Is this a contradicting definition for the symmetrical difference $\Delta$? The book I'm studying with has the following definition of the symmetrical difference:
$$ M_1\Delta M_2 = \{ x | ( x\in M_1 \lor x\in M_2 ) \land \lnot(x\in M_1 \land x\in M_2)\} $$
However, when I try to expand the the negated conjunction in the latter half of this definition, I arrive at the following contradictory defintion:
$$ M_1\Delta M_2 = \{ x | ( x\in M_1 \lor x\in M_2 ) \land (x\notin M_1 \lor x\notin M_2)\} $$
So, is that definition simply wrong or am I not even supposed to expand parts of definitions for some reason? Thanks!
 A: Note that $x \in M_1 \vee x \in M_2$ and $x \notin M_1 \vee x \notin M_2$ are not actually contradictory. In particular, if $x \in M_1$ and $x \notin M_2$, then $x \in M_1 \vee x \in M_2$ is true and $x \notin M_1 \vee x \notin M_2$ is also true (and similarly for $x \notin M_1$ and $x \in M_2$).
In general, $x \in M_1 \vee x \in M_2$ is true whenever $x$ is in $M_1$ or $M_2$ and $x \notin M_1 \vee x \notin M_2$ is true whenever $x$ is not contained in one of $M_1$ or $M_2$.
So, pretty much, everything is ok!
A: You distributed the symbol $\lnot$ correctly. In fact, you can go further and write 
$$ 
\begin{align*}
M_1\Delta M_2 &= \{ x : (( x\in M_1 \lor x\in M_2 ) \land x\notin M_1) 
\lor ( ( x\in M_1 \lor x\in M_2 ) \land x\notin M_2)\} \\ 
&= \{ x : x\in (M_2\setminus M_1)  
\lor x\in (M_1 \setminus M_2) \}. 
\end{align*} 
$$
I added the parentheses in the second equality to make the sets more clear. Furthermore, you can draw a Venn diagram to have a visual understanding of the symmetric difference. 
A: There is no contradiction in 
Person P  (2) is either a man or a woman and (2) person P either is not a man OR is not a woman. 
That is : person P (1) belongs to the union of the sets M and W, but (2) does not belong to the intersection of these two sets. 
Or, if you prefer : (1) there is at least one set to which person P belongs, AND (2) there is at least one set to which person P does not belong. 
The fact that the  formula  : (M OR W) & ( ~M OR ~W) is not a contradiction is shown by the fact that it's truth value is not " false" in all possible cases. It is false only when M and W are both false ( which makes " M OR W" false) , or both true ( which makes the secund conjunct false). 

These 2 cases of falsehood correspond exactly to the 2 cases where an object does not belong to the symmetric diffference of two sets : $x\notin A\Delta B$ if and only if : 
(1) $x\notin A$ AND $\notin B$ 
OR 
(2) $x\in A$ & $\in B$. 
A: The definition simply says: $x$ is in $M_1$ or $M_2$ but not in both at the same time. So it's $(M_1 \cup M2) \setminus (M_1 \cap M_2)$ and can also be written as $(M_1 \setminus M_2) \cup (M_2 \setminus M_1)$, which consists of two disjoint areas in a Venn diagram, as can be seen here, e.g.
