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Proof step 4 from proof gcdnex-prime (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ ¬(∃ n. n ≠ 0 ∧ n ≠ 1 ∧ n | m ∧ n | m0) ↔ ¬(∃ n. prime(n) ∧ n ≠ 0 ∧ n ≠ 1 ∧ n | m ∧ n | m0), m = 1, m0 = 1

Rule argument

Additional information

Used simplifier rules: divi-one: ⊦ m | 1 ↔ m = 1;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%one: ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1; from specification nat-basic2%one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1