hcju/proofwikiqa huggingface dataset:
We need to show that $\RR_f$ is an equivalence relation. Checking in turn each of the criteria for equivalence: === Reflexive === $\RR_f$ is reflexive: :$\forall x \in S: \map f x = \map f x \implies x \mathrel {\RR_f} x$ {{qed|lemma}} === Symmetric === $\RR_f$ is symmetric: {{begin-eqn}} {{eqn | l = x       | o = \RR_f       | r = y       | c = by definition }} {{eqn | ll= \leadsto       | l = \map f x       | r = \map f y       | c = Definition of $f$ }} {{eqn | ll= \leadsto       | l = \map f x       | r = \map f y       | c = Equality is Equivalence Relation, and so Symmetric }} {{eqn | ll= \leadsto       | l = y       | o = \RR_f       | r = x       | c = Definition of $f$ }} {{end-eqn}} {{qed|lemma}} === Transitive === $\RR_f$ is transitive: {{begin-eqn}} {{eqn | l = x       | o = \RR_f       | r = y       | c =  }} {{eqn | lo= \land       | l = y       | o = \RR_f       | r = z       | c =  }} {{eqn | ll= \leadsto       | l = \map f x       | r = \map f y       | c = Definition of $f$ }} {{eqn | lo= \land       | l = \map f y       | r = \map f z       | c = Definition of $f$ }} {{eqn | ll= \leadsto       | l = \map f x       | r = \map f z       | c = Equality is Equivalence Relation, and so Transitive }} {{eqn | ll= \leadsto       | l = x       | o = \RR_f       | r = z       | c = Definition of $f$ }} {{end-eqn}} {{qed|lemma}} Thus $\RR_f$ is reflexive, symmetric and transitive, and is therefore an equivalence relation. {{Qed}}
From Canonical Injection is Injection we have that the canonical injections are in fact injective. It remains to prove the morphism property. Let $x, y \in \struct {S_1, \circ_1}$. Then: {{begin-eqn}} {{eqn | l = \map {\inj_1} {x \circ_1 y}       | r = \tuple {x \circ_1 y, e_2} }} {{eqn | r = \tuple {x \circ_1 y, e_2 \circ_2 e_2} }} {{eqn | r = \tuple {x, e_2} \circ \tuple {y, e_2} }} {{eqn | r = \map {\inj_1} x \circ \map {\inj_1} y }} {{end-eqn}} and the morphism property has been demonstrated to hold for $\inj_1$. Thus $\inj_1: \struct {S_1, \circ_1} \to \struct {S_1, \circ_1} \times \struct {S_2, \circ_2}$ has been shown to be an injective homomorphism and therefore a monomorphism. The same argument applies to $\inj_2$. {{Qed}}
Suppose $\phi: S \to T$ is a mapping such that: : $\forall x, y \in S: x \preceq_1 y \iff \phi \left({x}\right) \preceq_2 \phi \left({y}\right)$ Then, for all $x, y \in S$: {{begin-eqn}} {{eqn | l = \phi \left({x}\right)       | r = \phi \left({y}\right)       | c =  }} {{eqn | ll= \implies       | l = \phi \left({x}\right) \preceq_2 \phi \left({y}\right)       | o = \land       | r = \phi \left({y}\right) \preceq_2 \phi \left({x}\right)       | c = Ordering $\preceq_2$ is Reflexive }} {{eqn | ll= \implies       | l = x \preceq_1 y       | o = \land       | r = y \preceq_1 x       | c = Definition of Order Embedding }} {{eqn | ll= \implies       | l = x       | r = y       | c = Ordering $\preceq_1$ is Antisymmetric }} {{end-eqn}} So $\phi$ is an injection. {{qed}} Category:Order Embeddings Category:Injections 6m3u2exgapmlruwa7kj0g9sw1d5vtq4
=== Necessary Condition === Suppose $f$ is an order isomorphism. Then by definition $f$ is a bijection and so a surjection. Also by definition, $f$ is increasing, and so: :$\forall x, y \in S: x \preceq_1 y \implies \map f x \preceq_2 \map f y$ Also by definition $f^{-1}$ is also a bijection which is increasing, and so: :$\forall x, y \in S: \map f x \preceq_2 \map f y \implies x = \map {f^{-1} } {\map f x} \preceq_1 \map {f^{-1} } {\map f y} = y$ and so: :$\forall x, y \in S: x \preceq_1 y \iff \map f x \preceq_2 \map f y$ {{qed|lemma}} === Sufficient Condition === Suppose $f: S \to T$ is a mapping such that: :$f$ is a surjection; :$\forall x, y \in S: x \preceq_1 y \iff \map f x \preceq_2 \map f y$ From Order Embedding is Injection we have that $f$ is an injection. As, by hypothesis, it is also surjective, it follows that it is a bijection. Now, suppose $a, b \in T$. As $f$ is surjective: :$\exists x, y \in S: \map f x = a, \map f y = b$ As $f$ is bijective, then: :$x = \map {f^{-1} } a, y = \map {f^{-1} } b$ So by hypothesis: :$a \preceq_2 b \implies \map f x \preceq_2 \map f y \implies \map {f^{-1} } a = x \preceq_1 y = \map {f^{-1} } b$ {{qed|lemma}} Hence, by definition, $f$ is an order isomorphism. {{qed}}
Follows directly from Empty Set is Subset of All Sets and the definition of infimum. {{qed}} Category:Power Set Category:Empty Set Category:Order Theory cl07ut95vzftu3omo47fhrfz7esyt8x
By the definition of the power set: :$\forall X \in \mathcal P \left({S}\right): X \subseteq S$ The result then follows from the definition of supremum. {{qed}} Category:Power Set Category:Order Theory ivqfn0w9rl1glequblavmrjiz9mpv69
From Trivial Ordering is Universally Compatible and Diagonal Relation is Equivalence we know that the diagonal relation possesses these properties. We now need to show that it is the '''only''' relation on $S$ which possesses these properties. Both an equivalence and an ordering are reflexive and transitive. Also: : An equivalence relation is symmetric : An ordering is antisymmetric. The result follows from Relation is Symmetric and Antisymmetric iff Coreflexive. {{qed}}
From the definition, a cardinal is a set, so standard set theoretic results apply. So, checking in turn each of the criteria for an ordering: === Reflexivity === For any cardinal $S$, the identity mapping $I_S: S \to S$ is an injection. Thus: :$\forall S \in \mathbb S: S \preccurlyeq S$ So $\preccurlyeq$ is reflexive. {{qed|lemma}} === Transitivity === Let $S_1, S_2, S_3 \in \mathbb S$ such that $f: S_1 \to S_2$ and $g: S_2 \to S_3$ be injections. From Composite of Injections is Injection, $g \circ f$ is an injection and so $S_1 \preccurlyeq S_3$. So $\preccurlyeq$ is transitive. {{qed|lemma}} === Antisymmetry === Suppose $S \preccurlyeq T$ and $T \preccurlyeq S$. Then from the Cantor-Bernstein-Schröder Theorem, $S \sim T$ and, as $S$ and $T$ are cardinals, $S = T$ by definition. So $\preccurlyeq$ is antisymmetric on a set of cardinals. {{qed|lemma}} Hence the result. {{qed}}
{{WLOG}}, suppose $I$ is oriented along the $z$-axis. By definition of moment of inertia: :$I = \Sigma m_j \lambda_j^2$ :$I_0 = \Sigma m_j \lambda_j'^2$ where: :$\lambda_j$ is the position vector to the $j$th particle from the $z$-axis :$\lambda_j'$ is related to $\lambda_j$ by: ::$\lambda_j = \lambda_j' + R_\perp$ :$R_\perp$ is the perpendicular distance from $I$ to the center of mass of $B$. Therefore: :$I = \Sigma m_j \lambda_j^2 = \Sigma m_j \paren {\lambda_j'^2 + 2 \lambda_j' \cdot R_\perp + R_\perp^2}$ The middle term is: :$2 R_\perp \cdot \Sigma m_j \lambda_j' = 2 R_\perp \cdot \Sigma m_j \paren {\lambda_j - R_\perp} = 2 R_\perp \cdot M \paren {R_\perp - R_\perp} = 0$ Thus: :$I = \Sigma m_j \lambda_j^2 = \Sigma m_j \paren {\lambda_j'^2 + R_\perp^2} = I_0 + M l^2$ {{Qed}}
Consider $X = \left\{{a, b}\right\}$ where $a, b \in S$. By hypothesis, $X$ has a smallest element. So either $\min X = a$ or $\min X = b$. If $\min X = a$, then $a \preceq b$. If $\min X = b$, then $b \preceq a$. So either $a \preceq b$ or $b \preceq a$. That is, $a$ and $b$ are comparable. As this applies to all $a, b \in S$, the ordering $\preceq$ is total.
Suppose $T = \bigcup \mathbb S$ is well-orderable. Then we can create a well-ordering $\preceq$ on $T$ so as to make $\left({T, \preceq}\right)$ a well-ordered set. From the definition of well-ordered set, every subset of $T$ is itself well-ordered. From Subset of Union: General Result we have that $\forall S \in \mathbb S: S \subseteq T$. So every $S \in \mathbb S$ is well-ordered and Choice Function Exists for Set of Well-Ordered Sets applies. {{qed}}
We use the Principle of Mathematical Induction on the cardinality of $I$. === Basis for the Induction === Let $\left\vert{I}\right\vert = 1$. Let $j \in I$. Then $I = \left\{{j}\right\}$.  By the definition of a non-empty set, there exists an $x \in S_j$. Hence, there exists a mapping $f: I \to S_j$ such that $f \left({j}\right) = x$. Since $\forall i \in I: i = j$, the result follows. === Induction Hypothesis === Assume that the theorem holds for all $I$ with $\left\vert{I}\right\vert = n$. This is the induction hypothesis. === Induction Step === Now, suppose that $\left\vert{I}\right\vert = n + 1$.  Let $j \in I$. Let $J = I \setminus \left\{{j}\right\}$. By Cardinality Less One, $\left\vert{J}\right\vert = n$. By the induction hypothesis, there exists a mapping: :$\displaystyle g: J \to \bigcup_{i \mathop \in J} S_i$ such that: :$\forall i \in J: g \left({i}\right) \in S_i$ By the definition of a non-empty set, there exists a $y \in S_j$. Hence, there exists a mapping: :$\displaystyle f: I \to \bigcup_{i \mathop \in I} S_i$ such that: :$\displaystyle \forall i \in I: f \left({i}\right) = \begin{cases} g \left({i}\right) & : i \ne j \\ y & : i = j \end{cases}$ Then: :$\forall i \in I: f \left({i}\right) \in S_i$ as desired. {{qed}}
Every member of $\mathbb S$ is a well-ordered set.  Thus, for $S \in \mathbb S$, there is a minimal element $s$ for $S$ (with respect to the ordering of $S$).  By Well-Ordering Minimal Elements are Unique, $s$ is unique. Therefore, we can define $f$ by: :$\forall S \in \mathbb S: f \left({S}\right) = s$ {{qed}}
From Empty Set is Element of Power Set: :$\O \in \powerset S$ From Union with Empty Set: :$\forall A \subseteq S: A \cup \O = A = \O \cup A$ By definition of power set: :$A \subseteq S \iff A \in \powerset S$ So: :$\forall A \in \powerset S: A \cup \O = A = \O \cup A$ Thus we see that $\O$ acts as the identity for $\struct {\powerset S, \cup}$. {{qed}}
We note that by Set is Subset of Itself, $S \subseteq S$ and so $S \in \powerset S$ from the definition of the power set. From Intersection with Subset is Subset, we have: :$A \subseteq S \iff A \cap S = A = S \cap A$ By definition of power set: :$A \subseteq S \iff A \in \powerset S$ So: :$\forall A \in \powerset S: A \cap S = A = S \cap A$ Thus we see that $S$ acts as the identity element. {{qed}}
=== Necessary Condition === Let $x$ commute with $y$. Then: {{begin-eqn}} {{eqn | l = y^{-1} \circ x       | r = y^{-1} \circ \left({x \circ \left({y \circ y^{-1} }\right)}\right)       | c = Invertibility of $y$ }} {{eqn | r = y^{-1} \circ \left({\left({x \circ y}\right) \circ y^{-1} }\right)       | c = Associativity of $\circ$ }} {{eqn | r = y^{-1} \circ \left({\left({y \circ x}\right) \circ y^{-1} }\right)       | c = $x$ commutes with $y$ }} {{eqn | r = \left({y^{-1} \circ \left({y \circ x}\right)}\right) \circ y^{-1}       | c = Associativity of $\circ$ }} {{eqn | r = \left({\left({y^{-1} \circ y}\right) \circ x}\right) \circ y^{-1}       | c = Associativity of $\circ$ }} {{eqn | r = x \circ y^{-1}       | c = Invertibility of $y$ }} {{end-eqn}} So $x$ commutes with $y^{-1}$. {{Qed|lemma}} === Sufficient Condition === Now let $x$ commute with $y^{-1}$. From the above, it follows that $x$ commutes with $\left({y^{-1}}\right)^{-1}$. From Inverse of Inverse in Monoid: : $\left({y^{-1}}\right)^{-1} = y$ Thus $x$ commutes with $y$. {{Qed}}
=== Necessary Condition === Let $x$ commute with $y$. Then: {{begin-eqn}} {{eqn | l = x^{-1} \circ y^{-1}       | r = \paren {y \circ x}^{-1}       | c = Inverse of Product }} {{eqn | r = \paren {x \circ y}^{-1}       | c =  $x$ commutes with $y$ }} {{eqn | r = y^{-1} \circ x^{-1}       | c = Inverse of Product }} {{end-eqn}} So $x^{-1}$ commutes with $y^{-1}$. {{qed|lemma}} === Sufficient Condition === Now let $x^{-1}$ commute with $y^{-1}$. From the above, $\paren {x^{-1} }^{-1}$ commutes with $\paren {y^{-1} }^{-1}$. From Inverse of Inverse in Monoid, $\paren {x^{-1} }^{-1} = x$ and $\paren {y^{-1} }^{-1} = y$. Thus $x$ commutes with $y$. {{Qed}}
{{begin-eqn}} {{eqn | l = x \circ y       | r = y \circ x       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \struct {x \circ y}^{-1}       | r = \struct {y \circ x}^{-1}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \struct {x \circ y}^{-1}       | r = x^{-1} \circ y^{-1}       | c = Inverse of Product }} {{end-eqn}} {{Qed}}
As $\struct {S, \circ}$ is a monoid, it is by definition a semigroup. Therefore it is taken for granted that $\circ$ is associative, so we can dispense with parentheses. We also take for granted the fact that $x$ and $y$ are cancellable from Invertible Element of Monoid is Cancellable. So: {{begin-eqn}} {{eqn | l = x \circ y       | r = y \circ x       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x \circ y \circ x^{-1}       | r = y \circ x \circ x^{-1}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x \circ y \circ x^{-1}       | r = y       | c = {{Defof|Invertible Element}} }} {{end-eqn}} {{qed}} Category:Monoids Category:Commutativity Category:Conjugacy pbws5hifelx5gn6hxgu0ztzt6nsj1zi
As $\struct {S, \circ}$ is a monoid, it is by definition a semigroup. Therefore $\circ$ is associative, so we can dispense with parentheses. From Invertible Element of Monoid is Cancellable, we also have that $x, y, x^{-1}, y^{-1}$ are cancellable. So: {{begin-eqn}} {{eqn | l = x \circ y       | r = y \circ x       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x \circ y \circ x^{-1}       | r = y       | c = Conjugate of Commuting Elements }} {{eqn | ll= \leadstoandfrom       | l = x \circ y \circ x^{-1} \circ y^{-1}       | r = y \circ y^{-1}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x \circ y \circ x^{-1} \circ y^{-1}       | r = e_S       | c = Invertibility of $y$ }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | l = x \circ y       | r = y \circ x       | c =  }} {{eqn | ll= \leadstoandfrom       | l = y^{-1} \circ x \circ y       | r = x       | c = Conjugate of Commuting Elements }} {{eqn | ll= \leadstoandfrom       | l = x^{-1} \circ y^{-1} \circ x \circ y       | r = x^{-1} \circ x       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x^{-1} \circ y^{-1} \circ x \circ y       | r = e_S       | c = Invertibility of $x$ }} {{end-eqn}} {{qed}} Category:Monoids Category:Commutativity q5vqzejvpwtlpl1e4pm57x9a41v72r7
Consider the power set $\powerset S$ of $S$. {{AimForCont}} $\forall x \in \powerset S: x \in S$. Then the identity mapping $I_S: S \to \powerset S$ would be a surjection. But from Cantor's Theorem, there is no surjection $f: S \to \powerset S$. From this contradiction it follows that: :$\exists x \in \powerset S: x \notin S$ Hence the result. {{qed}} Category:Set Theory f1mbpl5yapgomxeyhgtvk1au0ncrq6m
From Intersection with Subset is Subset: :$S \subseteq T \iff S \cap T = S$ From Union with Superset is Superset: :$S \subseteq T \iff S \cup T = T$ That is: :$T \subseteq S \iff S \cup T = S$ Thus: :$\paren {S \cup T = S} \land \paren {S \cap T = S} \iff S \subseteq T \subseteq S$ By definition of set equality: :$S = T \iff S \subseteq T \subseteq S$ Hence the result. {{qed}}
Consider $R, S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the universe. {{begin-eqn}} {{eqn | l = \left({R \cap S}\right) \setminus T       | r = \left({R \cap S}\right) \cap \complement \left({T}\right)       | c = Set Difference as Intersection with Complement }} {{eqn | r = \left({R \cap S}\right) \cap \left({\complement \left({T}\right) \cap \complement \left({T}\right)}\right)       | c = Intersection is Idempotent }} {{eqn | r = \left({R \cap \complement \left({T}\right)}\right) \cap \left({S \cap \complement \left({T}\right)}\right)       | c = Intersection is Associative and Intersection is Commutative }} {{eqn | r = \left({R \setminus T}\right) \cap \left({S \setminus T}\right)       | c = Set Difference as Intersection with Complement }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = x       | o = \in       | r = \paren {R \cap S} \setminus T       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \in        | r = R \land x \in S       | c = {{Defof|Set Intersection}} }} {{eqn | lo= \land       | l = x       | o = \notin        | r = T       | c = {{Defof|Set Difference}} }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \in       | r = R \land x \in S       | c = Rule of Idempotence }} {{eqn | lo= \land       | l = x       | o = \notin       | r = T \land x \notin T       | c = Rule of Idempotence }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \in       | r = R       | c = Rule of Association }} {{eqn | lo= \land       | l = x       | o = \in       | r = x \in S \land x \notin T       | c =  }} {{eqn | lo= \land       | l = x       | o = \notin       | r = T       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \in       | r = R       | c =  }} {{eqn | lo= \land       | l = x       | o = \notin       | r = T \land x \in S       | c = Rule of Commutation }} {{eqn | lo= \land       | l = x       | o = \notin       | r = T       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \in       | r = R \land x \notin T       | c = Rule of Association }} {{eqn | lo= \land       | l = x       | o = \in       | r = S \land x \notin T       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \in       | r = \paren {R \setminus T}       | c = {{Defof|Set Difference}} }} {{eqn | lo= \land       | l = x       | o = \in       | r = \paren {S \setminus T}       | c = {{Defof|Set Difference}} }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \in       | r = \paren {R \setminus T} \cap \paren {S \setminus T}       | c = {{Defof|Set Intersection}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \left({R \cap T}\right) \setminus \left({S \cap T}\right)       | r = \left({\left({R \cap T}\right) \setminus S}\right) \cup \left({\left({R \cap T}\right) \setminus T}\right)       | c = De Morgan's Laws: Difference with Intersection }} {{eqn | r = \left({\left({R \cap T}\right) \setminus S}\right) \cup \varnothing       | c = Set Difference of Intersection with Set is Empty Set }} {{eqn | r = \left({R \cap T}\right) \setminus S       | c = Union with Empty Set }} {{eqn | r = \left({R \setminus S}\right) \cap T       | c = Intersection with Set Difference is Set Difference with Intersection }} {{end-eqn}} {{qed|lemma}} Then: {{begin-eqn}} {{eqn | l = R \cap \left({S \setminus T}\right)       | r = \left({S \setminus T}\right) \cap R       | c = Intersection is Commutative }} {{eqn | r = \left({S \cap R}\right) \setminus \left({T \cap R}\right)       | c = from above }} {{eqn | r = \left({R \cap S}\right) \setminus \left({R \cap T}\right)       | c = Intersection is Commutative }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = S \setminus \paren {S \cap T}       | r = \paren {S \setminus S} \cup \paren {S \setminus T}       | c = De Morgan's Laws: Difference with Intersection }} {{eqn | r = \O \cup \paren {S \setminus T}       | c = Set Difference with Self is Empty Set }} {{eqn | r = S \setminus T       | c = Union with Empty Set }} {{end-eqn}} {{qed}}
From Set Difference is Right Distributive over Set Intersection  we have: : $\left({R \cap S}\right) \setminus T = \left({R \setminus T}\right) \cap \left({S \setminus T}\right)$ Hence: {{begin-eqn}} {{eqn | l=\left({S \cap T}\right) \setminus S       | r=\left({S \setminus S}\right) \cap \left({T \setminus S}\right)       | c=Set Difference is Right Distributive over Set Intersection }} {{eqn | r=\varnothing \cap \left({T \setminus S}\right)       | c=Set Difference with Self is Empty Set }} {{eqn | r=\varnothing       | c=Intersection with Empty Set }} {{end-eqn}} {{qed|lemma}} {{begin-eqn}} {{eqn | l=\left({S \cap T}\right) \setminus T       | r=\left({T \cap S}\right) \setminus T       | c=Intersection is Commutative }} {{eqn | r=\varnothing       | c=from above }} {{end-eqn}} {{qed}} Category:Set Difference Category:Set Intersection Category:Empty Set f034uz6dd62nuhy5cvh2fodolvpnqzh
{{begin-eqn}} {{eqn | l = \left({S \setminus T}\right)       | o = \subseteq       | r = S       | c = Set Difference is Subset }} {{eqn | ll= \implies       | l = \left({S \setminus T}\right) \cap S       | r = S \setminus T       | c = Intersection with Subset is Subset }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \left({S \setminus T}\right) \cap S       | r = \left({S \cap S}\right) \setminus T       | c = Intersection with Set Difference is Set Difference with Intersection }} {{eqn | r = S \setminus T       | c = Intersection is Idempotent }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \paren {S \setminus T} \cap T       | r = \paren {S \cap T} \setminus \paren {T \cap T}       | c = Set Intersection Distributes over Set Difference }} {{eqn | r = \paren {S \cap T} \setminus T       | c = Intersection is Idempotent }} {{eqn | r = \O       | c = Set Difference of Intersection with Set is Empty Set }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \paren {S \setminus T} \cap T       | r = \paren {S \cap T} \setminus T       | c = Intersection with Set Difference is Set Difference with Intersection }} {{eqn | r = \O       | c = Set Difference of Intersection with Set is Empty Set }} {{end-eqn}} {{qed}}
Consider $S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the universe. {{begin-eqn}} {{eqn | l=\left({S \cup T}\right) \setminus T       | r=\left({S \cup T}\right) \cap \overline T       | c=Set Difference as Intersection with Complement }} {{eqn | r=\left({S \cap \overline T}\right) \cup \left({T \cap \overline T}\right)       | c=Intersection Distributes over Union }} {{eqn | r=\left({S \setminus T}\right) \cup \left({T \setminus T}\right)       | c=Set Difference as Intersection with Complement }} {{eqn | r=\left({S \setminus T}\right) \cup \varnothing       | c=Set Difference with Self is Empty Set }} {{eqn | r=S \setminus T       | c=Union with Empty Set }} {{end-eqn}} {{qed}} Category:Set Difference Category:Set Union 1lf8oyetz2hfs1uhcwmh7yf0hm54g3x
Consider $S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the universe. {{begin-eqn}} {{eqn | l = \paren {S \setminus T} \cup S       | r = \paren {S \cap \map \complement T} \cup S       | c = Set Difference as Intersection with Complement }} {{eqn | r = S       | c = Union Absorbs Intersection }} {{end-eqn}} {{qed}} Category:Set Difference Category:Set Union g3no2t302x6dxigj7hmgavwabeapekl
Consider $S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the universe. {{begin-eqn}} {{eqn | l = \left({S \setminus T}\right) \cup T       | r = \left({S \cap \complement \left({T}\right)}\right) \cup T       | c = Set Difference as Intersection with Complement }} {{eqn | r = \left({S \cup T}\right) \cap \left({\complement \left({T}\right) \cup T}\right)       | c = Union Distributes over Intersection }} {{eqn | r = \left({S \cup T}\right) \cap \mathbb U       | c = Union with Complement }} {{eqn | r = S \cup T       | c = Intersection with Universe }} {{end-eqn}} {{qed}} {{LEM|Union with Complement}}
Consider $R, S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the universe. {{begin-eqn}} {{eqn | l = \paren {R \cup T} \setminus \paren {S \cup T}       | r = \paren {R \cup T} \cap \overline {\paren {S \cup T} }       | c = Set Difference as Intersection with Complement }} {{eqn | r = \paren {R \cup T} \cap \paren {\overline S \cap \overline T}       | c = De Morgan's Laws: Complement of Union }} {{eqn | r = \paren {\paren {R \cup T} \cap \overline T} \cap \overline S       | c = Intersection is Associative and Intersection is Commutative }} {{eqn | r = \paren {\paren {R \cup T} \setminus T} \setminus S       | c = Set Difference as Intersection with Complement }} {{eqn | r = \paren {R \setminus T} \setminus S       | c = Set Difference with Union is Set Difference }} {{end-eqn}} {{qed|lemma}} Then: {{begin-eqn}} {{eqn | l = R \setminus \paren {S \cup T}       | r = R \cap \overline {\paren {S \cup T} }       | c = Set Difference as Intersection with Complement }} {{eqn | r = R \cap \paren {\overline S \cap \overline T}       | c = De Morgan's Laws: Complement of Union }} {{eqn | r = \paren {R \cap \overline S} \cap \overline T       | c = Intersection is Associative }} {{eqn | r = \paren {R \setminus S} \setminus  T       | c = Set Difference as Intersection with Complement }} {{end-eqn}} {{qed|lemma}} Then: {{begin-eqn}} {{eqn | l = R \setminus \paren {S \cup T}       | r = R \setminus \paren {T \cup S}       | c = Union is Commutative }} {{eqn | r = \paren {R \setminus T} \setminus S       | c = from above }} {{end-eqn}} {{qed}}