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Proof step 12 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 13:
⊦ m0 * m * n = (m0 * m) * n
Proof step 14:
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

Rule argument

spec: nat-mult name: mult-a seq: ⊦ m0 * m * n = (m0 * m) * n subst: [n, m, m0] → [n1, n, n0] flags: <all paths>

Additional information

Used terms: n1, n, n0