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

⊦ gcd(2 * n - n, n) = n

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: nat inst: name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [2 * n, n, n0] flags:

Additional information

Used simplifier rules: gcd-id: ⊦ gcd(m, m) = m;%gcd-a: ⊦ gcd(m, gcd(n, n0)) = gcd(gcd(m, n), n0);%numeral-25: m2 ≠ 0 ⊦ (* m2) * m = m + n0 ↔ (m2 - 1) * m = n0; from specification nat-mult%one: ⊦ m0 * 1 = m0; 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%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%twole-02: ⊦ m ≤ 2 * m; from specification nat-mult%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%twols: ⊦ ¬ 2 * n < n; from specification nat-mult%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: 2 * n, n, n0