hcju/proofwikiqa huggingface dataset:
Let $n \in \Z_{>0}$. For $n \le 3$ the cases can be examined in turn. :$\set {1, 3, 5}$ are not all prime, as $1$ is not classified as prime. :$\set {2, 4, 6}$ are not all prime, as $4$ and $6$ both have $2$ as a divisor. :$\set {3, 5, 7}$ is the set of three consecutive odd primes which is asserted as being the unique prime triplet. Suppose $n > 3$. If $n$ is even then it has $2$ as a divisor, and so the set $\set {n, n + 2, n + 4}$ contains only composite numbers. So, suppose now that $n$ is an odd integer such that $n > 3$. Any integer $n$ can be represented as either: {{begin-eqn}} {{eqn | l = n       | r = 3 k }} {{eqn | l = n       | r = 3 k + 1 }} {{eqn | l = n       | r = 3 k + 2 }} {{end-eqn}} If $n = 3 k$, then $n$ is not prime because $3 \divides 3 k$. If $n = 3 k + 1$, then $n + 2$ is not prime because $3 \divides 3 k + 3$. If $n = 3 k + 2$, then $n + 4$ is not prime because $3 \divides 3 k + 6$. Therefore no such $n$ exists for which $n$, $n + 2$, and $n + 4$ are all prime. {{Qed}}
By Upper Bound of Ordinal Sum: :$\displaystyle \sum_{i \mathop = 1}^n \paren {x^{a_i} \times b_i} \le x^{a_1} \times \paren {b_1 + 1}$ So: {{begin-eqn}} {{eqn | l = \paren {\sum_{i \mathop = 1}^n x^{a_i} \times b_i}^y       | o = \le       | r = \paren {x^{a_1} \times \paren {b_1 + 1} }^y       | c = Subset is Right Compatible with Ordinal Exponentiation }} {{eqn | r = \paren {x^{a_1} }^y \times \paren {b_1 + 1}       | c = Ordinal Multiplication via Cantor Normal Form/Limit Base }} {{eqn | r = x^{a_1 \mathop \times y} \times \paren {b_1 + 1}       | c = Ordinal Power of Power }} {{end-eqn}} {{qed}} {{explain|I can't see how $\paren {x^{a_1} \times \paren {b_1 + 1} }^y {{=}} \paren {x^{a_1} }^y \times \paren {b_1 + 1}$, from the link suggested or otherwise. I would have expected the RHS to be $\paren {x^{a_1} }^y \times \paren {b_1 + 1}^y$, unless $\paren {b_1 + 1} {{=}} \paren {b_1 + 1}^y$, which I can't get my head round without a specific link to it somewhere.}}
By Upper Bound of Ordinal Sum: :$\displaystyle \sum_{i \mathop = 1}^n \paren {x^{a_i} \times b_i} \le x^{a_1} \times \paren {b_1 + 1}$ Furthermore: :$\displaystyle x^{a_1} \le \sum_{i \mathop = 1}^n \paren {x^{a_i} \times b_i}$ It follows that: {{begin-eqn}} {{eqn | l = \paren {x^{a_1} }^y       | o = \le       | r = \paren {\sum_{i \mathop = 1}^n x^{a_i} \times b_i}^y       | c = Subset is Right Compatible with Ordinal Exponentiation }} {{eqn | o = \le       | r = \paren {x^{a_1} \times \paren {b_1 + 1} }^y       | c = Subset is Right Compatible with Ordinal Exponentiation }} {{eqn | r = x^{a_1 \times y}       | c = Ordinal Exponentiation of Terms }} {{end-eqn}} It follows that: {{begin-eqn}} {{eqn | l = x^{a_1 \mathop + y}       | o = \le       | r = \paren {\sum_{i \mathop = 1}^n x^{a_i} b_i}^y       | c = Ordinal Power of Power }} {{eqn | o = \le       | r = x^{a_1 \mathop + y}       | c = proven above }} {{eqn | ll= \leadsto       | l = \paren {\sum_{i \mathop = 1}^n x^{a_i} b_i}^y       | r = x^{a_1 \mathop + y}       | c = {{Defof|Set Equality|index = 2}} }} {{end-eqn}} {{qed}}
By the hypothesis, $x^y$ is a limit ordinal by Limit Ordinals Closed under Ordinal Exponentiation. The result follows from Ordinal Exponentiation via Cantor Normal Form/Limit Exponents. {{qed}}
For any set $X$. we have: {{begin-eqn}} {{eqn|l = C \operatorname{id}_X      |r = \operatorname{id}_X^{-1}      |c = Definition of inverse relation functor }} {{eqn|r = \left\{ {\left({x, x}\right): \left({x, x}\right) \in \operatorname{id}_X} \right\}      |c = Definition of inverse relation }} {{eqn|r = \operatorname{id}_X }} {{end-eqn}} For any two relations $\mathcal R \subseteq X \times Y$ and $\mathcal S \subseteq Y \times Z$, we have: {{begin-eqn}} {{eqn|l = C \left({\mathcal S \circ \mathcal R}\right)      |r = \left({\mathcal S \circ \mathcal R}\right)^{-1}      |c = Definition of inverse relation functor }} {{eqn|r = \mathcal R^{-1} \circ \mathcal S^{-1}      |c = Inverse of Composite Relation }} {{eqn|r = C \mathcal R \circ C \mathcal S      |c = Definition of inverse relation functor }} {{end-eqn}} Hence $C$ is shown to be a contravariant functor. {{qed}}
=== Necessary Condition === Suppose that $f$ is injective. Suppose further that we have mappings $g, h: Z \to X$ such that $g \ne h$. Then necessarily there exists some $z \in Z$ such that $g \left({z}\right) \ne h \left({z}\right)$ by Equality of Mappings. As $f$ is injective, it follows that: :$f \left({g \left({z}\right)}\right) \ne f \left({h \left({z}\right)}\right)$ which, again by Equality of Mappings, means that $g \circ f \ne h \circ f$. Hence $f$ is monic, by the Rule of Transposition. {{qed|lemma}} === Sufficient Condition === Suppose that $f: X \rightarrowtail Y$ is a monomorphism. By definition of injection, it will suffice to show that: :$x \ne x' \implies f \left({x}\right) \ne f \left({x'}\right)$ To this end, consider a singleton $\left\{{a}\right\}$, and define: :$\bar x: \left\{{a}\right\} \to X, \bar x \left({a}\right) := x$ :$\bar x': \left\{{a}\right\} \to X, \bar x' \left({a}\right) := x'$ In particular, $\bar x \ne \bar x'$, and so, $f$ being monic, we deduce: :$f \circ \bar x \ne f \circ \bar x'$ It follows that it must be that: {{begin-eqn}} {{eqn|l = f \left({x}\right)      |r = f \left({\bar x \left({a}\right)}\right) }} {{eqn|o = \ne      |r = f \left({\bar x' \left({a}\right)}\right)      |c = Equality of Mappings }} {{eqn|r = f \left({x'}\right) }} {{end-eqn}} Hence $f$ is injective. {{qed}}
$\omega$ is a set by the Axiom of Infinity. Thus by the Axiom of Replacement, the image of $\omega$ under $F$ is also a set. Since $T$ is the union of $\map F \omega$, it is thus a set by the axiom of unions. Furthermore: {{begin-eqn}} {{eqn | l = x       | o = \in       | r = T       | c = Assumption }} {{eqn | ll= \leadsto       | lo= \exists n:       | l = x       | o = \in       | r = \map F n       | c = {{Defof|Union of Family}} }} {{eqn | ll= \leadsto       | lo= \exists n:       | l = x       | o = \subseteq       | r = \bigcup \map F n       | c = Set is Subset of Union }} {{eqn | l = y       | o = \in       | r = x       | c = Assumption }} {{eqn | ll= \leadsto       | lo= \exists n:       | l = y       | o = \in       | r = \bigcup \map F n       | c =  }} {{eqn | ll= \leadsto       | lo= \exists n:       | l = y       | o = \in       | r = \map F {n + 1}       | c = Definition of $F$ }} {{eqn | ll= \leadsto       | l = y       | o = \in       | r = T       | c = Definition of $T$ }} {{end-eqn}} {{explain|domain of $n$}} By the above equations: :$x \in T \land y \in x \implies y \in T$ Thus by definition $T$ is transitive. {{qed|lemma}} We have that: :$S = \map F 0$ By Set is Subset of Union: :$\displaystyle S \subseteq \bigcup_{n \mathop \in \omega} \map F n$ So $S \subseteq T$. {{qed|lemma}} Finally, suppose that $S \subseteq R$ and $R$ is transitive. $T \subseteq R$ follows by finite induction: For all $n \in \omega$, let $\map P n$ be the proposition: :$\map F n \subseteq R$ === Basis for the Induction === $\map P 0$ is the case: :$\map F 0 \subseteq R$ which has been proved above. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\map F k \subseteq R$ Then we need to show: :$\map F {k + 1} \subseteq R$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | n = 1       | l = \map F k       | o = \subseteq       | r = R       | c = Hypothesis }} {{eqn | ll= \leadsto       | l = \bigcup \map F k       | o = \subseteq       | r = \bigcup R       | c = Set Union Preserves Subsets }} {{eqn | n = 2       | o = \subseteq       | r = R       | c = Class is Transitive iff Union is Subset }} {{eqn | ll= \leadsto       | l = \map F k \cup \bigcup \map F k       | o = \subseteq       | r = R       | c = by $(1)$, $(2)$, Set Union Preserves Subsets }} {{eqn | ll= \leadsto       | l = \map F {k + 1}       | o = \subseteq       | r = R       | c = Definition of $\map F {k + 1}$ }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\map F n \subseteq R$ for all $n \in \omega$ :$T \subseteq R$ {{qed}}
Suppose that $g, h \in \mathbf P_1$ are morphisms such that: :$f \circ g = f \circ h$ In particular then, $g$ and $h$ have equal domain and codomain. Since $\mathbf P$ is a preorder category, there is at most one morphism between any two objects. Thus necessarily $g = h$, and hence $f$ is monic. {{qed}}
Suppose that $g, h \in \mathbf P_1$ are morphisms such that: :$g \circ f = h \circ f$ In particular then, $g$ and $h$ have equal domain and codomain. Since $\mathbf P$ is a preorder category, there is at most one morphism between any two objects. Thus necessarily $g = h$, and hence $f$ is epic. {{qed}}
=== Necessary Condition === Suppose that $f$ is surjective. Suppose further that we have mappings $g, h: Y \to Z$ such that $g \ne h$. Then necessarily there exists some $y \in Y$ such that $g \left({y}\right) \ne h \left({y}\right)$ by Equality of Mappings. As $f$ is surjective, it follows that there is an $x \in X$ such that $f \left({x}\right) = y$. Hence, we conclude that: :$g \left({f \left({x}\right)}\right) \ne h \left({f \left({x}\right)}\right)$ which, again by Equality of Mappings, means that $g \circ f \ne h \circ f$. Therefore, $f$ is epic, by the Rule of Transposition. {{qed|lemma}} === Sufficient Condition === Suppose that $f: X \twoheadrightarrow Y$ is an epimorphism. By definition of surjection, it will suffice to show that: :$\forall y \in Y: \exists x \in X: f \left({x}\right) = y$ Let us reason by contradiction. So suppose $f$ were not surjective. Then there would be an $y_0 \in Y$ such that: :$\forall x \in X: f \left({x}\right) \ne y_0$ Consider the mappings defined by: :$g: Y \to Y \cup \left\{{Y}\right\}, g \left({y}\right) := y$ :$h: Y \to Y \cup \left\{{Y}\right\}, h \left({y}\right) := \begin{cases}y & \text{if $y \ne y_0$}\\ Y & \text{if $y = y_0$} \end{cases}$ The assumption on $f$ yields that $g \circ f = h \circ f$. Since $h \ne g$, it follows that $f$ cannot be epic. This contradiction shows that $f$ is necessarily surjective. {{qed}}
Let $\left({M, \circ}\right)$ be a monoid with identity $e$. Let $f, g: \Z \to M$ be monoid homomorphisms such that: :$f \circ \iota = g \circ \iota$ that is, by definition of inclusion, such that: :$\forall n \in \N: f \left({n}\right) = g \left({n}\right)$ Now $\iota$ will be epic if we can show that $f = g$. The morphism property of $f$ and $g$ yields that, for any $m > 0$: :$f \left({- m}\right) = f \left({-1}\right) \circ \cdots \circ f \left({-1}\right)$ :$g \left({- m}\right) = g \left({-1}\right) \circ \cdots \circ g \left({-1}\right)$ with on the right-hand side $m$ copies of $f \left({-1}\right)$ and $g \left({-1}\right)$, respectively. It thus is necessary and sufficient to show that $f \left({-1}\right) = g \left({-1}\right)$. To this end, compute: {{begin-eqn}} {{eqn|l = f \left({-1}\right)      |r = f \left({-1}\right) \circ e      |c = $e$ is an identity }} {{eqn|r = f \left({-1}\right) \circ g \left({0}\right)      |c = $g$ is a monoid homomorphism }} {{eqn|r = f \left({-1}\right) \circ g \left({1 + \left({-1}\right)}\right) }} {{eqn|r = f \left({-1}\right) \circ g \left({1}\right) \circ g \left({-1}\right)      |c = $g$ is a monoid homomorphism }} {{eqn|r = f \left({-1}\right) \circ f \left({1}\right) \circ g \left({-1}\right)      |c = By the assumption on $f$ and $g$ }} {{eqn|r = f \left({\left({-1}\right) + 1}\right) \circ g \left({-1}\right)      |c = $f$ is a monoid homomorphism }} {{eqn|r = f \left({0}\right) \circ g \left({-1}\right) }} {{eqn|r = e \circ g \left({-1}\right)      |c = $f$ is a monoid homomorphism }} {{eqn|r = g \left({-1}\right)      |c = $e$ is an identity }} {{end-eqn}} Hence the result. {{qed}}
Since $f$ is an isomorphism, it is a fortiori a split monomorphism. The result follows from Split Monomorphism is Monic. {{qed}}
Let $g \in \Stab x$. Then: {{begin-eqn}} {{eqn | l = g * x       | r = x       | c = {{Defof|Stabilizer}} }} {{eqn | ll= \leadsto       | l = g \circ x       | r = x       | c = {{Defof|Group Action}} (this particular one) }} {{eqn | ll= \leadsto       | l = g       | r = e       | c = {{Defof|Identity Element}} }} {{end-eqn}} Hence the result, by definition of trivial subgroup. {{qed}}
From the definition of centralizer: :$\map {C_G} x = \set {g \in G: g \circ x = x \circ g}$ Then: {{begin-eqn}} {{eqn | l = z       | o = \in       | r = \Stab x       | c =  }} {{eqn | ll= \leadstoandfrom       | l = z       | o = \in       | r = \set {g \in G: g \circ x \circ g^{-1} = x}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = z       | o = \in       | r = \set {g \in G: g \circ x = x \circ g}       | c =  }} {{end-eqn}} {{qed}}
From the definition of stabilizer: :$\Stab S = \set {g \in G: g * S = S}$ The result follows from the definition of the group action $*$ given. {{qed}}
From the definition of Stabilizer of Subset Product Action on Power Set: :$\Stab H = H = \set {g \in G: g * H = H}$ The result follows from Left Coset Equals Subgroup iff Element in Subgroup. {{qed}}
Since $f$ is an isomorphism, it is a fortiori a split epimorphism. The result follows from Split Epimorphism is Epic. {{qed}}
Let $g: D \to C$ be a morphism such that: :$f \circ g = \operatorname{id}_D$ which is guaranteed to exist by definition of split epimorphism. Suppose that $x, y: D \to E$ are morphisms such that: :$x \circ f = y \circ f$ Then necessarily also: :$x \circ f \circ g = y \circ f \circ g$ and hence, since $f \circ g = \operatorname{id}_D$, it follows that: :$x \circ \operatorname{id}_D = y \circ \operatorname{id}_D$ which yields the result by the definition of identity morphism. The situation is illustrated by the following commutative diagram: ::$\begin{xy} <0em,0em> *+{C} = "C", <0em,-4em>*+{D} = "D", <4em,0em> *+{D} = "D2", <8em,0em> *+{E} = "E", "D2"+/r.5em/+/^.25em/;"E"+/l.5em/+/^.25em/ **@{-} ?>*@{>} ?*!/_.6em/{x}, "D2"+/r.5em/+/_.25em/;"E"+/l.5em/+/_.25em/ **@{-} ?>*@{>} ?*!/^.6em/{y}, "C";"D2"  **@{-} ?>*@{>} ?*!/_.6em/{f}, "D";"D2" **@{-} ?>*@{>} ?*!/^.6em/{\operatorname{id}_C}, "D";"C" **@{-} ?>*@{>} ?*!/_.6em/{g}, \end{xy}$ {{qed}}
Let $g: D \to C$ be a morphism such that: :$g \circ f = \operatorname{id}_C$ which is guaranteed to exist by definition of split monomorphism. Suppose that $x, y: B \to C$ are morphisms such that: :$f \circ x = f \circ y$ Then necessarily also: :$g \circ f \circ x = g \circ f \circ y$ and hence, since $g \circ f = \operatorname{id}_C$, it follows that: :$\operatorname{id}_C \circ x = \operatorname{id}_C \circ y$ which yields the result by the definition of identity morphism. The situation is illustrated by the following commutative diagram: ::$\begin{xy} <-4em,0em>*+{B} = "B", <0em,0em> *+{C} = "C", <4em,0em> *+{D} = "D", <4em,-4em>*+{C} = "C2", "B"+/r.5em/+/^.25em/;"C"+/l.5em/+/^.25em/ **@{-} ?>*@{>} ?*!/_.6em/{x}, "B"+/r.5em/+/_.25em/;"C"+/l.5em/+/_.25em/ **@{-} ?>*@{>} ?*!/^.6em/{y}, "C";"D"  **@{-} ?>*@{>} ?*!/_.6em/{f}, "C";"C2" **@{-} ?>*@{>} ?*!/^.6em/{\operatorname{id}_C}, "D";"C2" **@{-} ?>*@{>} ?*!/_.6em/{g}, \end{xy}$ {{qed}}
Consider the identity morphism $\operatorname{id}_P: P \to P$. By definition of projective object, we obtain the following commutative diagram: ::$\begin{xy} <0em,0em>*+{P} = "P", <4em,0em>*+{P} = "P2", <4em,4em>*+{E} = "E", "P";"E" **@{-} ?>*@{>} ?*!/_.6em/{f}, "P";"P2" **@{-} ?>*@{>} ?*!/^.6em/{\operatorname{id}_P}, "E";"P2" **@{-} ?>*@2{>} ?<>(.7)*{\vee} ?*!/_.6em/{e}, \end{xy}$ where $f = \bar{\operatorname{id}_P}$. It follows that $e \circ f = \operatorname{id}_P$, and so $e$ is a split epimorphism. {{qed}}
=== Proof of $(1)$ === Let $x \in A$ and $y \in T$ such that $x \prec y$. Then by definition: :$x \in {\prec^{-1}} \left({y}\right)$ Moreover, since $y \in T$: :$\exists n \in \omega: y \in F \left({n}\right)$ Therefore: :$x \in F \left({n+1}\right)$ and so: :$x \in T$ {{qed|lemma}} === Proof of $(2)$ === Let $x \in S$. We have that: :$F \left({0}\right) = S$ Therefore: :$\exists n \in \omega: x \in F \left({n}\right)$ It follows that $x \in T$. By the definition of subset, it follows that: :$S \subseteq T$ {{qed|lemma}} === Proof of $(3)$ === Let $R$ be $\prec$-transitive. Let $S \subseteq R$. Then: :$F \left({n}\right) \subseteq R$ shall be proved by finite induction. For all $n \in \omega$, let $P \left({n}\right)$ be the proposition: :$F \left({n}\right) \subseteq R$ === Basis for the Induction === $P \left({0}\right)$ is the case: :$F \left({0}\right) \subseteq R$ which has been proved above. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k+1}\right)$ is true. So this is our induction hypothesis: :$F \left({k}\right) \subseteq R$ Then we need to show: :$F \left({k+1}\right) \subseteq R$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = F \left({ k }\right)       | o = \subseteq       | r = R       | c = Hypothesis $(1)$ }} {{eqn | ll= \implies       | l = { \prec^{-1} } \left({ F \left({ k }\right) }\right)       | o = \subseteq       | r = { \prec^{-1} } \left({ R }\right)       | c = Image Preserves Subsets }} {{eqn | l = A \cap { \prec^{-1} } \left({ R }\right)       | o = \subseteq       | r = R       | c = Definition of $\prec$-transitive (2) }} {{eqn | ll = \implies       | l = F \left({ k }\right) \cup \left({ A \cap { \prec^{-1} } \left({ F \left({ k }\right) }\right) }\right)       | o = \subseteq       | r = R       | c = by $(1)$, $(2)$, Set Union Preserves Subsets }} {{eqn | ll = \implies       | l = F \left({k+1}\right)       | o = \subseteq       | r = R       | c = Definition of $F \left({k+1}\right)$ }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$F \left({n}\right) \subseteq R$ for all $n \in \omega$ Then by Indexed Union Subset: :$T \subseteq R$ {{qed}}
We have that: :$\Stab H = \set {g \in G: g \circ H \circ g^{-1} = H}$ which is precisely how the normalizer is defined. {{Qed}}
$B$ is not empty. So $B$ has at least one element $x$. By Singleton of Element is Subset: : $\left\{{ x }\right\} \subseteq B$ By Relational Closure Exists for Set-Like Relation: : $\left\{ x \right\}$ has a $\prec$-relational closure. {{explain|Set-like relation: it needs to be shown that $\prec$ is one of those.}} This $\prec$-relational closure shall be denoted $y$. By Intersection is Subset: :$y \cap B \subseteq A$ As $x \in y$ and $x \in B$: :$y \cap B \ne \varnothing$ By the definition of foundational relation: :$(1): \quad \exists x \in \left({ y \cap B }\right): \forall w \in \left({ y \cap B }\right): w \not \prec x$ Suppose that $w \in B$ and $w \prec x$. Since $x \in y$, it follows that $w \in y$, so $w \in \left({ B \cap y }\right)$ by the definition of intersection. This contradicts $(1)$. Therefore: : $w \not \prec x$ and so $B$ has a $\prec$-minimal element. {{qed}}
Consider the following commutative diagram: ::$\begin{xy} <-4em,0em>*+{0} = "M", <0em,0em> *+{0'}= "N", <0em,-4em>*+{0} = "M2", <4em,-4em>*+{0'}= "N2", "M";"N"  **@{-} ?>*@{>} ?*!/_.6em/{u}, "M";"M2" **@{-} ?>*@{>} ?*!/^.6em/{\operatorname{id}_0}, "N";"M2" **@{-} ?>*@{>} ?*!/_.6em/{v}, "N";"N2" **@{-} ?>*@{>} ?*!/_1em/{\operatorname{id}_{0'}}, "M2";"N2"**@{-} ?>*@{>} ?*!/^.6em/{u}, \end{xy}$ It commutes as each of the morphisms in it originates from an initial object, and hence is unique. Thus, $v$ is an inverse to $u$, and so $u$ is an isomorphism. {{qed}}
Consider the following commutative diagram: ::$\begin{xy} <-4em,0em>*+{1} = "M", <0em,0em> *+{1'}= "N", <0em,-4em>*+{1} = "M2", <4em,-4em>*+{1'}= "N2", "M";"N"  **@{-} ?>*@{>} ?*!/_.6em/{u}, "M";"M2" **@{-} ?>*@{>} ?*!/^.6em/{\operatorname{id}_1}, "N";"M2" **@{-} ?>*@{>} ?*!/_.6em/{v}, "N";"N2" **@{-} ?>*@{>} ?*!/_1em/{\operatorname{id}_{1'}}, "M2";"N2"**@{-} ?>*@{>} ?*!/^.6em/{u}, \end{xy}$ It commutes as each of the morphisms in it points to a terminal object, and hence is unique. Thus, $v$ is an inverse to $u$, and so $u$ is an isomorphism. {{qed}}
Let $T$ be a set, and let $f: T \to S$ be a mapping. Then since for all $t \in T$, we have $f \left({t}\right) \in S$, it follows that: :$\forall t \in T: f \left({t}\right) = x$ By Equality of Mappings, there is precisely one such mapping $f: T \to S$. Hence the result, by definition of terminal object. {{qed}}
Let $\mathbf C$ be an object of $\mathbf{Cat}$, i.e. a small category. By Empty Mapping is Unique, there are unique mappings: :$F_0: \mathbf 0_0 = \varnothing \to \mathbf C_0$ :$F_1: \mathbf 0_1 = \varnothing \to \mathbf C_1$ making $F: \mathbf 0 \to \mathbf C$ a functor by vacuous truth. That $F_0$ and $F_1$ are actually mappings follows from $\mathbf C$ being a small category. Hence the result, by definition of initial object. {{qed}}
Let $\mathbf C$ be an object of $\mathbf{Cat}$, i.e. a small category. From Singleton is Terminal Object of Category of Sets, there exist unique mappings: :$F_0: \mathbf C_0 \to \mathbf 1_0 = \left\{{*}\right\}$ :$F_1: \mathbf C_1 \to \mathbf 1_1 = \left\{{\operatorname{id}_*}\right\}$ since the latter sets are singletons. It remains to verify that $F: \mathbf C \to \mathbf 1$ so defined, in fact is a functor. Trivially, $F$ preserves identity morphisms. That $F$ has the morphism property follows from: :$\operatorname{id}_* \circ \operatorname{id}_* = \operatorname{id}_*$ Hence $F: \mathbf C \to \mathbf 1$ constitutes a functor. The result follows by definition of terminal object. {{qed}}
Let $f: \O \to T$ be the empty mapping. Vacuously: :$\forall x \in \O: \exists y \in T: \tuple {x, y} \in f$ thus showing that $f$ is left-total. Also vacuously: :$\forall x \in \O: \forall y_1, y_2 \in T: \tuple {x, y_1} \in f \land \tuple {x, y_2} \in f \implies y_1 = y_2$ thus showing that $f$ is many-to-one. Hence the result, from the definition of a mapping. {{qed}}
By definition a mapping from $\O$ to $T$ is a subset of the cartesian product $\O \times T$: :$f: \O \to T \subseteq \O \times T$ From Empty Mapping is Mapping, we have that the empty mapping from $\O$ to $T$ exists. From Cartesian Product is Empty iff Factor is Empty it follows that the empty mapping equals the empty set: :$\O \times T = \O$ The result follows by Empty Set is Unique. {{qed}}
=== Sufficient Condition === Let $S = \O$. Then the null relation $\RR = \O \subseteq S \times T$ is a mapping from Empty Mapping is Mapping. {{qed|lemma}} === Necessary Condition === Suppose $S \ne \O$. From the definition of an empty set, $S \ne \O \implies \exists x \in S$. Thus: {{begin-eqn}} {{eqn | l = \RR       | r = \O }} {{eqn | ll= \leadsto       | lo= \exists x \in S: \neg \exists y \in T:       | l = \tuple {x, y}       | o = \in       | r = \RR       | c = {{Defof|Empty Set}} }} {{eqn | ll= \leadsto       | l = \RR       | o =       | r = \text{ is not a mapping}       | c = {{Defof|Mapping}} }} {{end-eqn}} {{qed}}
Let $y \in G$. Then: {{begin-eqn}} {{eqn | lo= \exists g \in G:       | l = y       | r = g \circ x       | c = Group has Latin Square Property }} {{eqn | ll= \leadsto       | l = y       | o = \in       | r = \Orb x       | c = {{Defof|Orbit (Group Theory)|Orbit}} }} {{end-eqn}} Hence the result. {{qed}}
Follows from the definition of the conjugacy class. {{explain|Needs expansion.}} {{qed}}
{{NotZFC}} By Epsilon is Foundational, $\Epsilon$, the epsilon relation, is a foundational relation on $B$. By Epsilon Relation is Proper, $\left({\mathbb U, \Epsilon}\right)$ is a proper relational structure, where $\mathbb U$ is the universal class. By Well-Founded Proper Relational Structure Determines Minimal Elements, $B$ has an $\Epsilon$-minimal element. {{qed}}