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Proof step 2 from proof gcd-div-gcd (previous step)

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

User applied rule apply lemma

And obtained the following new goals

Proof step 3:
⊦ gcd(i, j) = gcd(abs(i), abs(j))
Proof step 4:
gcd(i, j) = gcd(abs(i), abs(j)) ⊦ gcd(i / gcd(i, j), j / gcd(i, j)) = 1, i = 0

Rule argument

spec: <toplevel> name: gcd-abs-rew-02 seq: ⊦ gcd(i, j) = gcd(abs(i), abs(j)) subst: [i, j] → [i, j]

Additional information

Used terms: i, j