I have the following boolean expression that I want to simplify

$$B\cdot D+ \overline{A\cdot B\cdot D} + \overline{B}\cdot C\cdot \overline{D}$$

Here is what I have been able to due so far

$$B\cdot D+ \overline{A} + \overline{B} + \overline{D} + \overline{B} \cdot \overline{D}\cdot C$$

I know that the answer is suppose to be $$\overline{A} + \overline{B}\cdot B $$

How can I simplify my initial expression any further thank you very much any help.


Factorising $$\overline{A}+B\cdot D + \overline{B}+\overline{D} \cdot (1+\overline{B}\cdot C)$$ $$=\overline{A}+B\cdot D + \overline{B}+\overline{D} \cdot 1$$ $$=\overline{A}+B\cdot D + \overline{B}+\overline{D}$$ Then by the absorption law, $$B\cdot D + \overline{B}=D + \overline{B}$$ So this simplifies to $$=\overline{A}+D + \overline{B}+\overline{D}$$ $$=\overline{A} + \overline{B}+D+\overline{D}$$ $$=\overline{A} + \overline{B}+1$$ $$=1$$

  • $\begingroup$ I drew a Karnaugh map and all the squares were full so I agree with the answer of $1$. $\endgroup$ – poetasis Mar 22 '19 at 16:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.