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Proof step 44 from proof int2bytes-butlast (previous step)

i = ~256, i < 0, i ≤ ~256, i = ~256 ∨ i % 256 ≠ 0, i = ~256 ∨ i % 256 = 0 ⊦ int2bytes(i / 256) = int2bytes(i >> 8)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: int2bytes-byte: ~128 ≤ i, i < 128 ⊦ int2bytes((* i)) = i2b(i) ';%int2bytes-minus-one: ⊦ int2bytes(i) = i2b(~1) ' ↔ i = ~1;