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Proof step 1 from proof gcdnex (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ gcd(m, m0) = 1 ↔ m + m0 = 1 ∨ ¬(∃ n. n ≠ 0 ∧ n ≠ 1 ∧ n | m ∧ n | m0)

Rule argument

Additional information

Used simplifier rules: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s)