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Proof step 27 from proof int2bytes-rec-neg-03 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 28:
0 < i, j + 2 ^ (i * 8) < 0 ⊦ j % 2 ^ (i * 8) = 0, int2bytes(j / 2 ^ (i * 8) - 1) = int2bytes((j + 1 - 2 ^ (i * 8)) / 2 ^ (i * 8))

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ i < ~ j ↔ i + j < 0;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ bytes + bytes0 = bytes1 + bytes0 ↔ bytes = bytes1;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ 0 < i → i→n(i + 1) = i→n(i) + 1;% ⊦ m + n = n + m;