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Proof step 51 from proof int2bytes-rec-pos-any (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 52:
# 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) = int2bytes(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 = []

Rule argument

Additional information

Used simplifier rules: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 0 < i → (0 < i * j ↔ 0 < j);% ⊦ int2bits(j) ≠ [];% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ 0 < i → i→n(i + 1) = i→n(i) + 1;