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Proof step 14 from proof rightshift-8 (previous step)

i < 0 ∧ 8 ≤ 8 ⊦ (i - 255) / 256 >> (8 - 8) = bytes2int(int2bytes(i).butlast), # int2bytes(i) = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
i < 0 ⊦ # int2bytes(i) = 1, (i - 255) / 256 = bytes2int(int2bytes(i).butlast)

Rule argument

Additional information

Used simplifier rules: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i >> 0 = i;