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Proof step 9 from proof gcd-div-one (previous step)

n0 = dist(n0 * m * m0, n0 * n * n1), 1 = dist(m * m0, n * n1), gcd(m, n) = 1, m * n2 = n * n0 ⊦ m | n0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ m | n ↔ (∃ n0. m * n0 = n)
Proof step 11:
n0 = dist(n0 * m * m0, n0 * n * n1), 1 = dist(m * m0, n * n1), gcd(m, n) = 1, m * n2 = n * n0 ⊦ ∃ n1. m * n1 = n0

Rule argument

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

Additional information

Used terms: n0, m