Proving Validity of a Symbolic Argument Using Truth Tables Proving Validity of a Symbolic Argument Using Truth Tables
I am looking to determine whether the following argument is valid/invalid using the truth table method:
$$(P \to (Q \land\lnot Q)) \models \lnot P$$
I cannot yet embed the image of my truth table on here because I haven't earned enough points of my profile yet, so I will try to explain it the best I can.
I have used these column headers:
$$\begin{array}{|c:c|c:c:c|c|}\hline
P & Q & ¬Q & Q ∧ ¬Q & P → (Q ∧ ¬Q) & ¬P 
\\ \hline
 T & T & F & F & F & F 
\\
 T & F & T & F & F & F 
\\
 F & T & F & F & T & T 
\\
 F & F & T & F & T & T 
\\ \hline
\end{array}$$
When it comes to reading validity from the truth table, I am not 100% certain what columns I should be taking into account. At the moment I am only taking account of the 4th column i.e. $(P \to (Q \land \lnot Q)$ as the only premise whose truth value I should look at, from which I can read that there are no situations where a false conclusion is derived from true premises and hence the argument is valid.
I just would like to clarify whether this is the right way of approaching reading validity (as you can probably tell I am a beginner in logic). Thank you.
 A: Yes, you are correct.
In other words, $(P \to (Q \wedge \neg Q)) \to \neg P$ is false precisely when $P \to (Q \wedge \neg Q)$ is true but $\neg P$ is false. Hence, after drawing your truth table, you should look at whether there is any case where the former is true but the latter is false. This will tell you that your proposition is false. If this does not occur, then the proposition is true.
Just to add the truth table:
$$
\begin{array} {c|c|c|c|c} 
P & Q & \neg Q & Q \wedge \neg Q & P \to (Q \wedge \neg Q) & \neg P & (P \to (Q \wedge \neg Q)) \to \neg P \\
\hline
T & T & F & F & F & F & T\\
T & F & T & F & F & F & T\\
F & T & F & F & T & T & T\\
F & F & T & F & T & T & T\\
\hline 
\end{array}
$$
Thence, the conclusion holds. Alternatively, you can go for simplification of the following kind in boolean algebra:
$$
(p \implies (qq')) \implies p' = p' + (p \implies (qq'))' = (p(p \implies qq'))' 
$$
But $qq' = 0$, so $(p \implies qq') = p'+0 = p'$. Therefore, the above just becomes $(pp')' = 0' = 1$, which means the statement made is true always.
A: Comments
To be precise, the argument must be: $P \to (Q \land \lnot Q) \vDash \lnot P$;
i.e. "from the premise $P \to (Q \land \lnot Q)$, infer the conclusion $\lnot P$".
$(P \to (Q \land \lnot Q)) \to \lnot P$, instead, is a single formula.
We have valid arguments, and valid formulas: a valid formula of propositional calculus is called a tautology.
Thus, two ways:
(i) check if the formula $(P \to (Q \land \lnot Q)) \to \lnot P$ is a tautology (i.e. its truth table has only TRUE);
(ii) check if in every row where the premise $P \to (Q \land \lnot Q)$ is evaluated to TRUE also the conclusion $\lnot P$ is TRUE.

Note: as you can verify, the two "procedures" give the same result.
This is so because:


$\varphi \vDash \psi \text{ iff } \vDash \varphi \to \psi \text { ( i.e. } \varphi \to \psi \text { is a tautology).}$


