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Proof step 1 from proof rightshift-8-01 (conclusion)

⊦ i→n(j) < # int2bytes(i) → i >> 8 * j = bytes2int(firstn(# int2bytes(i) - i→n(j), int2bytes(i)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
i→n(j) < # int2bytes(i) ⊦ i >> 8 * j = bytes2int(firstn(# int2bytes(i) - i→n(j), int2bytes(i)))

Rule argument

Additional information

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