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Proof step 22 from proof int2bytes-first-zerobyte (previous step)

¬ abs(n→i(# int2bits(i))) < 8, n→i(# int2bits(i)) % 8 = 1, ~128 ≤ i, i ≤ 127, 0 < i, ∀ i0. i→n(i0) < i→n(i) → 0 ≤ i0 → (int2bytes(i0).first = i2b(0) ↔ n→i(# int2bits(i0)) % 8 = 1) ⊦ i = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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