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Proof step 3 from proof gcd-conq-one (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 4:
gcd(m, n) = 1, 1 < n ⊦ ∃ m0. (m * m0) % n = 1
Proof step 5:
∃ m0. (m * m0) % n = 1, gcd(m, n) = 1 ⊦ n = 1, n = 0, ∃ m0. n | dist(m * m0, 1)

Rule argument

spec: nat-prime inst: name: gcd-mod-one-01 seq: gcd(m, n) = 1, 1 < n ⊦ ∃ m0. (m * m0) % n = 1 subst: [m, n] → [m, n] flags: current, []

Additional information

Used simplifier rules: one-02: ⊦ m ≠ 0 → (¬ 1 < m ↔ m = 1); from specification nat-basic1 Used terms: m, n