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Proof step 3 from proof gcd-one-prime (previous step)

prime(i) ⊦ gcd(i, j) = 1, i | j

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: gcd-one seq: ⊦ gcd(i, j) = 1 ↔ ¬(∃ i1. i1 > 1 ∧ i1 | i ∧ i1 | j) subst: [j, i] → [j, i] flags: <all paths>

Additional information

Used terms: j, i