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Proof step 10 from proof int2bytes-rec-neg-06 (previous step)

¬ 2 ^ (i * 8) ≠ 0, j % 2 ^ (i * 8) = 0, 0 < i, j + 2 ^ (i * 8 - 1) < 0 ⊦ int2bytes(j / 2 ^ (i * 8)) = int2bytes((j + 1 - 2 ^ (i * 8)) / 2 ^ (i * 8))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
j % 0 = 0, 2 ^ (i * 8) = 0, j + 2 ^ (i * 8 - 1) < 0, 0 < i ⊦ int2bytes(j / 0) = int2bytes((1 + j) / 0)

Rule argument

Additional information

Used simplifier rules: ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ (* i) + i0 - (* j) = (i - j) + i0;% ⊦ i * j = j * i;