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Proof step 1 from proof gcd-div-gcd (conclusion)

⊦ i ≠ 0 → gcd(i / gcd(i, j), j / gcd(i, j)) = 1

User applied rule insert lemma

And obtained the following new goals

Proof step 2:
⊦ gcd(i, j) = gcd(abs(i), abs(j))
Proof step 3:
gcd(i, j) = gcd(abs(i), abs(j)) ⊦ i ≠ 0 → gcd(i / gcd(i, j), j / gcd(i, j)) = 1

Rule argument

spec: int-gcd inst: name: gcd-abs-rew-02 seq: ⊦ gcd(i, j) = gcd(abs(i), abs(j)) subst: [i, j] → [i, j] flags: current, []

Additional information

Used terms: i, j