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Proof step 4 from proof gcd-eq (previous step)

n ≠ 0 → gcd(n, m) ≤ n, gcd(m, n) = m, n < m ⊦ m = 0, n = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ gcd(m, n) = gcd(n, m)

Rule argument

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

Additional information

Used simplifier rules: fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%ref-1: ⊦ ¬ n < n; from specification nat-basic1 Used terms: n, m