How Can You Simplify This Boolean Equation Further?

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SUMMARY

The Boolean equation X = A\overline{D}+\overline{B}AC+\overline{B}\overline{D}\overline{C}+BA\overline{C} can be simplified using a Karnaugh map. The final simplified expression is A(B XOR C + D') + B'C'D', which effectively reduces the number of gate inputs. Techniques such as De Morgan's law, double negation, and the identification of tautologies are essential for this simplification process.

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  • Understanding of Boolean algebra
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  • Basic concepts of logic gates and their functions
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Homework Statement



X = A\overline{D}+\overline{B}AC+\overline{B}\overline{D}\overline{C}+BA\overline{C}

Homework Equations



I need to try to simplify or reduce that boolean equation to something as small as possible.

The Attempt at a Solution



It used to be worse and I used a karnaugh map to get to where I am now from a truth table. I am having trouble getting anything better. I can reduce it to get an XOR gate, but I think it can do better. Any tips?

Thanks
 
Last edited:
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de morgan's law, double negation, tautologies, and logic properties
http://www.facstaff.bucknell.edu/mastascu/eLessonsHTML/Logic/Logic2.html
 
Last edited:
You can reduce gate inputs by collecting terms, but this is as simplified as it gets. Draw the k-map. A(B xor C + D') + B'C'D'
 
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