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Proof step 4 from proof int2int-fillsgn (previous step)

n ≤ # int2bytes(i) ⊦ bytes2int(fillsgn(n, int2bytes(i))) = i

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ # bytes ≥ n → fillsgn(n, bytes) = bytes

Rule argument

name: fillsgn-longer rotate: F subst: [bytes, n] → [int2bytes(i), n]

Additional information

Used simplifier rules: int2int: ⊦ bytes2int(int2bytes(i)) = i;% ⊦ n ≥ m ↔ m ≤ n;% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ m ≤ n → (n < m ↔ false); Used terms: int2bytes(i), n