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Propositional Logic

  1. Aug 24, 2009 #1
    The book which i read for improving my logic sense~
    There is a theorem called REPLACEMENT ..

    ( P [tex]\rightarrow[/tex] Q ) [tex]\vee[/tex] [tex]\neg[/tex] ( P [tex]\rightarrow[/tex] Q)
    where (P[tex]\rightarrow[/tex] Q) is the second occurence of ( P [tex]\rightarrow[/tex] Q)

    But what if the replace the second occurrence with [tex]\neg[/tex] P[tex]\vee[/tex] Q!!
    And i try to check with the truth table it does not gv me the values !!
    Help~~
     
  2. jcsd
  3. Aug 24, 2009 #2
    I am not absolutely sure what you mean by the truth table not giving you the values. Are you saying that

    [tex](P \rightarrow Q) \vee \neg (P \rightarrow Q) \text{ and } (P \rightarrow Q) \vee \neg (\neg P \vee Q)[/tex]

    are not appearing to be logically equivalent?

    --Elucidus
     
  4. Sep 3, 2009 #3
    yup yup~~coz i cant prove this is tvalidity by using truth table..~~
     
  5. Sep 3, 2009 #4
    Here is the side-by-side truth table of the two expressions. The final values of each is in boldface.

    [tex]\begin{array}{c|c|cccc|cccc}
    P & Q & (P \rightarrow Q) & \vee & \neg & (P \rightarrow Q) & (P \rightarrow Q) & \vee & \neg & (\neg P \vee Q) \\
    \hline
    T & T & T & \bold{T} & F & T & T & \bold{T} & F & T \\
    T & F & F & \bold{T} & T & F & F & \bold{T} & T & F \\
    F & T & T & \bold{T} & F & T & T & \bold{T} & F & T \\
    F & F & T & \bold{T} & F & T & T & \bold{T} & F & T
    \end{array}[/tex]

    Additionally, any expression of the form [itex](A \vee \neg A)[/itex] (like the one on the left) is a tautology. Also the implication [itex](P \rightarrow Q)[/itex] is reducible to [itex](\neg P \vee Q)[/itex].

    --Elucidus
     
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