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Proof step 6 from proof divi-primmult (previous step)

prime(m), m | n * n0 ⊦ gcd(m, n) = 1, m | n0, m | n

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ gcd(m, m0) = 1 ↔ m + m0 = 1 ∨ ¬(∃ n. prime(n) ∧ 1 < n ∧ n | m ∧ n | m0)

Rule argument

spec: <toplevel> name: gcdnex-prime seq: ⊦ gcd(m, m0) = 1 ↔ m + m0 = 1 ∨ ¬(∃ n. prime(n) ∧ 1 < n ∧ n | m ∧ n | m0) subst: [m0, m] → [n, m] flags: <all paths>

Additional information

Used simplifier rules: div: ⊦ prime(m) ∧ n ≠ 1 → (n | m ↔ n = m);% ⊦ 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% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s) Used terms: n, m