# Proof involving different quantifiers?

• Mar 1st 2009, 12:05 PM
tshare1
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.
• Mar 1st 2009, 01:28 PM
Plato
$\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}$

$\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}$