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Proof step 19 from proof gcd-mult (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 20:
n0 * m = n3 + n0 * n, n ≤ m, n0 * n ≤ n0 * m, ∀ n1, m1, n2. n1 * m1 + n1 * n2 < n0 * m + n0 * n → true → gcd(n1 * m1, n1 * n2) = n1 * gcd(m1, n2) ⊦ gcd(n3, n0 * n) = n0 * gcd(m, n), n0 = 0

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [n0 * m, n0 * n, n3]

Additional information

Used simplifier rules: ⊦ n0 ≤ n → n * m - n0 * m = (n - n0) * m; (s)%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult% ⊦ m0 ≠ 0 → (m0 * m ≤ m0 * n ↔ m ≤ n); (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m ≤ n → (n < m ↔ false); (s)%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n0 * m, n0 * n, n3