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Proof step 4 from proof gcdnex (previous step)

gcd(m, m0) = 1, n | m0, n | m ⊦ n = 0, n = 1, m = 1, m0 = 1

User applied rule insert lemma

And obtained the following new goals

Proof step 5:
⊦ n0 | m ∧ n0 | n → n0 | gcd(m, n)

Rule argument

spec: inst: name: gcd-alldiv seq: ⊦ n0 | m ∧ n0 | n → n0 | gcd(m, n) subst: [n0, m, n] → [n, m, m0] flags: current, []

Additional information

Used simplifier rules: divi-one: ⊦ m | 1 ↔ m = 1; Used terms: n, m, m0