} {{end-eqn}} {{Qed}}
{{begin-eqn}} {{eqn | l=S \cup T = \varnothing       | o=\iff       | r=\neg \exists x: x \in \left ({S \cup T}\right)       | c=Definition of Empty Set }} {{eqn | o=\iff       | r=\forall x: \neg \left ({x \in \left ({S \cup T}\right)}\right)       | c=De Morgan's Laws (Predicate Logic) }} {{eqn | o=\iff       | r=\forall x: \neg \left ({x \in S \lor x \in T}\right)       | c=Definition of Set Union }} {{eqn | o=\iff       | r=\forall x: x \notin S \land x \notin T       | c=De Morgan's Laws: Conjunction of Negations }} {{eqn | o=\iff       | r=S = \varnothing \land T = \varnothing       | c=Definition of Empty Set }} {{end-eqn}} {{qed}}
Let $S \cup T = \varnothing$. We have: {{begin-eqn}} {{eqn | l=S       | o=\subseteq       | r=S \cup T       | c=Set is Subset of Union }} {{eqn | ll=\implies       | l=S       | o=\subseteq       | r=\varnothing       | c=by hypothesis }} {{end-eqn}} From Empty Set is Subset of All Sets: : $\varnothing \subseteq S$ So it follows by definition of set equality that $S = \varnothing$. Similarly for $T$. {{qed}}
{{begin-eqn}} {{eqn | l = S * \O       | r = \paren {S \cup \O} \setminus \paren {S \cap \O}       | c = {{Defof|Symmetric Difference|index = 2}} }} {{eqn | r = S \setminus \paren {S \cap \O}       | c = Union with Empty Set }} {{eqn | r = S \setminus \O       | c = Intersection with Empty Set }} {{eqn | r = S       | c = Set Difference with Empty Set is Self }} {{end-eqn}} {{qed}}
From Set Intersection Distributes over Set Difference, we have: :$\paren {R \setminus S} \cap T = \paren {R \cap T} \setminus \paren {S \cap T}$ So: {{begin-eqn}} {{eqn | l = \paren {R \cap T} * \paren {S \cap T}       | r = \paren {\paren {R \cap T} \setminus \paren {S \cap T} } \cup \paren {\paren {S \cap T} \setminus \paren {R \cap T} }       | c = {{Defof|Symmetric Difference}} }} {{eqn | r = \paren {\paren {R \setminus S} \cap T} \cup \paren {\paren {S \setminus R} \cap T}       | c = Set Intersection Distributes over Set Difference }} {{eqn | r = \paren {\paren {R \setminus S} \cup \paren {S \setminus R} } \cap T       | c = Intersection Distributes over Union }} {{eqn | r = \paren {R * S} \cap T       | c = {{Defof|Symmetric Difference}} }} {{end-eqn}} The second part of the proof is a direct consequence of the fact that Intersection is Commutative. {{qed}}
{{begin-eqn}} {{eqn | l = \paren {R \cup T} * \paren {S \cup T}       | r = \paren {\paren {R \cup T} \setminus \paren {S \cup T} } \cup \paren {\paren {S \cup T} \setminus \paren {R \cup T} }       | c = {{Defof|Symmetric Difference|index = 1}} }} {{eqn | r = \paren {\paren {R \setminus S} \setminus T} \cup \paren {\paren {S \setminus R} \setminus T}       | c = Set Difference with Union }} {{eqn | r = \paren {\paren {R \setminus S} \cup \paren {S \setminus R} } \setminus T       | c = Set Difference is Right Distributive over Union }} {{eqn | r = \paren {R * S} \setminus T       | c = {{Defof|Symmetric Difference|index = 1}} }} {{end-eqn}} {{Qed}} Category:Symmetric Difference Category:Set Union 2awpfjbdrubswknh7xks2pus30vn6r2
{{begin-eqn}} {{eqn | l = S       | o = \subseteq       | r = \mathbb U       | c = {{Defof|Universe (Set Theory)}} }} {{eqn | ll= \leadstoandfrom       | l = \mathbb U \cap S       | r = S       | c = Intersection with Subset is Subset }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = S       | o = \subseteq       | r = \mathbb U       | c = {{Defof|Universe (Set Theory)}} }} {{eqn | ll= \leadstoandfrom       | l = \mathbb U \cup S       | r = \mathbb U       | c = Union with Superset is Superset }} {{end-eqn}} {{qed}}
Substitute $\mathbb U$ for $S$ in $\relcomp S \O = S$ from Relative Complement of Empty Set. {{qed}}
Substitute $\mathbb U$ for $S$ in $\relcomp S S = \O$ from Relative Complement with Self is Empty Set. {{Qed}}
Substitute $\mathbb U$ for $S$ and $S$ for $T$ in $\relcomp S {\relcomp S T} = T$ from Relative Complement of Relative Complement. {{qed}}
Substitute $\mathbb U$ for $S$ and $S$ for $T$ in $T \cap \relcomp S T = \O$ from Intersection with Relative Complement is Empty. {{qed}}
Substitute $\mathbb U$ for $S$ and $S$ for $T$ in $T \cup \relcomp S T = S$ from Union with Relative Complement. {{qed}} {{LEM|Union with Relative Complement}}
Follows directly from Set with Relative Complement forms Partition: If $\varnothing \subset T \subset S$, then $\left\{{T, \complement_S \left({T}\right)}\right\}$ is a partition of $S$. {{Qed}} Category:Set Complement Category:Universe osay7sq601e55jilomfupzcdydrwx3i
This follows directly from Set Difference as Intersection with Relative Complement: :$A \setminus B = A \cap \relcomp S B$ Let $S = \Bbb U$. Since $A, B \subseteq \Bbb U$ by definition of the universe, the result follows. {{qed}}
{{begin-eqn}} {{eqn | l=\complement \left({S}\right) \setminus \complement \left({T}\right)       | r=\left\{ {x: x \in \complement \left({S}\right) \land x \notin \complement \left({T}\right)}\right\}       | c=Definition of Set Difference }} {{eqn | r=\left\{ {x: x \notin S \land x \in T}\right\}       | c=Definition of Complement }} {{eqn | r=\left\{ {x: x \in T \land x \notin S}\right\}       | c=Rule of Commutation }} {{eqn | r=T \setminus S       | c=Definition of Set Difference }} {{end-eqn}} {{qed}} Category:Set Difference Category:Set Complement o7i9en0eps0ze8bfgdd7olycgzkpgca
{{begin-eqn}} {{eqn | l = \map \complement S * \map \complement T       | r = \paren {\map \complement S \setminus \map \complement T} \cup \paren {\map \complement T \setminus \map \complement S}       | c = {{Defof|Symmetric Difference}} }} {{eqn | r = \paren {T \setminus S} \cup \paren {S \setminus T}       | c = Set Difference of Complements }} {{eqn | r = S * T       | c = {{Defof|Symmetric Difference}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \mathbb U * S       | r = \mathbb U \cup S \setminus \mathbb U \cap S       | c = {{Defof|Symmetric Difference|index = 2}} }} {{eqn | r = \mathbb U \cup S \setminus S       | c = Intersection with Universe }} {{eqn | r = \mathbb U \setminus S       | c = Union with Universe }} {{eqn | r = \map \complement S       | c = {{Defof|Set Complement}} }} {{end-eqn}} {{qed}}
This follows directly from Symmetric Difference of Equal Sets: :$S * T = \O \iff S = T$ substituting $S$ for $T$. {{Qed}}
{{begin-eqn}} {{eqn | l = S * \relcomp {} S       | r = \paren {S \cup \relcomp {} S} \setminus \paren {S \cap \relcomp {} S}       | c = {{Defof|Symmetric Difference|index = 2}} }} {{eqn | r = \paren {S \cup \relcomp {} S} \setminus \O       | c = Intersection with Complement }} {{eqn | r = \mathbb U \setminus \O       | c = Union with Complement }} {{eqn | r = \mathbb U       | c = Set Difference with Empty Set is Self }} {{end-eqn}} {{qed}}
We can directly expand the expressions for $R * \paren {S * T}$ and $\paren {R * S} * T$, and see that they come to the same thing. Expanding the {{RHS}}: {{begin-eqn}} {{eqn | o =        | r = \paren {R * S} * T       | c =  }} {{eqn | r = \paren {\paren {\paren {R \cap \overline S} \cup \paren {\overline R \cap S} } \cap \overline T} \cup \paren {\overline {R * S} \cap T}       | c = {{Defof|Symmetric Difference|index = 3}} }} {{eqn | r = \paren {\paren {\paren {R \cap \overline S} \cup \paren {\overline R \cap S} } \cap \overline T} \cup \paren {\overline {\paren {\paren {R \cup S} \cap \paren {\overline R \cup \overline S} } } \cap T}       | c = {{Defof|Symmetric Difference|index = 4}} }} {{eqn | r = \paren {R \cap \overline S \cap \overline T} \cup \paren {\overline R \cap S \cap \overline T} \cup \paren {\paren {\overline {\paren {R \cup S} } \cup \overline {\paren {\overline R \cup \overline S} } } \cap T}       | c = Intersection Distributes over Union and De Morgan }} {{eqn | r = \paren {R \cap \overline S \cap \overline T} \cup \paren {\overline R \cap S \cap \overline T} \cup \paren {\paren {\paren {\overline R \cap \overline S} \cup \paren {R \cap S} } \cap T}       | c = De Morgan }} {{eqn | r = \paren {R \cap \overline S \cap \overline T} \cup \paren {\overline R \cap S \cap \overline T} \cup \paren {\overline R \cap \overline S \cap T} \cup \paren {R \cap S \cap T}       | c = Intersection Distributes over Union }} {{end-eqn}} Expanding the {{LHS}}: {{begin-eqn}} {{eqn | o =        | r = R * \paren {S * T}       | c =  }} {{eqn | r = \paren {R \cap \overline {S * T} } \cup \paren {\overline R \cap \paren {\paren {S \cap \overline T} \cup \paren {\overline S \cap T} } }       | c = {{Defof|Symmetric Difference|index = 3}} }} {{eqn | r = \paren {R \cap \overline {\paren {\paren {S \cup T} \cap \paren {\overline S \cup \overline T} } } } \cup \paren {\overline R \cap \paren {\paren {S \cap \overline T} \cup \paren {\overline S \cap T} } }       | c = {{Defof|Symmetric Difference|index = 4}} }} {{eqn | r = \paren {R \cap \paren {\overline {\paren {S \cup T} } \cup \overline {\paren {\overline S \cup \overline T} } } } \cup \paren {\overline R \cap S \cap \overline T} \cup \paren {\overline R \cap \overline S \cap T}       | c = Intersection Distributes over Union and De Morgan }} {{eqn | r = \paren {R \cap \paren {\paren {\overline S \cap \overline T} \cup \paren {S \cap T} } } \cup \paren {\overline R \cap S \cap \overline T} \cup \paren {\overline R \cap \overline S \cap T}       | c = De Morgan }} {{eqn | r = \paren {R \cap \overline S \cap \overline T} \cup \paren {R \cap S \cap T} \cup \paren {\overline R \cap S \cap \overline T} \cup \paren {\overline R \cap \overline S \cap T}       | c = Intersection Distributes over Union }} {{end-eqn}} From Union is Commutative it is seen that the {{LHS}} and {{RHS}} are the same, and the result is proved. {{qed}}
Expanding the {{RHS}}: {{begin-eqn}} {{eqn | o =        | r = \paren {R * S} * T       | c =  }} {{eqn | r = \paren {\paren {\paren {R \cup S} \cap \paren {\overline R \cup \overline S} } \cup T} \cap \paren {\overline {R * S} \cup \overline T}       | c = {{Defof|Symmetric Difference|index = 4}} }} {{eqn | r = \paren {\paren {\paren {R \cup S} \cap \paren {\overline R \cup \overline S} } \cup T} \cap \paren {\overline {\paren {\paren {R \cap \overline S} \cup \paren {\overline R \cap S} } } \cup \overline T}       | c = {{Defof|Symmetric Difference|index = 3}} }} {{eqn | r = \paren {R \cup S \cup T} \cap \paren {\overline R \cup \overline S \cup T} \cap \paren {\paren {\overline {\paren {R \cap \overline S} } \cap \overline {\paren {\overline R \cap S} } } \cup \overline T}       | c = Union Distributes over Intersection and De Morgan }} {{eqn | r = \paren {R \cup S \cup T} \cap \paren {\overline R \cup \overline S \cup T} \cap \paren {\paren {\paren {\overline R \cup S} \cap \paren {R \cup \overline S} } \cup \overline T}       | c = De Morgan }} {{eqn | r = \paren {R \cup S \cup T} \cap \paren {\overline R \cup \overline S \cup T} \cap \paren {\overline R \cup S \cup \overline T} \cap \paren {R \cup \overline S \cup \overline T}       | c = Union Distributes over Intersection }} {{end-eqn}} Expanding the {{LHS}}: {{begin-eqn}} {{eqn | o =        | r = R * \paren {S * T}       | c =  }} {{eqn | r = \paren {R \cup \paren {\paren {S \cup T} \cap \paren {\overline S \cup \overline T} } } \cap \paren {\overline R \cup \overline {S * T} }       | c = {{Defof|Symmetric Difference|index = 4}} }} {{eqn | r = \paren {R \cup \paren {\paren {S \cup T} \cap \paren {\overline S \cup \overline T} } } \cap \paren {\overline R \cup \overline {\paren {\paren {S \cap \overline T} \cup \paren {\overline S \cap T} } } }       | c = {{Defof|Symmetric Difference|index = 3}} }} {{eqn | r = \paren {R \cup S \cup T} \cap \paren {R \cup \overline S \cup \overline T} \cap \paren {\overline R \cup \paren {\overline {\paren {S \cap \overline T} } \cap \overline {\paren {\overline S \cap T} } } }       | c = Union Distributes over Intersection and De Morgan }} {{eqn | r = \paren {R \cup S \cup T} \cap \paren {R \cup \overline S \cup \overline T} \cap \paren {\overline R \cup \paren {\paren {\overline S \cup T} \cap \paren {S \cup \overline T} } }       | c = De Morgan }} {{eqn | r = \paren {R \cup S \cup T} \cap \paren {R \cup \overline S \cup \overline T} \cap \paren {\overline R \cup \overline S \cup T} \cap \paren {\overline R \cup S \cup \overline T}       | c = Union Distributes over Intersection }} {{end-eqn}} From Intersection is Commutative, it can be seen that the {{LHS}} and {{RHS}} are the same, and the result is proved. {{qed}}
{{begin-eqn}} {{eqn | l = S \times T       | o = \ne       | r = \O }} {{eqn | ll= \leadstoandfrom       | l = \exists \tuple {s, t}       | o = \in       | r = S \times T       | c = {{Defof|Empty Set}} }} {{eqn | ll= \leadstoandfrom       | l = \exists s \in S       | o = \land       | r = \exists t \in T       | c = {{Defof|Cartesian Product}} }} {{eqn | ll= \leadstoandfrom       | l = S \ne \O       | o = \land       | r = T \ne \O       | c = {{Defof|Empty Set}} }} {{eqn | ll= \leadstoandfrom       | l = \neg \leftparen {S = \O}       | o = \lor       | r = \rightparen {T = \O}       | c = De Morgan's Laws: Conjunction of Negations }} {{end-eqn}} So by the Rule of Transposition: :$S = \O \lor T = \O \iff S \times T = \O$ {{Qed}}
Suppose $S \times T = T \times S$. Then: {{begin-eqn}} {{eqn | l = x \in S       | o = \land       | r = y \in T       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \tuple {x, y}       | o = \in       | r = S \times T       | c = {{Defof|Cartesian Product}} }} {{eqn | ll= \leadstoandfrom       | l = \tuple {x, y}       | o = \in       | r = T \times S       | c = by hypothesis }} {{eqn | ll= \leadstoandfrom       | l = x \in T       | o = \land       | r = y \in S       | c = {{Defof|Cartesian Product}} }} {{end-eqn}} Thus it can be seen from the definition of set equality that $S \times T = T \times S \implies S = T$. Note that if $S = \O$ or $T = \O$ then, from Cartesian Product is Empty iff Factor is Empty, $S \times T = T \times S = \O$ whatever $S$ and $T$ are, and the result does not hold. {{qed}}