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Proof step 1 from proof gcd-one-eq-01 (conclusion)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 2:
⊦ j2 / gcd(i2, j2) = 1 → gcd(i2, j2) = j2
Proof step 3:
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))

Rule argument

spec: int-gcd inst: name: gcd-one-eq seq: ⊦ j2 / gcd(i2, j2) = 1 → gcd(i2, j2) = j2 subst: [j2, i2] → [abs(i), j] flags: current, []

Additional information

Used terms: abs(i), j