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Proof step 16 from proof gcd-a (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 17:
⊦ gcd(m, gcd(n, n0)) | n
Proof step 18:
gcd(n0, gcd(m, n)) | m ⊦ gcd(gcd(n, m), n0) | gcd(m, gcd(n, n0)), gcd(gcd(n, m), n0) | m ∧ gcd(gcd(n, m), n0) | gcd(n, n0)

Rule argument

spec: <toplevel> name: gcd-div-03 seq: ⊦ gcd(m, gcd(n, n0)) | n subst: [m, n, n0] → [n0, m, n]

Additional information

Used terms: n0, m, n