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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 ⊦ 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, n0 | gcd(m, n2), m ≤ n, ∀ n1, m, n0. n1 < n → m < n1 ∧ n0 | gcd(m, n1) → n0 | n1 ∧ n0 | m ⊦ n0 | n ∧ n0 | m

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