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Proof step 20 from proof int2bytes-rec-neg-06 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
0 < i, j0 * 2 ^ (i * 8) + 2 ^ (i * 8 - 1) < 0, j0 < 0 ⊦ 2 ^ (i * 8) = 0, int2bytes(j0) = int2bytes(j0 + (1 - 2 ^ (i * 8)) / 2 ^ (i * 8))

Rule argument

Additional information

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