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Proof step 2 from proof int2int-pos (previous step)

true, ∀ i0. i→n(i0) < i→n(i) → true → 0 ≤ i0 → bytes2int(int2bytes(i0)) = i0 ⊦ 0 ≤ i → bytes2int(int2bytes(i)) = i

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: int2int: ⊦ bytes2int(int2bytes(i)) = i;