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Proof step 2 from proof gcd-mod-one (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
⊦ ∃ m0, n0. m * m0 + gcd(m, n) = n * n0 ∨ n * n0 + gcd(m, n) = m * m0
Proof step 4:
∃ m0, n0. m * m0 + gcd(m, n) = n * n0 ∨ n * n0 + gcd(m, n) = m * m0, gcd(m, n) = 1 ⊦ m = 0, n = 1, n = 0, ∃ m0. (m * m0) % n = 1

Rule argument

spec: nat-prime inst: name: gcdex seq: ⊦ ∃ m0, n0. m * m0 + gcd(m, n) = n * n0 ∨ n * n0 + gcd(m, n) = m * m0 subst: [m, n] → [m, n] flags: current, []

Additional information

Used terms: m, n