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Proof step 19 from proof maxinf-skolemf (previous step)

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

User applied rule insert elim lemma

And obtained the following new goals

Proof step 20:
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))

Rule argument

spec: nat inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n3 + n0, m] flags: []

Additional information

Used simplifier rules: z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%lez: ⊦ 0 ≤ m; from specification nat%ls-01: n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0; from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%leprf: ⊦ m ≤ m - 1 → m = 0; (forward) from specification nat%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n3 + n0, m