A problem in Logic If A,B and C are statements such that C is true only if exactly one of A and B is true.If C is
false then which of the following statement is true?
$1$.If A is false then B is false.
$2$.If A is true then B is false.
$3$.Both A and B are true.
$4$.Both A and C are false.
 A: It is just a matter of interpreting the given information correctly.
What we are given is that if $C$ is true, then exactly on of $A$ and $B$ is true.
Now, if $C$ is false, our hypothesis is not satisfied and we are not sure of anything about $A$ and $B$. Thus, $A$ and $B$ could be anything.
Do not get confused thinking that the given statement is saying that "if exactly one of $A$ and $B$ is true, then $C$ is true."
A: Hint
There are two cases if $C$ is false :
First case : $A$ true, $B$ true
Second case : $A$ false, $B$ false
C is called the XOR function.
I let you finish the exercise and check the answers.  
Solutions
1 is 

 true 

2 is

 false

3 is  

 false

4 is 

 false

A: OK, so assuming you mean an exclusive or (which I think is correct) and that $P$ only if $Q$ means $P \rightarrow Q$ (i.e. if $P$ then $Q$)...which is explained here, then you have:
\begin{align}
\left((A \wedge \neg B) \vee (\neg A \wedge B)\right) \rightarrow C =& \neg\left((A \wedge \neg B) \vee (\neg A \wedge B)\right) \vee C \\
= & \left(\neg(A \wedge \neg B) \wedge \neg(\neg A \wedge B)\right) \vee C\\
=& \left((\neg A \vee B) \wedge ( A \vee \neg B)\right) \vee C \\
=& \left((\neg A \wedge ( A \vee \neg B)) \vee (B \wedge ( A \vee \neg B))\right)\vee C \\
=& \left(((\neg A \wedge A) \vee (\neg A \wedge \neg B)) \vee ((B \wedge A) \vee (B \wedge \neg B))\right)\vee C \\
=& \left((\neg A \wedge \neg B) \vee (B \wedge A) \right)\vee C \\
\end{align}
I think that's about as far as we can go--either $C$ is true (regardless of $A$ and $B$'s values), both $A$ and $B$ are false, or both $A$ and $B$ are true (note that both of the last two conditions negate the exclusive or making the implication trivially true).
Assuming $C$ is false, then we are left with: $(\neg A \wedge \neg B) \vee (B \wedge A)$.  This means that whatever value $A$ has, then $B$ has and vice versa.  This makes the first statement true, the second statement definitively false, the third statement false (because $A$ and $B$ could both be false), and the fourth statement false (because $A$ could be true and $B$ could be true...and of course $C$ is false as this was our assumption).
...note that if we look at the case when $C$ is true, then all of these statements are trivially true (we don't need to check that case).
