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Proof step 5 from proof gcd-alldiv (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ gcd(m, n) = gcd(n, m)
Proof step 7:
n0 | m, n0 | n ⊦ m < n, n0 | gcd(n, m)

Rule argument

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

Additional information

Used terms: n, m