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Math Help - Proof involving different quantifiers?

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