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Proof step 46 from proof int2bytes-first-zerobyte (previous step)

¬ # int2bits(i % 256) ≤ 9, 0 < i, 0 ≤ i, 256 ≤ i, ∀ i0. i→n(i0) < i→n(i) → 0 ≤ i0 → (int2bytes(i0).first = i2b(0) ↔ n→i(# int2bits(i0)) % 8 = 1) ⊦ fillsgn(9, int2bits(i % 256)) = [], int2bytes(i / 256).first = i2b(0) ↔ (n→i(# fillsgn(9, int2bits(i % 256)) - 1) + n→i(# int2bits(i / 256))) % 8 = 1, n→i(# fillsgn(9, int2bits(i % 256)) - 1) % 8 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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