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Proof step 20 from proof int2bytes-rec-neg-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
i→n(i * 8) + 1 < # int2bits(j % 2 ^ (i * 8)), j + 2 ^ (i * 8) < 0, 0 < i, ~1 ≤ j % 2 ^ (i * 8), 0 < j % 2 ^ (i * 8) ⊦ bits2bytes(int2bits(j / 2 ^ (i * 8) - 1) + 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, # fillsgn(i→n(i * 8) + 1, int2bits(j % 2 ^ (i * 8))).rest % 8 = 0 ∧ fillsgn(i→n(i * 8) + 1, int2bits(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 → (i ≠ j ↔ j < i);