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Proof step 61 from proof int2bytes-rec-neg-02 (previous step)

¬(# int2bits(j % 2 ^ (i * 8)) ≤ i→n(i * 8) + 1 ∧ int2bits(j % 2 ^ (i * 8)) ≠ []), # 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 62:
# fillsgn(i→n(i * 8) + 1, int2bits(j % 2 ^ (i * 8))).rest % 8 = 0, i→n(i * 8) + 1 < # int2bits(j % 2 ^ (i * 8)), j + 2 ^ (i * 8) < 0, 0 < i ⊦ bits2bytes(fillsgn(i→n(i * 8) + 1, int2bits(j % 2 ^ (i * 8))).rest) = 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 = []

Rule argument

Additional information

Used simplifier rules: ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ int2bits(j) ≠ [];% ⊦ ¬ m ≤ n ↔ n < m;