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Proof step 26 from proof maxinf-skolemf (previous step)

(∃ n. np(n) ∧ (∀ n0. n < n0 → ¬ np(n0))) ∨ (∃ n. n3 + n0 - 1 < n ∧ np(n)), np(n0), ∀ n1, n2, np. n1 - n2 < n3 → np(n2) → (∃ n. np(n) ∧ (∀ n0. n < n0 → ¬ np(n0))) ∨ (∃ n0. n1 < n0 ∧ np(n0)) ⊦ np(n3 + n0), ∃ n1. n3 + n0 < n1 ∧ np(n1), ∃ n. np(n) ∧ (∀ n0. n < n0 → ¬ np(n0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 27:
np(n0), n3 + n0 - 1 < n, np(n), ∀ n1, n2, np. n1 - n2 < n3 → np(n2) → (∃ n. np(n) ∧ (∀ n0. n < n0 → ¬ np(n0))) ∨ (∃ n0. n1 < n0 ∧ np(n0)) ⊦ n = 0, np(n3 + n0), ∃ n1. n3 + n0 < n1 ∧ np(n1), ∃ n. np(n) ∧ (∀ n0. n < n0 → ¬ np(n0))

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1