hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | l = \paren {a \circ b} \circ \paren {a \circ b}^{-1}       | r = e       | c = {{Defof|Inverse Element}} }} {{eqn | ll= \leadsto       | l = a \circ \paren {b \circ \paren {a \circ b}^{-1} }       | r = e       | c = Group Axiom $G1$: Associativity }} {{eqn | ll= \leadsto       | l = b \circ \paren {a \circ b}^{-1}       | r = a^{-1}       | c = Group Product Identity therefore Inverses }} {{eqn | ll= \leadsto       | l = b^{-1} \circ b \circ \paren {a \circ b}^{-1}       | r = b^{-1} \circ a^{-1}       | c =  }} {{eqn | ll= \leadsto       | l = e \circ \paren {a \circ b}^{-1}       | r = b^{-1} \circ a^{-1}       | c = {{Defof|Inverse Element}} }} {{eqn | ll= \leadsto       | l = \paren {a \circ b}^{-1}       | r = b^{-1} \circ a^{-1}       | c = {{Defof|Identity Element}} }} {{end-eqn}} {{Qed}}
Let $A \ne B$. Then $B \subsetneq A$. Therefore, by Set Difference with Proper Subset: :$A \setminus B \ne \varnothing$ Then: {{begin-eqn}} {{eqn | l = A \setminus B \ne \varnothing       | o = \implies       | r = \exists x \in \left({A \setminus B}\right): \left({A \setminus B}\right) \cap A_x = \varnothing       | c = Proper Well-Ordering Determines Smallest Elements }} {{eqn | o = \implies       | r = \exists x \in \left({A \setminus B}\right): \left({A \cap A_x}\right) \subseteq B       | c = Set Difference with Superset is Empty Set }} {{end-eqn}} One direction of inclusion is proven. By the hypothesis: {{begin-eqn}} {{eqn | l = x \in A \land x \prec y \land y \in B       | o = \implies       | r = x \in B }} {{end-eqn}} But $x \in A \land x \notin B$, so: {{begin-eqn}} {{eqn | l = y \in B       | o = \implies       | r = \neg x \prec y       | c = Modus Tollendo Tollens and other propositional manipulations }} {{eqn | o = \implies       | r = y \prec x       | c = $\prec$ is totally ordered and $y \ne x$}} {{end-eqn}} Therefore: :$B \subseteq A_x$ and so  $B: \subseteq A \cap A_x$ {{qed}}
Let $x \in S$ be invertible, where $y$ is an inverse of $x$. Then: :$x \circ y = e = y \circ x$ by definition. {{qed}}
Let $g \in G$. Then: {{begin-eqn}} {{eqn | l = g g^{-1}       | r = e       | c = {{Defof|Inverse Element}} }} {{eqn | ll= \leadsto       | l = g       | r = \paren {g^{-1} }^{-1}       | c = Group Product Identity therefore Inverses }} {{end-eqn}} {{qed}}
{{NotZFC}} Let $x$ be an ordinal. Then $\map F x = \map G {F \restriction x}$ and $\map G {F \restriction x} \in A$ by hypothesis. Therefore, $\map F x \in A$. This satisfies the first statement. Take two distinct ordinals $x$ and $y$. {{WLOG}}, assume $x \in y$ (we are justified in this by Ordinal Membership is Trichotomy). Then: {{begin-eqn}} {{eqn | l = x \in y       | o = \leadsto       | r = \map F x \in \Img y }} {{eqn | o = \leadsto       | r = \map F x \in \Img y \land \map F y \notin \Img y       | c = by hypothesis }} {{eqn | o = \leadsto       | r = \map F x \ne \map F y }} {{end-eqn}} Thus for distinct ordinals $x$ and $y$, $\map F x \ne \map F y$. Therefore, $F$ is injective. $F$ is injective and $F: \On \to A$. Therefore, if $A$ is a set, then $\On$ is a set. But by the Burali-Forti Paradox, this is impossible, so $A$ is not a set. Therefore, $A$ is a proper class. {{qed}}
{{NotZFC}} Set $B$ equal to the class of all ordinals $x$ such that $\left({A \setminus \operatorname{Im} \left({x}\right)}\right) \ne \varnothing$. Assume $B = \operatorname{On}$.   Then: {{begin-eqn}} {{eqn | l = B = \operatorname{On}       | o = \implies       | r = \forall x: F \left({x}\right) = G \left({F \restriction x}\right)       | c = by definition of $B$ }} {{eqn | o = \implies       | r = \forall x: G \left({F \restriction x}\right) \in \left({A \setminus \operatorname{Im}(F)}\right)       | c = by hypothesis }} {{end-eqn}} By Condition for Injective Mapping on Ordinals, $A$ is a proper class. This contradicts the fact that $A$ is a set. Therefore $B \subsetneq \operatorname{On}$. Because $B$ is bounded above, $\bigcup B \in \operatorname{On}$. By Union of Ordinals is Least Upper Bound, the union of ordinals is the least upper bound of $B$. Setting $\bigcup B = x$: :$(1): \quad \left({A \setminus \operatorname{Im} \left({x}\right)}\right) = \varnothing \land \forall y \in x: \left({A \setminus \operatorname{Im} \left({y}\right)}\right) \ne \varnothing$ The first condition is satisfied. In addition: :$(2): \quad A \subseteq \operatorname{Im} \left({x}\right)$ Take any $y \in \operatorname{Im} \left({x}\right)$. Then: {{begin-eqn}} {{eqn | l = y \in \operatorname{Im} \left({x}\right)       | o = \implies       | r = \exists z \in x: \left({y = F \left({z}\right)}\right)       | c = Definition of Image of Element }} {{eqn | o = \implies       | r = \exists z: \left({y = F \left({z}\right) \land \left({A \setminus \operatorname{Im} \left({z}\right)}\right) \ne \varnothing}\right)       | c = Equation $(1)$ }} {{eqn | o = \implies       | r = \exists z: \left({y = F \left({z}\right) \land F \left({z}\right) \in  \left({A \setminus \operatorname{Im} \left({z}\right)}\right)}\right)       | c = by hypothesis }} {{eqn | o = \implies       | r = y \in A }} {{end-eqn}} This means that: :$\operatorname{Im} \left({x}\right) \subseteq A$ Combining with $(2)$: :$\operatorname{Im} \left({x}\right) = A$ $F$ is a mapping, so $\left({F \restriction x}\right)$ is a mapping.  Take any $y, z \in x$ such that $y$ and $z$ are distinct. {{WLOG}}, allow $y \in z$ (justified by Ordinal Membership Trichotomy). {{begin-eqn}} {{eqn | l = y \in z \land z \in x       | o = \implies       | r = F \left({y}\right) \in \operatorname{Im} \left({z}\right) \land F \left({z}\right) \in \left({A \setminus \operatorname{Im} \left({z}\right)}\right)       | c = by hypothesis }} {{eqn | o = \implies       | r = F \left({y}\right) \in \operatorname{Im} \left({z}\right) \land F \left({z}\right) \notin \operatorname{Im} \left({z}\right)       | c = Definition of Set Difference }} {{eqn | o = \implies       | r = F \left({y}\right) \ne F \left({z}\right)       | c =  }} {{end-eqn}} From this, we may conclude that $F$ is injective. {{qed}}
The proof will proceed by the Principle of Mathematical Induction on $\N$. Let $T$ be the set defined as: :$\displaystyle T := \set {n \in \N: \bigoplus_{k \mathop = 1}^n a_k = \oplus^n a}$ First, recall the definition of the composite of $\tuple {a_1, a_2, \ldots, a_n}$ for $\oplus$: :$\displaystyle \bigoplus_{k \mathop = 1}^n a_k = \begin{cases} a: & n = 1 \\ \map {\oplus_m} {a_1, \ldots, a_m} \oplus a_{m + 1}: & n = m + 1 \end{cases}$ Secondly, recall the definition of the $n$th power of $a$ under $\oplus$: :$\forall n \in \N_{>0}: \oplus^n a = \begin{cases} a & : n = 1 \\ \paren {\oplus^m a} \oplus a & : n = m + 1 \end{cases}$ === Basis for the Induction === We have that: {{begin-eqn}} {{eqn | l = \bigoplus_{k \mathop = 1}^1 a_k       | r = a_1       | c = {{Defof|Composite (Abstract Algebra)|Composite}} of $\paren {a_1}$ }} {{eqn | r = a       | c = Definition of $\tuple {a_1, a_2, \ldots, a_n}$: $\forall k \in \N_n: a_k = a$ }} {{eqn | r = \oplus^1 a       | c = {{Defof|Power of Element of Semigroup}} }} {{end-eqn}} So $1 \in T$. This is our basis for the induction. === Induction Hypothesis === It is to be shown that, if $j \in T$ where $j \ge 1$, then it follows that $j + 1 \in T$. This is the induction hypothesis: :$\displaystyle \bigoplus_{k \mathop = 1}^j a_k = \oplus^j a$ It is to be demonstrated that it follows that: :$\displaystyle \bigoplus_{k \mathop = 1}^{j + 1} a_k = \oplus^{j + 1} a$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \bigoplus_{k \mathop = 1}^{j + 1} a_k       | r = \map {\oplus_j} {a_1, \ldots, a_j} \oplus a_{j + 1}       | c = Definition of Composite of $\left({a_1, \ldots, a_j, a_{j+1} }\right)$ }} {{eqn | r = \paren {\bigoplus_{k \mathop = 1}^j a_k} \oplus a_{j + 1}       | c = {{Defof|Composite (Abstract Algebra)|Composite}} of $\tuple {a_1, \ldots, a_j}$ }} {{eqn | r = \paren {\oplus^j a} \oplus a_{j + 1}       | c = Induction Hypothesis }} {{eqn | r = \paren {\oplus^j a} \oplus a       | c = Definition of $\tuple {a_1, a_2, \ldots, a_{j + 1} }$: $\forall k \in \N_{j + 1}: a_k = a$ }} {{eqn | r = \oplus^{j + 1} a       | c = {{Defof|Power of Element of Semigroup}} }} {{end-eqn}} So $k \in T \implies k + 1 \in T$ and the result follows by the Principle of Mathematical Induction: :$\displaystyle \forall n \in \N: \bigoplus_{k \mathop = 1}^n a_k = \oplus^n a$
The existence of $x$ and $f$ follows from Woset is Isomorphic to Unique Ordinal. The uniqueness of $x$ follows from Woset is Isomorphic to Unique Ordinal. The uniqueness of $f$ follows from Order Isomorphism between Wosets is Unique. {{qed}}
Since $S \subset \operatorname{On}$, $\left({S, \in}\right)$ is a strict well-ordering. {{explain|Link to result justifying this statement.}} The result follows directly from Strict Well-Ordering Isomorphic to Unique Ordinal under Unique Mapping. {{qed}}
Define the mapping $F: \operatorname{On} \to \operatorname{On} \times \operatorname{On}$ as: : $\forall x \in \operatorname{On}: F \left({x}\right) = \left({\varnothing, x}\right)$ Then, $F: \operatorname{On} \to \left({\operatorname{On} \times \operatorname{On}}\right)_{\left({1, \varnothing}\right)}$, since $\varnothing < 1$. {{explain|The above statement does not make linguistic sense. The entity $F: \operatorname{On} \to \left({\operatorname{On} \times \operatorname{On} }\right)_{\left({1, \varnothing}\right)}$ is a noun, and there is no verb in the "then" clause.}} That is, $F$ is a mapping from the class of all ordinals to the initial segment of $\left({1, \varnothing}\right)$ with respect to lexicographic order. By Equality of Ordered Pairs, $F$ is injective. But since $\operatorname{On}$ is a proper class by the Burali-Forti Paradox, the initial segment of $\left({1, \varnothing}\right)$ is a proper class as well. {{qed}}
Let $z = \max \left({x, y}\right)^+$. Let $\left({v, w}\right) R_0 \left({x, y}\right)$. Then: {{begin-eqn}} {{eqn | l = \max \left({v, w}\right)       | o = \le       | r = \max \left({x, y}\right) }} {{eqn | o = \lt       | r = z }} {{end-eqn}} Thus, the initial segment: : $\left({\operatorname{On} \times \operatorname{On} }\right)_{\left({x, y}\right)} \subseteq \left({z \times z}\right)$ {{explain|Clarify the sentence flow of the above statement.}} By Axiom of Subsets Equivalents, the initial segment of $\left({x, y}\right)$ is a set. {{qed}}
Using Transfinite Induction on $y$: === Base Case === {{begin-eqn}} {{eqn|l = x      |o = \in      |r = \operatorname{On}      |c =  }} {{eqn|ll = \implies      |l = \left({x + \varnothing}\right)      |o = \in      |r = \operatorname{On}      |c = Definition of Ordinal Addition }} {{end-eqn}} === Inductive Case === {{begin-eqn}} {{eqn|l = \left({x+y}\right)      |o = \in      |r = \operatorname{On}      |c =  }} {{eqn|ll = \implies      |l = \left({x+y}\right)^+      |o = \in      |r = \operatorname{On}      |c = Successor Set of Ordinal is Ordinal }} {{eqn|ll = \implies      |l = \left({x+y^+}\right)      |o = \in      |r = \operatorname{On}      |c = Definition of Ordinal Addition }} {{end-eqn}} === Limit Case === {{begin-eqn}} {{eqn|lo = \forall z \in y:      |l = \left({x + z}\right)      |o = \in      |r = \operatorname{On}      |c =  }} {{eqn|ll = \implies      |l = \bigcup_{z \mathop \in y} \left({x + z}\right)      |o = \in      |r = \operatorname{On}      |c = Union of Ordinals is Ordinal }} {{eqn|ll = \implies      |l = \left({x + z}\right)      |o = \in      |r = \operatorname{On}      |c = Definition of Ordinal Addition }} {{end-eqn}} {{qed}} {{MissingLinks|The meanings of "base case", "inductive case" and "limit case".}}
{{begin-eqn}} {{eqn|l = x \in \bigcup_{z \mathop \in y} F \left({z}\right)      |o = \implies      |r = \exists z \in y: x \in F \left({z}\right)      |c = Definition of Set Union }} {{eqn|o = \implies      |r = x \in \operatorname{On}      |c = by hypothesis }} {{end-eqn}} So: : $\displaystyle \bigcup_{z \mathop \in y} F \left({z}\right) \subseteq \operatorname{On}$ {{begin-eqn}} {{eqn|l = x \in \bigcup_{z \mathop \in y} F \left({z}\right)      |o = \implies      |r = \exists z \in y: x \in F \left({z}\right)      |c = Definition of Set Union }} {{eqn|o = \implies      |r = \exists z \in y: x \subseteq F \left({z}\right)      |c = Ordinals are Transitive }} {{eqn|o = \implies      |r = x \subseteq \bigcup_{z \mathop \in y} F \left({z}\right)      |c = Union Preserved Under Subset Relation }} {{end-eqn}} So $\displaystyle \bigcup_{z \mathop \in y} F \left({z}\right)$ is a transitive set. Therefore $\displaystyle \bigcup_{z \mathop \in y} F \left({z}\right)$ is an ordinal. {{begin-eqn}} {{eqn|l = \bigcup_{z \mathop \in y} F \left({z}\right)      |r = \bigcup \operatorname{Im} \left({y}\right)      |c = }} {{end-eqn}} Let $U$ denote the universal class. {{begin-eqn}} {{eqn|l = y \in V      |o = \implies      |r = \operatorname{Im} \left({y}\right) \in V      |c = Axiom of Replacement Equivalents }} {{eqn|o = \implies      |r = \bigcup \operatorname{Im} \left({y}\right) \in V      |c = Axiom of Unions Equivalents }} {{eqn|o = \implies      |r = \bigcup_{z \mathop \in y} F \left({z}\right) \in V      |c = Equality given above }} {{end-eqn}} Therefore $\displaystyle \bigcup_{z \mathop \in y} F \left({z}\right)$ is a set, so it is a member of the ordinal class. {{qed}} Category:Ordinals fofy4ax4wx02wa3nh17e3rx0ddwmep2
By definition of ordinal addition, it is immediate that: :$x + \varnothing = x$ {{qed|lemma}} We shall use Transfinite Induction on $x$ to prove $\left({\varnothing + x}\right) = x$ === Base Case === The induction basis $x = \varnothing$ comes down to: :$\varnothing + \varnothing = \varnothing$ This follows by the above. {{qed|lemma}} === Inductive Case === For the induction step, suppose that $\varnothing + x = x$. Then, also: {{begin-eqn}} {{eqn|l = x^+      |r = \left({\varnothing + x}\right)^+      |c = Substitutivity of Equality }} {{eqn|r = \varnothing + x^+      |c = Definition of ordinal addition }} {{end-eqn}} {{qed|lemma}} === Limit Case === Finally, the limit case. So let $x$ be a limit ordinal, and suppose that: :$\forall y \in x: \varnothing + y = y$ Now we have: {{begin-eqn}} {{eqn|l = x      |r = \bigcup_{y \mathop \in x} y      |c = Union of Limit Ordinal }} {{eqn|r = \bigcup_{y \mathop \in x} \left({\varnothing + y}\right)      |c = Indexed Union Equality }} {{eqn|r = \varnothing + x      |c = Definition of ordinal addition }} {{end-eqn}} {{qed|lemma}} Hence the result, by Transfinite Induction. {{qed}}
{{ProofWanted|Complete the Proof (use Union of Ordinals is Least Upper Bound)}}
By induction on the well-formed parts of $\map P x$. The proof shall use $\implies$ and $\neg$ as the primitive connectives. === Atoms === :$x = y \implies \paren {x \in z \iff y \in z}$ by Substitution of Elements. :$x = y \implies \paren {z \in x \iff z \in y}$ by definition of Set Equality. === Inductive Step for $\implies$ === Suppose that $\map P x$ is of the form $\map Q x \implies \map R x$ Furthermore, suppose that: :$x = y \implies \paren {\map Q x \iff \map Q y}$ and: :$x = y \implies \paren {\map R x \iff \map R y}$ It follows that: :$x = y \implies \paren {paren {\map Q x \implies \map R x} \iff \paren {\map Q y \implies \map R y} }$ :$x = y \implies \paren {\map P x \iff \map P y}$ === Inductive Step for $\neg$ === Suppose that $\map P x$ is of the form $\neg \map Q x$ Furthermore, suppose that: :$x = y \implies \paren {\map Q x \iff \map Q y}$ It follows that: :$x = y \implies \paren { \neg \map Q x \iff \neg \map Q y}$ :$x = y \implies \paren {\map P x \iff \map P y}$ === Inductive Step for $\forall x:$ === Suppose that $\map P x$ is of the form $\forall z: \map Q {x, z}$ If $x$ and $z$ are the same variable, then $x$ is a bound variable and the theorem holds trivially. If $x$ and $z$ are distinct, then: {{begin-eqn}} {{eqn | l = x = y       | o = \leadsto       | r = \paren {\map Q {x, z} \iff \map Q {y, z} }       | c = Inductive Hypothesis }} {{eqn | o = \leadsto       | r = \forall z: \paren {\map Q {x, z} \iff \map Q {y, z} }       | c = Universal Generalization }} {{eqn | o = \leadsto       | r = \paren {\forall z: \map Q {x, z} \iff \forall z: \map Q {y, z} }       | c = Predicate Logic manipulation }} {{eqn | o = \leadsto       | r = \paren {\map P x \iff \map P y}       | c = Definition of $P$ }} {{end-eqn}} {{qed}}
The proof proceeds by transfinite induction on $y$. In the proof, we shall use $\in$, $\subsetneq$, and $<$ interchangeably. We are justified in this by Transitive Set is Proper Subset of Ordinal iff Element of Ordinal. === Base Case === {{begin-eqn}} {{eqn | l = \neg x       | o <       | r = \O       | c = {{Defof|Empty Set}} }} {{end-eqn}} The conclusion: :$x < \O \implies \paren {z + x} < \paren {z + \O}$ follows from propositional logic. {{explain|... how?}} === Inductive Case === {{begin-eqn}} {{eqn | l = x < y       | o = \leadsto       | r = \paren {z + x} < \paren {z + y}       | c = Hypothesis }} {{eqn | l = x < y^+       | o = \leadsto       | r = x < y \lor x = y       | c = {{Defof|Successor Set}} }} {{eqn | l = \paren {z + y^+} = \paren {z + y}^+       | o =       | c = {{Defof|Ordinal Addition}} }} {{eqn | l = \paren {z + y} < \paren {z + y^+}       | o =       | c = Ordinal is Less than Successor }} {{eqn | l = x < y       | o = \leadsto       | r = \paren {z + x} < \paren {z + y^+}       | c = Hypothetical Syllogism }} {{eqn | l = x = y       | o = \leadsto       | r = \paren {z + x} < \paren {z + y^+}       | c = Substitutivity of Equality }} {{eqn | l = x < y^+       | o = \leadsto       | r = \paren {z + x} < \paren {z + y^+}       | c = Proof by Cases }} {{end-eqn}} === Limit Case === {{begin-eqn}} {{eqn| o =       | r = \forall w < y: \paren {x < w \implies \paren {z + x} < \paren {z + w} }      | c =  }} {{eqn| o = \leadsto      | r = \paren {\exists w < y: x < w \implies \exists w < y \paren {z + x} < \paren {z + w} }      | c = Predicate Logic Manipulation }} {{eqn| o = \leadsto      | r = \paren {x < y \implies \paren {z + x} < \bigcup_{w \mathop \in y} \paren {z + w} }      | c = Membership of Indexed Union }} {{eqn| o = \leadsto      | r = \paren {x < y \implies \paren {z + x} < \paren {z + y} }      | c = {{Defof|Ordinal Addition}} }} {{end-eqn}} {{qed}} {{MissingLinks|"base case", "inductive case" and "limit case".}}
For the proof, $<$, $\in$, and $\subsetneq$ will be used interchangeably. This is justified by Transitive Set is Proper Subset of Ordinal iff Element of Ordinal. Note that: {{begin-eqn}} {{eqn | l = x < y       | o = \implies       | r = \left({z + x}\right) < \left({z + y}\right)       | c = Membership is Left Compatible with Ordinal Addition }} {{eqn | l = y < x       | o = \implies       | r = \left({ z + y }\right) < \left({ z + x }\right)       | c = Membership is Left Compatible with Ordinal Addition }} {{end-eqn}} However: {{begin-eqn}} {{eqn | l = \left({z + x}\right) = \left({z + y}\right)       | o = \implies       | r = \left({z + x}\right) \not < \left({z + y}\right) \land \left({z + y}\right) \not < \left({z + x}\right)       | c = No Membership Loops }} {{end-eqn}} This contradicts the consequents of the first two equations, so: {{begin-eqn}} {{eqn | o = \implies       | r = x \not < y \land y \not < x       | c = Rule of Transposition }} {{eqn | o = \implies       | r = x = y       | c = Ordinal Membership is Trichotomy }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = x < \bigcup A       | o = \implies       | r = \exists z: \left({x < z \land z < A}\right)       | c = Definition of Set Union }} {{eqn | o = \implies       | r = \exists y \in B: \exists z: \left({x < z \land z \le y}\right)       | c = By Hypothesis }} {{eqn | o = \implies       | r = \exists y \in B: x < y }} {{eqn | o = \implies       | r = x < \bigcup B       | c = Union of Ordinals is Least Upper Bound }} {{end-eqn}} Therefore: : $\displaystyle \bigcup A \subseteq \bigcup B$ and: : $\displaystyle \bigcup A \le \bigcup B$ {{qed}}
The proof proceeds by transfinite induction on $z$. === Base Case === {{begin-eqn}} {{eqn | l = \left({x + \varnothing}\right)       | r = x       | c = Definition of Ordinal Addition }} {{eqn | l = \left({y + \varnothing}\right)       | r = y       | c = Definition of Ordinal Addition }} {{eqn | ll= \leadsto       | l = x \le y       | o = \implies       | r = \left({x + \varnothing}\right) \le \left({y + \varnothing}\right)       | c = Substitutivity of Equality }} {{end-eqn}} === Inductive Case === {{begin-eqn}} {{eqn | l = x       | o = \le       | r = y       | c =  }} {{eqn | ll= \implies       | l = \left({x + z}\right)       | o = \le       | r = \left({y + z}\right)       | c = Induction Hypothesis }} {{eqn | ll= \implies       | l = \left({x + z}\right)^+       | o = \le       | r = \left({y + z}\right)^+       | c = Successor Preserved under Subset }} {{eqn | ll= \implies       | l = \left({x + z^+}\right)       | o = \le       | r = \left({y + z^+}\right)       | c = Definition of Ordinal Addition }} {{end-eqn}} === Limit Case === {{begin-eqn}} {{eqn | lo= \forall w < z:       | l = x \le y       | o = \implies       | r = \left({x + w}\right) \le \left({y + w}\right)       | c = }} {{eqn | ll= \leadsto       | l = x \le y       | o = \implies       | r = \forall w < z: \left({x + w}\right) \le \left({y + w}\right)       | c = }} {{eqn | ll= \leadsto       | l = x \le y       | o = \implies       | r = \bigcup_{w \mathop \in z} \left({x + w}\right) \le \bigcup_{w \mathop \in z} \left({y + w}\right)       | c = Indexed Union Subset }} {{eqn | ll= \leadsto       | l = x \le y       | o = \implies       | r = \left({x + z}\right) \le \left({y + z}\right)       | c = Definition of Ordinal Addition }} {{end-eqn}} The result follows by transfinite induction. {{qed}}
{{begin-eqn}} {{eqn|l = x = y |o = \implies |r = x^+ = y^+ |c = Substitutivity of Equality }} {{end-eqn}} Conversely, {{begin-eqn}} {{eqn|l = x^+ = y^+ |o = \implies |r = \bigcup x^+ = \bigcup y^+ |c = Substitutivity of Equality }} {{eqn|o = \implies |r = x = y |c = Union of Successor Ordinal }} {{end-eqn}} {{qed}}