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

255 < i ⊦ int2bytes(i) = int2bytes(i / 256) + int2bytes(i % 256).last '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
255 < i, 0 < i, 0 ≤ i ⊦ bits2bytes(int2bits(i)) = bits2bytes(int2bits(i / 256)) + bits2bytes(int2bits(i % 256)).last '

Rule argument

Additional information

Used simplifier rules: int2bytes: ⊦ int2bytes(i) = bits2bytes(int2bits(i));% ⊦ (* i) < j ∧ 0 < i → 0 ≤ j ∧ 0 < j; (forward)