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September 10th, 2014, 08:31 PM   #1
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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.
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September 11th, 2014, 02:10 PM   #2
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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:

$\displaystyle
\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:

$\displaystyle
\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.
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