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Proof step 6 from proof gcd-conq-one (previous step)

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

User applied rule exists right

And obtained the following new goals

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

Rule argument

pos: right 3 terms: m0 abort: F discard: T computed: F

Additional information

Used terms: m0