How to convert expression to its NOR form The expression I am working on is $xy'+x'y$.
Is this the correct conversion to NOR?
$$(x'+y)'+(x+y')'$$
Just adding for clarification
the question is in an old logic notation ( think it was used by Lewis Carroll)


*

*the $'$ stands for negation

*the $+$ stands for or 

 A: By "Nor form" I'm assuming you mean create a circuit of only variable inputs, their negation, and "Nor" gates.  This is done sometimes with CMOS circuitry because of its simple low power nature.
First notice that if you have an expression of the form
$$(A \text{ or } B \text{ or } C...) \text{ and } (D \text{ or } E \text{ or } F...) \text{ or } ...$$
That it is equivalent to the expression:
$$(A \text{ nor } B \text{ nor } C...) \text{ nor } (D \text{ nor } E \text{ nor } F...) \text{ nor } ...$$
This can be seen using either demorgans, or just notice that the nots after the ors cancel with the nots before the ands.
The first expression above is called "conjunctive normal form".  So the first thing you want to do is convert your expression to conjunctive normal form.  There are general algorithms for doing this, but your XOR has an easy CNF:
$$x\bar y + \bar x y = x \text{ xor } y = (x \text{ or } y) \text{ and } (\bar x \text{ or } \bar y)$$
Then converting to NOR form is straightforward:
$$(x \text{ or } y) \text{ and } (\bar x \text{ or } \bar y) = (x \text{ nor } y) \text{ nor } (\bar x \text{ nor } \bar y)$$
The answer you've given above is actually the "or" of 2 "nor" gates (or the nand of 2 or gates).  While your answer is logically equivalent to your question, it isn't in a form directly made up of only nor gates.
A: If you want to see if you are right, write a truth table, with $L$ being left and $R$ being right side of your equation:
$$
x=0,y=0 \Rightarrow L=0, R=0\\
x=0,y=1 \Rightarrow L=1, R=1\\
\text{etc}
$$
Alternatively, take the complement of R, expand it, then take its complement.
