Answer
Yes, it can be proved if ∀x(P(x) ∨ Q(x)),
∀x(¬Q(x) ∨ S(x)), ∀x(R(x)→¬S(x)), and ∃x¬P(x)
are true, then ∃x¬R(x) is true using these rules of inference listed below
Existential instantiation
Universal instantiation
Disjunctive syllogism
Existential generalization
Work Step by Step
Lets assume that :
∀ x (P(x) ∨ Q (x)),
∀ x(-Q(x) ∨ S(x))
∀ x (R(x) →¬ S(x))
and ∃ x¬ P(x) are true
hence these four statements are the premises.
Let c be arbitrary.
1. ∀ x(P(x) ∨ Q(x)) premise
2. ∀ x(-Q(x)) ∨ S(x)) premise
3. ∀ x (R(x) →¬ S(x)) premise
4. ∃ x¬ P(x) premise
5. ¬ P(c) existential instantiation from (4)
6. P(c) ∨ Q(c) universal instantiation from (1)
7. Q(c) disjunctive syllogism from (5) and (6)
8. ¬Q(c) ∨ S(c) universal instantiation from (2)
9. S(c) disjunctive syllogism from (7) and (8)
10. R(c) ∨ ¬S(c) universal instantiation from (3)
11. ¬R(c) ∨ ¬S(c) logical equivalence (1) from (10)
12. ¬(¬S(c)) double negation law from (9)
13. ¬R(c) disjunctive syllogism from (11) and (12)
14. ∃ x¬ R(x) existential generalization from (13)
Hence we have shown that if the premises
∀ x(P(x) ∨ Q(x))
∀ x(-Q(x) ∨ S(x))
∀ x (R(x) →¬ S(x))
and ∃ x¬ P(x) are true ,
then the conclusion ∃ x¬ R(x) is also true.
