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Proof step 12 from proof gcd-div-both-h (previous step)

n0 | gcd(m, n2), n2 ≤ m, ∀ n1, m0, n0. n1 < n2 + m → m0 < n1 ∧ n0 | gcd(m0, n1) → n0 | n1 ∧ n0 | m0 ⊦ n2 = 0, n2 < m ∧ n0 | gcd(n2, m), n0 | n2 ∧ n0 | m

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 13:
⊦ gcd(m, n) = gcd(n, m)

Rule argument

spec: <toplevel> name: gcd-c seq: ⊦ gcd(m, n) = gcd(n, m) subst: [n, m] → [m, n2] flags: <all paths>

Additional information

Used simplifier rules: gcd-id: ⊦ gcd(m, m) = m;% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m ≤ n → (n ≤ m ↔ m = n); (s)% ⊦ m ≤ m; (s) Used terms: m, n2