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 it 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?
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;