Ackermann-Péter Function is Strictly Increasing on First Argument

Theorem

For all $x, y \in \N$:

$\map A {x + 1, y} > \map A {x, y}$

where $A$ is the Ackermann-Péter function.

General Result

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

\(\ds \map A {x + 1, y}\)

\(\ge\)

\(\ds \map A {x, y + 1}\)

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

\(\ds \)

\(>\)

\(\ds \map A {x, y}\)

Ackermann-Péter Function is Strictly Increasing on Second Argument

$\blacksquare$