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Proof step 7 from proof gcdnex-prime (previous step)

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

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 4 terms: n

Additional information

Used simplifier rules: ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1; (ws)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: n