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Proof step 25 from proof int2bytes-first-zerobyte (previous step)

0 < i, n→i(# int2bits(i)) % 8 = 1, ¬ i < 2 ^ (n→i(8) - 2), 0 < i, i ≤ 127, ~128 ≤ 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: i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j;% ⊦ ¬ i < j ↔ j ≤ i;