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Proof step 24 from proof int2bytes-rec-neg-03 (previous step)

¬ abs(1 - 2 ^ (i * 8)) < abs(2 ^ (i * 8)), 0 < i, 2 ^ (i * 8) * (j0 + 1) < 0, j0 < 0 ⊦ 2 ^ (i * 8) = 0, int2bytes(j0) = int2bytes(j0 + (1 - 2 ^ (i * 8)) / 2 ^ (i * 8))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
0 < i, 2 ^ (i * 8) * (j0 + 1) < 0, j0 < 0, 2 ^ (i * 8) ≤ abs(1 - 2 ^ (i * 8)) ⊦ 2 ^ (i * 8) = 0, int2bytes(j0) = int2bytes(j0 + (1 - 2 ^ (i * 8)) / 2 ^ (i * 8))

Rule argument

Additional information

Used simplifier rules: ⊦ i * j = j * i;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ abs(2 ^ i) = 2 ^ i;