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Proof step 11 from proof gcd-conq-one (previous step)

m * m0 = m1 * n + 1, gcd(m, n) = 1 ⊦ n = 1, n = 0, ∃ n0. n * n0 = dist(m * m0, 1)

User applied rule exists right

And obtained the following new goals

Proof step 12:
m * m0 = m1 * n + 1, gcd(m, n) = 1 ⊦ n = 1, n = 0, n * m1 = dist(m * m0, 1)

Rule argument

pos: right 3 terms: m1 abort: F discard: T computed: F

Additional information

Used terms: m1