hcju/proofwikiqa huggingface dataset:
Denote the implicit projections of the two products by: ::$\begin{xy}\xymatrix@C=2em@L+4mu{ &  A \times B   \ar@/_/[ddl]_*{\textrm{pr}_1}   \ar@/^/[ddr]^*{\textrm{pr}_2} \\ &  A \times' B   \ar[dl]^*{\textrm{pr}'_1}   \ar[dr]_*{\textrm{pr}'_2} \\  A & &  B }\end{xy}$ By the UMP of $A \times' B$, there is a unique $u: A \times B \to A \times' B$ fitting the dashed arrow in: ::$\begin{xy}\xymatrix@C=2em@L+4mu{ &  A \times B   \ar@/_/[ddl]_*{\textrm{pr}_1}   \ar@/^/[ddr]^*{\textrm{pr}_2}   \ar@{-->}[d] \\ &  A \times' B   \ar[dl]^*{\textrm{pr}'_1}   \ar[dr]_*{\textrm{pr}'_2} \\  A & &  B }\end{xy}$ By the UMP of $A \times B$, there is a further unique $v: A \times' B \to A \times B$ fitting the dashed arrow in: ::$\begin{xy}\xymatrix@C=2em@L+4mu{ &  A \times B   \ar@/_/[ddl]_*{\textrm{pr}_1}   \ar@/^/[ddr]^*{\textrm{pr}_2}   \ar[d]^*{u} \\ &  A \times' B   \ar[dl]^*{\textrm{pr}'_1}   \ar[dr]_*{\textrm{pr}'_2}   \ar@{-->}[d] \\  A &  A \times B   \ar[l]^*{\textrm{pr}_1}   \ar[r]_*{\textrm{pr}_2} &  B }\end{xy}$ We can now form the following diagram: ::$\begin{xy}\xymatrix@C=2em@R=3em@L+4mu{  & A \times B   \ar[dl]_*{\textrm{pr}_1}   \ar[dr]^*{\textrm{pr}_2}   \ar@{-->}[d] \\  A &  A \times B   \ar[l]^*{\textrm{pr}_1}   \ar[r]_*{\textrm{pr}_2} &  B }\end{xy}$ where we see both the identity morphism $\operatorname{id}_{A \times B}$ and $v \circ u$ fit the dashed arrow. The UMP of $A \times B$ however states that there is a unique fitting morphism. Hence $v \circ u = \operatorname{id}_{A \times B}$. That $u \circ v = \operatorname{id}_{A \times' B}$ follows, mutatis mutandis, by interchanging $A \times B$ and $A \times' B$. Hence there is a unique isomorphism $u: A \times B \to A \times' B$, as desired. {{qed}}
{{improve|wrong use of links}} Let $\On$ denote the class of ordinals. By definition of subset, $S \subseteq \On$. But by Ordinal Class is Ordinal $\On$ is an ordinal. Therefore $\On$ is well-ordered by $\in$. This means that $S$ is also well-ordered by $\in$. {{qed}} Category:Ordinals b2ykdmc9n2644i77ponqyuon8ozzs8s
For the proof: :the ordering relation $\le$ for ordinals and :the subset relation $\subseteq$ shall be used interchangeably. Let $f: T \to \card T$ be a bijection. It follows that $f \restriction_S : S \to \card T$ is an injection. The image of $S$ under $f$ is a subset of $\card T$ and thus is a subset of an ordinal. By Unique Isomorphism between Ordinal Subset and Unique Ordinal, there is a unique mapping $\phi$ and a unique ordinal $x$ such that $\phi: x \to f \sqbrk S$ is an order isomorphism. It follows that $S \sim x$ by the definition of order isomorphism. Furthermore, $\phi$ is a strictly increasing mapping from ordinals to ordinals. {{begin-eqn}} {{eqn | l = y       | o = \in       | r = x       | c =  }} {{eqn | ll= \leadsto       | l = y       | o = \le       | r = \map \phi y       | c = Strictly Increasing Ordinal Mapping Inequality }} {{eqn | o = \in       | r = f \sqbrk  S       | c = Definition of $\phi$ }} {{eqn | o = \subseteq       | r = \card T       | c = Image Preserves Subsets }} {{eqn | ll= \leadsto       | l = y       | o = \in       | r = \card T       | c = Cardinal Number is Ordinal }} {{end-eqn}} Therefore, $y \in x \implies y \in \card T$ and $x \le \card T$ by the definition of subset. But $\card S \le x$ by Cardinal Number Less than Ordinal. So $\card S \le \card T$ by the fact that Subset Relation is Transitive. {{qed}}
Since $x$ is an ordinal, it follows that $x \sim \left|{x}\right|$ by Ordinal Number Equivalent to Cardinal Number. This satisfies the hypothesis for Subset implies Cardinal Inequality. Therefore: : $\left|{S}\right| \le \left|{x}\right|$ {{qed}}
Let $y \in T$. Define the mapping $f : S \to S \times T$ as follows: :$f\left({x}\right) = \left({ x,y }\right)$ If $f\left({x_1}\right) = f\left({x_2}\right)$, then $\left({ x_1 , y }\right) = \left({ x_2 , y }\right)$ by the definition of $f$. It follows that $x_1 = x_2$ by Equality of Ordered Pairs. Thus, $f : S \to S \times T$ is an injection. By Injection implies Cardinal Inequality, it follows that $\left|{S}\right| \le \left|{S \times T}\right|$ {{qed}}
Let $f \sqbrk S$ denote the image of $S$ under $f$. {{begin-eqn}} {{eqn | l = S       | o = \sim       | r = f \sqbrk S       | c = Set is Equivalent to Image under Injection }} {{eqn | o = \subseteq       | r = T       | c = Image Preserves Subsets }} {{eqn | ll= \leadsto       | l = \card S       | r = \card {f \sqbrk S}       | c = Equivalent Sets have Equal Cardinal Numbers }} {{eqn | o = \le       | r = \card T       | c = Subset implies Cardinal Inequality }} {{end-eqn}} {{qed}} Category:Cardinals tua6h54zbpnzh0nqtbpd8wt33rgftl8
By Restriction of Mapping to Image is Surjection, the mapping: :$f: S \to \Img f$ is a surjection. Let $h$ be a mapping such that: :$h: \card S \to S$ is a bijection. By Composite of Surjections is Surjection: :$f \circ h: \card S \to \Img f$ is a surjection. Construct a set $R$ such that: :$R = \set {x \in \card S: \forall y \in x: \map f {\map h x} \ne \map f {\map h y} }$ It follows that: :$R \subseteq \card S$ By Subset of Ordinal implies Cardinal Inequality: :$\card R \le \card S$ Suppose: :$x \in R$ :$y \in R$ :$\map f {\map h x} = \map f {\map h y}$ Then, $x$ and $y$ are ordinals and by Ordinal Membership is Trichotomy: :$x < y \lor x = y \lor y < x$ If $x < y$, then: :$\map f {\map h x} \ne \map f {\map h y}$ by the definition of $R$. Similarly, $y < x$ implies that: :$\map f {\map h x} \ne \map f {\map h y}$ Therefore, $x = y$. It follows that the restriction $f \circ h \restriction_R : R \to \Img f$ is an injection. {{qed|lemma}} Finally, by the definition of surjection, for all $x \in \Img f$, there is some $y \in \card S$ such that: :$\map f {\map h y} = x$ But since this is true for some $y$, the set: :$\set {\map y \in \card S: \map f {\map h y} = x}$ has a minimal element. For this minimal element $y$, it follows that: :$\forall z \in y: \map f {\map h z} \ne \map f {\map h y}$ since if it were equal, this would contradict the fact that $y$ is a $\in$-minimal element. It follows that the restriction $f \circ h \restriction_R : R \to \Img f$ is a bijection. By the definition of set equivalence, $R \sim \Img f$. So: {{begin-eqn}} {{eqn | l = \Img f       | r = \card R       | c = Equivalent Sets have Equal Cardinal Numbers }} {{eqn | o = \le       | r = \card S       | c = Subset implies Cardinal Inequality }} {{end-eqn}} {{qed}}
=== Necessary Condition === Suppose that $\card S \le \card T$. Let $g : S \to \card S$ be a bijection and $h: \card T \to T$ be a bijection. It follows that $g: S \to \card T$ is an injection by the fact that $\card T \le \card S$. Then from Composite of Injections is Injection, $h \circ g: S \to T$ is an injection. {{qed|lemma}} === Sufficient Condition === The other direction follows from Injection implies Cardinal Inequality. {{qed}} Category:Cardinals g0oty2vpdzs9newer6xbwhqpc386hyu
=== Necessary Condition === Suppose $f: S \to T$ is a surjection. Then $\Img f = T$ by definition. From Cardinality of Image of Mapping not greater than Cardinality of Domain: :$\card T \le \card S$ Furthermore, if $S$ is non-empty, then $\map f x \in T$ for some $x \in S$. Thus, $T$ is non-empty and $0 < \card T$ by Cardinality of Empty Set. {{qed|lemma}} === Sufficient Condition === Suppose that $0 < \card T \le \card S$. By Injection iff Cardinal Inequality, it follows that $g: T \to S$ for some injection $g$. Take an arbitrary $y \in T$. Define the function $f: S \to T$ as follows: :$\map f x = \begin{cases} \map {g^{-1} } x & : x \in \Img g \\ y & : x \notin \Img g \end{cases}$ For any $z \in T$, $\map g z \in \Img g$. Thus, $\map f x = z$ for some $x \in S$. It follows that $f: S \to T$ is a surjection. {{qed}}
Let $x \in S$ be any element of $S$. From Right Inverse for All is Left Inverse we have that $x_R \circ x = e_R$. Then: {{begin-eqn}} {{eqn | l = e_R \circ x       | r = \paren {x \circ x_R} \circ x       | c = {{Defof|Right Inverse Element}} }} {{eqn | r = x \circ \paren {x_R \circ x}       | c = as $\circ$ is associative }} {{eqn | r = x \circ e_R       | c = $x_R$ is a Right Inverse }} {{eqn | r = x       | c = {{Defof|Right Identity}} }} {{end-eqn}} So $e_R$ behaves as a left identity as well as a right identity. That is, by definition, $e_R$ is an identity element. {{qed}}
From Left Inverse for All is Right Inverse we have that $x \circ x_L = e_L$. Then: {{begin-eqn}} {{eqn | l = x \circ e_L       | r = x \circ \paren {x_L \circ x}       | c = {{Defof|Left Inverse Element}} }} {{eqn | r = \left({x \circ x_L}\right) \circ x       | c = as $\circ$ is associative }} {{eqn | r = e_L \circ x       | c = $x_L$ is a Right Inverse }} {{eqn | r = x       | c = {{Defof|Left Identity}} }} {{end-eqn}} So $e_L$ behaves as a right identity as well as a left identity. That is, by definition, $e_L$ is an identity element. {{qed}}
Let $y = x_R \circ x$. Then: {{begin-eqn}} {{eqn | l = y \circ e_R       | r = y \circ \paren {y \circ y_R}       | c = {{Defof|Right Inverse Element}} }} {{eqn | r = \paren {y \circ y} \circ y_R       | c = as $\circ$ is associative }} {{eqn | r = y \circ y_R       | c = Product of Semigroup Element with Right Inverse is Idempotent: $y = x_R \circ x$ }} {{eqn | r = e_R       | c = {{Defof|Right Inverse Element}} }} {{end-eqn}} So $x_R \circ x = e_R$, and $x_R$ behaves as a left inverse as well as a right inverse with respect to the right identity. {{qed}}
We have that a group is a monoid, all of whose elements are invertible. The result follows from Inverse of Product/Monoid/General Result. {{Qed}}
Proof by induction: For all $n \in \N_{> 0}$, let $\map P n$ be the proposition: :$\paren {a_1 \circ a_2 \circ \cdots \circ a_n}^{-1} = a_n^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$ $\map P 1$ is (trivially) true, as this just says: :$\paren {a_1}^{-1} = a_1^{-1}$ === Basis for the Induction === $\map P 2$ is the case: :$\paren {a_1 \circ a_2}^{-1} = a_2^{-1} \circ a_1^{-1}$ which has been proved in Inverse of Group Product. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\paren {a_1 \circ a_2 \circ \cdots \circ a_k}^{-1} = a_k^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$ Then we need to show: :$\paren {a_1 \circ a_2 \circ \cdots \circ a_{k + 1} }^{-1} = a_{k + 1}^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \paren {a_1 \circ a_2 \circ \cdots \circ a_k \circ a_{k + 1} }^{-1}       | r = \paren {\paren {a_1 \circ a_2 \circ \cdots \circ a_k} \circ a_{k + 1} }^{-1}       | c = General Associativity Theorem }} {{eqn | r = a_{k + 1}^{-1} \circ \paren {a_1 \circ a_2 \circ \cdots \circ a_k}^{-1}       | c = Basis for the Induction }} {{eqn | r = a_{k + 1}^{-1} \circ \paren {a_k^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1} }       | c = Induction Hypothesis }} {{eqn | r = a_{k + 1}^{-1} \circ a_k^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}       | c = General Associativity Theorem }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \N_{> 0}: \paren {a_1 \circ a_2 \circ \cdots \circ a_n}^{-1} = a_n^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$ {{qed}}
{{WLOG}}, suppose that $r > s$. From $g^r = g^s$ it follows from Powers of Group Elements that: : $g^{r - s} = g^r g^{-s} = g^s g^{-s} = e$ Thus there exists $t \in \Z_{>0}$ such that $g^t = e$. Let $m \in \Z_{>0}$ be the smallest such that $g^m = e$. Suppose $0 \le i < j < m$ such that $g^i = g^j$. Then $g^{j - i} = e$ and $0 < j - i < m$. This contradicts the choice of $m$ as the smallest $m \in \Z_{>0}$ such that $g^m = e$. Hence the result. {{qed}}
If $z \in \C$, then by the definition of the complex exponential function, $\exp$ is a mapping $\C \to \C_{\ne 0}$. Let $z_1, z_2 \in \C$. By Exponential of Sum: :$\map \exp {z_1 + z_2} = \map \exp {z_1} \, \map \exp {z_2}$ Therefore $\exp: \struct {\C, +} \to \struct {\C_{\ne 0}, \times}$ is a group homomorphism. {{qed}}
Let $F$ be a group. By Direct Product of Group Homomorphisms is Homomorphism, given group homomorphisms: :$g: F \to G, h: F \to H$ their direct product $g \times h: F \to G \times H$ is a group homomorphism. From Projections on Direct Product of Group Homomorphisms, the following diagram is commutative: ::$\begin{xy}\xymatrix@L+3mu@R=3em{ &  F   \ar@/_/[dl]_*{g}   \ar[d]^*{\quad g \times h}   \ar@/^/[dr]^*{h} \\  G &  G \times H   \ar[l]^*{\mathrm{pr}_1}   \ar[r]_*{\mathrm{pr}_2} &  H }\end{xy}$ By Cartesian Product is Set Product, $g \times h$ is the only mapping $F \to G \times H$ that could fit into the diagram. Since it is also a group homomorphism, we see that the UMP for the binary product is satisfied. Thus the direct product $G \times H$ indeed is a categorical product for $G$ and $H$ in $\mathbf{Grp}$. {{qed}}
Let $F: \mathbf A \to \mathbf C$ and $G: \mathbf A \to \mathbf D$ be morphisms in $\mathbf{Cat}$, i.e. functors. Suppose $X: \mathbf A \to \mathbf C \times \mathbf D$ is a functor such that $\pr_1 \circ X = F$ and $\pr_2 \circ X = G$: ::$\begin{xy}\xymatrix@C=3em@R=3em@L+3mu{ &  \mathbf A   \ar[dl]_*{F}   \ar[d]^*{X}   \ar[dr]^*{G} \\  \mathbf C &  \mathbf C \times \mathbf D   \ar[l]^*{\pr_1}   \ar[r]_*{\pr_2} &  \mathbf D }\end{xy}$ Here, $\pr_1$ and $\pr_2$ are projection functors. Thus, for all objects $A$ of $\mathbf A$, it is required that: :$\map {\pr_1} {X A} = F A$ :$\map {\pr_2} {X A} = G A$ meaning that $X A = \tuple {F A, G A}$ by definition of the objects of the product category $\mathbf C \times \mathbf D$. Similarly, for all morphisms $f$ of $\mathbf A$, it is required that: :$\map {\pr_1} {X f} = F f$ :$\map {\pr_2} {X f} = G f$ meaning that $X f = \tuple {F f, G f}$ by definition of the morphisms of the product category $\mathbf C \times \mathbf D$. This uniquely determines the functor $X$, if it exists. That this definition actually yields a functor is shown on Functor to Product Category. Hence $\mathbf C \times \mathbf D$ is a binary product of $\mathbf C$ and $\mathbf D$ in $\mathbf{Cat}$. {{qed}}
Suppose that there are morphisms $l \to p$ and $l \to q$ in $\mathbf P$. That is to say, suppose $l \preceq p$ and $l \preceq q$. Then $l$ is a lower bound for $\left\{{p, q}\right\}$. By definition of infimum, we then have: :$l \preceq r = \inf \left\{{p, q}\right\}$ By definition of $\mathbf P$ as an order category, this means there is a morphism: :$l \to r$ By definition of $\mathbf P$, this morphism is necessarily unique. Thus $r$ satisfies the UMP for the binary product of $p$ and $q$. The result follows. {{qed}}
Let $x_1$ and $x_2$ be distinct elements of $S$. Let $y_1$ and $y_2$ be distinct elements of $T$. Define the mapping $f : S \times T \to S \cup T$ as follows: :$\map f {x, y} = \begin {cases} y &: x = x_1 \\ x_1 &: x = x_2 \land y = y_1 \\ x &: \text {otherwise} \end {cases}$ If $x \in S$, then we have that either $x = x_1$ or $x \ne x_1$. If $x \ne x_1$, then $\map f {x, y_2} = x$ by the definition of $f$. If $x = x_1$, then $\map f {x_2, y_1} = x_1$ by the definition of $f$. If $y \in S$, then we have that: :$\map f {x_1, y} = y$ by the definition of $f$. Therefore, it follows that $f : S \times T \to S \cup T$ is a surjection. Thus by Surjection iff Cardinal Inequality: :$\card {S \cup T} \le \card {S \times T}$ {{qed}}
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: :$\forall x \in S, y \in T: F \left({x, y}\right) = \left|{S}\right| \cdot g \left({y}\right) + f \left({x}\right)$ Suppose $F\left({x_1,y_1}\right) = F\left({x_2,y_2}\right)$. {{begin-eqn}} {{eqn|ll =       |l = \left\vert{ S }\right\vert \cdot g\left({y_1}\right) + f\left({x_1}\right)      |o = =      |r = \left\vert{ S }\right\vert \cdot g\left({y_2}\right) + f\left({x_2}\right)      |c =  }} {{eqn|ll = \implies      |l = f\left({x_1}\right) , f\left({x_2}\right)      |o = \in      |r = \left\vert{ S }\right\vert      |c = definition of the mapping $f$ }} {{eqn|ll = \implies      |l = f\left({x_1}\right) = f\left({x_2}\right)      |o = \land      |r = g\left({y_1}\right) = g\left({y_2}\right)      |c = by the Division Theorem for Ordinals }} {{eqn|ll = \implies      |l = x_1 = x_2      |o = \land      |r = y_1 = y_2      |c = by the fact that $f$ and $g$ are bijections. }} {{end-eqn}} It follows that $F: S \times T \to \left|{S}\right| \cdot \left|{T}\right|$ is a injection. {{qed|lemma}} Furthermore, take any $z \in \left|{ S }\right| \cdot \left|{ T }\right|$. By the Division Theorem for Ordinals, it follows that: :$z = \left|{ S }\right| \cdot a + b$ for some unique $a$ and $b \in \left|{ S }\right|$. Moreover, $a \in \left|{ T }\right|$ since otherwise, this would contradict the fact that $z \in \left|{ S }\right| \cdot \left|{ T }\right|$. Therefore, $a = f\left({x}\right)$ for some $x \in S$ and $b = g\left({y}\right)$ for some $y \in T$. Thus, $F : S \times T \to \left|{ S }\right| \cdot \left|{ T }\right|$ is a surjection. {{qed|lemma}} Therefore, $F : S \times T \to \left|{ S }\right| \cdot \left|{ T }\right|$ is a bijection by the definition of bijection. By Condition for Set Equivalent to Cardinal Number, it follows that $S \times T \sim \left|{S}\right| \cdot \left|{T}\right|$. {{qed}} Category:Cardinals Category:Ordinal Arithmetic k5bj0hcoqzqbjyzfb8rv7fias5dxfe1
Let $\card S$ denote the cardinal number of $S$. Let $\cdot$ denote ordinal multiplication. By Cardinal Product Equinumerous to Ordinal Product, it follows that $S \times T \sim \card S \cdot \card T$. But then $\card S$ and $\card T$ are members of the minimal infinite successor set. Therefore, $\card S \cdot \card T \in \omega$ by Natural Number Multiplication is Closed. Since $S \times T$ is equinumerous to a member of the minimal infinite successor set, it follows that $S \times T$ is finite. {{qed}}
By the definition of Cartesian product: :$S \times T = \set {\tuple {s, t}: s \in S, t \in T}$ Then by definition of set union: :$S \times T = \displaystyle \bigcup_{s \mathop \in S} \set s \times T$ Also, the mapping $g_s: \set s \times T \to T$ defined by: :$\map {g_s} {s, t} = t$ is a bijection. Therefore, since $T$ is finite, so is $\set s \times T$ for all $s \in S$. Since $S$ is finite, the result follows from Finite Union of Finite Sets is Finite. {{qed}}
The proof proceeds by induction. Let $S$ be a finite set with cardinality $n$. If $n = 0$ then $S = \varnothing$, so $\bigcup S = \varnothing$, which is finite. Suppose that the union of any finite set of finite sets with cardinality $n$ has a finite union. {{explain|The above statement is tautological. Suggest the final sentence be rethought.}} Let $S$ have cardinality $n^+$. Then there is a bijection $f: n^+ \to S$. Then: : $\displaystyle \bigcup S = \bigcup_{k \mathop \in n^+} f \left({k}\right) = \bigcup_{k \mathop \in n} f \left({k}\right) \cup f \left({n}\right)$ {{explain|Explain in detail.}} By Union of Finite Sets is Finite, $\displaystyle \bigcup S$ is finite. {{qed}}
Let $S = \left\{{A_1, \ldots, A_n}\right\}$ such that $A_k$ is finite $\forall k = 1, \ldots, n$. Set: : $m = \max \left\{{ \left|{A_1}\right|, \ldots, \left|{A_n}\right|}\right\}$ Then: : $\displaystyle \left|{ \bigcup_{k \mathop = 1}^n A_k}\right| \le \sum_{k \mathop = 1}^n \left|{A_k}\right| \le \sum_{k \mathop = 1}^n m = n m$ {{explain|needs to invoke a link to result that shows $\left\vert{ \bigcup_{k \mathop {{=}} 1}^n A_k}\right\vert \le \sum_{k \mathop {{=}} 1}^n \left\vert{A_k}\right\vert$}} Hence the result. {{qed}}
Note that $\card {S \cup T} \le \card {S \times T}$ by Cardinal of Union Less than Cardinal of Cartesian Product. The theorem follows from the fact that $S \times T$ is finite by Product of Finite Sets is Finite. {{qed}}
If $S$ or $T$ is empty, the result is trivial. Otherwise, let $f: \N_{<n} \to S$ and $g: \N_{<m} \to T$ be bijections, where $\N_{<n}$ is an initial segment of $\N$. Now define $h: \N_{< n + m} \to S \cup T$ by: :$\map h i = \begin{cases} \map f i : & \text {if $i < n$} \\ \map g {i - n} : & \text{if $i \ge n$} \end{cases}$ By Set Finite iff Surjection from Initial Segment of Natural Numbers, it suffices to show that $h$ is surjective. Let $s \in S$. Then: {{begin-eqn}} {{eqn | l = \map h {\map {f^{-1} } s}       | r = \map f {\map {f^{-1} } s}       | c = as $\map {f^{-1} } s \in \N_{<n}$ }} {{eqn | r = s       | c = Definition of $f^{-1}$ }} {{end-eqn}} Next let $t \in T$.  We have that: :$n + \map {g^{-1} } t < n + m$ so that: :$n + \map {g^{-1} } t \in \N_{< n + m}$ Then: {{begin-eqn}} {{eqn | l = \map h { n + \map {g^{-1} } t}       | r = \map g {n + \map {g^{-1} } t - n}       | c = as $\map {g^{-1} } t \ge 0$ and so $n + \map {g^{-1} } t \ge n$ }} {{eqn | r = \map g {\map {g^{-1} } t} }} {{eqn | r = t       | c = Definition of $g^{-1}$ }} {{end-eqn}} Thus it has been shown that for all $s \in S$ and $t \in T$, there exists an element of $\N_{< n + m}$ which maps to it. So by definition of set union, for all $x \in S \cup T$: :$\exists y \in \N_{< n + m}: \map h y = x$. Hence, by definition, $h$ is a surjection. Therefore $S \cup T$ is finite. {{qed}}
$x$ is finite {{iff}} $x \sim \N_n$ for some $n \in \N$, by definition. But $x$ is an ordinal, and by definition, it is equal to its initial segment. By definition of the von Neumann construction of natural numbers, it follows that $x \sim n$ for some $n$. By Finite Ordinal is equal to Natural Number, it follows that $x$ is equal to $n$. Thus, $x$ is an element of the minimal infinite successor set. {{qed}}
=== Sufficient Condition === By Cardinal Number Less than Ordinal, it follows that $\left|{x}\right| \le x$. So if $x < \left|{T}\right|$, then $\left|{x}\right| < \left|{T}\right|$. {{qed|lemma}} === Necessary Condition === Suppose $\left|{T}\right| \le x$. By Subset of Ordinal implies Cardinal Inequality: : $\left|{\left({\left|{T}\right|}\right)}\right| \le \left|{x}\right|$. Therefore, by Cardinal of Cardinal Equal to Cardinal: : $\left|{T}\right| \le \left|{x}\right|$ By the Rule of Transposition: : $\neg \left|{T}\right| \le \left|{x}\right| \implies \neg \left|{T}\right| \le x$ By Ordinal Membership is Trichotomy: :$x < \left|{T}\right| \iff \left|{x}\right| < \left|{T}\right|$ {{qed}}
Since $x > 1$, then $0 < x$ and $1 < x$. Define the function $f : x + 1 \to x \times x$ as follows: :$f\left({y}\right) = \begin{cases} \left({ y , 0 }\right) &: y < x \\ \left({ 0 , 1 }\right) &: y = x \end{cases}$ If $f\left({y}\right) = f\left({z}\right)$, then $y = z$ by cases. === Case 1: $y = x$ === If $y = x$, then $f\left({y}\right) = \left({ 0,1 }\right)$ by the definition of $f$. So if $z < x$, then $f\left({y}\right) \ne f\left({z}\right)$. This contradicts the hypothesis, so $z = x$ and therefore $z = y$. === Case 2: $y < x$ === If $y < x$, then $f\left({y}\right) = \left({ y,0 }\right)$ by the definition of $f$. $z = x$ yields the contradictory statement $\left({ y,0 }\right) = \left({ 0,1 }\right)$. Therefore, $z < x$ and $\left({ y,0 }\right) = \left({ z,0 }\right)$. By Equality of Ordered Pairs, it follows that $y = z$. It follows that $f$ is an injection. By Injection iff Cardinal Inequality, it follows that $\left|{ x + 1 }\right| \le \left|{ x \times x }\right|$. {{qed}}
=== Necessary Condition === {{ProofWanted}} === Sufficient Condition === Recall the definition of cyclic permutation. Let: :$\rho = \begin {bmatrix} i & \map \rho i & \ldots & \map {\rho^{k - 1} } i \end{bmatrix}$ Note that the orbit $\set {i, \map \rho i, \ldots, \map {\rho^{k - 1} } i}$ is the only non-trivial orbit. (If $k = 1$, then the above-mentioned orbit is also trivial.) {{explain|A page to define a trivial orbit}} {{qed|lemma}}