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Math Help - logic: expressing "or" in terms of "implies not"

  1. #1
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    Question logic: expressing "or" in terms of "implies not"

    Hi,

    I have to express "p | q" (p or q) by using only =>~ (implies not)

    I posted this yesterday night, this morning I looked at it again and I came up with a solution:

    Let ->~ be *

    p|q is equivalent to (( ((P*Q)*P) * ( (P*Q)*Q))*(P*Q) ) * (((p*q)*(p*q))*q)

    My first approach was using truth tables, finding new ones and trying to combine them to get the result I am looking for. This worked in the end, I was able to find a working combination with those truth tables I created, but there must be better ways to solve this kind of problems and to make sure one gets as simple a solution as possible, mine seems too long to be the simplest one.

    Anyone knows the right way to solve this?


    Thank you
    Last edited by Roboris; September 28th 2007 at 02:55 AM.
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  2. #2
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    It is very hard to follow what you have written.
    However, here are some standard equivalencies.

    \begin{array}{l}<br />
 \left[ {P \Rightarrow Q} \right] \Leftrightarrow \left[ {\neg P \vee Q} \right] \\ <br />
 \left[ { P \vee Q} \right] \Leftrightarrow \left[ {\neg P \Rightarrow Q} \right] \Leftrightarrow \left[ {\neg Q \Rightarrow P} \right] \\ <br />
 \end{array}
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  3. #3
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    If I understand this right, P*P has the same truth table as \neg P. So P\vee Q is equivalent to (P*P)*(Q*Q).
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  4. #4
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    Quote Originally Posted by Opalg View Post
    If I understand this right, P*P has the same truth table as \neg P. So P\vee Q is equivalent to (P*P)*(Q*Q).
    Yes, you are totally right, I even had \neg P to express in terms of *. Now I see for sure that my answer, even though it works, was way too long.

    Thank you
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