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