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Proof step 41 from proof int2bytes-rec-neg-06 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 42:
~(2 ^ (i * 8)) % 2 ^ (i * 8) = 0, 0 < i, j + 2 ^ (i * 8 - 1) < 0, j < ~1, j < 0 ⊦ j % 2 ^ (i * 8) = 0, int2bytes(j / 2 ^ (i * 8) - 1) = int2bytes(~(2 ^ (i * 8)) / 2 ^ (i * 8) + (j + 1) / 2 ^ (i * 8))

Rule argument

Additional information

Used simplifier rules: ⊦ i < (* j) ∧ j < 0 → i < 0; (forward)% ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j);% ⊦ 0 < 2 ^ i ↔ 0 ≤ i;% ⊦ ~ i < 0 ↔ 0 < i;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + (* j) < (* j0) ↔ i < j0 - j;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * ~1 = ~ i;