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Proof step 6 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)) = 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 7:
0 < i, j + 2 ^ (i * 8 - 1) < 0, 0 < j + 2 ^ (i * 8) ⊦ j % 2 ^ (i * 8) = 0, bits2bytes(int2bits(j)) = int2bytes(j / 2 ^ (i * 8) - 1) + fix(i→n(i) + 1, int2bytes(j % 2 ^ (i * 8))).rest

Rule argument

Additional information

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