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Proof step 24 from proof gcd-mod-one (previous step)

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

User applied rule exists right

And obtained the following new goals

Proof step 25:
n * n0 + 1 = m * m0, gcd(m, n) = 1 ⊦ m = 0, n = 1, n = 0, (m * m0) % n = 1

Rule argument

pos: right 4 terms: m0 abort: F discard: T computed: F

Additional information

Used terms: m0