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Proof step 70 from proof rightshift-8-01-01 (previous step)

i→n(j) < # int2bytes(i) ⊦ bytes2int(firstn(# int2bytes(i) - i→n(j), int2bytes(i))) = bytes2int(firstn(# int2bytes(i), fillsgn(i→n(j) + # int2bytes(i), int2bytes(i))))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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