zemzela Posted March 20, 2012 Share Posted March 20, 2012 Could someone help me , how to simplify this expression not c xor (not b or c) Please help me Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted March 20, 2012 Share Posted March 20, 2012 Could you tell us how you've started, what you've tried, and what problems you've encountered? Link to comment Share on other sites More sharing options...
zemzela Posted March 20, 2012 Author Share Posted March 20, 2012 First I try to solve this with de Morgan's laws. For the first time I work with this laws, and I don't know if my result is true? c' xor (b' + c) And I don't know how to continue with solving the expression Link to comment Share on other sites More sharing options...
John Posted March 21, 2012 Share Posted March 21, 2012 (edited) What you might do first is express the xor in terms of and and or. Your problem is saying that either the left side is true or the right side is true, but not both, so [math]\neg c \oplus (\neg b \vee c)[/math] is the same as [math](\neg c \wedge \neg (\neg b \vee c)) \vee (\neg (\neg c) \wedge (\neg b \vee c))[/math]. You can then use DeMorgan's law on the [math]\neg (\neg b \vee c)[/math] part and go from there. Does that help at all? As for verifying your result, you can construct a truth table (if you've learned how to do so) and see whether the values for your result match the values for the original statement. In fact, constructing a truth table for the original statement will make it clear what the simplified statement will be. Edited March 21, 2012 by John Link to comment Share on other sites More sharing options...
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