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Thread: 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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    $\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} $
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