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Proof step 7 from proof gcd-multdiv (previous step)

gcd(m, n) * n1 = m ⊦ gcd(m, n) | (gcd(m, n) * n1) * n0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ m | n ↔ (∃ n0. m * n0 = n)
Proof step 9:
gcd(m, n) * n1 = m ⊦ ∃ n2. gcd(m, n) * n2 = (gcd(m, n) * n1) * n0

Rule argument

spec: nat-prime inst: name: Divides seq: ⊦ m | n ↔ (∃ n0. m * n0 = n) subst: [n, m] → [(gcd(m, n) * n1) * n0, gcd(m, n)] flags: rewrite, current

Additional information

Used terms: (gcd(m, n) * n1) * n0, gcd(m, n)