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Proof step 15 from proof maxinf-skolemf (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
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))

Rule argument

Additional information

Used simplifier rules: lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat