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Proof step 3 from proof gcd-same (previous step)

⊦ i ≠ 0 → gcd(abs(i), abs(i)) = abs(i)

User applied rule insert lemma

And obtained the following new goals

Proof step 4:
⊦ gcd(i, j) | i
Proof step 5:
gcd(abs(i), abs(i)) | abs(i) ⊦ i ≠ 0 → gcd(abs(i), abs(i)) = abs(i)

Rule argument

spec: int-gcd inst: name: gcd-div1 seq: ⊦ gcd(i, j) | i subst: [i, j] → [abs(i), abs(i)] flags: current, []

Additional information

Used terms: abs(i), abs(i)