# Simplification of the expression using De-Morgan's rule

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

          =¬(AD)+¬(¬(B+¬C))


is the solution correct?

Many thanks!

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

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