Show that $P$ is prime iff the following holds: For any $A\in \mathcal P\left(S\right)$ either $A\in P$ or $A^{c}=S\setminus A\in P$ but not both. 
Let $S$ be an infinite set and consider the ring $\left\langle  \mathcal P\left(S\right),\triangle,\cap\right\rangle $.
(a) Show that the collection $I$ of finite subsets of $S$ is an ideal of $ \mathcal P\left(S\right)$.

Proof. We show that $A\triangle B\in I$ and $A\cap B\in I$ for $A,B\in I$. Let $A,B\in I$ then $A$ and $B$ are finite. $A\triangle B=\left(A\backslash B\right)\cup\left(B\backslash A\right)$.The union of finite subsets is also finite, hence, $A\triangle B\in I$. The intersection of finite group $B$ with any group $A$ is finite, hence, $A\cap B\in I$. Thus, $I$ is an ideal of $ \mathcal P\left(S\right)$.

(b) Let $P$ be an ideal of $ \mathcal P\left(S\right)$. Show that $P$ is prime iff the following holds: For any $A\in  \mathcal P\left(S\right)$ either $A\in P$ or $A^{c}=S\setminus A\in P$ but not both.

Proof. $\left(\implies\right)$ Suppose $P$ is prime. For $A\cap B\in P$, then $A\in P$ or $B\in P$. If $A\in P$, then we are done. If $A\not\in P$ then $B\in P$. Moreover, $B\subseteq S-A$ .......,,...i got nothing
In $a$, is $A\triangle B\in I$ sufficient? From what I have read, to show that $I$ is an ideal $A+B,A-B\in I$ (in different notation). How can I write the "$-$" in the operation given?
In $b$, that's all I have. How about the converse?
 A: First, as a short note, your proof of (a) is a bit off - it seems like you properly say this later, but at the start you state that you want to show $A\Delta B \in I$ for $A,B\in I$ and $A\cap B\in I$ for $A,B\in I$ - but you actually want the stronger condition that $A\cap B\in I$ whenever $A\in I$ and $B\in P(S)$ - and you also use the word "group" to refer to sets, which is really not a good idea in this context.
It is not necessary to handle differences because if you know an ideal is closed under sums and multiplication by arbitrary elements, you know that if $a,b\in I$ then $(-1)\cdot b \in I$ and therefore $a+(-b)=a-b \in I$. Also, in this ring, addition and subtraction are the same operation because $1=-1$ (they're both the whole set $S$) - if you wanted to show that something is a subring, you would check that it contains $0$ (and also $1$ if you require rings to be unital), is closed under addition, multiplication, and negation, but the stronger condition of ideals lets one accomplish this in fewer steps.
For (b), your proof doesn't really work out, but it's close; you say "For $A\cap B\in P$" but this doesn't really mean anything because neither $A$ nor $B$ is fixed - and then you later say $B\subseteq S-A$ without ever assuming anything of $B$. It's good to consider what happens when $A\cap B$ is in $P$, since that all you have to work with, but it's not useful to write that as the first step in a proof! That proof should definitely begin something like:

Let $P$ be a prime ideal of $P(S)$ and $A$ be an arbitrary element of $P(S)$. We wish to show that either $A\in P$ or $S-A \in P$.

This makes it very clear what you are given and what variable ($A$) is fixed. Then, you should make an advantageous choice of $B$ to make use of the expression $A\cap B$ - there's a hint in your proof that setting $B=S-A$ could work. So, your proof could proceed:

Consider the expression $A\cap (S-A)$. This intersection is empty, so is the additive identity of $P(S)$ and thus must be in $P$. However, since $P$ is prime, either $A\in P$ or $S-A\in P$.

For the other direction, you can start with your given:

Let $P$ be any ideal such that for every set $A\in P(S)$ either $A\in P$ or $S-A\in P$. We wish to show that for any $B,C\in P(S)$ if $B\cap C \in P$ then either $B\in P$ or $C\in P$.

Then, it's a bit more difficult, but we can just use our givens. It's fairly typical to prove this sort of statement by noting that if $B\in P$, we are done - so we can proceed as follows:

If $B\in P$, we are done. Otherwise, assume $B\not\in P$. We wish to show $C\in P$.

Then, we should spot that $B\not\in P$ tells us, by our condition that $S-B$ is in $P$. However, then we know that $B\cap C$ and $S-B$ are both in $P$ and must show that $C$ is - but $C$ is a linear combination of $B\cap C$ and $S-A$ by $C = (B\cap C) \Delta ((S-B)\cap C)$ so we can finish:

Note that $S-B$ is in $P$. Since $C = (B\cap C) \Delta ((S-B)\cap C)$ and $P$ is an ideal, this implies that $C\in P$.

Lastly, you should just show a simple lemma to handle the "not both" part of this:

We also note that if there were any set $A$ such that $S-A$ and $A$ were both in an ideal $P$, then $P=P(S)$ since $A\Delta (S-A) = S = 1_{P(S)}$ and conversely that if $P$ were the whole ring, any set $A$ would have that both $A$ and $S-A$ were in $P$.

(Though, it's worth noting that some authors admit the whole ring as a prime ideal since, as we see here, the proof of $P$ being proper and the proof of $ab\in P\Rightarrow a\in P \text{ or }b\in P$ are largely unrelated)
A: Here are some hints:
First let's see that if $P$ satisfies the desired property, then it is prime. 
If $A \cap B \in P$, then:
If $A \in P$, we are done. 
If $A \not \in P$, then $A^c \in P$, and so $A \cap B$ and $A^c$ are both in $P$. Can you use this to show $B \in P$, keeping in mind $B$ is the disjoint union of $A^c$ and $A \cap B$?
For the other direction, consider $P$ prime, and notice $A$ and $A^c$ cannot both be in $P$, as otherwise $A + A^c = S \in P$ and so $P$ would be the whole ring. 
But now $A \cap A^c = \emptyset \in P$. Can you finish it from here?

Good luck! I hope this helps ^_^
