# Use basic logical equivalences to show?

• Oct 21st 2009, 05:55 PM
Mitchell
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...
• Oct 21st 2009, 09:38 PM
tonio
Quote:

Originally Posted by Mitchell
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
• Oct 22nd 2009, 10:51 AM
Mitchell
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).
• Oct 22nd 2009, 11:39 AM
tonio
Quote:

Originally Posted by Mitchell
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
• Oct 25th 2009, 07:04 AM
oldguynewstudent
Quote:

Originally Posted by Mitchell
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)]

=[(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.
• Oct 25th 2009, 03:43 PM
Plato
Quote:

Originally Posted by Mitchell
Use basic logical equivalences to verify that: (pqr)∨(¬pqr)=(qr).

Why make is look so hard?
$\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}$
• Oct 25th 2009, 04:14 PM
oldguynewstudent
Quote:

Originally Posted by Plato
Why make is look so hard?
$\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}$

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!