Ackermann-Péter Function is Strictly Increasing on Second Argument/General Result

Theorem
For all $x, y, z \in \N$ such that:
 * $y < z$

we have:
 * $\map A {x, y} < \map A {x, z}$

where $A$ is the Ackermann-Péter function.

Proof
Let $z$ be expressed as:
 * $z = y + k$

for some $k \in \N_{>0}$.

Proceed by induction on $k$.

Basis for the Induction
Suppose $k = 1$.

Then:

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

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

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