Probability Proof about A and B I have to formally prove that:
$$P(A) = P(A\wedge \neg B) + P(A\wedge B)$$
so I did like this:
$$P(A\wedge \neg B) + P(A\wedge B)$$
$$=P(A\wedge \neg B) + P(A)\cdot P(B)$$
$$=P(A)\cdot P(\neg B) + P(A)\cdot P(B)$$
$$=P(A)\cdot (1-P(B)) + P(A)\cdot P(B)$$
$$=P(A)-(P(A)\cdot P(B)) + (P(A)\cdot P(B))$$
$$=P(A)-0$$
$$=P(A)$$
Is this a valid formal proof? Did I commit any errors?
Edit: 
$$P(A\wedge \neg B) + P(A\wedge B)$$
$$= P((A\wedge \neg B) \lor (A\wedge B))$$
$$= P(A\wedge (B\lor \neg B))$$
$$= P(A)$$
I'm not sure where to go from here? Does $(B\lor \neg B) = 0?$
 A: When talking about probability, you usually start with a probability space $(\Omega, \mathcal{F}, P)$, where:
1. $\Omega$ is an arbitrary set, called the sample space. It is the set of outcomes, for example the sample space of a dice roll is $\Omega =\{1, 2, 3, 4, 5, 6\}.$
2. $\mathcal{F}$ is a collection of subsets of $\Omega$, called events. This collection also satisfies certain properties or axioms that (a) involve unions and complements: if $A,B$ are events, then $A\cup B, A^c, B^c$ are also events (the union also holds for countable collections of events) and (b) an important subset that always belong to $\mathcal{F}$ (namely $\Omega$). It is what is called a $\sigma$-algebra. In the dice roll case, you take the power set $\mathcal{P}(\Omega)$ as $\mathcal{F}$, and an example of an event $A \in \mathcal{F}$, is that the dice roll is an even number, so $A= \{2\} \cup \{ 4\} \cup \{6\} =\{2,4,6\} \subset \Omega$, so $A \in \mathcal{F}.$
Using these axioms and some properties of the set operations you can see that in every probability space: (a)$\, \emptyset \in \mathcal{F}$ (the empty set), and (b) if $A,B \in \mathcal{F}$, then $A \cap B \in \mathcal{F}.$
3. P is a function that sets a value between $0$ and $1$ to each event $ A \in \mathcal{F}$, so $P: \mathcal{F} \to [0,1]$. It is called a probability measure, and also satisfy certain axioms: (a) $P(\emptyset)=0$, (b) $P(\Omega)=1$ and (c) $P(A_1 \cup A_2 \cup \dots \cup A_n) = P(A_1) + P(A_2) + \dots + P(A_n)$ if the sets $A_k$ are mutually disjoint, this is $A_1 \cap A_2 = \emptyset$, for example. (This is the hint they gave you, it is also satisfied if you take a countable collection of events). For the event $A$ in 2., for example, $P(A)=P(\{2,4,6\})= P(\{2\}) + P(\{4 \}) + P(\{6\})$, since $A$ is a union of mutually disjoint sets, because the outcome of a dice roll can't be $2$ and $4$ at the same time, etc... 
As you can see, when talking about probability, your basic objects of study are sets ($\Omega, A, B$) and functions ($P$), so it is important to learn some properties about them and their operations. The most important operations between sets are the union $A \cup B$, intersection $A \cap B$, difference $A - B$, and complement $A^c = \Omega - A$ (what you write as $\neg B$), and their properties (for example, associative, commutative, distributive, De Morgan Laws, etc...) so you can be able to manipulate them. 
On the comments and your edit, for example, the last question $B \lor \neg B = 0$?, made me think that maybe you don't know some of these important facts, but since you showed interest and work I decided to make this long post so you have some basic definitions. I leave to you the task of researching more about them. 
The complete statement and proof of your theorem should be as follows:
Theorem: Let ($\Omega, \mathcal{F}, P$) be a probability space, and $A,B \in \mathcal{F}$. Then 
$$P(A) = P(A \cap B^c) + P(A \cap B).$$
Proof: Since $A,B \in \mathcal{F}$, by 2., $B^c \in \mathcal{F}$ and also by 2., $A \cap B \in \mathcal{F}$ and $ A \cap B^c \in \mathcal{F}$ (This is to ensure that these are in fact events, so we can talk about their probabilities).
$Afirmation \, 1: (A \cap B) \cap (A \cap B^c)= \emptyset.$ (that is, this events are mutually disjoint)
$Proof:$ Since $\cap$ is associative and commutative then
$$(A \cap B) \cap (A \cap B^c) = A \cap B \cap A \cap B^c = (A \cap A) \cap (B \cap B^c) = A \cap ((\Omega - B) \cap B) = A \cap \emptyset = \emptyset,$$
hence $(A \cap B)$ and  $(A \cap B^c)$ are mutually disjoint.
Now that we now this fact we may continue with the main proof. 
Using 3.(c) for these mutually disjoint events we get
$$P(A \cap B) + P(A \cap B^c) = P((A \cap B) \cup (A \cap B^c)),$$
but by the distributive property of the intersection over the union we have
$$P((A \cap B) \cup (A \cap B^c))= P(A \cap (B \cup B^c))= P(A \cap (B \cup (\Omega - B))= P(A \cap \Omega)$$
Since $A \in \mathcal{F}$ is a subset of $\Omega$ by definition, we have 
$$P(A \cap \Omega) = P(A)$$
Hence $P(A \cap B) + (A \cap B^c) = P(A) \, \blacksquare$ 
