# Exponent Not Equal to Zero

## Theorem

Let $x$ and $y$ be ordinals.

Let $x \ne 0$.

Then:

$x^y \ne 0$

## Proof

The proof shall proceed by Transfinite Induction on $y$.

### Basis for the Induction

$x^0 = 1$ by the definition of ordinal exponentiation.

Therefore, $x^0 \ne 0$.

This proves the basis for the induction.

### Induction Step

The inductive hypothesis supposes that $x^y \ne 0$.

 $\displaystyle x^{y^+}$ $=$ $\displaystyle x^y \times x$ definition of Definition:Ordinal Exponentiation $\displaystyle x^y$ $\ne$ $\displaystyle 0$ Inductive Hypothesis $\displaystyle x$ $\ne$ $\displaystyle 0$ Hypothesis $\displaystyle x^y \times x$ $\ne$ $\displaystyle 0$ by Ordinals have No Zero Divisors

This proves the induction step.

### Limit Case

The inductive hypothesis says that:

$\forall z \in y: x^z \ne 0$

 $\displaystyle \forall z \in y: \ \$ $\displaystyle x^z$ $\ne$ $\displaystyle 0$ Inductive Hypothesis $\displaystyle \implies \ \$ $\displaystyle 0$ $\in$ $\displaystyle x^z$ by Ordinal Membership is Trichotomy $\displaystyle \implies \ \$ $\displaystyle 0$ $\in$ $\displaystyle \bigcup_{z \in y} x^z$ by the fact that limit ordinals are nonempty $\displaystyle \implies \ \$ $\displaystyle 0$ $\in$ $\displaystyle x^y$ definition of ordinal exponentiation $\displaystyle \implies \ \$ $\displaystyle x^y$ $\ne$ $\displaystyle 0$ definition of empty set

This proves the limit case.

$\blacksquare$