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Proof step 54 from proof int2bytes-rec-pos-any (previous step)

# fillsgn(i→n(i * 8) + 1, int2bits(j % 2 ^ (i * 8))).rest % 8 = 0, 2 ^ (i * 8) ≤ j, 0 < i ⊦ bits2bytes(int2bits(j / 2 ^ (i * 8))) + bits2bytes(fillsgn(i→n(i * 8) + 1, int2bits(j % 2 ^ (i * 8))).rest) = bits2bytes(int2bits(j / 2 ^ (i * 8))) + fix(i→n(i) + 1, int2bytes(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 55:
# fillsgn(i→n(i * 8) + 1, int2bits(j % 2 ^ (i * 8))).rest % 8 = 0, 2 ^ (i * 8) ≤ j, 0 < i ⊦ fillsgn(i→n(i * 8) + 1, int2bits(j % 2 ^ (i * 8))).rest = [], 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: ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ bytes + bytes0 = bytes + bytes1 ↔ bytes0 = bytes1;% ⊦ i * j = j * i;