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Proof step 1 from proof maxinf-skolemf (conclusion)

⊦ ∃ n∞. (n∞ = ∞ ⊃ (∀ n. ∃ n0. n < n0 ∧ np(n0)) ∨ (∀ n. ¬ np(n));(∃ n. np(n∞.n)) ∧ (∀ m. np(m) → ¬ n∞ < ⌜m⌝ ))

User applied rule exists right

And obtained the following new goals

Proof step 2:
⊦ (∞ = ∞ ⊃ (∀ n. ∃ n0. n < n0 ∧ np(n0)) ∨ (∀ n. ¬ np(n));(∃ n. np(∞.n)) ∧ (∀ m. np(m) → ¬ ∞ < ⌜m⌝ )), ∃ n∞. (n∞ = ∞ ⊃ (∀ n. ∃ n0. n < n0 ∧ np(n0)) ∨ (∀ n. ¬ np(n));(∃ n. np(n∞.n)) ∧ (∀ m. np(m) → ¬ n∞ < ⌜m⌝ ))

Rule argument

pos: right 1 terms: ∞ abort: F discard: F computed: T

Additional information

Used terms: ∞