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Proof step 49 from proof int2bytes-rec-neg-06 (previous step)

j = i0 * 2 ^ (i * 8) + j0 ∧ j0 ≤ 0 ∧ abs(j0) < 2 ^ (i * 8) ∧ i0 ≤ 0 ∧ 0 < 2 ^ (i * 8), 0 < i, j + 2 ^ (i * 8 - 1) < 0, j < ~1, j < 0 ⊦ 2 ^ (i * 8) = 0, j0 = 0, int2bytes(i0 - 1) = int2bytes((j + 1) / 2 ^ (i * 8) - 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 50:
0 < i, i0 * 2 ^ (i * 8) + j0 < ~1, i0 ≤ 0, i0 < i, i0 < 2 ^ (i * 8) + j0, (i0 * 2 ^ (i * 8) + j0) + 2 ^ (i * 8 - 1) < 0, i0 * 2 ^ (i * 8) + j0 < 0, j0 < 0, 0 < 2 ^ (i * 8) + j0 ⊦ 2 ^ (i * 8) = 0, j0 = 0, int2bytes(i0 - 1) = int2bytes(((i0 * 2 ^ (i * 8) + j0) + 1) / 2 ^ (i * 8) - 1)

Rule argument

Additional information

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