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Proof step 9 from proof int2bytes-rec-neg-05 (previous step)

0 < 2 ^ (i * 8) ∧ j < 0 ∧ abs(j) < 2 ^ (i * 8), 0 < i, j + 2 ^ (i * 8 - 1) < 0, 0 < j + 2 ^ (i * 8) ⊦ j % 2 ^ (i * 8) = 0, bits2bytes(int2bits(j)) = int2bytes(0 - 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 10:
0 < i, j + 2 ^ (i * 8 - 1) < 0, 0 < j + 2 ^ (i * 8), 0 < 2 ^ (i * 8) + j, j < 0 ⊦ j = 0, bits2bytes(int2bits(j)) = i2b(~1) ' + fix(i→n(i) + 1, int2bytes(j)).rest

Rule argument

Additional information

Used simplifier rules: int2bytes-byte: ~128 ≤ i, i < 128 ⊦ int2bytes((* i)) = i2b(i) ';% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ ~ i < j ↔ 0 < j + i;% ⊦ i < 0 → abs(i) = ~ i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j);% ⊦ 0 < 2 ^ i ↔ 0 ≤ i;% ⊦ abs(i) < j → i % j = i;% ⊦ i * j = j * i;