Algebra - Prove this is a Ring Let $S$ be a nonempty set. Prove that the power set $P(S)$ whose elements are all subsets of $S$, forms ring under the following operations:
\begin{align*}
a + b &= (a \cup b) \setminus (a \cap b)\\
a \cdot b &= a \cap b
\end{align*}
For proving it's an abelian group first, I've gotten that the identity is $\emptyset$. 
\begin{align*}
      a + \emptyset
          &= (a \cup \emptyset) \setminus (a \cap \emptyset)\\
          &= a \setminus \emptyset\\
          &= a
     \end{align*}
Each element is it's own inverse. 
\begin{align*}
      a + a
        &= (a \cup a) \setminus (a \cap a)\\
        &= a \setminus a\\
        &= \emptyset
     \end{align*}
It's certainly commutative.
\begin{align*}
      a + b  
        &= (a \cup b) \setminus (a \cap b)\\ 
        &= (b \cup a) \setminus (b \cap a)\\
        &= b + a
     \end{align*}
But I can't get associative. That is, $(a + b) + c = a + (b + c)$. Anyone want to aid?
The properties for multiplication aren't too bad either. The only one that is throwingme a curve ball is the left/right distribution. That is, 
$a(b + c) = ab + ac$ and $(a + b)c = ac + bc$.
While this questions is asked elsewhere on this site and has answers to most parts in proving that it is a ring, it does not explain or show the associativity portion for addition in detail. That is $(a + b) + c = a + (b + c)$ is left very vague.
 A: It's much easier if you compute the truth table for $+$:
$$
\begin{array}{cc|c}
a & b & a+b \\
\hline
T & T & F \\
T & F & T \\
F & T & T \\
F & F & F
\end{array}
$$
Then the truth table for $(a+b)+c$ is
$$
\begin{array}{ccc|c}
a & b & c & a+b & (a+b)+c \\
\hline
T & T & T & F & T \\
T & T & F & F & F \\
T & F & T & T & F \\
T & F & F & T & T \\
F & T & T & T & F \\
F & T & F & T & T \\
F & F & T & F & T \\
F & F & F & F & F
\end{array}
$$
You can clearly see that a return of $T$ corresponds to an odd number of $T$’s in the data and $F$ to an even number of $T$. This is independent on the order $a$, $b$ and $c$ are considered in. Hence the truth table for $(b+c)+a$ is the same, which means
$$
(a+b)+c=(b+c)+a=a+(b+c)
$$
using the obvious commutativity of $+$.
You can do similarly for the distributive property:
$$
\begin{array}{ccc|c}
a & b & c & a+b & (a+b)c \\
\hline
T & T & T & F & F \\
T & T & F & F & F \\
T & F & T & T & T \\
T & F & F & T & F \\
F & T & T & T & T \\
F & T & F & T & F \\
F & F & T & F & F \\
F & F & F & F & F
\end{array}
$$
and computing the one for $ac+bc$.
A different strategy is to consider indicator functions. The indicator function for $a$ is the map $\chi_a\colon S\to\{0,1\}$ defined by $\chi_a(x)=1$ if $x\in a$, $\chi_a(x)=0$ if $x\notin a$. Two subsets of $S$ are equal if and only if they have the same indicator function.
If we use modulo $2$ arithmetic,  the indicator function of $a+b$ is easily seen to be $\chi_{a}+\chi_b$ (pointwise sum). Thus the indicator function for $(a+b)+c$ is $\chi_a+\chi_b+\chi_c$.
For the distributive property, observe that the indicator function of $ab$ is $\chi_a\chi_b$ (pointwise product).
A: Hint: $a+b=\ldots =a\backslash b\cup b\backslash a$
Second hint: $x\backslash y=x\cap y^c$
Now write out both $(a+b)+c$ and $a+(b+c)$ independently in terms of $\cap$, $\cup$ and compare.
A: Here is an answer using Damian's hints, if you want to go that route for some reason. We have:
\begin{align}a + b &= (a \cup b)\setminus(a\cap b) \\
&= (a\cup b) \cap (a\cap b)^c \\
&= (a\cup b) \cap (a^c \cup b^c) \\
&= (a\cap b^c) \cup (b\cap a^c) \\
&= (a\setminus b) \cup (b\setminus a)\end{align}
So, in excruciating detail, we see that:
\begin{align}
(a+b)+c &= ((a\setminus b) \cup (b\setminus a)) + c \\
&= [((a\setminus b)\cup (b\setminus a))\setminus c] \cup [c \setminus ((a\setminus b)\cup (b\setminus a))] \\
&= [((a\cap b^c)\cup (b\cap a^c))\cap c^c] \cup [c\cap ((a\cap b^c) \cup (b\cap a^c))^c] \\
&= [(a\cap b^c \cap c^c) \cup (b\cap a^c \cap c^c)]\cup [c\cap (a^c \cup b)\cap (b^c \cup a)] \\
&=(a\cap b^c \cap c^c) \cup (b\cap a^c\cap c^c) \cup (c\cap a^c \cap b^c) \cup (c\cap b \cap a) \\
&= (c\cap a^c \cap b^c) \cup (b\cap a^c \cap c^c) \cup (a \cap b^c \cap c^c) \cup (a\cap b \cap c)\\
&= [(c\cap a^c \cap b^c) \cup (b\cap a^c \cap c^c)]\cup [a \cap (c^c \cup b)\cap (b^c \cup c)] \\
&= [((c\cap b^c)\cup (b\cap c^c))\cap a^c] \cup [a\cap ((c^c\cap b)\cup (b^c\cap c))^c] \\
&= [((c\setminus b) \cup (b\setminus c))\setminus a]\cup [a\setminus ((c\setminus b) \cup (b\setminus c))] \\
&= a + ((c\setminus b) \cup (b\setminus c)) \\
&= a + (b + c)
\end{align}
For distribution property, we have, in less detail
\begin{align}
a(b + c) &= a\cap ((b\setminus c) \cup (c\setminus b))\\
&= ((a\cap b)\setminus c) \cup ((a\cap c)\setminus b) \\
&= ((a\cap b)\setminus (a\cap c))\cup ((a\cap c)\setminus (a\cap b)) \\
&= (a\cap b) + (a\cap c) \\
&= ab + ac
\end{align}
and similarly for the other way around.
