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.

  • $\begingroup$ I think that the column corersponding to $P \to (Q \land \lnot Q)$ is the 5th... $\endgroup$ – Mauro ALLEGRANZA Nov 30 '17 at 9:36
  • $\begingroup$ Thank you for your help $\endgroup$ – user508231 Nov 30 '17 at 9:49
  • $\begingroup$ @smiles47 Aha! I am glad to see you have learned not to teat the helper columns as premises, very good! But I would still urge you to use Mathjax, please. $\endgroup$ – Bram28 Nov 30 '17 at 12:01
  • $\begingroup$ Asking three very similar very elementary questions suggests that you need to go away and have a look at the presentation of truth-tables in a few text books. There are literally dozens of excellent elementary logic texts, any one of which would have got you to the point where you didn't need to trouble people here. Many are available online. For a legally available one, Paul Teller's A Modern Formal Logic Prime is reliable and very clear. tellerprimer.ucdavis.edu $\endgroup$ – Peter Smith Nov 30 '17 at 13:41


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:

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

  • $\begingroup$ I am unsure how to check whether a symbolic argument is a tautology, however if I look at column 5 of rows 3 and 4 where the premise is evaluated is true the conclusion in column 6 is also true. So from this I suppose I could determine the argument's validity (whether or not I know that is it a tautology) $\endgroup$ – user508231 Nov 30 '17 at 9:41
  • $\begingroup$ @smiles47 $P\to(Q\land\lnot Q)\models \lnot P$ is a valid argument. $(P\to(Q\land\lnot Q))\to \lnot P$ is a tautology. $\endgroup$ – Graham Kemp Nov 30 '17 at 12:21

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.

  • $\begingroup$ Thank you that is really informative. $\endgroup$ – user508231 Nov 30 '17 at 9:49
  • $\begingroup$ But of course, you are welcome! Also +1. Do visit this site more often! $\endgroup$ – астон вілла олоф мэллбэрг Nov 30 '17 at 9:50

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