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

n3 + n0 = m + 1 ∧ n3 + n0 ≠ 0, np(n0), ∀ n1, n2, np. n1 - n2 < n3 → np(n2) → (∃ n. np(n) ∧ (∀ n0. n < n0 → ¬ np(n0))) ∨ (∃ n0. n1 < n0 ∧ np(n0)) ⊦ m - n0 < n3, 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 21:
n3 + n0 = m + 1, n3 ≤ m - n0, np(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-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat