hcju/proofwikiqa huggingface dataset:
Aiming for a contradiction, assume that there is a $l$ such that: :$\forall m \in \N: m \nrightarrow \left({l}\right)_r^n$ Let $\hat{K_i}$ denote a hypergraph on $i$ vertices where all possible $n$-subsets of the vertices are the hyperedges. Let $G$ be a hypergraph with vertices $V = \left\{ {v_i: i \in \N}\right\}$. Let the hyperedges of $G <$ be enumerated by: :$E = \left\{ {E_i: E_i \subset \N, \left\vert{E_i}\right\vert = n}\right\}$ {{explain|What does $G <$ mean?}} We construct a (rooted) tree $T$ as follows: $(1): \quad$ First introduce a root node $r t$. $(2): \quad$ Each node is allowed to have at most $r<$ children which correspond to the $r$-colors, subject to it satisfying the criteria: {{explain|What does $r <$ mean?}} :::A child is always labeled by one among the $r$-colors. :::Let the colors be denoted $c_1, c_2, \ldots, c_r$. $(3): \quad$ A child $c_i$ is '''permitted''' {{iff}} its introduction creates a path of some finite length $k$ starting from the root. :::So, if the hyperedges $E_1, E_2, \ldots, E_k$ are colored by the colors used in the path in the same order, then the corresponding subgraph of $G$ does not contain a monochromatic $\hat{K_l}$. :::For example: if the introduction of a child $c_i$ creates the $k$ length path $r t, c_a, c_b, \ldots, c_i$ and the hyperedges $E_1, E_2, \ldots, E_k$ when colored $c_a, c_b, \ldots, c_i$ do not contain a monochromatic $\hat{K_l}$, the child $c_i$ is permitted to be added to $T$. Note that for all $m$, there always exists a coloring of $\hat{K_m}$ such that no monochromatic $\hat{K_l}$ exists within. Thus the situation that a child cannot be added to any vertex at a given level $k$ cannot arise. For we can always take a coloring of $\hat{K_{k+n}}$ containing no monochromatic $\hat{K_l}$. Since any $k$ hyperedges in it would yield a sequence of colors already existing in $T$, we know which vertex to add the child to. We give the child the color corresponding to any other edge. Hence we can forever keep adding children and so $T$ is infinite. It is also obvious that each level $k$ of $T$ has at most $r^k<$ vertices. So each level is finite. By König's Tree Lemma there will be an infinite path $P$ in $T$. $P$ provides a $r$-coloring of $G$ that contains no monochromatic $\hat{K_i}$. Hence $P$ contains no monochromatic infinite hypergraph. This contradicts the Infinite Ramsey's Theorem. The result follows by Proof by Contradiction. {{qed}} Category:Ramsey Theory 3umvqy1sfhkc24t7gy141uavbr51b6h
By definition of the product morphism $\operatorname{id}_C \times \operatorname{id}_D$, it is the unique morphism making: ::$\begin{xy}\xymatrix@+1em@L+5px{  C   \ar[d]_*+{\operatorname{id}_C} &  C \times D   \ar[l]_*+{\operatorname{pr}_1}   \ar[r]^*+{\operatorname{pr}_2}   \ar@{-->}[d]^*+{\hskip{2.2em} \operatorname{id}_C \times \operatorname{id}_D} &  D   \ar[d]^*+{\operatorname{id}_D} \\  C &  C \times D   \ar[l]^*+{\operatorname{pr}_1}   \ar[r]_*+{\operatorname{pr}_2} &  D }\end{xy}$ a commutative diagram. It is immediate by the behaviour of identity morphisms that $\operatorname{id}_{\left({C \mathop \times D}\right)}$ is the unique morphism sought. {{qed}} Category:Category Theory 5p7uglo9unav3u087bc1zlr3fehq6qm
Since $S$ and $T$ are equivalent, there exists a bijection $f: S \to T$. Let $g: T \to S$ be the inverse of $f$; its existence is assured by Bijection iff Left and Right Inverse. Define $\hat f: R \times S \to R \times T$ by: :$\hat f \left({r, s}\right) := \left({r, f \left({s}\right)}\right)$ Next, define $\hat g: R \times T \to R \times S$ by: :$\hat g \left({r, t}\right) := \left({r, g \left({t}\right)}\right)$ Then, for all $\left({r, s}\right) \in R \times S$: {{begin-eqn}} {{eqn|l = \hat g \circ \hat f \left({r, s}\right)      |r = \hat g \left({r, f \left({s}\right)}\right) }} {{eqn|r = \left({r, g \left({f \left({s}\right)}\right)}\right) }} {{eqn|r = \left({r, s}\right)      |c = $g$ is the inverse of $f$ }} {{end-eqn}} Similarly, for all $\left({r, t}\right) \in R \times T$: {{begin-eqn}} {{eqn|l = \hat f \circ \hat g \left({r, s}\right)      |r = \hat f \left({r, g \left({s}\right)}\right) }} {{eqn|r = \left({r, f \left({g \left({s}\right)}\right)}\right) }} {{eqn|r = \left({r, s}\right)      |c = $g$ is the inverse of $f$ }} {{end-eqn}} Thus, it follows that $\hat g$ is a inverse for $\hat f$. From Bijection iff Left and Right Inverse, we conclude $\hat f$ is a bijection. Hence $R \times S \sim R \times T$, by definition of set equivalence. Mutatis mutandis, the other relation also follows. {{qed}} Category:Set Theory 6g87mvgqee3iiu13jl1u791e7u76ib1
The situation is efficiently captured in the following commutative diagram: ::$\begin{xy} <-5em,0em>*+{A}            = "A", <0em,0em>*+{A \times A'}   = "P", <5em,0em>*+{A'}            = "A2", <-5em,-5em>*+{B}           = "B", <0em,-5em>*+{B \times B'}  = "Q", <5em,-5em>*+{B'}           = "B2", <-5em,-10em>*+{C}          = "C", <0em,-10em>*+{C \times C'} = "R", <5em,-10em>*+{C'}          = "C2", "P";"A"  **@{-} ?>*@{>} ?*!/^.8em/{\operatorname{pr}_1}, "P";"A2" **@{-} ?>*@{>} ?*!/_.8em/{\operatorname{pr}_2}, "Q";"B"  **@{-} ?>*@{>} ?*!/^.8em/{\operatorname{pr}_1}, "Q";"B2" **@{-} ?>*@{>} ?*!/_.8em/{\operatorname{pr}_2}, "R";"C"  **@{-} ?>*@{>} ?*!/_.8em/{\operatorname{pr}_1}, "R";"C2" **@{-} ?>*@{>} ?*!/^.8em/{\operatorname{pr}_2}, "A";"B"   **@{-} ?>*@{>}  ?*!/^.8em/{f}, "A2";"B2" **@{-} ?>*@{>}  ?*!/_.8em/{f'}, "P";"Q"   **@{--} ?>*@{>} ?*!/_1.4em/{f \times f'}, "B";"C"   **@{-} ?>*@{>}  ?*!/^.8em/{g}, "B2";"C2" **@{-} ?>*@{>}  ?*!/_.8em/{g'}, "Q";"R"   **@{--} ?>*@{>} ?*!/_1.4em/{g \times g'}, \end{xy}$ The result is immediate from deleting the $B$s: ::$\begin{xy} <-5em,0em>*+{A}            = "A", <0em,0em>*+{A \times A'}   = "P", <5em,0em>*+{A'}            = "A2", <-5em,-10em>*+{C}          = "C", <0em,-10em>*+{C \times C'} = "R", <5em,-10em>*+{C'}          = "C2", "P";"A"  **@{-} ?>*@{>} ?*!/^.8em/{\operatorname{pr}_1}, "P";"A2" **@{-} ?>*@{>} ?*!/_.8em/{\operatorname{pr}_2}, "R";"C"  **@{-} ?>*@{>} ?*!/_.8em/{\operatorname{pr}_1}, "R";"C2" **@{-} ?>*@{>} ?*!/^.8em/{\operatorname{pr}_2}, "A";"C"   **@{-} ?>*@{>}  ?*!/^1.1em/{g \circ f}, "A2";"C2" **@{-} ?>*@{>}  ?*!/_1.3em/{g' \circ f'}, "P";"R"   **@{--} ?>*@{>} ?*!/_2em/{\left({g \circ f}\right) \times \\ \left({g' \circ f'}\right)}, \end{xy}$ The uniqueness of $\left({g \circ f}\right) \times \left({g' \circ f'}\right)$ implies it equals $\left({g \times g'}\right) \circ \left({f \times f'}\right)$. {{qed}} Category:Product Categories Category:Category Theory n2g9d4g2ru4tgwly9wfqww8e1ixxqql
Let $x$ denote $\max \left({ \left|{ S }\right| , \left|{ T }\right| }\right)$. Then by Cartesian Product Preserves Cardinality: :$S \times T \sim \left|{ S }\right| \times \left|{ T }\right|$ Let $f : S \times T \to \left|{ S }\right| \times \left|{ T }\right|$ be a bijection. It follows that $f : S \times T \to x \times x$ is an injection. Hence: {{begin-eqn}} {{eqn | l = \left\vert{ S \times T }\right\vert       | o = \le       | r = \left\vert{ x \times x }\right\vert       | c = Injection iff Cardinal Inequality }} {{eqn | r = \left\vert{ x }\right\vert       | c = Non-Finite Cardinal is equal to Cardinal Product }} {{eqn | o = \le       | r = x       | c = Cardinal Number Less than Ordinal: Corollary }} {{end-eqn}} Therefore: :$\left|{ S \times T }\right| \le x$ {{qed|lemma}} Conversely: :$x = \left|{ S }\right|$ if $\left|{ T }\right| \le \left|{ S }\right|$ and: :$x = \left|{ T }\right|$ if $\left|{ S }\right| \le \left|{ T }\right|$ By Relation between Two Ordinals: :$x = \left|{ S }\right|$ or $x = \left|{ T }\right|$ It follows by Set Less than Cardinal Product that: : $x \le \left|{ S \times T }\right|$ {{qed|lemma}} Combining the two lemmas, it follows that: :$x = \left|{ S \times T }\right|$ {{qed}}
Let $x$ denote $\max \left({ \left|{ S }\right| , \left|{ T }\right| }\right)$. $x = \left|{ S }\right|$ if $\left|{ T }\right| \le \left|{ S }\right|$. $x = \left|{ T }\right|$ if $\left|{ S }\right| \le \left|{ T }\right|$ By Relation between Two Ordinals, $x = \left|{ S }\right|$ or $x = \left|{ T }\right|$. In either case, it follows that $x \le \left|{ S \cup T }\right|$ by Subset of Union. {{qed|lemma}} If $\left|{ T }\right| = 1$ or $\left|{ T }\right| = 0$, it follows that $\left|{ S \cup T }\right| = \left|{ S }\right| = x$ by the definition of an infinite set. If $1 < \left|{ T }\right|$, then $\left|{ S \cup T }\right| \le \left|{ S \times T }\right|$ by Cardinal of Union Less than Cardinal of Cartesian Product. But by Cardinal Product Equal to Maximum, it follows that $\left|{ S \cup T }\right| \le x$. {{qed|lemma}} Combining the two lemmas, it follows that $x = \left|{ S \cup T }\right|$ {{qed}}
{{NotZFC}} By definition of the cardinal class: :$\mathcal N = \left\{ x \in \operatorname{On} : \exists y: x = \left|{ y }\right| \right\}$ Every element of $\mathcal N$ is thus an element of $\operatorname{On}$. {{qed}}
=== Necessary Condition === Suppose $x = \left|{ x }\right|$. Then $x = \left|{ y }\right|$ for some $y$ by Existential Generalisation. By definition of the cardinal class: :$\mathcal N = \left\{ x \in \operatorname{On} : \exists y: x = \left|{ y }\right| \right\}$ It follows that $x \in \mathcal N$. {{qed|lemma}} === Sufficient Condition === {{begin-eqn}} {{eqn|ll =       |l = x      |o = \in      |r = \mathcal N      |c = Hypothesis }} {{eqn|ll = \implies      |l = x      |o = =      |r = \left\vert{ y }\right\vert      |c = definition of cardinal class }} {{eqn|ll = \implies      |l = \left\vert{ x }\right\vert      |o = =      |r = \left\vert{ y }\right\vert      |c = by Cardinal of Cardinal Equal to Cardinal }} {{eqn|ll = \implies      |l = x      |o = =      |r = \left\vert{ x }\right\vert      |c = by Equality is Transitive }} {{end-eqn}} {{qed}}
{{NotZFC}} Suppose $n \in \omega$. By Cardinal of Finite Ordinal, $n = \left|{ n }\right|$. By Cardinal of Cardinal Equal to Cardinal/Corollary, $n \in \mathcal N$. {{qed}}
=== $x$ is an ordinal === $x$ is clearly a subset of the class of all ordinals. Moreover, suppose $y \in x$ and $z \in y$. Then $z \subseteq y$ by the fact that $y$ is an ordinal. $y \in x$ means that $f : y \to S$ for some injective mapping $f$ by the definition of dominance. But then, $f \restriction_z : z \to S$ is also an injection by Restriction of Injection is Injection. Thus, $z \in x$ by the definition of $x$, so $x$ is transitive. Therefore, $x$ is an ordinal, since it is a transitive subset of the ordinal class. {{qed|lemma}} === $x$ is a cardinal === {{AimForCont}} $x$ is not a cardinal. It follows that $\size x \in x$ and $x \sim \set x$ by corollary to Cardinal of Cardinal Equal to Cardinal and Ordinal Number Equivalent to Cardinal Number. Let $g: x \to \size x$ be a bijection. Let $f: \size x \to S$ be an injection. Then $f \circ g : x \to S$ is an injection by Composite of Injections is Injection. It follows that $x \preccurlyeq S$, so $x \in x$. This means that $x \subsetneq x$ by Transitive Set is Proper Subset of Ordinal iff Element of Ordinal, which is a contradiction. Therefore, $x$ must be an element of the class of cardinals. {{qed|lemma}} === No injection $f: x \to S$ === {{AimForCont}} that there is an injection $f: x \to S$. By the definition of $x$, it follows that $x \in x$. This means that $x \subsetneq x$ by Transitive Set is Proper Subset of Ordinal iff Element of Ordinal, which is a contradiction. Therefore, no injection can exist. {{qed}}
{{NotZFC}} Suppose $\mathcal N$ is a small class. By Cardinal Class is Subset of Ordinal Class: :$\mathcal N \subseteq \operatorname{On}$ Therefore, $\bigcup \mathcal N$ is an ordinal by Union of Subset of Ordinals is Ordinal. Take $x = \left\{{y \in \operatorname{On}: y \preccurlyeq \bigcup \mathcal N}\right\}$. By Cardinal Equal to Collection of All Dominated Ordinals, $x$ is a cardinal number and there is no injection $f : x \to \bigcup \mathcal N$. But since $x$ is a cardinal number, $x \in \mathcal N$, so $x \subseteq \bigcup \mathcal N$ by Set is Subset of Union/General Result. Therefore, the identity map $I_x: x \to \bigcup \mathcal N$ is an injection. This is a contradiction. Therefore, $\mathcal N$ is not a small class. {{qed}}
{{NotZFC}} Assume $\mathcal N’$ is a small class. By Union of Small Classes is Small, $\mathcal N’ \cup \omega$ is a small class. By the definition of the infinite cardinal class, $\mathcal N \subseteq \mathcal N’ \cup \omega$. But by Axiom of Subsets Equivalents, this means that $\mathcal N$ is a small class. This contradicts Cardinal Class is Proper Class. Therefore, $\mathcal N’$ is not a small class. {{qed}}
By the definition of the aleph mapping, $\aleph_y$ is an element of the infinite cardinal class. By Cardinal Inequality implies Ordinal Inequality, it follows that: :$x \in \aleph_y \iff \left|{ x }\right| \in \aleph_y$ {{qed}}
{{begin-eqn}} {{eqn | l = \left\vert{\aleph_x \times \aleph_x }\right\vert       | r = \left\vert{\aleph_x}\right\vert       | c = Non-Finite Cardinal is equal to Cardinal Product }} {{eqn | r = \aleph_x       | c = Aleph is Infinite Cardinal }} {{end-eqn}} {{qed}}
By the definition of the aleph mapping: : $\aleph : \operatorname{On} \to \mathcal N'$ {{explain|$\operatorname{On}$ and $\mathcal N'$}} The theorem statement is an immediate consequence of this fact. {{qed}} Category:Aleph Mapping h42qe2cr691hetujzbdfzo0avumsniu
:$y < \aleph_{x+1}$, then $y < \aleph_x \lor y \sim \aleph_x$ by Ordinal Less than Successor Aleph. In either case, $\left|{ y }\right| \le \aleph_x$ by Ordinal in Aleph iff Cardinal in Aleph and Equivalent Sets have Equal Cardinal Numbers. The existence of the surjection follows from Surjection iff Cardinal Inequality. {{qed}}
=== Sufficient Condition === {{begin-eqn}} {{eqn|ll =       |l = y      |o = <      |r = \aleph_{x+1}      |c =  }} {{eqn|ll = \iff      |l = \left\vert{ y }\right\vert      |o = <      |r = \aleph_{x+1}      |c = by Ordinal in Aleph iff Cardinal in Aleph }} {{end-eqn}} But $\left|{ y }\right|$ is a cardinal number, so it is either finite or an element of the infinite cardinal class. It is a finite set iff $y \in \omega$ by Ordinal is Finite iff Natural Number. If $y \in \omega$, then $y \in \aleph_x$ since $\aleph_x$ is an infinite set by Aleph is Infinite. If it is an element of the infinite cardinal class, then $y = \aleph_z$ for some ordinal $z$ by the definition of the aleph mapping. This means that: {{begin-eqn}} {{eqn|ll =       |l = \aleph_z      |o = <      |r = \aleph_{x+1}      |c = Hypothesis }} {{eqn|ll = \implies      |l = z      |o = <      |r = x+1      |c = since $\aleph$ is an order isomorphism }} {{eqn|ll = \implies      |l = z < x      |o = \lor      |r = z = x      |c = definition of successor }} {{end-eqn}} ==== Case 1: $z < x$ ==== In the first case $z < x$: {{begin-eqn}} {{eqn|ll =       |l = z      |o = <      |r = x      |c =  }} {{eqn|ll = \implies      |l = \aleph_z      |o = <      |r = \aleph_x      |c = definition of aleph mapping }} {{eqn|ll = \implies      |l = \left\vert{ y }\right\vert      |o = <      |r = \aleph_x      |c = definition of $\aleph_z$ }} {{eqn|ll = \implies      |l = y      |o = <      |r = \aleph_x      |c = by Ordinal in Aleph iff Cardinal in Aleph }} {{end-eqn}} ==== Case 2: $z = x$ ==== {{begin-eqn}} {{eqn|ll =       |l = z      |o = =      |r = x      |c =  }} {{eqn|ll = \implies      |l = \aleph_z      |o = =      |r = \aleph_x      |c = by Substitutivity of Class Equality }} {{eqn|ll = \implies      |l = y      |o = \sim      |r = \aleph_x      |c = by definition of $\aleph_z$ }} {{end-eqn}} Therefore, $y < \aleph_x \lor y \sim \aleph_x$. {{qed|lemma}} === Necessary Condition === Suppose that $y < \aleph_x \lor y \sim \aleph_x$. ==== Case 1: $y < \aleph_x$ ==== Suppose that $y < \aleph_x$. It follows that $y < \aleph_{x+1}$ since $\aleph$ is strictly monotone by the definition of the aleph mapping. ==== Case 2: $y \sim \aleph_x$ ==== Suppose that $y \sim \aleph_x$. Then $\left|{ y }\right| = \aleph_x$ by Equivalent Sets have Equal Cardinal Numbers. But $\aleph_x < \aleph_{x+1}$ since $\aleph$ is strictly monotone. Therefore, $\left|{ y }\right| < \aleph_{x+1}$ and $y < \aleph_{x+1}$ by Ordinal in Aleph iff Cardinal in Aleph. Category:Aleph Mapping 0pr2ix4obha5er6bf8uz7kyxv83za30
Since $0 \le x$, it follows that $\aleph_0 \le \aleph_x$ by definition of the aleph mapping. But $\aleph_0 = \omega$ by Aleph-Null. Therefore, $\omega \le \aleph_x$. {{qed}} Category:Aleph Mapping ac1bym14fscxt2vpvqawgpgqbf3le87
For all $n \in \omega$, $n \notin \NN'$ by the definition of the infinite cardinal class. Therefore, $\omega \le \aleph_0$. {{qed|lemma}} Moreover, $\omega \in \NN'$ by Minimal Infinite Successor Set is Infinite Cardinal. Therefore, $\aleph_x = \omega$ for some ordinal $x$. It follows that $\aleph_0 \le \aleph_x$ since $0 \le x$ and the definition of the aleph mapping, so $\aleph_0 \le \omega$. Thus, $\aleph_0 = \omega$. {{qed}} Category:Aleph Mapping r0o075wc8ay2tgjozsuu2k8k1hx195l
By Cardinal Number Less than Ordinal: Corollary: :$\left\vert{\omega}\right\vert \le \omega$ Moreover, for any $n \in \omega$, by Cardinal of Finite Ordinal: :$\left\vert{n}\right\vert < \left\vert{n+1}\right\vert \le \left\vert{\omega}\right\vert$ Thus by Cardinal of Finite Ordinal: : $n \in \left\vert{\omega}\right\vert$ Therefore: : $\omega = \left\vert{\omega}\right\vert$ By Cardinal of Cardinal Equal to Cardinal: Corollary: : $\omega \in \mathcal N’$ {{qed}} Category:Cardinals Category:Minimal Infinite Successor Set encs8nywgu8r7y5ek5ygnq7v3r0hw6l
The set of all mappings $S^T$ is equal to the collection of all mappings $f : S \to T$. Each of these mappings $f$ is a subset of $S \times T$. Thus, $S^T \subseteq \mathcal P \left({ S \times T }\right)$. Therefore, by Cartesian Product is Small and the axiom of powers, $S^T$ is a small class. {{qed}}
Define the mapping $F : \left({ R^S }\right)^T \to R^{S \times T}$ as follows: : $F\left({ f }\right) \left({ x,y }\right) = \left({f\left({x}\right)}\right) \left({y}\right)$ for all $x \in S , y \in T$. Suppose $F\left({f_1}\right) = F\left({f_2}\right)$. Then $\left({f_1\left({x}\right)}\right)\left({y}\right) = \left({f_2\left({x}\right)}\right)\left({y}\right)$ for all $x \in S , y \in T$ by the definition of $F$. Therefore, $f_1\left({x}\right) = f_2\left({x}\right)$ and $f_1 = f_2$ by Equality of Mappings. It follows that $F$ is an injection. Take any $g \in R^{S \times T}$. Define a function $f$ as $\left({f\left({x}\right)}\right)\left({y}\right) = g\left({x,y}\right)$. It follows that: {{begin-eqn}} {{eqn|l = F\left({f}\right) \left({x,y}\right)      |r = \left({ f\left({x}\right) }\right) \left({y}\right)      |c = definition of $F$ }} {{eqn|l =       |r = g\left({x,y}\right)      |c = definition of $f$ }} {{end-eqn}} Therefore, $F$ is a surjection. Thus, $F$ is a bijection. It follows that $ R^{S \times T} \sim \left({ R^S }\right)^T$ by the definition of set equivalence. {{qed}}
For brevity and readability, let us write $\hom$ for $\hom_\mathbf C$. For any object $D$ of $\mathbf C$ and morphism $f: C \to D \in \hom \paren {C, D}$ we have: :$\hom \paren {C, \operatorname{id}_D} \paren f = \operatorname{id}_D \circ f = f$ and so $\hom \paren {C, \operatorname{id}_D} = \operatorname{id}_{\hom \paren {C, D} }$. Thus $\hom \paren {C, \cdot}$ preserves identity morphisms. For composable morphisms $g$ and $f$ of $\mathbf C$ we furthermore have: {{begin-eqn}} {{eqn | l = \hom \paren {C, g \circ f} \paren x       | r = \paren {g \circ f} \circ x }} {{eqn | r = g \circ \paren {f \circ x}       | c = Composition of morphisms is associative }} {{eqn | r = \hom \paren {C, g} \circ \hom \paren {C, f} \paren x }} {{end-eqn}} for any $x \in \hom \paren {C, \operatorname{dom} f}$ composable with $f$. Hence the result, by Equality of Mappings. {{qed}}
Since $f$ is an isomorphism, it admits at least one inverse morphism. Now let $g, g': D \to C$ be two inverse morphisms for $f$. Then: {{begin-eqn}} {{eqn|l = g      |r = g \circ \operatorname{id}_D      |c = Axiom $(C2)$ for metacategories }} {{eqn|r = g \circ \left({f \circ g'}\right)      |c = $g'$ is an inverse morphism for $f$ }} {{eqn|r = \left({g \circ f}\right) \circ g'      |c = Axiom $(C3)$ for metacategories }} {{eqn|r = {\operatorname{id}_C} \circ g'      |c = $g$ is an inverse morphism for $f$ }} {{eqn|r = g'      |c = Axiom $(C2)$ for metacategories }} {{end-eqn}} In conclusion, $g = g'$. Hence the result. {{qed}}
{{MissingLinks|to $\le$, mostly. Some of the results have their own pages already}} Each of the conditions for cofinal ordinals shall be verified: :$x \le x$ follows by Set is Subset of Itself. The mapping $f: x \to x$ can simply be the identity mapping $I_x$: :$I_x: x \to x$ Moreover, $a < b \implies I_x \left({a}\right) < I_x \left({b}\right)$ by the definition of the identity mapping. Therefore: :$I_x$ is strictly increasing. Finally, $I_x \left({a}\right) \ge a$ by definition of identity mapping. It follows by Existential Generalisation that: :$\forall a \in x: \exists b \in x: I_x \left({b}\right) \ge a$ {{qed}}
The conditions for $z$ being cofinal with $x$ shall be verified individually. Let $\le$ denote the subset relation. Since $x \le y$ and $y \le z$, it follows that $x \le z$ by Subset Relation is Transitive Let $f : y \to x$ be the strictly increasing mapping that satisfies: :$\forall a \in x: \exists b \in y: f \left({b}\right) \ge a$ Let $g : z \to y$ be the strictly increasing mapping that satisfies: :$\forall a \in y: \exists b \in z: g \left({b}\right) \ge a$ Then $f \circ g : z \to x$ is a strictly increasing mapping. Moreover, take any $a \in x$. There is some $b \in y$ such that $f \left({b}\right) \ge a$. But $b \in y$, so there is some $c \in z$ such that: :$g \left({c}\right) \ge b$. {{begin-eqn}} {{eqn | l = f\left({g \left({c}\right)}\right)       | o = \ge       | r = f \left({b}\right)       | c = $f$ is strictly increasing }} {{eqn | l =        | o = \ge       | r = a       | c = Inequality above }} {{end-eqn}} Therefore, $f \circ g \left({c}\right) \ge a$ for some $c \in z$. {{qed}}
=== $\operatorname{cof} \left({ x,0 }\right) \implies x = 0$ === If $\operatorname{cof} \left({ x,0 }\right)$, then there is a function $f : x \to 0$. If $x \ne 0$, then $x$ has an element $a$. But then, $f\left({a}\right) \in 0$, which contradicts the definition of the empty set. Therefore, $x = 0$. {{qed|lemma}} === $x = 0 \implies \operatorname{cof} \left({ 0,x }\right)$ === Follows directly from Cofinal Ordinal Relation is Reflexive. {{qed|lemma}} === $\operatorname{cof} \left({ 0,x }\right) \implies \operatorname{cof} \left({ x,0 }\right)$ === If $\operatorname{cof} \left({ 0,x }\right)$, then $x \le 0$ by the definition of cofinal. By Subset of Empty Set, $x = 0$. $\operatorname{cof} \left({ x,0 }\right)$ follows directly from Cofinal Ordinal Relation is Reflexive. {{qed}}
Both $x$ and $y$ are non-empty, so by the definition of a limit ordinal: :$x = z^+$ for some $z$. :$y = w^+$ for some $w$. :$\bigcup z^+ \le \bigcup w^+$ follows by Set Union Preserves Subsets/General Result. :$z \le w$ follows by Union of Successor Ordinal. Define the function $f : x \to y$ as follows: :$\map f a = \begin{cases} a &: a \ne z \\ w &: a = z \end{cases}$ $a < b \le w$, so $f \left({a}\right) = a$. Take any $a,b \in x$ such that $a < b$. :$\map f a < \map f b$ shall be proven by cases: === Case 1: $b \ne z$ === If $b \ne z$: :$\map f a < \map f b$ is simply a restatement of $a < b$. === Case 2: $b = z$ === If $b = z$, then $\map f b = w$ by the definition of $f$. Since $a < z \le w$, $\map f a < \map f b$. It follows that $f$ is strictly increasing. {{qed|lemma}} Moreover, since $\bigcup y = w$ is the least upper bound, $f\left({z}\right) \ge a$ for all $a \in y$. {{qed}}
Since $1 = 0^+$, $1$ is not a limit ordinal. Let $\le$ denote the subset relation. It follows that $0 < 1$ by Ordinal is Less than Successor. Moreover, $1 \le x$ follows by the fact that $0 < x$ and Successor of Element of Ordinal is Subset. Thus we have $0 < 1 \le x$ and so by Condition for Cofinal Nonlimit Ordinals: :$\operatorname{cof} \left({x, 1}\right)$ {{qed}}