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Proof step 3 from proof gcd-div-one (previous step)

gcd(m, n) = 1, m | n * n0 ⊦ m | n0

User applied rule insert lemma

And obtained the following new goals

Proof step 4:
⊦ ∃ m0, n0. gcd(m, n) = dist(m * m0, n * n0)
Proof step 5:
gcd(m, n) = 1, m | n * n0, ∃ m0, n0. gcd(m, n) = dist(m * m0, n * n0) ⊦ m | n0

Rule argument

spec: <toplevel> name: gcdex-dist seq: ⊦ ∃ m0, n0. gcd(m, n) = dist(m * m0, n * n0) subst: [m, n] → [m, n]

Additional information

Used terms: m, n