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Proof step 5 from proof gcd-div-one (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 6:
n0 = dist(n0 * m * m0, n0 * n * n1), 1 = dist(m * m0, n * n1), gcd(m, n) = 1, m | n * n0 ⊦ m | n0
Proof step 18:
1 = dist(m * m0, n * n1), gcd(m, n) = 1, m | n * n0 ⊦ n0 = dist(n0 * m * m0, n0 * n * n1), m | n0

Rule argument

fma: n0 = dist(n0 * m * m0, n0 * n * n1)

Additional information

Used formulas: n0 = dist(n0 * m * m0, n0 * n * n1)