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Proof step 87 from proof int2bytes-rec-neg-05 (previous step)

j ≤ ~(2 ^ (i * 8)) ∧ 0 < i * 8 ∧ j % 2 ^ (i * 8) ≠ 0, 0 < i, j + 2 ^ (i * 8 - 1) < 0 ⊦ j % 2 ^ (i * 8) = 0, bits2bytes(int2bits(j / 2 ^ (i * 8) - 1) + fillsgn(i→n(i * 8 + 1), int2bits(j % 2 ^ (i * 8))).rest) = int2bytes(j / 2 ^ (i * 8) - 1) + fix(i→n(i) + 1, int2bytes(j % 2 ^ (i * 8))).rest

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 88:
0 < i, j + 2 ^ (i * 8 - 1) < 0, j + 2 ^ (i * 8) ≤ 0, j + 2 ^ (i * 8) < i ⊦ j % 2 ^ (i * 8) = 0, bits2bytes(int2bits(j / 2 ^ (i * 8) - 1) + fillsgn(i→n(i * 8) + 1, int2bits(j % 2 ^ (i * 8))).rest) = int2bytes(j / 2 ^ (i * 8) - 1) + fix(i→n(i) + 1, int2bytes(j % 2 ^ (i * 8))).rest

Rule argument

Additional information

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