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Proof step 6 from proof gcd-trans (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 7:
⊦ n0 | m ∧ m | n → n0 | n
Proof step 8:
n0 | n, gcd(m, n0) | n0 ∧ n0 | n → gcd(m, n0) | n ⊦ gcd(m, n0) | n, gcd(m, n0) | gcd(m, n)

Rule argument

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

Additional information

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