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Proof step 31 from proof int2bytes-first-zerobyte (previous step)

255 < i, 0 ≤ i, ∀ i0. i→n(i0) < i→n(i) → 0 ≤ i0 → (int2bytes(i0).first = i2b(0) ↔ n→i(# int2bits(i0)) % 8 = 1) ⊦ (int2bytes(i / 256) + int2bytes(i % 256).last ').first = i2b(0) ↔ n→i(# int2bits(i)) % 8 = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 32:
0 < i, 0 ≤ i, 255 < i, ∀ i0. i→n(i0) < i→n(i) → 0 ≤ i0 → (int2bytes(i0).first = i2b(0) ↔ n→i(# int2bits(i0)) % 8 = 1) ⊦ (int2bytes(i / 256) + int2bytes(i % 256).last ').first = i2b(0) ↔ n→i(# int2bits(i)) % 8 = 1

Rule argument

Additional information

Used simplifier rules: i0 ≠ 0, i0 < i ⊦ (* i) < j → (* i0) < j;% ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward)% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)