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Proof step 3 from proof int2int-fillsgn (previous step)

¬ # int2bytes(i) < n ⊦ bytes2int(fillsgn(n, int2bytes(i))) = i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
n ≤ # int2bytes(i) ⊦ bytes2int(fillsgn(n, int2bytes(i))) = i

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ m < n ↔ n ≤ m;