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Proof step 17 from proof gcd-div-one (previous step)

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

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: dist(n0 * m0, n2 * n1)

Additional information

Used terms: dist(n0 * m0, n2 * n1)