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Proof step 14 from proof gcd-div-one (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 15:
⊦ m * n = n * m
Proof step 16:
m * n2 = n * n0, 1 = dist(m * m0, n * n1), gcd(m, n) = 1, n0 = dist(n0 * m * m0, (n * n0) * n1) ⊦ ∃ n1. m * n1 = n0

Rule argument

spec: nat-mult name: mult-c seq: ⊦ m * n = n * m subst: [n, m] → [n, n0] flags: <all paths>

Additional information

Used simplifier rules: mult-c: ⊦ m * n = n * m; from specification nat-mult Used terms: n, n0