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Is there any way to simplify a combination of XOR and XNOR gates in the following expression? I have tried multiple boolean theorems and I have not been able to simplify this any further:

The simplified version is XOR(A,B)*XNOR(C,D) + XOR(C,D)*XNOR(A,B)

The actual expression is: (A'B + AB')(CD + C'D') + (AB + A'B')(C'D + CD')

According to my lab, this expression can be simplified, but I don't know where to start.

I guess you could look at each element, such as (A'B + AB') as a single variable, such as X and then the equation would be XY' + X'Y, but I still don't see how to simplify that.

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3 Answers 3

up vote 3 down vote accepted

You're on the right track. Note that $X\bar Y + \bar X Y$ is exactly $X\oplus Y$ (where $\oplus$ is XOR), so you can simplify to $$ (A\oplus B) \oplus (C\oplus D)$$ Since $\oplus$ is associative and commutative, this simplifies further to $$ A\oplus B\oplus C\oplus D $$ which is particularly nice because it's a linear function of the inputs in the associated Boolean ring (whose multiplication is conjunction and addition is $\oplus$).

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@miracle173: Good catch. I have no idea where I got the idea that $Y$ would be $\overline{C\oplus D}$ rather than just $C\oplus D$. –  Henning Makholm Sep 20 '13 at 21:43

The definition (xor, xnor) of XOR(A,B)is A'B + AB' and of XNOR(C,D) is CD+C'D'. So XOR(A,B)*XNOR(C,D) + XOR(C,D)*XNOR(A,B) is a direct translation of (A'B + AB')(CD + C'D') + (AB + A'B')(C'D + CD').

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In plain boolean algebra: XOR = (a || b) && !(a && b)

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