# Thread: Proof involving different quantifiers?

1. ## Proof involving different quantifiers?

I am confused how to use (or if it is even possible) to use existential instantiation and universal generalization in a proof. How would I solve the following proof and would I need these rules?

Premise 1: for all x in D [ P(x) -> Q(x) ]
Premise 2: for all x in D [ not P(x) -> R(x) ]
Premise 3: there exists an x in D [ not Q(x) ]

Solution: there exists an x in D [ R(x) ]

Any help is appreciated. Thanks.

2. $\displaystyle \begin{array}{llr} 1 & {\left( {\forall x} \right)\left[ {P(x) \to Q(x)} \right]} & {} \\ 2 & {\left( {\forall x} \right)\left[ {\neg P(x) \to R(x)} \right]} & {} \\ 3 & {\left( {\exists x} \right)\left[ {\neg Q(x)} \right]} & {} \\ \hline 4 & {\neg Q(a)} & 3 \\ \end{array}$
$\displaystyle \begin{array}{llr} 5 & {P(a) \to Q(a)} & 1 \\ 6 & {\neg P(a)} & {5,4} \\ 7 & {\neg P(a) \to R(a)} & 2 \\ 8 & {R(a)} & {6,7} \\ 9 & {\left( {\exists x} \right)\left[ {R(x)} \right]} & 8 \\ \end{array}$