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Math Help - Use basic logical equivalences to show?

  1. #1
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    Use basic logical equivalences to show?

    Use basic logical equivalences to verify that: (pqr)∨(pqr)=(qr).

    ie. associativity, commutativity, distributivity, etc.
    I know its probably something really simple but I just don't see it...
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  2. #2
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    Quote Originally Posted by Mitchell View Post
    Use basic logical equivalences to verify that: (pqr)∨(pqr)=(qr).

    ie. associativity, commutativity, distributivity, etc.
    I know its probably something really simple but I just don't see it...

    As you can see it's impossible to understand what you meant. Please check with "preview post" before you submit anything.

    Tonio
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  3. #3
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    I'm not sure I understand what's unclear. I want to prove that (p and q and r) or ((not p) and q and r) = q and r:

    (pqr)∨(pqr)=(qr).
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    Quote Originally Posted by Mitchell View Post
    I'm not sure I understand what's unclear. I want to prove that (p and q and r) or ((not p) and q and r) = q and r:

    (pqr)∨(pqr)=(qr).

    I wonder whether it is only my puter: between p and q and r only appear little blank squares, not one single logical connective!

    Anyway you already wrote it: you want to show that (p \wedge q \wedge r)\vee (\neg p \wedge q \wedge r)\equiv q \wedge r

    Ok, so why don't you build a truth-false table for the left side, another one for the right side, and show that for any possible values you assign to the variables you get the same truth value?

    Tonio
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  5. #5
    Member oldguynewstudent's Avatar
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    Quote Originally Posted by Mitchell View Post
    I'm not sure I understand what's unclear. I want to prove that (p and q and r) or ((not p) and q and r) = q and r:

    (pqr)∨(pqr)=(qr).
    This is rather tedious and I had trouble with something similar on my first quiz this year:

    (p∧q∧r)V(p∧q∧r)=[pV(p∧q∧r)]∧[qV(p∧q∧r)]∧[rV(p∧q∧r)]

    =[(pVp)∧(pVq)∧(pVr)]∧[(qV∧p)∧(qVq)∧(qVr)]∧[(rVp)∧(rVq)∧(rVr)]

    =[T∧(pVq)∧(pVr)]∧[(qVp)∧q∧(qVr)]∧[(rVp)∧(rVq)∧r]

    =(pVq)∧(pVr)∧(qVp)∧q∧(qVr)∧(rVp)∧(rVq)∧r

    =[(qVp)∧(qVp)]∧[(rVp)∧(rVp)]∧[(qVr)∧(qVr)]∧q∧r

    =[q]∧[r]∧[qVr]∧q∧r =

    Hope you can finish it from here, the main trick in the previous step was that [(qVp)∧(qVp)]=q.
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  6. #6
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    Quote Originally Posted by Mitchell View Post
    Use basic logical equivalences to verify that: (pqr)∨(pqr)=(qr).
    Why make is look so hard?
    \begin{array}{rc}<br />
    \equiv  & {\left[ {p \wedge \left( {q \wedge r} \right)} \right] \vee \left[ {\neg p \wedge \left( {q \wedge r} \right)} \right]}  \\<br />
    \equiv  & {\left[ {p \vee \neg p} \right] \wedge \left( {q \wedge r} \right)}  \\<br />
    \equiv  & {T \wedge \left( {q \wedge r} \right)}  \\<br />
    \equiv  & {\left( {q \wedge r} \right)}  \\<br /> <br />
 \end{array}
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  7. #7
    Member oldguynewstudent's Avatar
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    Quote Originally Posted by Plato View Post
    Why make is look so hard?
    \begin{array}{rc}<br />
\equiv & {\left[ {p \wedge \left( {q \wedge r} \right)} \right] \vee \left[ {\neg p \wedge \left( {q \wedge r} \right)} \right]} \\<br />
\equiv & {\left[ {p \vee \neg p} \right] \wedge \left( {q \wedge r} \right)} \\<br />
\equiv & {T \wedge \left( {q \wedge r} \right)} \\<br />
\equiv & {\left( {q \wedge r} \right)} \\<br /> <br />
\end{array}
    WOW! Why didn't I think of that? Because I have a very poor instructor. My whole class is having immense problems. Thanks for the tip, I made need it on my final!
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