Simplify this boolean expression Z=(AB+C)(B'D+C'E')+(AB+C)'

Question

I need help on how to simplify the following expression:

Z=(AB+C)(B'D+C'E')+(AB+C)' I find this example to be very difficult, for other examples I used many rules of boolean algebra such as DeMorgan's Theorem, but I don't any thing about it for this expression (AB+C)(B'D+C'E')+(AB+C)' it's a big problem to me, perhaps it cannot be simplified?

Answer 1

It's almost like algebra for numerical expressions; just work things out, and then you can usually get rid of or simplify some of the terms:

$$(AB+C)(B'D+C'E')+(AB+C)'= \text{ (DeMorgan)}$$

$$(AB+C)(B'D+C'E')+(AB)'C'= \text{ (DeMorgan)}$$

$$(AB+C)(B'D+C'E')+(A'+B')C'= \text{ (Distribution)}$$

$$ABB'D+ABC'E'+B'CD+CC'D+A'C'+B'C'= \text{ (Complement)}$$

$$0+ABC'E'+B'CD+0+A'C'+B'C'= \text{ (Identity)}$$

$$ABC'E'+B'CD+A'C'+B'C'= \text{ (Distribution)}$$

$$(ABE'+A')C'+(CD+C')B'= \text{ (Reduction)}$$

$$(BE'+A')C'+(D+C')B'= \text{ (Distribution)}$$

$$BC'E'+A'C'+B'D+B'C'= \text{ (Distribution)}$$

$$(BE'+B')C'+A'C'+BD'= \text{ (Reduction)}$$

$$(E'+B')C'+A'C'+BD'= \text{ (Distribution)}$$

$$E'C'+B'C'+A'C'+BD'$$

... and that's about as simple as you can get, unless you find that combining the first three terms to $(ABE+C)'$ makes it 'simpler'. But the result above is in DNF, which is what most books are looking for.

Answer 2

Lets deal with the term $(AB+C)'$ first ... two aplications of DeMorgan's Law gives \begin{eqnarray*} (AB+C)'=(AB)^{'} C^{'} = (A'+B')C'=A'C'+B'C' \end{eqnarray*} Now use distributivity & $BB'=0$ we have \begin{eqnarray*} Z=(AB+C)(B'D+C'E')+A'C'+B'C'=ABB'D+ABC'E'+CB'D+CC'E+A'C'+B'C'=\color{red}{ABC'E'+CB'D+A'C'+B'C'}. \end{eqnarray*}

EDIT: ((A && B ) ||C) && ((not(B)&&D) || (not(C) && not(E))) || not((A && B ) ||C) , https://www.wolframalpha.com/input/?i=((A+%26%26+B+)+%7C%7CC)+%26%26+((not(B)%26%26D)+%7C%7C+(not(C)+%26%26+not(E)))+%7C%7C+not((A+%26%26+B+)+%7C%7CC)

(A && B && not(C) && not(E)) || (C && not(B) && D) || (not(A) && (not(C)) || (not(B) && not(C)) https://www.wolframalpha.com/input/?i=(A+%26%26+B+%26%26+not(C)+%26%26+not(E))+%7C%7C+(C+%26%26+not(B)+%26%26+D)+%7C%7C+(not(A)+%26%26+(not(C))+%7C%7C+(not(B)+%26%26+not(C))

Answer 3

The given formula has the form $\alpha \beta + \alpha'$, which immediately simplifies to $\beta + \alpha'$ or $B'D + C'E' + (AB+C)'$.