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Proof step 1 from proof int2bytes-rec-neg-256-01 (conclusion)

⊦ i < ~256 ∧ i % 256 ≠ 0 → int2bytes(i) = int2bytes(i / 256 - 1) + int2bytes(i % 256).last '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
i < ~256, i < 0 ⊦ int2bytes(i) = int2bytes(i / 256 - 1) + int2bytes(i % 256).last ', i % 256 = 0

Rule argument

Additional information

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