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

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

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

where $A$ is the Ackermann-Péter function.

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

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

Proceed by induction on $k$.

Basis for the Induction
Suppose $k = 1$.

Then:
 * $\map A {x, z} < \map A {x + 1, z}$

by Ackermann-Péter Function is Strictly Increasing on First Argument.

This is the basis for the induction.

Induction Hypothesis
The induction hypothesis is:
 * $\map A {x, z} < \map A {x + k, z}$

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

Induction Step
Thus, the induction step is satisfied.

Thus, by Principle of Mathematical Induction, the result holds forall $x, y, z \in \N$.