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Proof step 62 from proof int2bytes-rec-pos-any (previous step)

¬ 0 < j % 2 ^ (i * 8), # fillsgn(i→n(i * 8) + 1, int2bits(j % 2 ^ (i * 8))).rest % 8 = 0, i→n(i * 8) + 1 < # int2bits(j % 2 ^ (i * 8)), 0 < i, 2 ^ (i * 8) ≤ j ⊦ bits2bytes(fillsgn(i→n(i * 8) + 1, int2bits(j % 2 ^ (i * 8))).rest) = fix(i→n(i) + 1, bits2bytes(int2bits(j % 2 ^ (i * 8)))).rest, 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 63:
# fillsgn(i→n(i * 8) + 1, int2bits(j % 2 ^ (i * 8))).rest % 8 = 0, i→n(i * 8) + 1 < # int2bits(j % 2 ^ (i * 8)), 0 < i, 2 ^ (i * 8) ≤ j, j % 2 ^ (i * 8) ≤ 0, j % 2 ^ (i * 8) < i ⊦ bits2bytes(fillsgn(i→n(i * 8) + 1, int2bits(j % 2 ^ (i * 8))).rest) = fix(i→n(i) + 1, bits2bytes(int2bits(j % 2 ^ (i * 8)))).rest, fillsgn(i→n(i * 8) + 1, int2bits(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;