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Proof step 5 from proof gcd-conq-one (previous step)

gcd(m, n) = 1 ⊦ n = 0, n = 1, ∃ m0. n | dist(m * m0, 1)

User applied rule insert lemma

And obtained the following new goals

Proof step 6:
gcd(m, n) = 1, 1 < n ⊦ ∃ m0. (m * m0) % n = 1
Proof step 7:
gcd(m, n) = 1, ∃ m0. (m * m0) % n = 1 ⊦ n = 1, n = 0, ∃ m0. n | dist(m * m0, 1)

Rule argument

spec: <toplevel> name: gcd-mod-one-01 seq: gcd(m, n) = 1, 1 < n ⊦ ∃ m0. (m * m0) % n = 1 subst: [m, n] → [m, n]

Additional information

Used simplifier rules: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s) Used terms: m, n