Second Hyperoperation is Multiplication Operation

Theorem
The $2$nd hyperoperation is the multiplication operation restricted to the positive integers:


 * $\forall x, y \in \Z_{\ge 0}: H_2 \left({x, y}\right) = x \times y$

Proof
By definition of the hyperoperation sequence:


 * $\forall n, x, y \in \Z_{\ge 0}: H_n \left({x, y}\right) = \begin{cases}

y + 1 & : n = 0 \\ x & : n = 1, y = 0 \\ 0 & : n = 2, y = 0 \\ 1 & : n > 2, y = 0 \\ H_{n - 1} \left({x, H_n \left({x, y - 1}\right)}\right) & : n > 0, y > 0 \end{cases}$

Thus the $2$nd hyperoperation is defined as:


 * $\forall x, y \in \Z_{\ge 0}: H_2 \left({x, y}\right) = \begin{cases}

0 & : y = 0 \\ H_1 \left({x, H_2 \left({x, y - 1}\right)}\right) & : y > 0 \end{cases}$

From First Hyperoperation is Addition Operation:


 * $(1): \quad \forall x, y \in \Z_{\ge 0}: H_2 \left({x, y}\right) = \begin{cases}

0 & : y = 0 \\ x + H_2 \left({x, y - 1}\right) & : y > 0 \end{cases}$

The proof proceeds by induction.

For all $y \in \Z_{\ge 0}$, let $P \left({y}\right)$ be the proposition:
 * $\forall x \in \Z_{\ge 0}: H_2 \left({x, y}\right) = x \times y$

Basis for the Induction
$P \left({0}\right)$ is the case:

Thus $P \left({0}\right)$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown 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 the induction hypothesis:
 * $\forall x \in \Z_{\ge 0}: H_2 \left({x, k}\right) = x \times k$

from which it is to be shown that:
 * $\forall x \in \Z_{\ge 0}: H_2 \left({x, k + 1}\right) = x \times \left({k + 1}\right)$

Induction Step
This is the induction step:

So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall x, y \in \Z_{\ge 0}: H_2 \left({x, y}\right) = x \times y$