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Proof step 3 from proof gcd-trans (previous step)

n0 | n ⊦ gcd(m, n0) | gcd(m, n)

User applied rule insert lemma

And obtained the following new goals

Proof step 4:
⊦ n0 | m ∧ n0 | n → n0 | gcd(m, n)
Proof step 5:
n0 | n, gcd(m, n0) | m ∧ gcd(m, n0) | n → gcd(m, n0) | gcd(m, n) ⊦ gcd(m, n0) | gcd(m, n)

Rule argument

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

Additional information

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