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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) ⊦ gcd(n0 * m, n3) = n0 * gcd(m, n - m), n0 = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 15:
n0 * n = n3 + n0 * m, n = n4 + m, m < n, m ≤ n, ∀ n1, m1, n2. n1 * m1 + n1 * n2 < n0 * m + n0 * n → true → gcd(n1 * m1, n1 * n2) = n1 * gcd(m1, n2) ⊦ gcd(n0 * m, n3) = n0 * gcd(m, n4), n0 = 0

Rule argument

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

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < n → ¬ n < m; (s)%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n, m, n2