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Proof step 16 from proof maxinf-skolemf (previous step)

np(n0), n0 ≤ n, ∀ n1, n2, np. n1 - n2 < n - n0 → np(n2) → (∃ n. np(n) ∧ (∀ n0. n < n0 → ¬ np(n0))) ∨ (∃ n0. n1 < n0 ∧ np(n0)) ⊦ np(n), ∃ n0. n < n0 ∧ np(n0), ∃ n. np(n) ∧ (∀ n0. n < n0 → ¬ np(n0))

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 17:
n = n3 + n0, np(n0), n0 ≤ n, ∀ n1, n2, np. n1 - n2 < n3 → np(n2) → (∃ n. np(n) ∧ (∀ n0. n < n0 → ¬ np(n0))) ∨ (∃ n0. n1 < n0 ∧ np(n0)) ⊦ np(n), ∃ n0. n < n0 ∧ np(n0), ∃ n. np(n) ∧ (∀ n0. n < n0 → ¬ np(n0))

Rule argument

spec: nat inst: name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [n, n0, n3] flags:

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n, n0, n3