Use basic logical equivalences to verify that: (p∧q∧r)∨(¬p∧q∧r)=(q∧r).
ie. associativity, commutativity, distributivity, etc.
I know its probably something really simple but I just don't see it...
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 $\displaystyle (p \wedge q \wedge r)\vee (\neg p \wedge q \wedge r)\equiv q \wedge r$
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
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)]
=[(pV¬p)∧(pVq)∧(pVr)]∧[(qV∧¬p)∧(qVq)∧(qVr)]∧[(rV¬p)∧(rVq)∧(rVr)]
=[T∧(pVq)∧(pVr)]∧[(qV¬p)∧q∧(qVr)]∧[(rV¬p)∧(rVq)∧r]
=(pVq)∧(pVr)∧(qV¬p)∧q∧(qVr)∧(rV¬p)∧(rVq)∧r
=[(qVp)∧(qV¬p)]∧[(rVp)∧(rV¬p)]∧[(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)∧(qV¬p)]=q.
Why make is look so hard?
$\displaystyle \begin{array}{rc}
\equiv & {\left[ {p \wedge \left( {q \wedge r} \right)} \right] \vee \left[ {\neg p \wedge \left( {q \wedge r} \right)} \right]} \\
\equiv & {\left[ {p \vee \neg p} \right] \wedge \left( {q \wedge r} \right)} \\
\equiv & {T \wedge \left( {q \wedge r} \right)} \\
\equiv & {\left( {q \wedge r} \right)} \\
\end{array} $