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Proof step 10 from proof gcdnex (previous step)

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

User applied rule exists right

And obtained the following new goals

Proof step 11:
⊦ m0 = 1, gcd(m, m0) = 1, m0 = 0 ∧ m = 1, gcd(m, m0) ≠ 0 ∧ gcd(m, m0) ≠ 1 ∧ gcd(m, m0) | m ∧ gcd(m, m0) | m0, ∃ n. n ≠ 0 ∧ n ≠ 1 ∧ n | m ∧ n | m0

Rule argument

pos: right 4 terms: gcd(m, m0) abort: F discard: F computed: T

Additional information

Used terms: gcd(m, m0)