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Proof step 15 from proof gcd-div-one (previous step)

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

User applied rule insert equation

And obtained the following new goals

Proof step 16:
n0 = dist(n0 * m * m0, (m * n2) * n1), 1 = dist(m * m0, n * n1), gcd(m, n) = 1 ⊦ ∃ n1. m * n1 = n0

Rule argument

pos: left 4 flags: rotate, discard

Additional information