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Proof step 58 from proof int2bytes-rec-neg-02 (previous step)

# fillsgn(i→n(i * 8) + 1, int2bits(j % 2 ^ (i * 8))).rest % 8 = 0 ∧ int2bits(j / 2 ^ (i * 8) - 1) ≠ [] ∧ fillsgn(i→n(i * 8) + 1, int2bits(j % 2 ^ (i * 8))).rest ≠ [], 0 < i, j + 2 ^ (i * 8) < 0 ⊦ j % 2 ^ (i * 8) = 0, bits2bytes(int2bits(j / 2 ^ (i * 8) - 1)) + bits2bytes(fillsgn(i→n(i * 8) + 1, int2bits(j % 2 ^ (i * 8))).rest) = bits2bytes(int2bits(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 59:
# fillsgn(i→n(i * 8) + 1, int2bits(j % 2 ^ (i * 8))).rest % 8 = 0, 0 < i, j + 2 ^ (i * 8) < 0 ⊦ fillsgn(i→n(i * 8) + 1, int2bits(j % 2 ^ (i * 8))).rest = [], j % 2 ^ (i * 8) = 0, bits2bytes(fillsgn(i→n(i * 8) + 1, int2bits(j % 2 ^ (i * 8))).rest) = fix(i→n(i) + 1, int2bytes(j % 2 ^ (i * 8))).rest

Rule argument

Additional information

Used simplifier rules: ⊦ int2bits(j) ≠ [];% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ bytes + bytes0 = bytes + bytes1 ↔ bytes0 = bytes1;% ⊦ i * j = j * i;