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Proof step 18 from proof gcd-div-one (previous step)

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

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: dist-mult: ⊦ dist(m * n, m * n0) = m * dist(n, n0);%eq-same: ⊦ m * n = m ↔ ¬ ¬(m = 0 ∨ n = 1); from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult