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Proof step 25 from proof maxinf-skolemf (previous step)

n3 + n0 = m + 1, n0 < n, np(n), np(n0), n0 ≤ m + 1, 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), ∃ n1. m + 1 < n1 ∧ np(n1), ∃ n. np(n) ∧ (∀ n0. n < n0 → ¬ np(n0))

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 2 terms: n

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1); (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m < n ↔ ¬ n ≤ m; (ss)%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%le-one: ⊦ m ≤ n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)% ⊦ m ≤ m; (s)% ⊦ m + m0 ≤ m ↔ m0 = 0; (s)% ⊦ m < n → (n ≤ m ↔ false); (s)%n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n; (s)% ⊦ m = n + 1 → ¬ m ≤ n; (s) Used terms: n