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Proof step 6 from proof gcd-div-both-h (previous step)

m < n, n0 | gcd(m, n - m), ∀ n1, m, n0. n1 < n → m < n1 ∧ n0 | gcd(m, n1) → n0 | n1 ∧ n0 | m ⊦ n = 0, n0 | n ∧ n0 | m

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
n = n2 + m ∧ m ≤ n, m < n, n0 | gcd(m, n2), ∀ n1, m, n0. n1 < n → m < n1 ∧ n0 | gcd(m, n1) → n0 | n1 ∧ n0 | m ⊦ n = 0, n0 | n ∧ n0 | m

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-03: ⊦ m < n → m ≤ n; from specification nat%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n, m, n2