Ordinal Exponentiation of Terms

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Theorem

Let $x$, $y$, and $z$ be ordinals.

Let $n$ be a finite ordinal.

Let $x$ be a limit ordinal.

Let $y$, $z$ and $n$ be greater than $0$.


Then:

$\left({ x^y \times n }\right) ^z = x^{y \mathop \times z} \times n$ if $z$ is not a limit ordinal
$\left({ x^y \times n }\right) ^z = x^{y \mathop \times z}$ if $z$ is a limit ordinal


Proof

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 $z = 1$.


\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \left({ x^y \times n }\right) ^1\) \(=\) \(\displaystyle \) \(\) \(\displaystyle \) \(\displaystyle x^y \times n\) \(\displaystyle \) \(\displaystyle \)          definition of ordinal exponentiation          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\) \(\displaystyle \) \(\displaystyle x^{y \mathop \times 1} \times n\) \(\displaystyle \) \(\displaystyle \)          Ordinal Multiplication by One          

This proves the basis for the induction.

$\Box$

Induction Step

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \left({ x^y \times n }\right) ^z\) \(=\) \(\displaystyle \) \(\) \(\displaystyle \) \(\displaystyle x^{y \mathop \times z} \times n\) \(\displaystyle \) \(\displaystyle \)          Inductive Hypothesis          
\(\displaystyle \) \(\displaystyle \implies\) \(\displaystyle \) \(\displaystyle \left({ x^y \times n }\right) ^{z^+}\) \(=\) \(\displaystyle \) \(\) \(\displaystyle \) \(\displaystyle \left({ x^y \times n }\right) ^z \times x^y \times n\) \(\displaystyle \) \(\displaystyle \)          definition of ordinal exponentiation          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\) \(\displaystyle \) \(\displaystyle x^{y \mathop \times z} \times n \times x^y \times n\) \(\displaystyle \) \(\displaystyle \)          Inductive Hypothesis          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\) \(\displaystyle \) \(\displaystyle x^{y \mathop \times z} \times x^y \times n\) \(\displaystyle \) \(\displaystyle \)          Finite Ordinal Times Ordinal          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\) \(\displaystyle \) \(\displaystyle x^{y \mathop \times z + y} \times n\) \(\displaystyle \) \(\displaystyle \)          Ordinal Sum of Powers          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\) \(\displaystyle \) \(\displaystyle x^{y \mathop \times z^+} \times n\) \(\displaystyle \) \(\displaystyle \)          definition of ordinal multiplication          

This proves the induction step.

$\Box$

Limit Case

Suppose that this statement holds for all $w \in z$ where $z$ is a limit ordinal.


Then:

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle x^{y \mathop \times z}\) \(=\) \(\displaystyle \) \(\) \(\displaystyle \) \(\displaystyle \left({x^y}\right)^z\) \(\displaystyle \) \(\displaystyle \)          Ordinal Power of Power          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\le\) \(\displaystyle \) \(\) \(\displaystyle \) \(\displaystyle \left({x^y \times n}\right)^z\) \(\displaystyle \) \(\displaystyle \)          Subset Right Compatible with Ordinal Exponentiation          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\) \(\displaystyle \) \(\displaystyle \bigcup_{w \in z} \left({x^y \times n}\right)^w\) \(\displaystyle \) \(\displaystyle \)          definition of ordinal exponentiation          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\le\) \(\displaystyle \) \(\) \(\displaystyle \) \(\displaystyle \bigcup_{w \in z} x^{y \mathop \times w} \times n\) \(\displaystyle \) \(\displaystyle \)          Proof by Cases where $w \in K_{I}$ or $w \in K_{II}$          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\le\) \(\displaystyle \) \(\) \(\displaystyle \) \(\displaystyle \bigcup_{w \in z} x^{y \mathop \times w + 1}\) \(\displaystyle \) \(\displaystyle \)          Membership is Left Compatible with Ordinal Exponentiation          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\le\) \(\displaystyle \) \(\) \(\displaystyle \) \(\displaystyle \bigcup_{w \in z} x^{y \mathop \times w^+}\) \(\displaystyle \) \(\displaystyle \)          Membership is Left Compatible with Ordinal Multiplication and Membership is Left Compatible with Ordinal Exponentiation          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\) \(\displaystyle \) \(\displaystyle \bigcup_{w \in z} \left({ x^y }\right)^{w^+}\) \(\displaystyle \) \(\displaystyle \)          Ordinal Power of Power          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\) \(\displaystyle \) \(\displaystyle \left({ x^y }\right)^z\) \(\displaystyle \) \(\displaystyle \)          definition of ordinal exponentiation          

This proves the limit case.

$\blacksquare$


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