Discrete Math - Determine if the argument is valid

Question

Can you guys please check my work and syntax.

Question: Determine if the argument is valid.

p $\rightarrow $ q
$\underline{\urcorner{q}}$
$\therefore \urcorner$p

Answer:
T $\rightarrow $ T
$\underline{{F}}$
$\therefore$ F

Answer 1

Step 1 : Definition of a valid argument

An argument is valid if and only if its conclusion is never false while its premises are true.

Step 2 : Building the truth table

We follow the standard method, enumerating all possible binary cases for the sentence letters $p$ and $q$, and apply the usual boolean operators to find the truth values of the premises of the argument (columns 3 and 4) and of its conclusion (column 5).

\begin{array}{cc|ccc}p&q&p\to q&\neg q&\neg p\\\hline T&T&T&F&F\\T&F&F&T&F\\F&T&T&F&T\\F&F&T&T&T\end{array}

Step 3 : Interpretation of the truth table and conclusion

The only assignment of truth values to the sentence letters $p$ and $q$ that make both premises true is given in the last line of the truth table, i.e. when both $p$ and $q$ are false. In this configuration, the conclusion is also true.

We have shown that it is never the case that the premises are true while the conclusion is false, therefore, by definition, the argument is valid.

Answer 2

You seem confused about the very definition of a valid argument:

An argument is valid iff whenever its premises are true, so is its conclusion.

In your approach, you took $p$ and $q$ to be the premises. However, this is not quite true. In the given form:

$A$
$\underline{B}$
$C$

$A$ and $B$ are the premises, and $C$ is the conclusion. In the present case, we need to find all possible truth values for $p$ and $q$ so that both the premises $p \to q$ and $\neg q$ are true. Subsequently we need to check if the conclusion, in casu $\neg p$, is also true.

Basically, the only way that this could possibly fail would be if we took $q = F$ false (so that $\neg q$ is true) and at the same time $\neg p$ would be false, i.e. $p = T$.

However, if $p = T$ and $q = F$, then the first premise $p \to q$ is false.

Therefore, we are led to the conclusion that it is impossible to derive a false conclusion from true premises using this argument. I.e., that the argument is valid.