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Math Help - Please check my Truth Table 2!

  1. #1
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    SOLVED!! Please check my Truth Table 2!

    ( P AND Q ) AND ( ~ P OR ~ Q ) AND ( P AND Q ) AND ( ~ P OR Q )

    Is this a contradiction?

    Work: Well, lol, I got it is a contradiction. There is always one false in each row. So I assume it is a contradiction.
    Last edited by aeubz; July 15th 2009 at 10:25 PM.
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  2. #2
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    Quote Originally Posted by aeubz View Post
    ( P AND Q ) AND ( ~ P OR ~ Q ) AND ( P AND Q ) AND ( ~ P OR Q )
    Is this a contradiction?
    I think that you mised a 'not'.
    ]( P AND Q ) AND ( ~ P OR ~ Q ) AND ( P AND ~ Q ) AND ( ~ P OR Q )
    Did you?
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  3. #3
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    nope its typed correctly
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  4. #4
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    If this is it: ( P AND Q ) AND ( ~ P OR ~ Q ) AND ( P AND Q ) AND ( ~ P OR Q )

    Then drop a redundancy ( ~ P OR ~ Q ) AND ( P AND Q ) AND ( ~ P OR Q )
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  5. #5
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    it is a contradiction, is it not?
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  6. #6
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    Hello, aeubz!

    If it is typed correctly, it should read: . (P \wedge Q) \wedge (\sim\!P \:\vee \sim\!Q) \wedge (\sim\!P \vee Q)

    And we need a few theorems:
    . . \begin{array}{cc}(1) & R \:\wedge \sim\!R \:=\:f \\ (2) & R \vee f \:=\:R \\ (3) & R \wedge f \:=\:f \end{array}


    I would prove it like this:

    \begin{array}{ccc}(P \wedge Q) \wedge (\sim\!P\:\vee \sim\!Q) \wedge (\sim\!P \vee Q) & \text{given} \\ <br />
(P \wedge Q) \wedge \left[\sim\!P \vee (\sim\!Q \wedge Q)\right] & \text{distr.} \\ <br /> <br />
(P \wedge Q) \wedge \left[\sim\!P \vee f\right] & (1) \\ <br />
(P \wedge Q) \:\wedge \sim\!P & (2) \\ <br />
(P \:\wedge \sim\!P) \wedge Q & \text{comm/assoc.} \\ <br /> <br />
f \wedge P & (1) \\ <br />
f & (3)<br />
\end{array}


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  7. #7
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    TY All for your responses !! it is contradiction.. by what Soroban wrote. even though i did it the long way by using truth tables but his way is much neater and much better
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