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Proof step 34 from proof rightshift-8-01-01 (previous step)

# int2bytes(i) ≠ 0, i < 0, # int2bytes(i) ≤ i→n(j) ⊦ j = 0, i >> 8 * j = ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 35:
i < 0, # int2bytes(i) ≤ i→n(j) ⊦ j = 0, i >> 8 * j = ~1

Rule argument

Additional information

Used simplifier rules: int2bytes-empty: ⊦ int2bytes(i) ≠ [];% ⊦ # bytes = 0 ↔ bytes = [];