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Proof step 1 from proof int2bytes-first-zerobyte (conclusion)

⊦ 0 ≤ i → (int2bytes(i).first = i2b(0) ↔ n→i(# int2bits(i)) % 8 = 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
0 ≤ i ⊦ int2bytes(i).first = i2b(0) ↔ n→i(# int2bits(i)) % 8 = 1

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)