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Proof step 2 from proof fillsgn-short (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
⊦ bytes ≠ [] → n ≤ # fillsgn(n, bytes)
Proof step 4:
int2bytes(bytes2int(bytes)) ≠ [] → # bytes ≤ # fillsgn(# bytes, int2bytes(bytes2int(bytes))) ⊦ bytes = [], fillsgn(2, int2bytes(bytes2int(bytes))) = bytes ↔ ¬ # bytes < 2 ∧ (2 < # bytes → int2bytes(bytes2int(bytes)) = bytes)

Rule argument

name: fillsgn-length subst: [bytes, n] → [int2bytes(bytes2int(bytes)), # bytes]

Additional information

Used terms: int2bytes(bytes2int(bytes)), # bytes