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Proof step 40 from proof int2bytes-first-zerobyte (previous step)

¬(n→i(# fillsgn(9, int2bits(i % 256)).rest) % 8 = 0 ∧ 0 ≤ n→i(# int2bits(i / 256)) ∧ 0 ≤ n→i(# fillsgn(9, int2bits(i % 256)).rest) ∧ 0 < 8), 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) ⊦ int2bytes(i / 256).first = i2b(0) ↔ (n→i(# int2bits(i / 256)) + n→i(# fillsgn(9, int2bits(i % 256)).rest)) % 8 = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ 0 ≤ n→i(n);