hcju/proofwikiqa huggingface dataset:
=== Formulation 2 implies Formulation 1 === {{begin-eqn}} {{eqn | n = 1       | o =        | r = \forall y: y = y       | c = Equality is Reflexive }} {{eqn | n = 2       | o =        | r = \neg \exists y: y \ne y       | c = From $(1)$ }} {{eqn | n = 3       | o =        | r = x := \set { {y: y \ne y} }       | c =  }} {{eqn | n = 4       | o =        | r = \neg \exists y: y \in x       | c = From $(2)$ and $(3)$ }} {{end-eqn}} {{qed|lemma}} === Formulation 1 implies Formulation 2 === {{begin-eqn}} {{eqn | n = 1       | o =        | r = \forall y: \paren {\neg \paren {y \in x} }       | c =  }} {{eqn | n = 2       | o =        | r = \forall y: \neg \paren {y \in x} \land \paren {y \in x} \implies y \ne y       | c = Rule of Explosion }} {{eqn | n = 3       | o =        | r = \forall y: \paren {\neg \paren {y \in x} \implies \paren {y \in x \implies y \ne y} }       | c = Rule of Exportation, from $(2)$ }} {{eqn | n = 4       | o =        | r = \forall y: \paren {y \in x \implies y \ne y}       | c = Modus Ponendo Ponens, from $(1)$ and $(3)$ }} {{end-eqn}} {{qed}}
Since $p$ is the smallest element of $\mathbf P_0$, we have: :$\forall q \in \mathbf P_0: p \le q$ that is, for every object $q$ of $\mathbf P$ there is a unique morphism $p \to q$. That is, $p$ is initial. {{qed}}
{{NotZFC}} === Necessary Condition === Let $x \subseteq \operatorname{rank} \left({ S }\right)$. Then by Von Neumann Hierarchy Comparison: :$S \in V \left({x}\right) \implies S \in V \left({\operatorname{rank} \left({S}\right)}\right)$ But by Ordinal Equal to Rank: :$S \notin V \left({\operatorname{rank} \left({S}\right) }\right)$ By contraposition: :$S \notin V \left({x}\right)$ {{qed|lemma}} === Sufficient Condition === Let $S \notin V \left({x}\right)$. Then: {{begin-eqn}} {{eqn | l = S       | o = \in       | r = V \left({\operatorname{rank} \left({ S }\right) + 1 }\right)       | c = Ordinal Equal to Rank }} {{eqn | ll= \implies       | l = V \left({ \operatorname{rank} \left({ S }\right) + 1 }\right)       | o = \not \subseteq       | r = V \left({ x }\right)       | c = Rule of Transposition }} {{eqn | ll= \implies       | l = \operatorname{rank} \left({ S }\right) + 1       | o = \not \subseteq       | r = x       | c = Von Neumann Hierarchy Comparison }} {{eqn | ll= \implies       | l = x       | o = \in       | r = \operatorname{rank} \left({ S }\right) + 1       | c = Transitive Set is Proper Subset of Ordinal iff Element of Ordinal and Ordinal Membership is Trichotomy }} {{eqn | ll= \implies       | l = x       | o = \subseteq       | r = \operatorname{rank} \left({ S }\right)       | c = Definition of Successor Set }} {{end-eqn}} {{qed}}
{{NotZFC}} By Ordinal Equal to Rank: :$T \in V\left({ \operatorname{rank} \left({ T }\right) + 1 }\right)$ By the definition of rank: :$T \subseteq V\left({ \operatorname{rank} \left({ T }\right) }\right)$ Since $S \in T$: :$S \in V\left({ \operatorname{rank} \left({ T }\right) }\right)$ By Ordinal is Subset of Rank of Small Class iff Not in Von Neumann Hierarchy: :$\operatorname{rank} \left({ T }\right) \not \subseteq \operatorname{rank} \left({ S }\right)$ Therefore by Ordinal Membership is Trichotomy and Transitive Set is Proper Subset of Ordinal iff Element of Ordinal: :$\operatorname{rank} \left({ S }\right) < \operatorname{rank} \left({ T }\right)$ {{qed}}
Since $p$ is the greatest element of $\mathbf P_0$, we have: :$\forall q \in \mathbf P_0: q \le p$ i.e., for every object $q$ of $\mathbf P$ there is a unique morphism $q \to p$. That is, $p$ is terminal. {{qed}}
Let $f: B \to C$ be an object of $\mathbf C \mathop / C$. Then there is a morphism $a: f \to \operatorname{id}_C$ {{iff}}: :$f = \operatorname{id}_C \circ a = a$ Thus, $f$ itself defines the unique morphism $f \to \operatorname{id}_C$ in $\mathbf C \mathop / C$. We therefore have the following commutative diagram in $\mathbf C$: ::$\begin{xy} <-3em,0em>*+{B} = "X", <3em,0em>*+{C} = "X2", <0em,-4em>*+{C} = "C", "X";"X2" **@{--} ?>*@{>} ?*!/_1em/{f}, "X";"C" **@{-} ?>*@{>} ?<>(.3)*!/^1em/{f}, "X2";"C" **@{-} ?>*@{>} ?<>(.3)*!/_1em/{\operatorname{id}_C}, \end{xy}$ Hence the result, by definition of terminal object. {{qed}}
Let $f: C \to D$ be an object of $C \mathop / \mathbf C$. Then there is a morphism $a: \operatorname{id}_C \to f$ iff: :$f = a \circ \operatorname{id}_C = a$ Thus, $f$ itself defines the unique morphism $\operatorname{id}_C \to f$ in $C \mathop / \mathbf C$. We therefore have the following commutative diagram in $\mathbf C$: ::$\begin{xy} <-3em,0em>*+{C} = "X", <3em,0em>*+{D} = "X2", <0em,4em>*+{C} = "C", "X";"X2" **@{--} ?>*@{>} ?*!/^1em/{f}, "C";"X" **@{-} ?>*@{>} ?*!/^.6em/{\operatorname{id}_C}, "C";"X2" **@{-} ?>*@{>} ?<>(.6)*!/_1em/{f'}, \end{xy}$ Hence the result, by definition of initial object. {{qed}}
{{NotZFC}} Let: :$T = \bigcap \left\{ x \in \operatorname{On} : \forall y \in S: \operatorname{rank} \left({ y }\right) < x \right\}$ Let $y \in S$. Then by Membership Rank Inequality: :$\operatorname{rank} \left({ x }\right) < \operatorname{rank} \left({ S }\right)$ {{explain|$x$ or $y$?}} Therefore: :$T \subseteq \operatorname{rank} \left({ S }\right)$ Conversely, take any $x \in T$. By the definition of $T$: :$\forall y \in S: \operatorname{rank} \left({ y }\right) < x$ From Ordinal is Subset of Rank of Small Class iff Not in Von Neumann Hierarchy: : $\forall y \in S: y \in V \left({ x }\right)$ where $V \left({x}\right)$ denotes the von Neumann hierarchy. Therefore by the definition of subset: :$S \subseteq V \left({x}\right)$ By the definition of rank. :$\operatorname{rank} \left({ S }\right) \le x$ Therefore: : $\operatorname{rank} \left({ S }\right) = T$ {{qed}}
{{NotZFC}} The proof shall proceed by Transfinite Induction (Strong Induction) on $x$. Suppose $\forall y \in x: \map {\operatorname {rank} } y = y$. Then: {{begin-eqn}} {{eqn | l = \map {\operatorname {rank} } x       | r = \bigcap \set {z \in \On: \forall y \in x: y < z}       | c = Rank of Set Determined by Members }} {{eqn | r = \bigcap \set {z \in \On: x \subseteq z}       | c = {{Defof|Subset}} }} {{eqn | r = x       | c =  }} {{end-eqn}} {{qed}}
{{NotZFC}} Let $V$ denote the von Neumann hierarchy. Then: {{begin-eqn}} {{eqn | ll= \forall x \in S:       | l = \operatorname{rank} \left({x}\right)       | o = \le       | r = y       | c = by hypothesis }} {{eqn | ll= \implies       | l = x       | o = \in       | r = V \left({y + 1}\right)       | c = Ordinal is Subset of Rank of Small Class iff Not in Von Neumann Hierarchy and Ordinal Equal to Rank }} {{eqn | ll= \implies       | l = S       | o = \subseteq       | r = V \left({y + 1}\right)       | c = Definition of Subset }} {{end-eqn}} Therefore, by Axiom of Subsets Equivalents, it follows that $S$ is a small class. {{qed}}
Let $E$ be the total number of subsets of $S$ whose cardinality is even. From Cardinality of Set of Subsets, the number of subsets of $S$ with $m$ elements is $\displaystyle \binom n m$: :$\displaystyle \binom n m = \frac{n!}{m! \left({n-m}\right)}$ where $\displaystyle \binom n m$ is a binomial coefficient. Thus the total number of subsets of $S$ whose cardinality is even is given by: :$\displaystyle E = \binom n 0 + \binom n 2 + \binom n 4 + \cdots = \sum_{j \mathop \in \Z} \binom n {2j}$ Note the loose limits of the summation sign: the expression truly ranges over all $\Z$. This is because, when $2 j < 0$ and $2 j > n$, $\displaystyle \binom n {2j} = 0$ by definition of binomial coefficient. The result then follows from Sum of Even Index Binomial Coefficients: :$\displaystyle \sum_{j \mathop \ge 0} \binom n {2 j} = 2^{n-1}$ {{qed}}
Proof by induction: For all $n \in \N$, let $P \left({n}\right)$ be the proposition: :The number of subsets of $S$ whose cardinality is even is $2^{n-1}$, where $\left|{S}\right| = n$. === Basis for the Induction === When $n = 1$ we have from Cardinality of Power Set of Finite Set that $S$ has $2^1 = 1$ subsets: $\varnothing$ and $S$ itself. We have that $\left|{S}\right| = 1$ and $\left|{\varnothing}\right| = 0$. So there is indeed $2^{1-1} = 2^0 = 1$ subset of $S$ whose cardinality is even, that is $\varnothing$. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true. So this is our induction hypothesis: :The number of subsets of $S$ whose cardinality is even is $2^{k-1}$, where $\left|{S}\right| = k$. Then we need to show: :The number of subsets of $S$ whose cardinality is even is $2^k$, where $\left|{S}\right| = k+1$. === Induction Step === This is our induction step: Let $\left|{S}\right| = k+1$. Let $x \in S$. Consider the set $S\,' = S \setminus \left\{{x}\right\}$. We see that $\left|{S\,'}\right| = k$. Let $A \subseteq S$. If $x \notin A$ then $A \subseteq S\,'$. If $x \in A$ then there exists a unique $A' \subseteq S\,'$ such that $A' \cup \left\{{x}\right\} = A$. In fact, $A' = A \setminus \left\{{x}\right\}$. Let $K$ be the number of subsets of $S\,'$ with an even number of elements. From the induction hypothesis we have that $K = 2^{k-1}$. Now adjoin $x$ to all the subsets of $S\,'$. Suppose $E\,' \subseteq S\,'$, where $E\,'$ has an even number of elements. Then $E\,' \cup \left\{{x}\right\}$ has an odd number of elements. Similarly, $O\,' \subseteq S\,'$, where $O\,'$ has an odd number of elements. Then $O\,' \cup \left\{{x}\right\}$ has an even number of elements. So the number of subsets of $S$ with an even number of elements is: :$K$ subsets of $S$ that have an even number of elements and are subsets of $S\,'$ and: :$K$ subsets of $S$ of the form $O\,' \cup \left\{{x}\right\}$ We have that subset $A$ of $S$ has a unique expression of the form $A'$ or $A' \cup \left\{{x}\right\}$ with $A' \subseteq S\,'$. Thus there are exactly $2 K$ subsets of $S$ with an even number of elements. But as $K = 2^{k-1}$ it follows that $2 K = 2^k$. That is, there are $2^k$ subsets of $S$ with an even number of elements. So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: ::The number of subsets of $S$ whose cardinality is even is $2^{n-1}$, where $\left|{S}\right| = n$. {{qed}}
Let $F$ be the total number of subsets of $S$ whose cardinality is odd. From Cardinality of Set of Subsets, the number of subsets of $S$ with $m$ elements is $\displaystyle \binom n m$: :$\displaystyle \binom n m = \frac{n!}{m! \left({n-m}\right)}$ where $\displaystyle \binom n m$ is a binomial coefficient. Thus the total number of subsets of $S$ whose cardinality is Odd is given by: :$\displaystyle E = \binom n 1 + \binom n 3 + \binom n 5 + \cdots = \sum_{j \mathop \in \Z} \binom n {2j+1}$ Note the loose limits of the summation sign: the expression truly ranges over all $\Z$. This is because, when $2 j + 1 < 0$ and $2 j + 1> n$, $\displaystyle \binom n {2j+1} = 0$ by definition of binomial coefficient. The result then follows from Sum of Odd Index Binomial Coefficients: :$\displaystyle \sum_{j \mathop \ge 0} \binom n {2 j + 1} = 2^{n-1}$ {{qed}}
Proof by induction: For all $n \in \N$, let $\map P n$ be the proposition: :The number of subsets of $S$ whose cardinality is odd is $2^{n - 1}$, where $\card S = n$. === Basis for the Induction === When $n = 1$ we have from Cardinality of Power Set of Finite Set that $S$ has $2^1 = 2$ subsets: $\O$ and $S$ itself. We have that $\card S = 1$ and $\card \O = 0$. So there is indeed $2^{1 - 1} = 2^0 = 1$ subset of $S$ whose cardinality is odd, and that is $S$. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :The number of subsets of $S$ whose cardinality is odd is $2^{k - 1}$, where $\card S = k$. Then we need to show: :The number of subsets of $S$ whose cardinality is odd is $2^k$, where $\card S = k + 1$. === Induction Step === This is our induction step: Let $\card S = k + 1$. Let $x \in S$. Consider the set $S' = S \setminus \set x$. We see that $\card {S'} = k$. Let $A \subseteq S$. If $x \notin A$ then $A \subseteq S'$. If $x \in A$ then there exists a unique $A' \subseteq S'$ such that $A' \cup \set x = A$. In fact, $A' = A \setminus \set x$. Let $K$ be the number of subsets of $S'$ with an odd number of elements. From the induction hypothesis we have that $K = 2^{k - 1}$. Now adjoin $x$ to all the subsets of $S'$. Suppose $E' \subseteq S'$, where $E'$ has an even number of elements. Then $E' \cup \set x$ has an odd number of elements. Similarly, $O' \subseteq S'$, where $O'$ has an odd number of elements. Then $O' \cup \set x$ has an even number of elements. So the number of subsets of $S$ with an odd number of elements is: :$K$ subsets of $S$ that have an odd number of elements and are subsets of $S'$ and: :$K$ subsets of $S$ of the form $O' \cup \set x$. We have that subset $A$ of $S$ has a unique expression of the form $A'$ or $A' \cup \set x$ with $A' \subseteq S'$. Thus there are exactly $2 K$ subsets of $S$ with an odd number of elements. But as $K = 2^{k - 1}$ it follows that $2 K = 2^k$. That is, there are $2^k$ subsets of $S$ with an odd number of elements. So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: ::The number of subsets of $S$ whose cardinality is odd is $2^{n - 1}$, where $\card S = n$. {{qed}}
{{NotZFC}} Let $\prec' = \left({B \times B}\right) \cap {\prec}$. by Restriction of Foundational Relation is Foundational, $\prec'$ is a foundational relation By the lemma: :$B$ has a $\prec'$-minimal element $m$. By Minimal WRT Restriction, $m$ is $\prec$-minimal in $B$. {{qed}}
By Singleton Ordered Set is Terminal Object, we have that any ordered set with a singleton as underlying set is terminal in $\mathbf{OrdSet}$. Let $1$ be such a singleton ordered set. To show that $\mathbf{OrdSet}$ has enough constants, it is to be shown that if: :$f: P \to Q \ne g: P \to Q$ then there exists an $x: 1 \to P$ such that $f \circ x \ne g \circ x$. Now by definition of $\mathbf{OrdSet}$, $f$ and $g$ are increasing mappings. In particular, they are mappings, and by Equality of Mappings, there must be a $p \in P$ such that: :$f \left({p}\right) \ne g \left({p}\right)$ Define $\bar p: 1 \to P$ by $\bar p \left({*}\right) = p$ (where $*$ is the unique element of $1$). If this is a morphism in $\mathbf{OrdSet}$, we are evidently done. This follows from Mapping is Constant iff Increasing and Decreasing. Hence $\mathbf{OrdSet}$ has enough constants. {{qed}}
From Trivial Monoid is Terminal Element, we obtain that a constant of $M$ is a morphism $f: \left\{{e}\right\} \to M$. By Trivial Monoid is Initial Element, there is precisely one such morphism. Hence $M$ has one constant. {{qed}}
{{AimForCont}} $A \not \subseteq B$. Then $A \setminus B \not = 0$. By Well-Founded Relation is Strongly Well-Founded, $A \setminus B$ must have some $\prec$-minimal element. Then: :$\exists x \in \paren {A \setminus B}: \paren {A \setminus B} \cap \prec^{-1} \sqbrk x = \O$ so: :$A \cap \prec^{-1} \sqbrk x \subseteq B$ Since $\prec^{-1} \sqbrk x \subseteq A$: :$\prec^{-1} \sqbrk x \subseteq B$ Thus, by hypothesis $(1)$: :$x \in B$ But this contradicts the fact that: :$x \in \paren {A \setminus B}$ By Proof by Contradiction it follows that: :$\paren {A \setminus B} = \O$ and so: :$A \subseteq B$ Therefore: :$A = B$ {{qed}}
If $S \sim x$, then $\left\{{x \in \operatorname{On}: S \sim x}\right\}$ is a nonempty set of ordinals. It follows that this set has a minimal element, its intersection. {{explain|Needs a link to the fact that $\left\{ {x \in \operatorname{On}: S \sim x}\right\}$ is the subset of a well-ordered set, and also that the intersection is that minimal element.}} This minimal element is the cardinal number of $S$, by the definition of cardinal number. Thus, it is an ordinal. {{qed}}
=== $2 \implies 1$ === If $\exists x \in \On: S \sim x$, then by Ordinal Class is Strongly Well-Ordered by Subset there is a smallest ordinal $x$ such that $S \sim x$. This smallest ordinal $x$ is the cardinal number of $S$, by definition. {{qed|lemma}} === $3 \implies 2$ === Suppose that $y \subseteq x$ and $S \sim y$ for some ordinal $x$. Since $y \subseteq x$, it follows that $y \sim z$ for some $z \in \On$ by Unique Isomorphism between Ordinal Subset and Unique Ordinal. Therefore, by Set Equivalence is Equivalence Relation, $S \sim z$. {{qed|lemma}} === $1 \implies 3$ === Suppose that $3$ is not true. It follows that $S \not \sim x$ for any ordinal $x$. {{begin-eqn}} {{eqn | l = \bigcap \set {x \in \On : S \sim x}       | r = \bigcap \O       | c =  }} {{eqn | r = \mathbb U       | c = by Intersection of Empty Set }} {{end-eqn}} But $S \not \sim \mathbb U$, so $S \not \sim \card S$ by the definition of cardinal number. The result follows by contraposition. {{qed}}
By Condition for Set Equivalent to Cardinal Number, it follows that: If $\exists x \in \operatorname{On}: S \sim x$, then: :$S \sim \left\vert{S}\right\vert$ If $\neg \exists x \in \operatorname{On}: S \sim x$, then: {{begin-eqn}} {{eqn | l = \bigcap \left\{ {x \in \operatorname{On} : S \sim x}\right\}       | r = \bigcap \varnothing       | c =  }} {{eqn | r = \mathbb U       | c = by Intersection of Empty Set }} {{end-eqn}} And thus by the definition of cardinal number: :$\left\vert{S}\right\vert = \mathbb U$ {{qed}}
From Set Equivalence is Equivalence Relation: :$x \sim x$ Therefore, $x$ is equivalent to some ordinal. By Condition for Set Equivalent to Cardinal Number: :$x \sim \card x$ {{qed}}
Since $S \sim x$, it follows that: : $x \in \left\{{y \in \operatorname{On} : S \sim y}\right\}$ By Intersection is Subset: General Result, it follows that: : $\displaystyle \bigcap \left\{{y \in \operatorname{On} : S \sim y}\right\} \subseteq x$ {{explain|Link to the fact that $\le$ is isomorphic to $\subseteq$ on ordinals.}} Therefore $\left|{S}\right| \le x$ by the definition of cardinal number. {{qed}}
Let $x$ be an arbitrary set that is an ordinal: {{begin-eqn}} {{eqn | l = S       | o = \sim       | r = T       | c = By hypothesis }} {{eqn | ll= \leadsto       | l = S \sim x       | o = \iff       | r = T \sim x       | c = Set Equivalence is Equivalence Relation }} {{eqn | ll= \leadsto       | l = \set {x \in \On : S \sim x}       | r = \set {x \in \On : T \sim x}       | c = {{Defof|Class Equality}} }} {{eqn | ll= \leadsto       | l = \bigcap \set {x \in \On : S \sim x}       | r = \bigcap \set {x \in \On : T \sim x}       | c = Substitutivity of Class Equality }} {{eqn | ll= \leadsto       | l = \card S       | r = \card t       | c = {{Defof|Cardinal Number}} }} {{end-eqn}} {{qed}}
=== Necessary Condition === Let $S \cup T \sim \left|{S \cup T}\right|$. By definition of set equivalence, there exists a bijection: : $f: S \cup T \to \left|{S \cup T}\right|$ Since $f$ is a bijection, it follows that: :$S$ is equivalent to the image of $S$ under $f$. {{explain|Link to a result proving the above.}} This, in turn, is a subset of the ordinal $\left|{S \cup T}\right|$. {{explain|Link to a result proving the above.}} $\left|{S \cup T}\right|$ is an ordinal by Cardinal Number is Ordinal. By Condition for Set Equivalent to Cardinal Number, it follows that: : $S \sim \left|{S}\right|$ Similarly: :$T \sim \left|{T}\right|$ {{qed|lemma}} === Sufficient Condition === Suppose $S \sim \left|{S}\right|$ and $T \sim \left|{T}\right|$. Let $f: S \to \left|{S}\right|$ and $g: T \to \left|{T}\right|$ be bijections. By definition of set equivalence, these bijections are known to exist. Define the mapping $F : S \cup T \to \left|{S}\right| + \left|{T}\right|$ to be: :$F \left({x}\right) = \begin{cases} f \left({x}\right) & : x \in S \\ \left|{S}\right| + g \left({x}\right) & : x \notin S \end{cases}$ Suppose $F \left({x}\right) = F \left({y}\right)$. Let $x, y \in S$. Then: : $f \left({x}\right) = f\left({y}\right)$ Since $f$ is a bijection, it follows that: : $x = y$ Let $x \in S, y \in T$. Then: : $f \left({x}\right) = \vert S \vert + g\left({x}\right)$ But this is a contradiction, since $f\left({x}\right)$ has to be an element of $\left\vert{S}\right\vert$. Finally, let $x, y \in T$. Then: : $\left\vert{S}\right\vert + g \left({x}\right) = \left\vert{S}\right\vert + g\left({y}\right)$ From Ordinal Addition is Left Cancellable: :$g \left({x}\right) = g\left({y}\right)$ By the definition of a bijection: :$x = y$ It follows that $F: S \cup T \to \left|{S}\right| + \left|{T}\right|$ is an injection, where $\left|{S}\right| + \left|{T}\right|$ denotes ordinal addition. Therefore, $S \cup T$ is equivalent to some subset of the ordinal $\left|{S}\right| + \left|{T}\right|$. By Condition for Set Equivalent to Cardinal Number, it follows that $S \cup T \sim \left|{S \cup T}\right|$. {{qed}}
=== Necessary Condition === If $S \times T \sim \left|{S \times T}\right|$, then there is a mapping $f$ such that: :$f : S \times T \to \left|{S \times T}\right|$ is a bijection. Since $f$ is a bijection, it follows that: :$S$ is equivalent to the image of $S \times \left\{{x}\right\}$ under $f$ where $x \in T$. This, in turn, is a subset of the ordinal $\left|{S \times T}\right|$. $\left|{S \times T}\right|$ is an ordinal by Cardinal Number is Ordinal. By Condition for Set Equivalent to Cardinal Number, it follows that $S \sim \left|{S}\right|$. Similarly, $T \sim \left|{T}\right|$. {{qed|lemma}} === Sufficient Condition === Suppose $f: S \to \left|{S}\right|$ is a bijection and $g: T \to \left|{T}\right|$ is a bijection. Let $\cdot$ denote ordinal multiplication, while $\times$ shall denote the Cartesian product. Define the function $F$ to be: :$\forall x \in S, y \in T: F \left({x, y}\right) = \left|{S}\right| \cdot g \left({y}\right) + f \left({x}\right)$ It follows that $F: S \times T \to \left|{S}\right| \cdot \left|{T}\right|$ is a injection. By Condition for Set Equivalent to Cardinal Number, it follows that $S \times T \sim \left|{S \times T}\right|$. {{qed}}
By Condition for Set Equivalent to Cardinal Number: : $S \sim \left|{S}\right|$ Therefore, by Equivalent Sets have Equal Cardinal Numbers: : $\left|{S}\right| = \left|{ \left({ \left|{S}\right| }\right) }\right|$ {{qed}}
Since $n$ is an ordinal, it follows that $\left|{n}\right| \le n$ by Cardinal Number Less than Ordinal: Corollary. {{MissingLinks|Predecessor of Finite Ordinal is Finite Ordinal/Element of Finite Ordinal is Finite Ordinal}} Hence, $\left|{n}\right|$ is also a finite ordinal. Since $n$ is an ordinal, it also follows that $n \sim \left|{n}\right|$ by Ordinal Number Equivalent to Cardinal Number. {{link wanted|Equality of Finite Ordinals?}} By Equality of Natural Numbers, it follows that $n = \left|{n}\right|$. {{qed}}
Let $n \ne x$. Then either $n < x$ or $x < n$ by Ordinal Membership is Trichotomy. {{explain|Link to the result which states that $\le$ is isomorphic to $\subseteq$ on ordinals.}} If $x < n$, then by Subset of Finite Set is Finite both $x$ and $n$ are finite. Therefore by No Bijection between Finite Set and Proper Subset: : $x \not \sim n$ Suppose $n < x$. Then $x \sim n$ implies that $x$ is finite by definition. By No Bijection between Finite Set and Proper Subset: : $x \not \sim n$ Therefore by Rule of Transposition: : $n \sim x \implies n = x$ {{qed}}