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Proof step 34 from proof int2bytes-first-zerobyte (previous step)

int2bytes(i / 256) ≠ [], 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).first = i2b(0) ↔ n→i(# int2bits(i)) % 8 = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 35:
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).first = i2b(0) ↔ n→i(# int2bits(i)) % 8 = 1

Rule argument

Additional information

Used simplifier rules: int2bytes-empty: ⊦ int2bytes(i) ≠ [];