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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 + m = 0, n2 = 0, n2 < m ∧ n0 | gcd(n2, m), n0 | n2 ∧ n0 | m

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: nat-prime inst: name: gcd-c seq: ⊦ gcd(m, n) = gcd(n, m) subst: [n, m] → [m, n2] flags: rewrite, current

Additional information

Used simplifier rules: gcd-id: ⊦ gcd(m, m) = m;%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%anti: ⊦ m ≤ n → (n ≤ m ↔ m = n); from specification nat%two-02: ⊦ m + m = 2 * m; from specification nat-mult%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%zero-06: ⊦ m ≠ 0 → (m * n = 0 ↔ n = 0); from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%ler: ⊦ m ≤ m; from specification nat Used terms: m, n2