September 10th, 2014, 08:31 PM  #1 
Newbie Joined: Feb 2014 Posts: 15 Thanks: 0  Simplifying Logics Help
Â¬ (p ^ (q âˆ¨ r) ^ ((p ^ q) > r)) Simplify > Â¬ (p ^ (q âˆ¨ r) ^ (Â¬ (p ^ q) âˆ¨ r)) DE Morganâ€™s Law Â¬p âˆ¨ Â¬ (q âˆ¨ r) âˆ¨ Â¬ (Â¬ (p ^ q) âˆ¨ r) DE Morganâ€™s Law Â¬p âˆ¨ Â¬q ^ Â¬r âˆ¨ (p ^ q) ^ Â¬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) ^ Â¬r , but I'm not sure if this is allowed. Any help would be great. 
September 11th, 2014, 02:10 PM  #2 
Member Joined: Feb 2013 Posts: 80 Thanks: 8 
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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