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Proof step 64 from proof rightshift-8-01-01 (previous step)

2 ^ (8 * abs(j)) ≠ 0, ∃ j0. i = j0 * 2 ^ (8 * abs(j)), 0 ≤ 2 ^ (8 * abs(j)) + i, i < 0, 2 ^ (8 * abs(j) - 1) + i < 0 ⊦ j = 0, int2bytes(i).first = i2b(~1), i / 2 ^ (8 * abs(j)) = ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 65:
0 ≤ 2 ^ (8 * abs(j)) * (j0 + 1), 2 ^ (8 * abs(j) - 1) + j0 * 2 ^ (8 * abs(j)) < 0, j0 * 2 ^ (8 * abs(j)) < 0 ⊦ j0 = ~1, int2bytes(j0 * 2 ^ (8 * abs(j))).first = i2b(~1), j = 0, 2 ^ (8 * abs(j)) = 0

Rule argument

Additional information

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