How can $P \lor Q,~ P \vdash \lnot Q$ be invalid? While studying refutation tree and I came across this example:  $P \lor Q,~ P \vdash \lnot Q$ in Outline of Logic- Schaum's series. 
The solution invalidates the argument. 
But, when we cross examine this argument by constructing the truth-table we find that two premises: $P \lor Q$ and $P$ are true alone with thee Conclusion $ \lnot Q$ in 2nd line of  truth-table. 

For $P = {\rm T},~ Q = {\rm F}$
For $P \lor Q$ is ${\rm T}$, $P$ is ${\rm T}$, and $\lnot Q$ is ${\rm T}$
So, how can this form be invalid?
Here, 
Premises 1) $P \lor Q$ and 2) $P$ are true and 
Conclusion $\lnot Q$ is true as well 
When $P$ is ${\rm T}$ and $Q$ is ${\rm F}$.
 A: It is obviously invalid.
You have to check the truth assignment $v$ such that :

$v(P)=v(Q)= \text T$.

With it, we have that $v(P)=v(P \lor Q)= \text T$, while $v(\lnot Q)= \text F$.
Recall the definition of valid argument :

A deductive argument is said to be valid if and only if it takes a form that makes it impossible for the premises to be true and the conclusion to be false. 

In terms of propositional logic, this means that, in the truth table for the argument, the conclusion must have $\text T$ in every row where all premises are $\text T$. 
A: First of all, the argument is obviously invalid.
Suppose I claim that I am thinking of a number, and you know that the number is divisible by $2$ (represented by $P$) or by $3$ (represented by $Q$), or both (so we know $P \lor Q$).  Then I tell you the number is divisible by $2$ (so we know $P$).  Does this allow you to infer $\neg Q$, i.e. that it is divisible by $3$?  Of course not; my number could be $6$.  So, this argument is invalid.
Now, you might say: 
But wait, what if your number was $4$?  Then $P \lor Q$ and $P$ would be true, and $Q$ would be false. So, $\neg Q$ could be true.
Well yes, $\neg Q$ could be true, but that is not what makes an argument valid. For an argument to be valid, the conclusion has to be true, given the truth of the premises. The fact that the conclusion coulld be true is ont enough for the argument to be valid.
In general, an argument is valid if the truth of the premises forces the conclusion to be true, meaning that whenever the premises are true, the conclusion is true as well. In terms of truth-tables: it needs to be the case that for every row where the premises are true, the conclusion is true.
What you have pointed out is that there is some row where the premises are true and the conclusion is true as well. However, that does not mean that the conclusion is true for all rows where the premises are true. And as soon as there is a row with all true premises and a false conclusion, the argument is invalid.
Think about this argument:
$P$
$\therefore Q$
Intuitively this should of course be an invalid argument: the conclusion is completely unrelated to the premise.  So, the fact that there is a row where the premise and the conclusion are both true (namely with $v(P)=v(Q)=T$ means nothing. Indeed, there is a different valuation ($v(P)=T$ and $v(Q)=F$) that makes the premise True and the conclusion false, meaning that this is indeed an invalid argument.
Something similar is going on in those refutation trees you are working with: some branches of the tree may close ... but in order for the argument to be valid, all branches need to close. If some branches close but other branches end up open, then the argument is invalid, period.  And there is no such thing as 'sometimes valid, sometimes invalid'.
A: P or Q allows for the possibility of both P and Q to be true at the same time, whereas
P -> not Q does NOT allow for both P and Q to be true at the same time. 
