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Proof step 5 from proof gcd-le-both-h (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
n = n1 + m ∧ m ≤ n, m < n, ∀ n0, m. n0 < n → m < n0 → n0 = 0 ∨ m = 0 ∨ ¬ m < gcd(m, n0) ∧ ¬ n0 < gcd(m, n0) ⊦ n = 0, m = 0, ¬ m < gcd(m, n1) ∧ ¬ n < gcd(m, n1)

Rule argument

spec: nat inst: name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [n, m, n1] 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, n1