I need to show the validity of the below arguments by using a truth table I need to show the validity of 
$P \rightarrow Q$
$P \rightarrow R$ 
$\therefore  P \rightarrow (R \wedge Q)$
Can i just show the truth table for $P \rightarrow Q$ and the truth table for $P\rightarrow R$ then the truth table for $P \rightarrow (R \wedge Q)$
I think the truth table for $P \rightarrow (R \wedge Q)$ is:
T
T
F
T
F
T
 A: Use three columns, on the left most part of the truth-table, to keep track of the truth value assignments for $p, r, q$, one column per variable. So you need $2^3 = 8$ rows in your truth table, below the headings. Instead of creating three truth tables, create one truth-table, using a column for each premise, $p \rightarrow q$ and $p\rightarrow r$, and a final column for the conclusion: $p \rightarrow (q \land r)$. 
For your argument to be valid, it must be the case that whenever all the premises are true, the conclusion must be true. So your truth table needs to show that whenever the premises $(p \rightarrow q)$ and $(p \rightarrow r)$ are both true, then so is be the conclusion $p \rightarrow (r \wedge q)$ true.
The following truth-table was generated by Wolfram Alpha. Be sure to check whether the condition of validity is met.

A: An argument is valid if whenever all the premises are true, the conclusion must be true. So, the easiest thing is to compare the truth table of $(p \rightarrow q) \wedge (p \rightarrow r)$ with $p \rightarrow (r \wedge q)$.
Wolfram alpha can check your work: here are the two premises and here is the conclusion. I strongly encourage you to do the truth table yourself, and then use this as a check. Remember that when the premises are true, the conclusion must be true if the argument is valid.
A: $p \rightarrow q \equiv \neg p \vee q$
$p \rightarrow r \equiv \neg p \vee r$
$(p \rightarrow q ) \wedge ( p \rightarrow r ) \equiv ( \neg p \vee q ) \wedge ( \neg p \vee r)$  [reverse distribution property]
$(\neg p \vee (q \wedge r))$
$p \rightarrow  (q \wedge r)$
