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Proof step 1 from proof gcd-conq-one (conclusion)

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

User applied rule cut formula

And obtained the following new goals

Proof step 2:
gcd(m, n) = 1 ⊦ n = 0, ∃ m0. n | dist(m * m0, 1)

Rule argument

fma: n = 0

Additional information

Used simplifier rules: gcd-zero-02: ⊦ gcd(n, 0) = n;%gcd-a: ⊦ gcd(m, gcd(n, n0)) = gcd(gcd(m, n), n0);%divi-zero-01: ⊦ 0 | m ↔ m = 0;%dist-eq: ⊦ dist(m, n) = 0 ↔ m = n;%one-01: ⊦ m0 * m = 1 ↔ m0 = 1 ∧ m = 1; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult Used formulas: n = 0