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Proof step 12 from proof gcdnex (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 13:
⊦ gcd(m, n) = gcd(n, m)
Proof step 14:
⊦ m0 = 1, gcd(m0, m) = 1, m0 = 0 ∧ m = 1, ∃ n. n ≠ 0 ∧ n ≠ 1 ∧ n | m ∧ n | m0, gcd(m0, m) ≠ 0 ∧ gcd(m0, m) | m0

Rule argument

spec: nat-prime inst: name: gcd-c seq: ⊦ gcd(m, n) = gcd(n, m) subst: [n, m] → [m0, m] flags: rewrite, current

Additional information

Used terms: m0, m