truth tables and validity of arguments 
*

*$ p $

*$ p \to q $

*$ \lnot q \lor  r$

*$ \therefore r$


In order to prove validity with truth tables, do 1) 2) and 3) have to be true in order for the conclusion to be true? 
 A: A valid argument is an argument IF the premises are true, then the conclusion must be true. That is, for an argument to be valid, we must have that the truth of the premises guarantee/entail the truth of the conclusion. 
An argument can be valid even if one or more of the premises are false. Provided the premises logically entail the conclusion, the argument is valid. If we know the premises to be true, in a valid argument, we then know the conclusion is true. If we don't know the truth values of the premises, we can show such an argument is valid if we can show that, "if the premises are/ were true, then the conclusion is necessarily true." 
Whether or not the premises are true is another question: for example, a sound argument requires that the argument be valid and that the premises are all true.
A: The problem is an easy one and you do not necessarly have to "write down" all the truth table.
In order to prove with truth tables that 

$\Gamma \vDash r$

where $\Gamma = \{p, p \rightarrow q, \lnot q \lor r \}$ we apply the definition :

$\Gamma$ tautologically implies $\alpha$ (written $\Gamma \vDash \alpha$) iff every 
  truth assignment for the sentence symbols in $\Gamma$ and $\alpha$ that satisfies every member of $\Gamma$ also satisfies $\alpha$. 

In this case, we can "play a little bit" with the definition. We rephrase it in the equivalent form :

there is no truth assignment for the sentence symbols in $\Gamma$ and $\alpha$ that satisfies every member of $\Gamma$ and does not satisfy $\alpha$. 

Proving this with truth-tables amount to showing that there is no rows in the tt where all the formulas in $\Gamma$ are evaluated to true and the formula $r$ is evaluated to false.
We can take benefit of the fact that $\lnot q \lor r$ is equivalent to $q \rightarrow r$.
Let we start: we assume a truth assignment $v$ such that, for all $\varphi \in \Gamma$, $v(\varphi)=true$.
We have $v(p)=true$, and by truth-functional property of $\rightarrow$, also $v(q)=true$, in order $p \rightarrow q$ to be true.
So $v(r)=true$, for the same reason (we have assumed that, for the truth assignment $v$, $\lnot q \lor r$, i.e. $q \rightarrow r$, is true).
In conclusion, the truth assignment $v$, such that all the premises (the formulas in $\Gamma$) are true, boils down to the row of the tt corresponding to : t, t, t.
For this truth assignment $v$, $r$ is true; but $v$ is a generic truth assignment that satisfies all the premises, so we cannot have a truth assignment that satisfy all the premises and that evaluates $r$ to false.
In conclusion, we have proved that :


$\{p, p \rightarrow q, \lnot q \lor r \} \vDash r$.


A: Yes, by definition $\Gamma\vDash p$, if $p$ is true whenever $\forall \gamma\in\Gamma$, $\gamma$ is true.
Here, $\Gamma$ contains (1), (2) and (3).
A: The question calls for a look at the truth table proving the validity of the argument set out, so here it is.
It reveals that propositions 1) and 2) in the original argument can be false when the conclusion "r" is true. So in those cases the answer to the question is no.
Interestingly though, when proposition 3) is false, so is r. So 3) has to be true for r to be true.

A: No, 1., 2., and 3. do not have to be true in order to prove validity with truth tables.
You can prove the valid argument
(p $\land$ $\lnot$ p)
∴
q
using truth tables as follows:
  p   q   ¬p  (p ∧ ¬ p)  q
  F   F   T      F       F
  F   T   T      F       T
  T   F   F      F       F
  T   T   F      F       T

Consequently, you could prove the following argument:


*

*(p $\land$ $\lnot$ p)

*r

*s
∴
q
with truth tables.
