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Proof step 20 from proof maxinf-skolemf (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
n3 + n0 = m + 1, np(n0), n3 ≤ m - n0, ∀ n1, n2, np. n1 - n2 < n3 → np(n2) → (∃ n. np(n) ∧ (∀ n0. n < n0 → ¬ np(n0))) ∨ (∃ n0. n1 < n0 ∧ np(n0)) ⊦ np(m + 1), ∃ n. np(n) ∧ (∀ n0. n < n0 → ¬ np(n0)), ∃ n1. m + 1 < n1 ∧ np(n1)

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)