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Proof step 3 from proof gcdnex (previous step)

gcd(m, m0) = 1 ⊦ m + m0 = 1 ∨ ¬(∃ n. n ≠ 0 ∧ n ≠ 1 ∧ n | m ∧ n | m0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
gcd(m, m0) = 1, n | m0, n | m ⊦ n = 0, n = 1, m = 1, m0 = 1

Rule argument

Additional information

Used simplifier rules: divi-one: ⊦ m | 1 ↔ m = 1;%gcd-one: ⊦ gcd(m, 1) = 1;%gcd-a: ⊦ gcd(m, gcd(n, n0)) = gcd(gcd(m, n), n0);%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%one: ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1; from specification nat-basic2