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  • 1 Post By anakhronizein
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September 10th, 2014, 08:31 PM   #1
Joined: Feb 2014

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Simplifying Logics Help

¬ (p ^ (q ∨ r) ^ ((p ^ q) > r)) Simplify >
¬ (p ^ (q ∨ r) ^ (¬ (p ^ q) ∨ r)) DE Morgan’s Law
¬p ∨ ¬ (q ∨ r) ∨ ¬ (¬ (p ^ q) ∨ r) DE Morgan’s Law
¬p ∨ ¬q ^ ¬r ∨ (p ^ q) ^ ¬r

So I've reached this last step and I'm not sure how to simplify this further. The only thing I see is a Distributive Property of (p ^ q) ^ ¬r , but I'm not sure if this is allowed. Any help would be great.
extreme112 is offline  
September 11th, 2014, 02:10 PM   #2
Joined: Feb 2013

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How I would tackle this problem is to turn the original phrase into a truth table. Then from that I get the DNF in terms of when the proposition is false. That form is:

\neg (p \wedge q \wedge r) \vee \neg (p \wedge \neg q \wedge r)

Which you can then distribute the negation through the parentheses, then after that doing the distributive law for the following:

\equiv (\neg p \vee \neg q \vee \neg r) \wedge (\neg p \vee q \vee \neg r)

Which give you a long messy expansion. But you can simplify stuff like:

$\displaystyle \neg p \wedge \neg p \equiv \neg p$, or
$\displaystyle \neg q \wedge q \equiv F$.

Or further on you might see the following thing:

$\displaystyle (\neg r \wedge \neg q) \vee (\neg r \wedge q) $
$\displaystyle \equiv \neg r \wedge (\neg q \vee q)$
which we note very naturally that
$\displaystyle \equiv \neg r \wedge T$
$\displaystyle \equiv \neg r$

And we can get rid of any repetitions, hence you end up with the simplified DNF of

$\displaystyle \neg p \vee \neg r$

Of course you could probably do that on your version but this is just another route. You can come up with your own way as they often do in mathematics.
Thanks from extreme112
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