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Proof step 9 from proof gcd-conq-one (previous step)

m * m0 = m1 * n + 1, gcd(m, n) = 1 ⊦ n = 1, n = 0, n | dist(m * m0, 1)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ m | n ↔ (∃ n0. m * n0 = n)
Proof step 11:
m * m0 = m1 * n + 1, gcd(m, n) = 1 ⊦ n = 1, n = 0, ∃ n0. n * n0 = dist(m * m0, 1)

Rule argument

spec: nat-prime inst: name: Divides seq: ⊦ m | n ↔ (∃ n0. m * n0 = n) subst: [n, m] → [dist(m * m0, 1), n] flags: rewrite, current

Additional information

Used terms: dist(m * m0, 1), n