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Proof step 4 from proof int2bytes-rec-pos-03 (previous step)

255 < i, 0 ≤ i % 256, 0 < i % 256, 0 < i, 128 ≤ i % 256, 255 < i ⊦ int2bytes(i / 256) + int2bytes(i % 256).last ' = int2bytes(i / 256) + int2bytes(i % 256).rest

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
0 ≤ i % 256, 0 < i % 256, 0 < i, 128 ≤ i % 256, 255 < i ⊦ int2bytes(i % 256).last ' = int2bytes(i % 256).rest

Rule argument

Additional information

Used simplifier rules: ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ bytes + bytes0 = bytes + bytes1 ↔ bytes0 = bytes1;