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Proof step 4 from proof rightshift-8 (previous step)

¬(i < 0 ∧ 8 ≤ 8) ⊦ i >> 8 = bytes2int(int2bytes(i).butlast), # int2bytes(i) = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
0 ≤ i ⊦ i >> 8 = bytes2int(int2bytes(i).butlast), # int2bytes(i) = 1

Rule argument

Additional information

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