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Proof step 9 from proof gcdnex-prime (previous step)

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

User applied rule exists right

And obtained the following new goals

Proof step 10:
n | m0, n | m ⊦ m + m0 = 1, n = 0, n = 1, ∃ n. prime(n) ∧ n ≠ 0 ∧ n ≠ 1 ∧ n | m ∧ n | m0, prime(n) ∧ n ≠ 0 ∧ n ≠ 1 ∧ n | m ∧ n | m0

Rule argument

pos: right 4 terms: n abort: F discard: F computed: T

Additional information

Used terms: n