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Proof step 1 from proof fillsgn-short (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ m ≤ n ↔ n < m;