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Proof step 17 from proof rightshift-8 (previous step)

¬(i % 256 ≠ 0 ∧ i < 0), i < 0 ⊦ # int2bytes(i) = 1, (i - 255) / 256 = bytes2int(int2bytes(i).butlast)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
i % 256 = 0, i < 0 ⊦ # int2bytes(i) = 1, (i - 255) / 256 = bytes2int(int2bytes(i).butlast)

Rule argument

Additional information

Used simplifier rules: ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ ¬ i < j ↔ j ≤ i;