Increasing Second Argument of Ackermann-Péter Function is Not Greater than Increasing First Argument

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

where $A$ is the Ackermann-Péter function.

Proof
Proceed by induction on $y$.

Basis for the Induction
Suppose $y = 0$.

Then:

Therefore:
 * $\map A {x + 1, 0} \ge \map A {x, 1}$

which is the basis for the induction.

Induction Hypothesis
This is the induction hypothesis:
 * $\map A {x + 1, y} \ge \map A {x, y + 1}$

We want to show that:
 * $\map A {x + 1, y + 1} \ge \map A {x, y + 2}$

Induction Step
First, observe:

Thus, we have shown:
 * $\paren 1 \quad \map A {x + 1, y} \ge y + 2$

Now, we have:

This completes the induction step.

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