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Proof step 61 from proof rightshift-8-02 (previous step)

1 < # int2bytes(i) ⊦ bytes2int(int2bytes(i).butlast) = bytes2int(fillsgn(2, int2bytes(i)).butlast)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 62:
⊦ bytes2int(int2bytes(i).butlast) = bytes2int(fillsgn(2, int2bytes(i)).butlast), # int2bytes(i) = 1

Rule argument

Additional information

Used simplifier rules: int2bytes-empty: ⊦ int2bytes(i) ≠ [];% ⊦ # bytes = 0 ↔ bytes = [];% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);