hcju/proofwikiqa huggingface dataset:
Define the functor $F: \mathbf{Set}_* \to 1 \mathbin / \mathbf{Set}$ by: :$F \left({C, c}\right) := \bar c: 1 \to C$ :$F f := f$ where $\bar c: 1 \to C$ is defined by $\bar c (*) = c$. Further, define $G: 1 \mathbin / \mathbf{Set} \to \mathbf{Set}_*$ by: :$G \left({x: 1 \to C}\right) := \left({C, x (*)}\right)$ :$G f := f$ === $F$ is a functor === The definition of $F$ on morphisms is admissible, since for any pointed mapping $f: \left({C, c}\right) \to \left({D, d}\right)$: {{begin-eqn}} {{eqn|l = f \circ \bar c (*)      |r = f (c)      |c = Definition of $\bar c$ }} {{eqn|r = d      |c = $f$ is a pointed mapping }} {{eqn|r = \bar d (*)      |c = Definition of $\bar d$ }} {{end-eqn}} Thus by Equality of Mappings, $f \circ \bar c = \bar d$. So indeed $F f = f$ is a morphism $\bar c \to \bar d$, as desired. The identity morphisms of both $\mathbf{Set}_*$ and $1 \mathbin / \mathbf{Set}$ are the identity mappings, which $F$ thus preserves. That $F$ preserves composition is also trivial, since $\mathbf{Set}_*$ and $1 \mathbin / \mathbf{Set}$ both have composition of mappings as their $\circ$. In conclusion, $F$ is a functor. === $G$ is a functor === Let $x: 1 \to C$ and $y: 1 \to D$ be objects of the coslice $1 \mathbin / \mathbf{Set}$. Let $f: C \to D$ be a morphism $x \to y$. That is, let $f \circ x = y$. Thus, in particular: :$f \left({x (*)}\right) = y (*)$ showing that $f$ is a pointed mapping $\left({C, x(*)}\right) \to \left({D, y(*)}\right)$. Observe that the composition and the identity morphisms of $1 \mathbin / \mathbf{Set}$ and $\mathbf{Set}_*$ are identical. Because $G$ is the identity on morphisms, it is thus trivially a functor. === $F$ is an isomorphism === Because $F$ and $G$ are the identity on morphisms, it will suffice to show that: :$F G (x) = x$ for all objects $x$ of $1 \mathbin / \mathbf{Set}$ :$G F \left({C, c}\right) = \left({C, c}\right)$ for all objects $\left({C, c}\right)$ of $\mathbf{Set}_*$ Explicitly: {{begin-eqn}} {{eqn|l = F G \left({x: 1 \to C}\right)      |r = F \left({C, x (*)}\right) }} {{eqn|r = \overline{x (*)} }} {{end-eqn}} Now $\overline{x (*)}: 1 \to C$ is defined by $\overline{x (*)} (*) = x (*)$. Hence $\overline{x (*)} = x$ by Equality of Mappings. For the other equality: {{begin-eqn}} {{eqn|l = G F \left({C, c}\right)      |r = G \left({\bar c: 1 \to C}\right) }} {{eqn|r = \left({C, \bar c (*)}\right) }} {{eqn|r = \left({C, c}\right) }} {{end-eqn}} where the last equality follows by definition of $\bar c$. Thus $F$ is shown to be an isomorphism, and hence: :$\mathbf{Set}_* \cong 1 \mathbin / \mathbf{Set}$ {{qed}}
The proof shall proceed by Transfinite Induction on $z$. === Basis for the Induction === If $z = \varnothing$, then $x^z = 1$ {{begin-eqn}} {{eqn | l = x^z       | r = 1       | c = Definition of Ordinal Exponentiation }} {{eqn | l = y^z       | r = 1       | c = Definition of Ordinal Exponentiation }} {{eqn | l = x^z       | o = \le       | r = y^z       | c = Set is Subset of Itself }} {{end-eqn}} This proves the basis for the induction. {{qed|lemma}} === Induction Step === The inductive hypothesis states that $x^z \le y^z$ for $y$. Then: {{begin-eqn}} {{eqn | l = x^{z^+}       | r = x^z \times x       | c = Definition of Ordinal Exponentiation }} {{eqn | o = \le       | r = x^z \times y       | c = Membership is Left Compatible with Ordinal Multiplication }} {{eqn | o = \le       | r = y^z \times y       | c = Subset is Right Compatible with Ordinal Multiplication }} {{eqn | r = y^{z^+}       | c = Definition of Ordinal Exponentiation }} {{end-eqn}} This proves the induction step. {{qed|lemma}} === Limit Case === The inductive hypothesis for the limit case states that: :$\forall w \in z: x^w \le y^w$ where $z$ is a limit ordinal. {{begin-eqn}} {{eqn | lo= \forall w \in z:       | l = x^w       | o = \subseteq       | r = y^w       | c = Inductive Hypothesis }} {{eqn | ll = \implies       | l = \bigcup_{w \mathop \in z} x^w       | o = \subseteq       | r = \bigcup_{w \mathop \in z} y^w       | c = by Indexed Union Subset }} {{eqn | ll = \implies       | l = x^z       | o = \subseteq       | r = y^z       | c = Definition of Ordinal Exponentiation }} {{end-eqn}} This proves the limit case. {{qed}}
=== Sufficient Condition === Suppose that $x < y$. By Subset is Right Compatible with Ordinal Exponentiation: :$x^w \le y^w$ Then: {{begin-eqn}} {{eqn | l = x^z       | r = x^w \times x       | c = Definition of Ordinal Exponentiation }} {{eqn | l = y^z       | r = y^w \times y       | c = Definition of Ordinal Exponentiation }} {{eqn | l = x^w \times x       | o = \le       | r = y^w \times x       | c = Subset is Right Compatible with Ordinal Exponentiation }} {{eqn | l =        | o = <       | r = y^w \times y       | c = Membership is Left Compatible with Ordinal Exponentiation }} {{end-eqn}} Thus the sufficient condition is satisfied. {{qed}} === Necessary Condition === Suppose $x^z < y^z$. From Subset is Right Compatible with Ordinal Exponentiation: :$y \le x \implies y^z \le x^z$ This contradicts: : $x^z < y^z$ so: : $y \not \le x$ By Ordinal Membership is Trichotomy: : $x < y$ {{qed}}
The proof shall proceed by Transfinite Induction on $y$. === Basis for the Induction === If $y = 0$, then $0 \le x^y$ by Empty Set is Subset of All Sets. This proves the basis for the induction. {{qed|lemma}} === Induction Step === The inductive hypothesis states that: :$y \le x^y$ {{begin-eqn}} {{eqn | l = y       | o = \le       | r = x^y       | c = Inductive Hypothesis }} {{eqn | o = <       | r = x^y \times x       | c = Membership is Left Compatible with Ordinal Multiplication }} {{eqn | ll= \leadsto       | l = y \cup \left\{ {y}\right\}       | o = \le       | r = x^{y^+}       | c = Ordinal is Transitive }} {{eqn | ll= \leadsto       | l = y^+       | o = \le       | r = x^{y^+}       | c = {{Defof|Successor Set}} }} {{end-eqn}} This proves the induction step. {{qed|lemma}} === Limit Case === The inductive hypothesis for the limit case states that: :$\forall z \in y: z \le x^z$ and $y$ is a limit ordinal {{begin-eqn}} {{eqn | lo= \forall z \in y:       | l = z       | o = \le       | r = x^z       | c = Inductive Hypothesis }} {{eqn | ll= \leadsto       | l = \bigcup_{z \mathop \in y} z       | o = \le       | r = \bigcup_{z \mathop \in y} x^z       | c = Indexed Union Subset }} {{eqn | ll= \leadsto       | l = y       | o = \le       | r = x^y       | c = {{Defof|Ordinal Exponentiation}} and Union of Limit Ordinal }} {{end-eqn}} This proves the limit case. {{qed}}
=== Existence of $z$ === By Lower Bound for Ordinal Exponentiation, $y ≤ x^y$ so $y < x^{y^+}$. Therefore, $y$ is bounded above by $x^w$ for some $w$, so there is a smallest $w$ such that: :$y < x^w$ by Subset of Ordinals has Minimal Element. Moreover, if $w$ is a limit ordinal then there is some $v ∈ w$ such that $y < x^v$, contradicting the fact that $w$ is minimal. Therefore, $w$ is not a limit ordinal and therefore must be the successor of some ordinal. So $w = z^+$ for some $z$. Since $z < z^+$, it follows that $x^z ≤ y$ and $y < x^{z^+}$. {{qed|lemma}} === Uniqueness of $z$ === Assume that $x^z ≤ y$ and $y < x^{z^+}$ for some $z = z_1$ and $z = z_2$. Then $x^{z_1} < x^{z_2^+}$ and $x^{z_2} < x^{z_1^+}$. By Membership is Left Compatible with Ordinal Exponentiation $z_1 < z_2^+$ and $z_1 ≤ z_2$. Similarly $z_2 ≤ z_1$. By definition of set equality: : $z_1 = z_2$ {{qed}}
Suppose $x > 1$. Suppose also that $x^y$ is the successor of some ordinal $w$. By definition of ordinal multiplication: :$\displaystyle x^y = \bigcup_{z \mathop \in y} x^z$  Then: {{begin-eqn}} {{eqn|l = w      |o = \in      |r = x^y      |c = Ordinal is Less Than Successor }} {{eqn|ll = \exists z \in y:      |l = w      |o = \in      |r = x^z      |c = by definition of ordinal exponentiation }} {{eqn|l = w^+      |o = \subseteq      |r = x^z      |c = Successor of Element of Ordinal is Subset }} {{eqn|o = \in      |r = x^{z^+}      |c = Membership is Left Compatible with Ordinal Exponentiation }} {{end-eqn}} But $z^+ \in y$ by Successor in Limit Ordinal. So $w^+ \in x^y$ and $w^+ \in w^+$. This creates a membership loop and thus is a contradiction by No Membership Loops. {{qed|lemma}} For the second part, since $x$ is not the empty set it follows that $x = z^+$ for some $z$ or that $x$ is a limit ordinal. If $x$ is a limit ordinal, then $y^x$ is a limit ordinal by the first part. If $x$ is the successor of $z$, then: :$y^x = y^w × y$ by the definition of ordinal exponentiation. Then, $y^x$ is a limit ordinal by Limit Ordinals Preserved Under Ordinal Multiplication. {{qed}}
{{begin-eqn}} {{eqn|l = x \in y      |o = \iff      |r = x^+ \in y^+      |c = Successor is Less than Successor }} {{eqn|o = \iff      |r = \left({ x^+ \in y \lor x^+ = y }\right)      |c = definition of successor set }} {{eqn|o = \iff      |r = \left({ x^+ \subset y \lor x^+ = y }\right)      |c = Transitive Set is Proper Subset of Ordinal iff Element of Ordinal }} {{eqn|o = \iff      |r = x^+ \subseteq y      |c =  }} {{end-eqn}} {{qed}} Category:Ordinals 3ol827ogfd95sjoajp3m84i9j72uxcw
First, to prove that $\struct {S^*, *}$ is a semigroup. That is, to prove $*$ is associative. Let $s, s', s'' \in S^*$ be sequences of lengths $n, n', n''$, respectively. Then: {{begin-eqn}} {{eqn | l = \map {s * \paren {s' * s''} } i       | r = \begin {cases} \map s i & \text {if $1 \le i \le n$} \\ \map {s' * s''} {i - n} & \text {if $n < i \le n + n' + n''$} \end {cases} }} {{eqn | r = \begin {cases} \map s i & \text {if $1 \le i \le n$} \\ \map {s'} {i - n} & \text {if $n < i \le n + n'$} \\ \map {s''} {i - n - n'} & \text {if $n + n' < i \le n + n' + n''$} \end {cases} }} {{eqn | l = \map {\paren {s * s'} * s''} i       | r = \begin {cases} \map {s * s'} i & \text {if $1 \le i \le n + n'$} \\ \map {s''} {i - n - n'} & \text {if $n + n' < i \le n + n' + n''$} \end {cases} }} {{eqn | r = \begin {cases} \map s i & \text {if $1 \le i \le n$} \\ \map {s'} {i - n} & \text {if $n < i \le n + n'$} \\ \map {s''} {i - n - n'} & \text {if $n + n' < i \le n + n' + n''$} \end {cases} }} {{end-eqn}} Hence, by Equality of Mappings: :$s * \paren {s' * s''} = \paren {s * s'} * s''$ that is, $*$ is associative. Now, to prove $\struct {S^*, *}$ has an identity $e$. It follows immediately from the length of a concatenation that $e$ must have length $0$. That is, the only choice for $e$ is the empty sequence. Now, for any $s \in S^*$: :$\map {e * s} i = \begin {cases} \map e i & \text {if $1 \le i \le 0$} \\ \map s i & \text {if $0 < i < 0 + n$} \end {cases}$ from which we see that $e * s = s$. Also: :$\map {s * e} i = \begin {cases} \map s i & \text {if $1 \le i \le n$} \\ \map e i & \text {if $n < i < n + 0$} \end {cases}$ which shows $s * e = s$. So indeed the empty sequence is an identity element of $\struct {S^*, *}$. Hence $\struct {S^*, *}$ is a monoid. {{qed}}
By Kleene Closure is Monoid, $S^*$ is a monoid. It remains to verify the universal property. === Existence === Let $\left({N, \circ}\right)$ be a monoid with identity $e_N$, and let $\left\vert{N}\right\vert$ be its underlying set. Let $f: S \to \left\vert{N}\right\vert$ be a mapping. Define $\bar f: S^* \to N$ by: :$\bar f \left({-}\right) := e_N$ :$\bar f \left({\left\langle{s_1, \ldots, s_n}\right\rangle}\right) := f \left({s_1}\right) \circ \cdots \circ f \left({s_n}\right)$ The first condition is precisely saying that $\bar f$ preserves identity elements, and by: {{begin-eqn}} {{eqn|l = \bar f \left({\left\langle{s_1, \ldots, s_n}\right\rangle * \left\langle{t_1, \ldots, t_m}\right\rangle}\right)      |r = \bar f \left({\left\langle{s_1, \ldots, s_n, t_1, \ldots t_m}\right\rangle}\right)      |c = Definition of concatenation }} {{eqn|r = f \left({s_1}\right) \circ \cdots \circ f \left({s_n}\right) \circ f \left({t_1}\right) \cdots \circ f \left({t_m}\right)      |c = Definition of $\bar f$ }} {{eqn|r = \left({f \left({s_1}\right) \circ \cdots \circ f \left({s_n}\right)}\right) \circ \left({f \left({t_1}\right) \circ \cdots \circ f \left({t_m}\right)}\right)      |c = $\circ$ is associative }} {{eqn|r = \bar f \left({\left\langle{s_1 \ldots, s_n}\right\rangle}\right) \circ \bar f \left({\left\langle{t_1, \ldots, t_m}\right\rangle}\right)      |c = Definition of $\bar f$ }} {{end-eqn}} it follows that $\bar f$ is a monoid homomorphism. It also satisfies: :$\bar f \circ i \left({s}\right) = \bar f \left({\left\langle{s}\right\rangle}\right) = f \left({s}\right)$ and hence meets the set requirements. {{qed|lemma}} === Uniqueness === It remains to verify the uniqueness of $\bar f$. So let $g: S^* \to N \in \mathbf{Mon}_1$ be a monoid homomorphism such that for all $s \in S$: $g \circ i \left({s}\right) = f \left({s}\right)$ Then for any $\left\langle{s_1, \ldots, s_n}\right\rangle \in S^*$, compute: {{begin-eqn}} {{eqn|l = g \left({\left\langle{s_1, \ldots, s_n}\right\rangle}\right)      |r = g \left({\left\langle{s_1}\right\rangle * \cdots * \left\langle{s_n}\right\rangle}\right)      |c = Definition of concatenation }} {{eqn|r = g \left({\left\langle{s_1}\right\rangle}\right) \circ \cdots \circ g \left({\left\langle{s_n}\right\rangle}\right)      |c = $g$ is a homomorphism }} {{eqn|r = f \left({s_1}\right) \circ \cdots \circ f \left({s_n}\right)      |c = By assumption }} {{eqn|r = \bar f \left({\left\langle{s_1, \ldots, s_n}\right\rangle}\right)      |c = Definition of $\bar f$ }} {{end-eqn}} {{qed|lemma}} Hence $\left({S^*, i}\right)$ is a free monoid for $S$. {{qed}}
By the (categorial) definition of free monoid, we have the following commutative diagram: ::$\begin{xy} <-4em,4em>*{\mathbf{Mon} :}, <-4em,1em>*{\mathbf{Set} :}, <0em,4em>*+{N} = "N", <4em,4em>*+{M} = "M", <8em,4em>*+{N} = "N2", "N";"M" **@{.} ?>*@{>} ?*!/_1em/{\bar i}, "M";"N2" **@{.} ?>*@{>} ?*!/_1em/{\bar j}, <0em,1em>*+{\left\vert{N}\right\vert} = "NN", <4em,1em>*+{\left\vert{M}\right\vert} = "MM", <8em,1em>*+{\left\vert{N}\right\vert} = "NN2", <4em,-3em>*+{S} = "S", "NN";"MM" **@{-} ?>*@{>} ?*!/_1em/{\left\vert{\bar i}\right\vert}, "MM";"NN2" **@{-} ?>*@{>} ?*!/_1em/{\left\vert{\bar j}\right\vert}, "S";"NN" **@{-} ?>*@{>} ?*!/_1em/{j}, "S";"MM" **@{-} ?>*@{>} ?<>(.7)*!/_.6em/{i}, "S";"NN2" **@{-} ?>*@{>} ?*!/^1em/{j}, \end{xy}$ The outer rim of these diagrams also needs to satisfy the universal property. Since obviously $\operatorname{id}_N$ fits in place of $\bar j \circ \bar i$, we conclude these are equal. Exchanging the rôles of $\struct {M, i}$ and $\struct {N, j}$ proves that $\bar i \circ \bar j = \operatorname{id}_M$ as well. Hence $\bar i$ is the sought monoid isomorphism. Its uniqueness follows from the universal property of the free monoid. {{qed}}
The proof shall proceed by finite induction on $n$: For all $n \in \N_{\ge 0}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle \sum_{i \mathop = 1}^n x^{b_i} a_i < x^y$ === Basis for the Induction === :$P \left({0}\right)$ says that: :$\displaystyle \sum_{i \mathop = 1}^0 x^{b_i} a_i < x^y$ {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^0 x^{b_i} a_i       | r = 0       | c = Definition of Ordinal Sum }} {{eqn | l =        | o = <       | r = x^y       | c = Exponent Not Equal to Zero }} {{end-eqn}} This is our basis for the induction. {{qed|lemma}} === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k+1}\right)$ is true. So this is our induction hypothesis: :$\displaystyle \sum_{i \mathop = 1}^k x^{b_i} a_i < x^y$ Then we need to show: :$\displaystyle \sum_{i \mathop = 1}^{k \mathop + 1} x^{b_i} a_i < x^y$ === Induction Step === This is our induction step: By the inductive hypothesis: :$\displaystyle \sum_{i \mathop = 1}^k x^{b_{i + 1}} a_{i + 1} < x^{b_1}$ since $b_1$ is greater than all $b_i$ for $i > 1$. Then: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^{k+1} x^{b_i} a_i        | r = x^{b_1} a_1 + \sum_{i \mathop = 1}^k x^{b_{i + 1} } a_{i + 1}       | c = General Associative Law for Ordinal Sum }} {{eqn | o = <       | r = x^{b_1} a_1 + x^{b_1}       | c = Membership is Left Compatible with Ordinal Addition }} {{eqn | r = x^{b_1} a_1^+       | c = Ordinal Multiplication is Left Distributive }} {{eqn | o = \le       | r = x^{b_1^+}       | c = Membership is Left Compatible with Ordinal Multiplication and Successor of Element of Ordinal is Subset }} {{eqn | o = \le       | r = x^y       | c = Membership is Left Compatible with Ordinal Exponentiation and Successor of Element of Ordinal is Subset }} {{end-eqn}} {{explain|In the above, particularly the last couple of lines, it is not clear how the results invoked correspond to the operations performed. In any case, it is worth expanding those steps out to show how the individual steps are performed. ProofWiki prefers to perform steps individually.}} So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \sum_{i \mathop = 1}^n a_i x^{b_i} < x^y$ {{qed}}
The proof shall proceed by induction on $x$. === Basis for the Induction === If $x = 0$, then: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^{0 + 1} a_i       | r = \sum_{i \mathop = 1}^0 a_i + a_1       | c = definition of ordinal sum }} {{eqn | r = a_1       | c = by Ordinal Addition by Zero }} {{eqn | r = a_1 + \sum_{i \mathop = 1}^0 a_i       | c = by Ordinal Addition by Zero }} {{end-eqn}} This proves the basis for the induction. {{qed|lemma}} === Induction Step === Suppose that: : $\displaystyle \sum_{i \mathop = 1}^{x + 1} a_i = a_1 + \sum_{i \mathop = 1}^x a_{i + 1}$ Then: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^{x + 2} a_i       | r = \sum_{i \mathop = 1}^{x + 1} a_i + a_{i + 2}       | c = Definition of Ordinal Sum }} {{eqn | r = \left({a_1 + \sum_{i \mathop = 1}^x a_{i + 1} }\right) + a_{i + 2}       | c = by Inductive Hypothesis }} {{eqn | r = a_1 + \left({\sum_{i \mathop = 1}^x a_{i + 1} + a_{i + 2} }\right)       | c = Ordinal Addition is Associative }} {{eqn | r = a_1 + \sum_{i \mathop = 1}^{x + 1} a_{i + 1}       | c = Definition of Ordinal Sum }} {{end-eqn}} This proves the induction step. {{qed}}
From Ordinal Addition is Associative we have that: :$\forall a, b, c \in \operatorname{On}: a + \left({b + c}\right) = \left({a + b}\right) + c$ The result follows directly from the General Associativity Theorem. {{qed}}
The proof shall proceed by Transfinite Induction (Strong Induction) on $y$. The inductive hypothesis states that for all $v < y$, there exist unique finite sequences of ordinals: :$\sequence {c_i}, \sequence {d_i}$ both of unique length $n$ such that: :$(1): \quad \sequence {c_i}$ is strictly decreasing sequence for $1 \le i \le n$ :$(2): \quad 0 < d_i < x$ for all $1 \le i \le n$. :$(3): \quad \ds v = \sum_{i \mathop = 1}^n x^{c_i} d_i$ Since $x > 1$, it follows by Unique Ordinal Exponentiation Inequality that: :$x^z \le y < x^{z + 1}$ for some unique $z$. By the Division Theorem for Ordinals: :$y = x^z w + v$ and $v < x^z$ for some unique $z$, $w$, and $v$. Thus: {{begin-eqn}} {{eqn | l = v       | o = <       | r = x^z       | c = Division Theorem for Ordinals, as shown above }} {{eqn | o = \le       | r = y       | c = Unique Ordinal Exponentiation Inequality, as shown above }} {{end-eqn}} So $v < y$. If $v = 0$, then the statement is proven. If $v \ne 0$, then we may apply the inductive hypothesis. By the inductive hypothesis: :$\ds v = \sum_{i \mathop = 1}^n x^{c_i} d_i$ Therefore: :$\ds y = x^z w + \sum_{i \mathop = 1}^n x^{c_i} d_i$ {{qed|lemma}} Set: :$a_1 = z$ :$a_{i + 1} = c_i$ for $1 \le i \le n$ :$b_1 = w$ :$b_{i + 1} = d_i$ for $1 \le i \le n$ Since $w \ne 0$ and $d_i \ne 0$, it follows that $b_i \ne 0$ for all $1 \le i \le n+1$. {{qed|lemma}} Moreover, since $z > c_1$ and $\sequence {c_i}$ is strictly decreasing, it follows that also $\sequence {a_i}$ is strictly decreasing. {{qed|lemma}} The equation for the first lemma can be rewritten: :$\ds y = x^{a_1} b_1 + \sum_{i \mathop = 1}^n x^{a_{i + 1} } b_{i + 1}$ By General Associative Law for Ordinal Sum, it follows that: :$\ds y = \sum_{i \mathop = 1}^{n + 1} x^{a_i} b_i$ Thus, existence is proven. {{qed|lemma}} Furthermore, since $z$ and $\sequence {c_i}$ are unique, and $w$ and $\sequence {a_i}$ are unique, then $\sequence {a_i}$ and $\sequence {b_i}$ are unique. {{qed}}
It follows that: {{begin-eqn}} {{eqn | l = x^{a_1}       | o = \le       | r = \sum_{i \mathop = 1}^n \left({x^{a_i} b_i}\right)       | c = Ordinal is Less than Sum }} {{eqn | o = <       | r = x^{a_1 + 1}       | c = Upper Bound of Ordinal Sum }} {{end-eqn}} By multiplying the inequalities by $x^y$ on the left: {{begin-eqn}} {{eqn | l = x^{a_1} \times x^y       | o = \le       | r = \sum_{i \mathop = 1}^n \left({x^{a_i} b_i}\right) \times x^y       | c = Subset is Right Compatible with Ordinal Multiplication }} {{eqn | o = \le       | r = x^{a_1 \mathop + 1} \times x^y       | c =  }} {{end-eqn}} Solving for both sides of the inequality: {{begin-eqn}} {{eqn | l = x^{a_1} \times x^y       | r = x^{a_1 \mathop + y}       | c = Ordinal Sum of Powers }} {{eqn | l = x^{a_1 \mathop + 1} \times x^y       | r = x^{\left({a_1 \mathop + 1}\right) \mathop + y}       | c = Ordinal Sum of Powers }} {{eqn | r = x^{a_1 \mathop + \left({1 \mathop + y}\right)}       | c = Ordinal Addition is Associative }} {{eqn | r = x^{a_1 \mathop + y}       | c = Finite Ordinal Plus Transfinite Ordinal }} {{end-eqn}} Therefore: {{begin-eqn}} {{eqn | l = x^{a_1 \mathop + y}       | o = \le       | r = \sum_{i \mathop = 1}^n \left({x^{a_i} b_i}\right) \times x^y       | c = Substitutivity of Class Equality }} {{eqn | o = \le       | r = x^{a_1 \mathop + y}       | c = Substitutivity of Class Equality }} {{eqn | ll= \implies       | l = \sum_{i \mathop = 1}^n \left({x^{a_i} b_i}\right) \times x^y       | r = x^{a_1 \mathop + y}       | c = Definition of Set Equality }} {{end-eqn}} {{qed}}
The proof shall proceed by Transfinite Induction on $z$. === Basis for the Induction === The hypothesis requires that $z \ne 0$, so the induction starts at $z = 1$. {{begin-eqn}} {{eqn | l = \left({ x^y \times n }\right) ^1       | r = x^y \times n       | c = Definition of Ordinal Exponentiation }} {{eqn | r = x^{y \mathop \times 1} \times n       | c = Ordinal Multiplication by One }} {{end-eqn}} This proves the basis for the induction. {{qed|lemma}} === Induction Step === {{begin-eqn}} {{eqn | l = \left({ x^y \times n }\right) ^z       | r = x^{y \mathop \times z} \times n       | c = Inductive Hypothesis }} {{eqn | ll = \implies       | l = \left({ x^y \times n }\right) ^{z^+}       | r = \left({ x^y \times n }\right) ^z \times x^y \times n       | c = Definition of Ordinal Exponentiation }} {{eqn | r = x^{y \mathop \times z} \times n \times x^y \times n       | c = Inductive Hypothesis }} {{eqn | r = x^{y \mathop \times z} \times x^y \times n       | c = Finite Ordinal Times Ordinal }} {{eqn | r = x^{y \mathop \times z + y} \times n       | c = Ordinal Sum of Powers }} {{eqn | r = x^{y \mathop \times z^+} \times n       | c = Definition of Ordinal Multiplication }} {{end-eqn}} This proves the induction step. {{qed|lemma}} === Limit Case === Suppose that this statement holds for all $w \in z$ where $z$ is a limit ordinal. Then: {{begin-eqn}} {{eqn | l = x^{y \mathop \times z}       | r = \left({x^y}\right)^z       | c = Ordinal Power of Power }} {{eqn | o = \le       | r = \left({x^y \times n}\right)^z       | c = Subset is Right Compatible with Ordinal Exponentiation }} {{eqn | r = \bigcup_{w \mathop \in z} \left({x^y \times n}\right)^w       | c = Definition of Ordinal Exponentiation }} {{eqn | o = \le       | r = \bigcup_{w \mathop \in z} x^{y \mathop \times w} \times n       | c = Proof by Cases where $w \in K_I$ or $w \in K_{II}$ }} {{eqn | o = \le       | r = \bigcup_{w \mathop \in z} x^{y \mathop \times w + 1}       | c = Membership is Left Compatible with Ordinal Exponentiation }} {{eqn | o = \le       | r = \bigcup_{w \mathop \in z} x^{y \mathop \times w^+}       | c = Membership is Left Compatible with Ordinal Multiplication and Membership is Left Compatible with Ordinal Exponentiation }} {{eqn | r = \bigcup_{w \mathop \in z} \left({ x^y }\right)^{w^+}       | c = Ordinal Power of Power }} {{eqn | r = \left({ x^y }\right)^z       | c = Definition of Ordinal Exponentiation }} {{end-eqn}} This proves the limit case. {{qed}}