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Proof step 18 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)) ⊦ np(n3 + n0), ∃ n1. n3 + n0 < n1 ∧ np(n1), ∃ n. np(n) ∧ (∀ n0. n < n0 → ¬ np(n0))

User applied rule apply induction

And obtained the following new goals

Proof step 19:
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))
Proof step 26:
(∃ n. np(n) ∧ (∀ n0. n < n0 → ¬ np(n0))) ∨ (∃ n. n3 + n0 - 1 < n ∧ np(n)), np(n0), ∀ n1, n2, np. n1 - n2 < n3 → np(n2) → (∃ n. np(n) ∧ (∀ n0. n < n0 → ¬ np(n0))) ∨ (∃ n0. n1 < n0 ∧ np(n0)) ⊦ np(n3 + n0), ∃ n1. n3 + n0 < n1 ∧ np(n1), ∃ n. np(n) ∧ (∀ n0. n < n0 → ¬ np(n0))

Rule argument

subst: [n1, n2, np] → [n3 + n0 - 1, n0, np]

Additional information

Used terms: n3 + n0 - 1, n0, np