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Proof step 18 from proof gcd-a (previous step)

gcd(n0, gcd(m, n)) | m ⊦ gcd(gcd(n, m), n0) | m ∧ gcd(gcd(n, m), n0) | gcd(n, n0), gcd(gcd(n, m), n0) | gcd(m, gcd(n, n0))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 19:
⊦ gcd(m, n) = gcd(n, m)
Proof step 20:
gcd(n0, gcd(m, n)) | m ⊦ gcd(n0, gcd(n, m)) | m ∧ gcd(n0, gcd(n, m)) | gcd(n, n0), gcd(n0, gcd(n, m)) | gcd(m, gcd(n, n0))

Rule argument

spec: nat-prime inst: name: gcd-c seq: ⊦ gcd(m, n) = gcd(n, m) subst: [n, m] → [n0, gcd(n, m)] flags: rewrite, current

Additional information

Used terms: n0, gcd(n, m)