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Proof step 2 from proof gcdnex-prime (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ gcd(m, m0) = 1 ↔ m + m0 = 1 ∨ ¬(∃ n. 1 < n ∧ n | m ∧ n | m0)
Proof step 4:
⊦ m + m0 = 1 ∨ ¬(∃ n. 1 < n ∧ n | m ∧ n | m0) ↔ m + m0 = 1 ∨ ¬(∃ n. prime(n) ∧ n ≠ 0 ∧ n ≠ 1 ∧ n | m ∧ n | m0)

Rule argument

spec: nat-prime inst: name: gcdnex seq: ⊦ gcd(m, m0) = 1 ↔ m + m0 = 1 ∨ ¬(∃ n. 1 < n ∧ n | m ∧ n | m0) subst: [m0, m] → [m0, m] flags: rewrite, current

Additional information

Used terms: m0, m