Can someone please help me with some basic boolean minimization ?

Question

Hey guys,

Sorry to be annoying, but I am doing a little bit of work at the moment, and am trying to simplify the following piece of boolean algebra so that I can construct the circuit :

A'.B'.C.D  +  A'.B.C.D'  +  A'.B.C.D  +  A.B'.C'.D +  A.B'.C.D  +  A.B.C'.D  +  A.B.C.D' + A.B.C.D 

So far I have gotten it to :

(C.D) + (B.C) + (A.C'.D)

Is this correct ?

...but I know there is more that I can do ! Can anyone help me out please ? I want to get the best possible minimization, I'm just finding it a little difficult to figure out.

The steps I have went through so far are :

A'.B'.C.D  +  A'.B.C.D'  +  A'.B.C.D  +  A+B'+C'+D +  A.B'+C+D  +  A.B.C'.D  +  A.B.C.D' + A.B.C.D 
= A.A'(B'.C.D)  +  A.A'(B.C.D')  +  A.A'(B.C.D)  +  B.B'(A.C'.D)
= (B.C.D) + (B'.C.D) + (B.C.D) + (B.C.D') + (A.C'.D)
= (C.D) + (B.C) + (A.C'.D)

Can I do any more ?

Thank you very much ! :)

Answer 1

The only thing I can see that you could possibly do is distribute the "C" in the left two terms:

(C).(B+D)+(A.C'.D)

Or you could distribute the "D":

(C+A.C').D + (B.C)

Response to Comment: The distributive law is described here: http://www.ee.surrey.ac.uk/Projects/Labview/boolalgebra/. See the information under heading "T3"

Answer 2

Assuming your equation is actually:

X = (A'.B'.C.D) + (A'.B.C.D') + (A'.B.C.D) + (A+B'+C'+D) + (A.B'+C+D) + (A.B.C'.D) + (A.B.C.D') + (A.B.C.D);

I just ran this through Logic Friday and it factored it down to:

X = 1;

So you might want to check your simplification work and/or check that you've given the correct equation.

However I suspect there may be typos in the original equation above, and perhaps it should be:

X = (A'.B'.C.D) + (A'.B.C.D') + (A'.B.C.D) + (A.B'.C'.D) + (A.B'.C.D) + (A.B.C'.D) + (A.B.C.D') + (A.B.C.D);

?

In which case Logic Friday simplifies it to:

X = B.C + A.D + C.D;

Answer 3

Here's another solution (found by brute force):

(a+c).(b+d).(c+d)