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Proof step 3 from proof gcd-one-eq-01 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ gcd(i, abs(j)) = gcd(i, j)
Proof step 5:
abs(i) / gcd(j, i) = 1 → gcd(j, i) = abs(i) ⊦ i ≠ 0 → (abs(i) / gcd(i, j) = 1 ↔ gcd(i, j) = abs(i))

Rule argument

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

Additional information

Used terms: i, j