# How to simplify the Boolean function $A'B'C + A'BC' + ABC + AB'C'$?

So the question I have asks to implement the circuit with $XOR$ gates.

So I am 3/4 through the problem when I am having problems simplifying the Boolean expressions below:

$$A'B'C + A'BC' + ABC + AB'C'$$

According to the professor this can be simplified to $A \,\,XOR \,\,B\,\, XOR \,\,C$.

The next one: $$A'B'C'D + A'B'CD' + A'BC'D' + A'BCD + ABC'D + ABCD' + AB'C'D' + AB'CD$$

Can be simplified to: $A\,\,XOR \,\,B \,\,XOR\,\,C \,\,XOR\,\,D$. Again I have no idea how to simplify it to that.

I have not the slightest clue how to even get to that. I have tried many MANY methods I must be looking at this the wrong way. Can anyone help?

• What is your definition of XOR? If you agree that $A\,\,XOR\,\,B$ equals $AB'+A'B$ and if you know that $A\,\,XOR\,\,B\,\,XOR\,C$ is independent of the parentheses (order of operations), then you should be able to generalize this.. – Peter Franek Nov 8 '14 at 20:58
• I thought this was the way to go: = A'(B'C + BC') + A(BC + B'C') = A'(B XOR C) + A(BC + (B + C)') Which is where I ended up getting stuck. – George Nehme Nov 8 '14 at 21:00

$$A′B′C+A′BC′+ABC+AB′C′$$ $$=A^{'}(B^{'}C + BC^{'}) + A(BC + B^{'}C^{'})$$

Now

$X \oplus Y = XY^{'} + X^{'}Y$

$(X \oplus Y)^{'}= XY + X^{'}Y^{'}$

Thus

$$=A^{'}(B \oplus C) + A(B \oplus C)^{'}$$ $$=A \oplus B \oplus C$$

Note that $(\oplus)^{'}$ is the logical expression XNOR

This is very easy! If exclusive or is $$A\oplus B=\bar AB+A\bar B,~$$ then coincidence is $$A\odot B$$

$$=AB+\bar A\bar B.~$$ At the same time, the two operations are obviously opposite to one

another; i.e., if A is the same as B, then they coincide, and their XOR yields false;

and vice-versa. In other words, $$A\odot B=\overline{A\oplus B}.~$$ Now, factoring as you already

have, $$~\bar A\Big(B\oplus C\Big)+A\Big(B\odot C\Big)=\bar A\Big(B\oplus C\Big)+A\cdot\overline{B\oplus C}=A\oplus\Big(B\oplus C\Big)=$$

$$=A\oplus B\oplus C$$.