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Proof step 4 from proof fillsgn-short (previous step)

int2bytes(bytes2int(bytes)) ≠ [] → # bytes ≤ # fillsgn(# bytes, int2bytes(bytes2int(bytes))) ⊦ bytes = [], fillsgn(2, int2bytes(bytes2int(bytes))) = bytes ↔ ¬ # bytes < 2 ∧ (2 < # bytes → int2bytes(bytes2int(bytes)) = bytes)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
# bytes ≤ # fillsgn(# bytes, int2bytes(bytes2int(bytes))) ⊦ bytes = [], fillsgn(2, int2bytes(bytes2int(bytes))) = bytes ↔ ¬ # bytes < 2 ∧ (2 < # bytes → int2bytes(bytes2int(bytes)) = bytes)

Rule argument

Additional information

Used simplifier rules: int2bytes-empty: ⊦ int2bytes(i) ≠ [];