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Proof step 18 from proof int2bytes-rec-neg-03 (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, 2 ^ (i * 8) * (j0 + 1) < 0 ⊦ 2 ^ (i * 8) = 0, int2bytes(j0) = int2bytes(((1 - 2 ^ (i * 8)) + j0 * 2 ^ (i * 8)) / 2 ^ (i * 8))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
0 ≤ j0, 2 ^ (i * 8) * (j0 + 1) < 0, 0 < i ⊦ int2bytes(j0) = int2bytes((j0 * 2 ^ (i * 8) + 1 - 2 ^ (i * 8)) / 2 ^ (i * 8)), 2 ^ (i * 8) = 0

Rule argument

Additional information

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