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Proof step 14 from proof gcd-mult (previous step)

n0 * n = n3 + n0 * m, m < 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(n0 * m, n3) = n0 * gcd(m, n - m)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 15:
n = n4 + m ∧ m ≤ n, n0 * n = n3 + n0 * m, m < 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(n0 * m, n3) = n0 * gcd(m, n4)

Rule argument

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

Additional information

Used simplifier rules: lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%ls-02: ⊦ m < n → ¬ n < m; from specification nat-basic1%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n, m, n2