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Proof step 11 from proof int2bytes-first-zerobyte (previous step)

127 < i ∧ i ≤ 255, 0 < i, 127 < i, i ≤ 255, 0 ≤ i, ∀ i0. i→n(i0) < i→n(i) → 0 ≤ i0 → (int2bytes(i0).first = i2b(0) ↔ n→i(# int2bits(i0)) % 8 = 1) ⊦ (i2b(0) ' + i2b(i - 256) ').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: ~128 ≤ i, i ≤ 127, ~128 ≤ j, j ≤ 127 ⊦ i2b((* i)) = i2b((* j)) ↔ i = j;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ (by ' + bytes).first = by;