# problem simplifying boolean algebra expression using consensus theorem

Please simplify this logic expression for me with helping boolean algebra :

A'C'D + A'BD + BCD + ABC + ACD'

I know that must use consensus theorem .

my solve :

STEP 1 : Terms 1 & 3 ---eliminate---> Term 2

STEP 2 : Terms 3 & 5 ---eliminate---> Term 4

STEP 3 : Terms 2 & 4 ---eliminate---> Term 3

But truth table said step 3 is incorrect . but why ?

please tell me why step 3 is Wrong ?

and tell me What is the simplest form of it ?

• Please refrain from telling us your question is urgent. It won't get you an answer any faster, and irritates many users because they feel you are demanding a rapid answer. – Alex Becker Nov 26 '13 at 21:39

I think you are confusing during the application of minimization by using Boolean algebra formula

First you apply consensus theorem for terms 1,2,3 by taking D as a common factor. This helps in removing the term 2. So the minimized expression is A'C'D + BCD + ABC + ACD'

Now combining the last three terms similarly leads to elimination of the term ABC

So the minimized expression is A'C'D + BCD + ACD'.

Here you don't have any terms for which consensus theorem can be applied i.e., so you question describes there were no more 2 and 4 terms to eliminate the term 3.

So the minimized expression is A'C'D + BCD + ACD'

given $$A'C'D+A'BD+BCD+ABC+ACD'$$ use distributive law $xyz+pqr=(xyz+p).(xyz+qr)$ $$\to (A'C'D+A'D).(A'C'D+B) +(BCD+BC).(BCD+A)+ACD'$$ $$\to A'C'D+B+BCD+A+ACD'$$ $$\to A'C'D+B(1+CD)+A(1+CD')$$ $$\to A'C'D+A+B$$ $$\to (A+A').(A+C'D) +B$$ $$\to A+B+C'D$$

• hi thank you so much . – mine wwe Nov 26 '13 at 16:33
• you are welcome. you can accept my answer by clicking the tick sign in left of my answer. – Suraj M S Nov 26 '13 at 16:36
• hi thank you so much . but the answer is false . please let a , c , d = 0 and b = 1 ----> if use your final function , f = 1 but ----> if use the original function the output is 0 (f=0) . – mine wwe Nov 26 '13 at 16:41

Using a truth table Let $X=A'C'D + A'BD + BCD + ABC + ACD'$. (We shall leave entries blank where their value is zero) \begin{eqnarray*} \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline A & B & C & D & & A'C'D & A'BD & BCD & ABC & ACD' & & X \\ \hline 0&0&0&0& & & & & & & & & \\ \hline 0&0&0&1& &1& & & & & &1& \\ \hline 0&0&1&0& & & & & & & & & \\ \hline 0&0&1&1& & & & & & & & & \\ \hline 0&1&0&0& & & & & & & & & \\ \hline 0&1&0&1& &1&1& & & & &1& \\ \hline 0&1&1&0& & & & & & & & & \\ \hline 0&1&1&1& & &1&1& & & &1& \\ \hline 1&0&0&0& & & & & & & & & \\ \hline 1&0&0&1& & & & & & & & & \\ \hline 1&0&1&0& & & & & &1& &1& \\ \hline 1&0&1&1& & & & & & & & & \\ \hline 1&1&0&0& & & & & & & & & \\ \hline 1&1&0&1& & & & & & & & & \\ \hline 1&1&1&0& & & & &1&1& &1& \\ \hline 1&1&1&1& & & &1&1& & &1& \\ \hline \end{array} \end{eqnarray*}

From the table above it is clear that we only need to "or" the first, third & fifth terms in order to construct $X$. So the minimal expression is $X=\color{red}{A'C'D + BCD + ACD'}$.