Proving a Conditional Probability Theorem (General Addition Rule) If $B$ is an event with $P(B) > 0$, prove that the set function $Q(A) = P(A|B)$
satisfies the axioms for a probability measure. ,
$$P(A ∪ C|B) = P(A|B) + P(C|B) − P(A ∩ C|B)$$
What i tried
Basically the question is asking to prove the above Probability axiom using the fact that $Q(A) = P(A|B)$ .
$P(A ∩ C|B)=-P(A ∪ C|B)+P(A)+P(C|B)$
Substituting this to the Probability axiom it becomes
$$P(A ∪ C|B) = P(A|B) + P(C|B)+P(A ∪ C|B)-P(A)-P(C|B)$$
Simplifying it it becomes
$$P(A|B) =P(A)$$
This shows that events $A$ and $B$ are independent of each other. Hence showing $Q(A) = P(A|B)$ Is my method correct? Could anyone explain . Thanks
 A: Between the Question and some of the Comments, there seems to
be some confusion that we need to dispel from the start:
First, if $A$ and $B$ are independent, then $P(A|B) = P(A)$, but
there seems no reason to assume that.
Second, in most formulations of probability, the first equation you
write is a theorem, not an axiom.

I will begin by showing that $Q$ satisfies some of the $axioms$
of probability. (There are a few gaps, which I hope you will have
no trouble filling in.)
1) Show $Q(S) = 1$, where $S$ denotes the space on which $P$ and $Q$ are defined.
$$Q(S) = P(S|B) = P(S \cap B)/P(B) = P(B)/P(B) = 1.$$
2) Show $Q(A) \ge 0$, where $A$ is an event in $S$.
$$Q(A) = P(A \cap B)/P(B) \ge 0,$$
because $P(A \cap B) \ge 0$ and $P(B) > 0,$
3) If $A_1$ and $A_2$ are disjoint events in $S$, then
$Q(A_1 \cup A_2) = Q(A_1)+Q(A_2).$
$$Q(A_1 \cup A_2) = \frac{P((A_1 \cup A_2)\cap B)}{P(B)}
=\frac{P((A_1 \cap B) \cup (A_2 \cap B))}{P(B)}\\
= \frac{P(A_1 \cap B) + P(A_2 \cap B)}{P(B)} = Q(A_1) + Q(A_2),$$
because $A_1 \cap B$ and $A_2 \cap B$ must be disjoint.
Notes: 
(a) I used $A_1$ and $A_2$ instead of $A$ and $C$ because I think
there is potential distraction owing to the alphabetical order
of the letters. [In order to keep sight of event $B$, some
books would write $Q$ as $P_B,$ but I did not do that.]
(b) Depending on the level of your course, there may be a fourth
axiom involving an infinite sequence of mutually disjoint  events $A_i$ and an
infinite sum of probabilities. The proof for $Q$ is similar to that
of axiom (3). 
(c) Depending on the level of your course, there may be
a definition of 'events' as subsets of $S$, which are assigned
probabilities.

Now for the specific $theorem$ about $Q$ you mentioned in your
problem.
If $A_1$ and $A_2$ are not necessarily disjoint, it follows that $$Q(A_1) + Q(A_2) = Q(A_1)+Q(A_2)-Q(A_1 \cap A_2).$$
If we have proved $Q$ obeys all of the probability axioms,
this is just the same as the proof that 
$$P(A_1) + P(A_2) = P(A_1)+P(A_2)-P(A_1 \cap A_2).$$
If you want to prove this relationship directly from the 
definition of $Q$ as a conditional probability given $B$,
then the proof is very similar to the one shown for axiom (3).
