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Proof step 1 from proof int2bytes-rec-pos-04 (conclusion)

256 ≤ i ⊦ int2bytes(i) = int2bytes(i / 256) + fix(i→n(2), int2bytes(i % 256)).rest

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
256 ≤ i, 0 < i, 0 ≤ i ⊦ int2bytes(i) = int2bytes(i / 256) + fix(2, int2bytes(i % 256)).rest

Rule argument

Additional information

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