Can this boolean algebra be reduced further? f=/cd+/ab+/a/c+a/bcd

 

Yes.

Double NOT the function (this doesn't change the function), so: f=//(/cd+/ab+/a/c+a/bcd).

Now think about DeMorgan's theorems.

Have a go and post up your efforts, I'll give you some feedback.

Dave

 

I did the boolean algebra again this time using k maps and realized I had made a mistake in reducing my equation. The reduced one now is /CD+/AB+/A/C+A/BD and I don't think this can be reduced further. Can it?

 

I did the boolean algebra again this time using k maps and realized I had made a mistake in reducing my equation. The reduced one now is /CD+/AB+/A/C+A/BD and I don't think this can be reduced further. Can it?

Sorry I missed this one. Feel free to PM to remind me in future.

Without your working out I cannot give you any feedback on your answer, but I simplified it to:

f = (ABCD)'

(Note I use ' to denote NOT/INVERTED)

I used the two techniques I told you about above. How?

f = (CD)' + (AB)' + (A'B') + (AB'CD)

Double NOT (this doesn't change the function):

f = ((CD)' + (AB)' + A'B' + AB'CD)''

Take out a common factor in the last two terms:

f = ((CD)' + (AB)' + B'( A' + ACD))''

Apply DeMorgan's theorem:

f = ((CD)''(AB)''(B'(A' + ACD))')'

(CD)'' = CD; (AB)'' = AB; so:

f = (CDAB(B'(A' + ACD))')'

Apply DeMorgan's theorem on (B'(A' + ACD))':

f = (CDAB(B'' + (A' + ACD)'))'

B'' = B

Apply DeMorgan's theorem on (A' + ACD)':

f = (CDAB(B + (A''(ACD)')))'

A'' = A

Apply DeMorgan's theorem on (ACD)':

f = (CDAB(B + (A(A' + C' + D'))))'

Multiply out (A(A' + C' + D'):

f = (CDAB(B + (AA' + AC' + AD')))'

AA' = 0

f = (CDAB(B + AC' + AD'))'

Multiply everything out:

f = (CDABB + CDABAC' + CDABAD')'

BB = B; CC' = 0; DD' = 0; therefore

f = (CDAB)' = (ABCD)'

Not easy eh?!

Dave