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Proof step 2 from proof gcd-div-one-02 (previous step)

gcd(m, n) = 1, m | n0 * n ⊦ m | n0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ m * n = n * m
Proof step 4:
gcd(m, n) = 1, m | n * n0 ⊦ m | n0

Rule argument

spec: nat-mult inst: name: mult-c seq: ⊦ m * n = n * m subst: [n, m] → [n, n0] flags: rewrite

Additional information

Used terms: n, n0