Basic boolean minimization

I am trying to simplify the following piece of boolean algebra so 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 i开发者_如何学Ct to :

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

Is this correct?

I want to get the best possible minimization.

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?

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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;

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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"

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

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

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for simplifying boolean expressions use karnaugh maps. i think it is very much useful if we less number of variables. but if we have more variables then we can follow methods because this method is not that preferable.