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Proof step 2 from proof gcdnex (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 3:
gcd(m, m0) = 1, ∃ n. n ≠ 0 ∧ n ≠ 1 ∧ n | m ∧ n | m0 ⊦ m + m0 = 1
Proof step 6:
m + m0 = 1 ∨ ¬(∃ n. n ≠ 0 ∧ n ≠ 1 ∧ n | m ∧ n | m0) ⊦ gcd(m, m0) = 1

Rule argument

pos: right 1

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