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Proof step 15 from proof gcd-zero-01-h (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ gcd(m, n) = gcd(n, m)

Rule argument

spec: <toplevel> name: gcd-c seq: ⊦ gcd(m, n) = gcd(n, m) subst: [n, m] → [m, n1] flags: <all paths>

Additional information

Used simplifier rules: gcd-id: ⊦ gcd(m, m) = m;%gcd-a: ⊦ gcd(m, gcd(n, n0)) = gcd(gcd(m, n), n0);% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m ≤ n → (n ≤ m ↔ m = n); (s) Used terms: m, n1