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Proof step 43 from proof rightshift-8-01-01 (previous step)

i < 0 ∧ int2bytes(i).first ≠ i2b(~1), i < 0, ~(2 ^ (8 * n→i(i→n(j)))) ≤ i, 2 ^ (8 * abs(j) - 1) + i < 0 ⊦ j = 0, i >> 8 * j = ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 44:
0 ≤ 2 ^ (8 * abs(j)) + i, i < 0, 2 ^ (8 * abs(j) - 1) + i < 0 ⊦ int2bytes(i).first = i2b(~1), i >> 8 * j = ~1, j = 0

Rule argument

Additional information

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