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



is the solution correct?

Many thanks!

  • $\begingroup$ 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

I am reading this as:

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

which by DeMorgan would be:

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


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

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

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