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Proof step 2 from proof gcd-one-eq (previous step)

j2 / gcd(i2, j2) = 1 ⊦ gcd(i2, j2) = j2

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
⊦ 0 ≤ gcd(i, j)
Proof step 4:
0 ≤ gcd(i2, j2), j2 / gcd(i2, j2) = 1 ⊦ gcd(i2, j2) = j2

Rule argument

spec: int-gcd inst: name: gcd-pos seq: ⊦ 0 ≤ gcd(i, j) subst: [i, j] → [i2, j2] flags: current, []

Additional information

Used terms: i2, j2