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Proof step 10 from proof int2bytes-rec-pos-04 (previous step)

255 < i ∧ i % 256 < 128, 256 ≤ i, 0 < i, 0 ≤ i ⊦ int2bytes(i / 256) + int2bytes(i % 256) = int2bytes(i / 256) + fix(2, int2bytes(i % 256)).rest

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
i % 256 < 128, 256 ≤ i, 0 < i, 0 ≤ i ⊦ int2bytes(i % 256) = fix(2, int2bytes(i % 256)).rest

Rule argument

Additional information

Used simplifier rules: i0 ≠ 0, i0 < i ⊦ (* i) ≤ j → (* i0) < j;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ bytes + bytes0 = bytes + bytes1 ↔ bytes0 = bytes1;