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

1 < i, 2 < j, 0 < i, i < j, prime(j), 0 < j, 1 < j ⊦ gcd(i, j) = 1

User applied rule insert rewrite lemma

And obtained the following new goals

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, i] flags: rewrite, current

Additional information

Used terms: j, i