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Proof step 14 from proof int2bytes-first-zerobyte (previous step)

0 ≤ i ∧ i ≤ 127, 0 ≤ i, i ≤ 255, ∀ i0. i→n(i0) < i→n(i) → 0 ≤ i0 → (int2bytes(i0).first = i2b(0) ↔ n→i(# int2bits(i0)) % 8 = 1) ⊦ i2b(i) '.first = i2b(0) ↔ n→i(# int2bits(i)) % 8 = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: j0 ≠ 0, j ≤ j0 ⊦ i ≤ (* j) → i ≤ (* j0);% ⊦ by '.first = by;