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Proof step 20 from proof fillsgn-short (previous step)

# bytes ≤ # fillsgn(# bytes, int2bytes(bytes2int(bytes))) ⊦ ¬ # bytes < 2, 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 21:
fillsgn(2, int2bytes(bytes2int(bytes))) = bytes, # bytes ≤ # fillsgn(# bytes, int2bytes(bytes2int(bytes))), # bytes < 2 ⊦ bytes = []

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (forward)% ⊦ m < n → (n ≤ m ↔ false);% ⊦ ¬ m < n ↔ n ≤ m;