Logical implication and valid arguments question The following is a valid argument: $[[p \lor (q\lor r)]\land \neg q] \rightarrow (p\lor r)$. Determine the rows of the table crucial for assessing the validity of the argument and which rows can be ignored. 
$$
\begin{array}{c|c|c|c|c|c|c|c}
p & q & r & \neg q & q\lor r & p\lor(q\lor r) & [[p \lor (q\lor r)]\land \neg q] & p\lor r \\
\hline
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 \\
0 & 1 & 0 & 0 & 1 & 1 & 0 & 0 \\
0 & 1 & 1 & 0 & 1 & 1 & 0 & 1 \\
1 & 0 & 0 & 1 & 0 & 1 & 1 & 1 \\
1 & 0 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & 0 & 0 & 1 & 1 & 0 & 1 \\
1 & 1 & 1 & \underbrace{0}_{\text{premise 2}} & 1 & \underbrace{1}_{\text{premise 1}} & 0 & \underbrace{1}_{\text{conclusion}} \\
\end{array}
$$
So rows 2, 5, and 6 are crucial for assessing the validity of the argument, since we have the premises and conclusion $1$ in these rows.

Following this exercise are some questions I have, as the text I'm using  doesn't really cover these details. 


*

*$[[p \lor (q\lor r)]\land \neg q] \rightarrow (p\lor r)$ is also a tautology. Is it necessary that an implication of the form $$(p_{1}\land p_{2} \land \dots \land p_{n})\rightarrow q$$ to be a tautology in order to be a valid argument -- is this correct? 

*So in this exercise, only rows 2,5, and 6 assessed the validity of the argument. So for the other rows, the argument is invalid. But how can the exercise state $[[p \lor (q\lor r)]\land \neg q] \rightarrow (p\lor r)$ as a valid statement? Doesn't it depends on the values of the premises and conclusion being 1?

*What does it take for premises to logically imply a conclusion? Must it be that the implication be a tautology? or must it be that the rows by which the premises and conclusion be 1?


Thank you very much in advance! :) 
 A: I agree with the above comment; we have to say that the question must be rephrased as :

show that : "if $p∨(q∨r)$ and $¬q$, therefore $(p∨r)$" is a valid argument. 

But as you say in 1) :

$\varphi_1, \varphi_2,\ldots,\varphi_n \vDash \psi$ iff $\vDash \varphi_1 \land \varphi_2 \land \ldots \land \varphi_n \to \psi$.

Thus, the "procedure" of truth-table verification used to establish the validity [i.e. "tautologuesness"] of the conditional it is enough to show the validity of the corersponding argument. 
For 2), the definition of valid argument is "formalized" with the relation of logical consequence that, for propositional logic is :

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

Here the set of premises $\Sigma$ is : $\{ p∨(q∨r), ¬q \}$, while the conclusion $\tau$ is $(p∨r)$ and the truth table show that in all rows where both premises are true, also the conclusion is.
Thus, the conclusion is tautologically implied by the premises, i.e. the argument is valid.
