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Proof step 6 from proof divi-primmult (
previous step
)
⊦ gcd(m, m0) = 1 ↔ m + m0 = 1 ∨ ¬(∃ n. prime(n) ∧ 1 < n ∧ n | m ∧ n | m0)
Lemma
gcdnex-prime