Ackermann-Péter Function is Strictly Increasing on Second Argument

Theorem
For all $x, y \in \N$:
 * $\map A {x, y + 1} > \map A {x, y}$

where $A$ is the Ackermann-Péter function.

Proof
Proceed by induction on $x$.

Basis for the Induction
Suppose $x = 0$.

Then:

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

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

Induction Step
By the Principle of Mathematical Induction, the result holds.