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Proof step 1 from proof int2bytes-rec-pos-01 (conclusion)

255 < i, i % 256 = 0 ⊦ int2bytes(i) = int2bytes(i / 256) + i2b(0) '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
i % 256 = 0, 0 < i, 255 < i ⊦ int2bytes(i) = int2bytes(i / 256) + i2b(0) '

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)