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

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ gcd(i, j) = 1 ↔ ¬(∃ i1. i1 > 1 ∧ i1 | i ∧ i1 | j)
Proof step 3:
0 < i, 0 < j ⊦ ∀ i1. ¬(i1 > 1 ∧ i1 | i / gcd(i, j) ∧ i1 | j / gcd(i, j))

Rule argument

spec: int-gcd inst: name: gcd-one seq: ⊦ gcd(i, j) = 1 ↔ ¬(∃ i1. i1 > 1 ∧ i1 | i ∧ i1 | j) subst: [j, i] → [j / gcd(i, j), i / gcd(i, j)] flags: rewrite, current

Additional information

Used terms: j / gcd(i, j), i / gcd(i, j)