Ackermann-Péter Function is Greater than Second Argument

Theorem
For all $x, y \in \N$:
 * $\map A {x, y} > y$

where $A$ is the Ackermann-Péter function.

Proof
First, perform induction on $x$.

Basis for the Induction
Suppose $x = 0$.

Then:

Induction Hypothesis $\paren 1$
Suppose that, for every $y \in \N$:
 * $\map A {x, y} > y$

We want to show that, for every $y \in \N$:
 * $\map A {x + 1, y} > y$

Induction Step
We will now perform another induction on $y$.

Basis for the Induction
Suppose $y = 0$.

Then:

Induction Hypothesis $\paren 2$
Suppose that:
 * $\map A {x + 1, y} > y$

We want to show that:
 * $\map A {x + 1, y + 1} > y + 1$

Induction Step
By the Principle of Mathematical Induction, the Induction Step $\paren 1$ holds.

By the Principle of Mathematical Induction, the result holds.