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Proof step 37 from proof int2bytes-butlast (previous step)

i < 0 ∧ 8 ≤ 8, i < 0, i ≤ ~256, i < ~256 ⊦ int2bytes(i / 256 - 1) = int2bytes((i - 255) / 256 >> (8 - 8)), i % 256 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 38:
i < 0, i ≤ ~256, i < ~256 ⊦ i % 256 = 0, int2bytes(i / 256 - 1) = int2bytes((i - 255) / 256)

Rule argument

Additional information

Used simplifier rules: ⊦ i >> 0 = i;