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Math Help - is this logical expression equivalent?

  1. #1
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    is this logical expression equivalent?

    p + q = (pVq) ^ -(p^q)
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by EquinoX View Post
    p + q = (pVq) ^ -(p^q)
    what does the + mean? have you tried doing a truth table?
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    it means an exclusive or
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    Quote Originally Posted by EquinoX View Post
    it means an exclusive or
    it is equivalent, wikipedia actually has this as an example. i figured you wanted to show it though. i'd use a truth table. do you know how to construct one?
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    yes I do know how, actually my question is to show

    -(p+q) and p biimplication q are logically equivalent, and I am trying to find out how
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    Quote Originally Posted by EquinoX View Post
    yes I do know how, actually my question is to show

    -(p+q) and p biimplication q are logically equivalent, and I am trying to find out how
    Here's how to show the first one, maybe you can figure out from this how to show the second. Truth tables are the conventional way to test whether two statements are logically equivalent. Look at the truth table below. You will notice that the blue columns are the same. One represents P + Q and the other represents (P v Q) ^ ~(P ^ Q). if two statements have the same truth values in every instant in a truth table, then they are logically equivalent
    Attached Thumbnails Attached Thumbnails is this logical expression equivalent?-truthtable.gif  
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  7. #7
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    my teacher asks me not to use the truth table this time but to change the expression into other expression
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  8. #8
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    Recall that  p \Leftrightarrow q\quad  \equiv \quad \left( { \sim p \wedge  \sim q} \right) \vee \left( {p \wedge q} \right).

    Thus
     \begin{array}{rcl}<br />
  \sim \left( {p + q} \right) & \equiv &  \sim \left[ {\left( {p \vee q} \right) \wedge  \sim \left( {p \wedge q} \right)} \right] \\ <br />
  & \equiv & \left[ { \sim \left( {p \vee q} \right) \vee  \left( {p \wedge q} \right)} \right] \\ <br />
\end{array}
     \begin{array}{rcl}<br />
 \sim \left( {p + q} \right) & \equiv & \left[ {\left( { \sim p \wedge  \sim q} \right) \vee \left( {p \wedge q} \right)} \right] \\ <br />
  \; & \equiv & p \Leftrightarrow q \\ <br />
 \end{array}.
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    how did you change the sign of ^ into V in proofing the - (p + q)
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by EquinoX View Post
    how did you change the sign of ^ into V in proofing the - (p + q)
    be specific as to what line you are talking about
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  11. #11
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    sorry it's posted below, I dont understand on how to get the ^ symbol changes into v
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  12. #12
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    Thus
     \begin{array}{rcl}<br />
  \sim \left( {p + q} \right) & \equiv &  \sim \left[ {\left( {p \vee q} \right) \wedge  \sim \left( {p \wedge q} \right)} \right] \\ <br />
  & \equiv & \left[ { \sim \left( {p \vee q} \right) \vee  \left( {p \wedge q} \right)} \right] \\ <br />
\end{array}
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  13. #13
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    <br />
\begin{array}{l}<br />
  \sim \left( {A \wedge B} \right) \equiv \left[ { \sim A \vee  \sim B} \right] \\ <br />
  \sim \left( {A \vee B} \right) \equiv \left[ { \sim A \wedge  \sim B} \right] \\ <br />
 \end{array}<br />
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  14. #14
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    thanks that clears me up
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