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Proof step 5 from proof gcdnex-prime (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 6:
¬(∃ n. n ≠ 0 ∧ n ≠ 1 ∧ n | m ∧ n | m0) ⊦ ¬(∃ n. prime(n) ∧ n ≠ 0 ∧ n ≠ 1 ∧ n | m ∧ n | m0), m = 1, m0 = 1
Proof step 8:
¬(∃ n. prime(n) ∧ n ≠ 0 ∧ n ≠ 1 ∧ n | m ∧ n | m0) ⊦ ¬(∃ n. n ≠ 0 ∧ n ≠ 1 ∧ n | m ∧ n | m0), m = 1, m0 = 1

Rule argument

pos: right 1

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