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Proof step 4 from proof int2bytes-rec-neg-256-01 (previous step)

i < ~(2 ^ (0 +1 * 8)) ∧ 0 < 0 +1 ∧ i % 2 ^ (0 +1 * 8) ≠ 0, i < ~256, i < 0 ⊦ int2bytes(i / 2 ^ (0 +1 * 8) - 1) + fix(i→n(0 +1 + 1), int2bytes(i % 2 ^ (0 +1 * 8))).rest = int2bytes(i / 256 - 1) + int2bytes(i % 256).last ', i % 256 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
i < ~256, i < 0 ⊦ fix(2, int2bytes(i % 256)).rest = int2bytes(i % 256).last ', i % 256 = 0

Rule argument

Additional information

Used simplifier rules: j0 ≠ 0, j ≤ j0 ⊦ i < (* j) → i < (* j0);% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ bytes + bytes0 = bytes + bytes1 ↔ bytes0 = bytes1;