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Given the following pseudocode for function AP(x, y: integer) which returns an integer,

function AP(x,y: integer):
    if x = 0 then return y+1
    else 
         if y = 0 then return AP(x-1,1)
         else return AP(x-1, AP(x,y-1))

I need to prove, $\forall x (AP(x, y) > y)$.


I have tried solving it using mathematical induction,
Basis: $AP(0, y) = y + 1$
Inductive step: Assuming, $AP(n, y) > y$ and $y > 0$
Prove that $AP(n + 1, y) = AP(n, AP(n + 1, y - 1)) > y$

By unrolling recursion, I get relations like,
$AP(n + 1, y) > AP(n + 1, y - 1) > AP(n, 1)$,
but I am not sure how to proceed from here.

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  • 2
    $\begingroup$ Use double induction. That is do an induction proof inside of your induction step. Or do the whole thing by strong induction on lexicographic ordering of pairs of integers. $\endgroup$ – Jake Dec 23 '16 at 6:01

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