1
$\begingroup$

Simplify the following expressions to the simplest expression using De Morgan’s theorem and Boolean algebra.

¬(A(¬(B+¬C))D)=¬(AD(¬(B+¬C))

          =¬(AD)+¬(¬(B+¬C))
          =¬(AD)+(B+¬C)
          =¬(AD)+B+¬C 

is the solution correct?

Many thanks!

$\endgroup$
  • $\begingroup$ Not clear... Maybe you have to show that LHS=RHS... You can go one step further : $¬(AD)=¬A+¬D$ $\endgroup$ – Mauro ALLEGRANZA Apr 2 at 14:27
  • $\begingroup$ Actually I'm suppose to simplify the expression i.e. ¬(A(¬(B+¬C)D)) $\endgroup$ – Jarvis Ferns Apr 2 at 14:34
  • $\begingroup$ If so, you need one step only : $¬(A(¬(B+¬C)D)) = ¬A + (B+¬C) + ¬D$. $\endgroup$ – Mauro ALLEGRANZA Apr 2 at 14:37
  • $\begingroup$ Actually the expression is ¬(A(¬(B+¬C))D).sorry about the mistake $\endgroup$ – Jarvis Ferns Apr 2 at 14:42
2
$\begingroup$

I am reading this as:

$\neg (A \land \neg (B \lor \neg C) \land D)$

which by DeMorgan would be:

$\neg A \lor (B \lor \neg C) \lor \neg D$

i.e.

$\neg A + B + \neg C + \neg D$

$\endgroup$
  • $\begingroup$ Thanks a lot!This was very helpful! $\endgroup$ – Jarvis Ferns Apr 2 at 16:21

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.