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Proof step 1 from proof gcd-a (conclusion)

⊦ gcd(m, gcd(n, n0)) = gcd(gcd(m, n), n0)

User applied rule insert lemma

And obtained the following new goals

Proof step 2:
m | n, n | m ⊦ m = n
Proof step 3:
⊦ gcd(m, gcd(n, n0)) | gcd(gcd(m, n), n0) ∧ gcd(gcd(m, n), n0) | gcd(m, gcd(n, n0)), gcd(m, gcd(n, n0)) = gcd(gcd(m, n), n0)

Rule argument

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

Additional information

Used terms: gcd(m, gcd(n, n0)), gcd(gcd(m, n), n0)