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Proof step 24 from proof maxinf-skolemf (previous step)

n3 + n0 = m + 1, n3 ≤ m - n0, np(n0), n0 ≤ m + 1, ∀ n1, n2, np. n1 - n2 < n3 → np(n2) → (∃ n. np(n) ∧ (∀ n0. n < n0 → ¬ np(n0))) ∨ (∃ n0. n1 < n0 ∧ np(n0)) ⊦ np(m + 1), ∃ n1. m + 1 < n1 ∧ np(n1), np(n0) ∧ (∀ n1. 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 25:
n3 + n0 = m + 1, n3 ≤ m - n0, np(n0), n0 ≤ m + 1, np(n), n0 < 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(m + 1), ∃ n. np(n) ∧ (∀ n0. n < n0 → ¬ np(n0)), ∃ n1. m + 1 < n1 ∧ np(n1)

Rule argument

Additional information

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