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Proof step 24 from proof gcd-mult (previous step)

n0 * m = n3 + n0 * n, ∀ n1, m1, n2. n1 * m1 + n1 * n2 < n0 * m + n0 * n → true → gcd(n1 * m1, n1 * n2) = n1 * gcd(m1, n2) ⊦ n0 = 0, gcd(n3, n0 * n) = n0 * gcd(m - n, n)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

spec: nat inst: name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [m, n, n2] flags:

Additional information

Used simplifier rules: lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%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%addle-01: ⊦ m * n1 = m * m0 + n0 ∧ m ≠ 0 → ¬ n1 < m0; from specification nat-mult%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: m, n, n2