hcju/proofwikiqa huggingface dataset:
Let $T = T_1 \cap T_2$ where $T_1, T_2$ are subsemigroups of $\left({S, \circ}\right)$. Then: {{begin-eqn}} {{eqn | o =       | r = a, b \in T }} {{eqn | o = \implies       | r = a, b \in T_1 \land a, b \in T_2       | c = {{Defof|Set Intersection}} }} {{eqn | o = \implies       | r = a \circ b \in T_1 \land a \circ b \in T_2       | c = Subsemigroups are closed }} {{eqn | o = \implies       | r = a \circ b \in T       | c = {{Defof|Set Intersection}} }} {{end-eqn}} Thus $\left({T, \circ}\right)$ is closed, and is therefore a semigroup from the Subsemigroup Closure Test. {{Qed}} The other results follow from this and Intersection is Subset. Category:Set Intersection Category:Subsemigroups ju0qecld0oxti00l2zeoc6hopffz2k8
Now let $C$ be the set of cancellable elements of $\struct {S, \circ}$. Let $x, y \in C$. Then $x$ and $y$ are both left cancellable and right cancellable. Thus by Left Cancellable Elements of Semigroup form Subsemigroup: :$x \circ y$ is left cancellable and by Right Cancellable Elements of Semigroup form Subsemigroup: :$x \circ y$ is right cancellable. Thus $x \circ y$ is both left cancellable and right cancellable, and therefore cancellable. Thus $x \circ y \in C$. Thus $\struct {C, \circ}$ is closed and is therefore by the Subsemigroup Closure Test a subsemigroup of $\struct {S, \circ}$. {{qed}}
Let $G$ be a cyclic group generated by $a$. Let $H$ be a subgroup of $G$. If $H = \set e$, then $H$ is a cyclic group subgroup generated by $e$. Let $H \ne \set e$. By definition of cyclic group, every element of $G$ has the form $a^n$. Then as $H$ is a subgroup of $G$, $a^n \in H$ for some $n \in \Z$. Let $m$ be the smallest (strictly) positive integer such that $a^m \in H$. Consider an arbitrary element $b$ of $H$. As $H$ is a subgroup of $G$, $b = a^n$ for some $n$. By the Division Theorem, it is possible to find integers $q$ and $r$ such that $n = m q + r$ with $0 \le r < m$. It follows that: :$a^n = a^{m q + r} = \paren {a^m}^q a^r$ and hence: :$a^r = \paren {a^m}^{-q} a^n$ Since $a^m \in H$ so is its inverse $\paren {a^m}^{-1}$ By {{GroupAxiom|0}}, so are all powers of its inverse. Now $a^n$ and $\paren {a^m}^{-q}$ are both in $H$, thus so is their product $a^r$, again by Group Axiom $G 0$: Closure. However: : $m$ was the smallest (strictly) positive integer such that $a^m \in H$ and: :$0 \le r < m$ Therefore it follows that: : $r = 0$ Therefore: :$n = q m$ and: :$b = a^n = \paren {a^m}^q$. We conclude that any arbitrary element $b = a^n$ of $H$ is a power of $a^m$. So, by definition, $H = \gen {a^m}$ is cyclic. {{qed}}
Let $G$ be a cyclic group generated by $a$. === Finite Group === Let $G$ be finite. By Bijection from Divisors to Subgroups of Cyclic Group there are exactly as many subgroups of $G$ as divisors of the order of $G$. As each one of these is cyclic by Subgroup of Finite Cyclic Group is Determined by Order, the result follows. {{qed|lemma}} === Infinite Group === By Morphism from Integers to Group, an infinite cyclic group is isomorphic to $\left({\Z, +}\right)$. So all we need to do is show that any subgroup of $\left({\Z, +}\right)$ is cyclic. Suppose $H$ is a subgroup of $\left({\Z, +}\right)$. From Subgroup of Integers is Ideal and Ring of Integers is Principal Ideal Domain, we have that: :$\exists m \in \Z_{> 0}: H = \left({m}\right)$ where $\left({m}\right)$ is the principal ideal of $\left({\Z, +, \times}\right)$ generated by $m$. But $m$ is also a generator of the subgroup $\left({m}\right)$ of $\left({\Z, +}\right)$, as: :$n \in \Z: n \circ m = n \cdot m \in \gen m$ Hence the result. {{qed}}
Let $G$ be a cyclic group generated by $a$. Let $H$ be a subgroup of $G$. By Cyclic Group is Abelian, $G$ is abelian. By Subgroup of Abelian Group is Normal, $H$ is normal in $G$. Let $G / H$ be the quotient group of $G$ by $H$. Let $q_H: G \to G / H$ be the quotient epimorphism from $G$ to $G / H$: :$\forall x \in G: \map {q_H} x = x H$ Then from Quotient Group Epimorphism is Epimorphism, $H$ is the kernel of $q_H$. From Kernel of Homomorphism on Cyclic Group, $H$ is a cyclic group. {{qed}}
By the definition of subset, $\O \subseteq S$ means: :$\forall x: \paren {x \in \O \implies x \in S}$ By the definition of the empty set: :$\forall x: \neg \paren {x \in \O}$ Thus $\O \subseteq S$ is vacuously true. {{qed}}
$S \subseteq T$ means: :'''every element of $S$ is also in $T$''' or, equivalently: :'''every element that is ''not'' in $T$ is not in $S$ either.''' Thus: {{begin-eqn}} {{eqn | o =        | r = S \subseteq T       | c =  }} {{eqn | o = \leadstoandfrom       | r = \forall x \in S: x \in T       | c = {{Defof|Subset}} }} {{eqn | o = \leadstoandfrom       | r = \neg \paren {\exists x \in S: \neg \paren {x \in T} }       | c = De Morgan's Laws (Predicate Logic) }} {{end-eqn}} which means '''there is no element in $S$ which is not also in $T$'''. There are no elements of $\O$, from the definition of the empty set. Therefore $\O$ has no elements that are not also in any other set. Thus, from the above, all elements of $\O$ are all (vacuously) in every other set. {{qed}}
A reflexive relation can not be antireflexive, and vice versa: :$\tuple {x, x} \in \RR \iff \neg \paren {\tuple {x, x} \notin \RR}$ By the definition of non-reflexive, a reflexive relation can not also be non-reflexive. So a reflexive relation is neither antireflexive nor non-reflexive. An antireflexive relation can be neither reflexive (see above) nor non-reflexive (by the definition of non-reflexive). By its own definition, if a relation is neither reflexive nor antireflexive, then it is non-reflexive. {{qed}}
{{begin-eqn}} {{eqn | l = x       | o = \in       | r = S \cup S       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x \in S       | o = \lor       | r = x \in S       | c = {{Defof|Set Union}} }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \in       | r = S       | c = Rule of Idempotence: Disjunction }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = x       | o = \in       | r = S \cap S       | c = }} {{eqn | ll= \leadstoandfrom       | l = x \in S       | o = \land       | r = x \in S       | c = {{Defof|Set Intersection}} }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \in       | r = S       | c = Rule of Idempotence: Conjunction }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = x \in S       | o = \leadsto       | r = x \in S \lor x \in T       | c = Rule of Addition }} {{eqn | o = \leadsto       | r = x \in S \cup T       | c = {{Defof|Set Union}} }} {{eqn | o = \leadsto       | r = S \subseteq S \cup T       | c = {{Defof|Subset}} }} {{end-eqn}} Similarly for $T$. {{qed}}
{{begin-eqn}} {{eqn | ll= \forall x:       | l = \leftparen {x \in S}       | o = \implies       | r = \rightparen {x \in S}       | c = Law of Identity:       | cc= a statement implies itself }} {{eqn | lll=\leadsto       | l = S       | o = \subseteq       | r = S       | c = {{Defof|Subset}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = s       | o = \in       | r = \set {x \in S: \map P x}       | c =  }} {{eqn | ll= \leadsto       | l = s       | o = \in       | r = \set {x \in S \land \map P x}       | c =  }} {{eqn | ll= \leadsto       | l = s       | o = \in       | r = \set {x \in S} \land \map P s       | c = {{Defof|Set Definition by Predicate}} }} {{eqn | ll= \leadsto       | l = s       | o = \in       | r = \set {x \in S}       | c = Rule of Simplification }} {{eqn | ll= \leadsto       | l = s       | o = \in       | r = S       | c = {{Defof|Element}} }} {{eqn | ll= \leadsto       | l = \set {x \in S: \map P x}       | o = \subseteq       | r = S       | c = {{Defof|Subset}} }} {{end-eqn}} {{qed}}
We can express $\forall x$ using its propositional expansion: :$\map P {\mathbf X_1} \land \map P {\mathbf X_2} \land \map P {\mathbf X_3} \land \ldots$ where $\mathbf X_1, \mathbf X_2, \mathbf X_3 \ldots{}$ is the complete set of the objects in the universe of discourse. The fact that any object we care to choose has the property in question means that they ''all'' must have this property. The result then follows by generalising the Rule of Conjunction. {{qed}}
This is an extension of Proof by Cases. The propositional expansion of $\exists x: P \left({x}\right)$ is: :$P \left({\mathbf X_1}\right) \lor P \left({\mathbf X_2}\right) \lor P \left({\mathbf X_3}\right) \lor \ldots$ We know that any arbitrarily selected $\mathbf a$ with the property $P$ implies $y$. From this we can infer that ''all'' such $\mathbf a$ which have that property imply $y$. This is equivalent to the step in Proof by Cases in which we need to prove that both disjuncts lead to the same conclusion. The fact that we only need one of them in fact to ''be'' true is quite enough to draw the conclusion that $y$ is true. In this context, we are assured by the statement $\exists x: P \left({x}\right)$ that at least one such disjunct in the above propositional expansion is true. Thus the conclusion follows, and the result is proved. {{qed}}
{{begin-eqn}} {{eqn | o =        | r = \set x \subseteq A       | c =  }} {{eqn | o = \leadstoandfrom       | r = \forall y: \paren {y \in \set x \implies y \in A}       | c = {{Defof|Subset}} }} {{eqn | o = \leadstoandfrom       | r = \forall y: \paren {y = x \implies y \in A}       | c = {{Defof|Singleton}} }} {{eqn | o = \leadstoandfrom       | r = x \in A       | c = Equality implies Substitution }} {{end-eqn}} {{qed}}
=== Necessary Condition === Let $x \in S$. We have: :$\set x = \set {y \in S: y = x}$ From Subset of Set with Propositional Function: :$\set {x \in S: \map P x} \subseteq S$ Hence: :$\set x \subseteq S$ {{qed|lemma}} === Sufficient Condition === Let $\set x \subseteq S$. From the definition of a subset: :$x \in \set x \implies x \in S$ {{qed}}
{{begin-eqn}} {{eqn | o =       | r = \paren {R \subseteq S} \land \paren {S \subseteq T}       | c = }} {{eqn | o = \leadsto       | r = \paren {x \in R \implies x \in S} \land \paren {x \in S \implies x \in T}       | c = {{Defof|Subset}} }} {{eqn | o = \leadsto       | r = \paren {x \in R \implies x \in T}       | c = Hypothetical Syllogism }} {{eqn | o = \leadsto       | r = R \subseteq T       | c = {{Defof|Subset}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l=S \ne T       | o=\iff       | r=\neg \left({S = T}\right)       | c= }} {{eqn | o=\iff       | r=\neg \left({\left({S \subseteq T}\right) \land \left({T \subseteq S}\right)}\right)       | c=Definition of Set Equality }} {{eqn | o=\iff       | r=\neg \left({S \subseteq T}\right) \lor \neg \left({T \subseteq S}\right)       | c=De Morgan's Laws: Disjunction of Negations }} {{eqn | o=\iff       | r=\left({S \nsubseteq T}\right) \lor \left({T \nsubseteq S}\right)       | c= }} {{end-eqn}} {{qed}} Category:Set Theory Category:Subsets ayqpjtr3s2pd3goy1x3ti823vrizx12
{{begin-eqn}} {{eqn | l = S \subseteq S       | o \land        | r = S \supseteq S       | c = Set is Subset of Itself }} {{eqn | ll= \leadsto       | l = S       | r = S       | c = {{Defof|Set Equality}} }} {{end-eqn}} {{qed}} Category:Set Equality l4v6a9jr0a9dowr9jyr9l6oq8u5cubj
=== Necessary Condition === From Union of Subsets is Subset: :$\paren {S_1 \subseteq T} \land \paren {S_2 \subseteq T} \implies \paren {S_1 \cup S_2} \subseteq T$ {{qed|lemma}} === Sufficient Condition === Let $\paren {S_1 \cup S_2} \subseteq T$: {{begin-eqn}} {{eqn | l = S_1       | o = \subseteq       | r = S_1 \cup S_2       | c = Set is Subset of Union }} {{eqn | l = \paren {S_1 \cup S_2}       | o = \subseteq       | r = T       | c = by hypothesis }} {{eqn | ll= \leadsto       | l = S_1       | o = \subseteq       | r = T       | c = Subset Relation is Transitive }} {{end-eqn}} Similarly for $S_2$: {{begin-eqn}} {{eqn | l = S_2       | o = \subseteq       | r = S_1 \cup S_2       | c = Set is Subset of Union }} {{eqn | l = \paren {S_1 \cup S_2}       | o = \subseteq       | r = T       | c = by hypothesis }} {{eqn | ll= \leadsto       | l = S_2       | o = \subseteq       | r = T       | c = Subset Relation is Transitive }} {{end-eqn}} That is: :$\paren {S_1 \cup S_2} \subseteq T \implies \paren {S_1 \subseteq T} \land \paren {S_2 \subseteq T}$ {{qed|lemma}} Thus from the definition of equivalence: :$\paren {S_1 \subseteq T} \land \paren {S_2 \subseteq T} \iff \paren {S_1 \cup S_2} \subseteq T$ {{qed}}
{{begin-eqn}} {{eqn | l = S       | o = \subseteq       | r = S       | c = Set is Subset of Itself }} {{eqn | l = \O       | o = \subseteq       | r = S       | c = Empty Set is Subset of All Sets }} {{eqn | ll= \leadsto       | l = S \cup \O       | o = \subseteq       | r = S       | c = Union is Smallest Superset }} {{eqn | ll= \leadsto       | l = S       | o = \subseteq       | r = S \cup \O       | c = Set is Subset of Union }} {{eqn | ll= \leadsto       | l = S \cup \O       | r = S       | c = Definition of Set Equality }} {{end-eqn}} {{qed}}
From Empty Set is Subset of All Sets: :$\O \subseteq S$ From Union with Superset is Superset: :$S \cup \O = S$ {{qed}}
{{begin-eqn}} {{eqn | l = x       | o = \in       | r = S \cap T }} {{eqn | ll= \leadsto       | l = x       | o = \in       | r = S \land x \in T       | c = {{Defof|Set Intersection}} }} {{eqn | ll= \leadsto       | l = x       | o = \in       | r = S       | c = Rule of Simplification }} {{eqn | ll= \leadsto       | l = S \cap T       | o = \subseteq       | r = S       | c = {{Defof|Subset}} }} {{end-eqn}} Similarly for $T$. {{qed}}
{{begin-eqn}} {{eqn | l = S \cap \O       | o = \subseteq       | r = \O       | c = Intersection is Subset }} {{eqn | l = \O       | o = \subseteq       | r = S \cap \O       | c = Empty Set is Subset of All Sets }} {{eqn | ll= \leadsto       | l = S \cap \O       | r = \O       | c = {{Defof|Set Equality|index = 2}} }} {{end-eqn}} {{qed}}
=== Sufficient Condition === From Set is Subset of Intersection of Supersets we have that: :$S \subseteq T_1 \land S \subseteq T_2 \implies S \subseteq T_1 \cap T_2$ {{qed|lemma}} === Necessary Condition === Let: :$S \subseteq T_1 \cap T_2$ From Intersection is Subset we have $T_1 \cap T_2 \subseteq T_1$ and $T_1 \cap T_2\subseteq T_2$. From Subset Relation is Transitive, it follows directly that $S \subseteq T_1$ and $S \subseteq T_2$. So $S \subseteq T_1 \cap T_2 \implies S \subseteq T_1 \land S \subseteq T_2$. {{qed|lemma}} From the above, we have: :$S \subseteq T_1 \land S \subseteq T_2 \implies S \subseteq T_1 \cap T_2$ :$S \subseteq T_1 \cap T_2 \implies S \subseteq T_1 \land S \subseteq T_2$ Thus $S \subseteq T_1 \land S \subseteq T_2 \iff S \subseteq T_1 \cap T_2$ from the definition of equivalence. {{Qed}}
{{begin-eqn}} {{eqn | l = S \cap T       | o = \subseteq       | r = S       | c = Intersection is Subset }} {{eqn | l = S       | o = \subseteq        | r = S \cup T       | c = Set is Subset of Union }} {{eqn | ll= \leadsto       | l = S \cap T       | o = \subseteq       | r = S \cup T       | c = Subset Relation is Transitive }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | o =        | r = x \in R \cup \paren {S \cap T} }} {{eqn | o = \leadstoandfrom       | r = x \in R \lor \paren {x \in S \land x \in T}       | c = {{Defof|Set Union}} and {{Defof|Set Intersection}} }} {{eqn | o = \leadstoandfrom       | r = \paren {x \in R \lor x \in S} \land \paren {x \in R \lor x \in T}       | c = Disjunction is Left Distributive over Conjunction }} {{eqn | o = \leadstoandfrom       | r = x \in \paren {R \cup S} \cap \paren {R \cup T}       | c = {{Defof|Set Union}} and {{Defof|Set Intersection}} }} {{end-eqn}} {{qed}}
From Intersection Distributes over Union: :$R \cap \paren {S \cup T} = \paren {R \cap S} \cup \paren {R \cap T}$ From the Duality Principle for Sets, exchanging $\cup$ for $\cap$ throughout, and vice versa, reveals the result: :$R \cup \paren {S \cap T} = \paren {R \cup S} \cap \paren {R \cup T}$ {{qed}}
{{begin-eqn}} {{eqn | o =        | r = x \in R \cap \paren {S \cup T} }} {{eqn | o = \leadstoandfrom       | r = x \in R \land \paren {x \in S \lor x \in T}       | c = {{Defof|Set Union}} and {{Defof|Set Intersection}} }} {{eqn | o = \leadstoandfrom       | r = \paren {x \in R \land x \in S} \lor \paren {x \in R \land x \in T}       | c = Conjunction is Left Distributive over Disjunction }} {{eqn | o = \leadstoandfrom       | r = x \in \paren {R \cap S} \cup \paren {R \cap T}       | c = {{Defof|Set Union}} and {{Defof|Set Intersection}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = x \in S \setminus T       | o = \leadsto       | r = x \in S \land x \notin T       | c = {{Defof|Set Difference}} }} {{eqn | o = \leadsto       | r = x \in S       | c = Rule of Simplification }} {{end-eqn}} The result follows from the definition of subset. {{Qed}}
{{begin-eqn}} {{eqn | l = S \setminus T       | r = S \cap \complement_S \left({T}\right)       | c = Set Difference as Intersection with Relative Complement }} {{eqn | o = \subseteq       | r = S       | c = Intersection is Subset }} {{end-eqn}} {{Qed}}
From Set Difference is Subset: :$S \setminus \O \subseteq S$ From the definition of the empty set: :$\forall x \in S: x \notin \O$ Let $x \in S$. Thus: {{begin-eqn}} {{eqn | l = x \in S       | o = \leadsto       | r = x \in S \land x \notin \O       | c = Rule of Conjunction }} {{eqn | o = \leadsto       | r = x \in S \setminus \O       | c = {{Defof|Set Difference}} }} {{eqn | o = \leadsto       | r = S \subseteq S \setminus \O       | c = {{Defof|Subset}} }} {{end-eqn}} Thus we have: :$S \setminus \O \subseteq S$ and: :$S \subseteq S \setminus \O$ So by definition of set equality: :$S \setminus \O = S$ {{qed}}
From Set is Subset of Itself: :$S \subseteq S$ From Set Difference with Superset is Empty Set we have: :$S \subseteq T \iff S \setminus T = \O$ Hence the result. {{qed}}
Assume $S, T \subseteq \Bbb U$ where $\Bbb U$ is a universal set. {{begin-eqn}} {{eqn | l = S \setminus T       | r = S }} {{eqn | ll= \leadstoandfrom       | l = S \cap \map \complement T       | r = S       | c = Set Difference as Intersection with Complement }} {{eqn | ll= \leadstoandfrom       | l = S       | o = \subseteq       | r = \map \complement T       | c = Intersection with Subset is Subset }} {{eqn | ll= \leadstoandfrom       | l = S \cap \map \complement {\map \complement T}       | r = \O       | c = Intersection with Complement is Empty iff Subset }} {{eqn | ll= \leadstoandfrom       | l = S \cap T       | r = \O       | c = Complement of Complement }} {{end-eqn}} {{qed}}
From Set Difference and Intersection form Partition: : $\paren {R \setminus S} \cup \paren {R \cap S} = R = \paren {R \setminus T} \cup \paren {R \cap T}$ : $\paren {R \cap S} \cap \paren {R \setminus S} = \O = \paren {R \cap T} \cap \paren {R \setminus T}$ whatever $R, S, T$ might be. Let $R \setminus S = R \setminus T$. Then: {{begin-eqn}} {{eqn | l = \paren {\paren {R \setminus S} \cup \paren {R \cap S} } \setminus \paren {R \setminus S}       | r = \paren {\paren {R \setminus T} \cup \paren {R \cap T} } \setminus \paren {R \setminus T} }} {{eqn | ll= \leadsto       | l = \paren {R \cap S} \setminus \paren {R \setminus S}       | r = \paren {R \cap T} \setminus \paren {R \setminus T}       | c = Set Difference with Union is Set Difference }} {{end-eqn}} Now, we have from Set Difference with Disjoint Set: :$S \cap T = \O \iff S \setminus T = S$ and so: :$\paren {R \cap S} \setminus \paren {R \setminus S} = R \cap S$ and: :$\paren {R \cap T} \setminus \paren {R \setminus T} = R \cap T$ So: :$R \cap S = R \cap T$ We can use exactly the same reasoning if we assume $R \cap S = R \cap T$: {{begin-eqn}} {{eqn | l = \paren {\paren {R \setminus S} \cup \paren {R \cap S} } \setminus \paren {R \cap S}       | r = \paren {\paren {R \setminus T} \cup \paren {R \cap T} } \setminus \paren {R \cap T} }} {{eqn | ll= \leadsto       | l = \paren {R \setminus S} \setminus \paren {R \cap S}       | r = \paren {R \setminus T} \setminus \paren {R \cap T}       | c = Set Difference with Union is Set Difference }} {{end-eqn}} and then because of Set Difference with Disjoint Set as above: :$\paren {R \setminus S} \setminus \paren {R \cap S} = R \setminus S$ and: :$\paren {R \setminus T} \setminus \paren {R \cap T} = R \setminus T$ So: :$R \setminus S = R \setminus T$ {{qed}}
Consider $R, S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the universe. Then: {{begin-eqn}} {{eqn | l = \left({R \setminus S}\right) \cap T       | r = \left({R \cap \complement \left({S}\right)}\right) \cap T       | c = Set Difference as Intersection with Complement }} {{eqn | r = \left({R \cap T}\right) \cap \complement \left({S}\right)       | c = Intersection is Commutative and Intersection is Associative }} {{eqn | r = \left({R \cap T}\right) \setminus S       | c = Set Difference as Intersection with Complement }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | o =        | r = x \in \paren {R \setminus S} \cap T       | c =  }} {{eqn | ll= \leadstoandfrom       | o =        | r = \paren {x \in R \land x \notin S} \land x \in T       | c = {{Defof|Set Intersection}} and {{Defof|Set Difference}} }} {{eqn | ll= \leadstoandfrom       | o =        | r = \paren {x \in R \land x \in T} \land x \notin S       | c = Rule of Commutation and Rule of Association }} {{eqn | ll= \leadstoandfrom       | o =        | r = x \in \paren {R \cap T} \setminus S       | c = {{Defof|Set Intersection}} and {{Defof|Set Difference}} }} {{end-eqn}} {{qed}}
Suppose you were a member of a club with $n + 1$ members (including you). Suppose it were time to elect a committee of $k$ members from that club. From Cardinality of Set of Subsets, there are $\dbinom {n + 1} k$ ways to select the members to form this committee. Now, you yourself may or may not be elected a member of this committee. Suppose that, after the election, you are not a member of this committee. Then, from Cardinality of Set of Subsets, there are $\dbinom n k$ ways to select the members to form such a committee. Now suppose you ''are'' a member of the committee. Apart from you, there are $k - 1$ such members. Again, from Cardinality of Set of Subsets, there are $\dbinom n {k - 1}$ ways of selecting the ''other'' $k - 1$ members so as to form such a committee. In total, then, there are $\dbinom n k + \dbinom n {k - 1}$ possible committees. Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = \paren {S \setminus T} \cup \paren {S \cap T}       | r = \paren {\paren {S \setminus T} \cup S} \cap \paren {\paren {S \setminus T} \cup T}       | c = Union Distributes over Intersection }} {{eqn | r = S \cap \paren {\paren {S \setminus T} \cup T}       | c = Set Difference Union First Set is First Set }} {{eqn | r = S \cap \paren {S \cup T}       | c = Set Difference Union Second Set is Union }} {{eqn | r = S       | c = Intersection Absorbs Union }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \left({S \setminus T}\right) \cup \left({S \cap T}\right)       | r = S \setminus \left({T \setminus T}\right)       | c = Set Difference with Set Difference is Union of Set Difference with Intersection }} {{eqn | r = S \setminus \varnothing       | c = Set Difference with Self is Empty Set }} {{eqn | r = S       | c = Set Difference with Empty Set is Self }} {{end-eqn}} {{qed}}
By Set Difference is Subset: :$S \setminus T \subseteq S$ By Intersection is Subset: :$S \cap T \subseteq S$ Hence from Union is Smallest Superset: :$\paren {S \setminus T} \cup \paren {S \cap T} \subseteq S$ Let $s \in S$. Either: :$s \in T$, in which case $s \in S \cap T$ by definition of set intersection or :$s \notin T$, in which case $s \in S \setminus T$ by definition of set difference. That is, by definition of set union: :$s \in \paren {S \setminus T} \cup \paren {S \cap T}$ and so by definition of subset: :$S \subseteq \paren {S \setminus T} \cup \paren {S \cap T}$ Hence the result by definition of set equality. {{qed}}
{{begin-eqn}} {{eqn | l = S \setminus \paren {S \setminus T}       | r = \paren {S \setminus S} \cup \paren {S \cap T}       | c = Set Difference with Set Difference is Union of Set Difference with Intersection }} {{eqn | r = \O \cup \paren {S \cap T}       | c = Set Difference with Self is Empty Set }} {{eqn | r = S \cap T       | c = Union with Empty Set }} {{end-eqn}} Interchanging $S$ and $T$: {{begin-eqn}} {{eqn | l = T \setminus \paren {T \setminus S}       | r = T \cap S }} {{eqn | r = S \cap T       | c = Intersection is Commutative }} {{end-eqn}} {{qed}}