Is $[p \land q] \rightarrow ( p \rightarrow q)$ a valid argument?

Question

I need some help reconciling this. Let
$p$: I'm stuck inside.
$q$: I'm reading.

The argument: "I'm stuck inside and I'm reading, therefore I'm reading because I'm stuck inside." seems invalid to me (at least intuitively). But, when I do a truth table for the statement of the argument, $$(p \land q) \rightarrow (p \rightarrow q),$$ it seems valid.

Am I doing the math wrong here or is my intuition misguided? Thanks!

Answer 1

Your truth table is correct: the statement is indeed a tautology. The problem is with your informal example of it: the implication really cannot be translated as because. For the $p$ and $q$ that you’ve chosen, a better translation of the formal expression would be If I’m stuck inside, and I’m reading, then if I’m stuck inside, then I’m reading. And this is true, because if I’m stuck inside and am reading, then I’m reading whether or not I’m stuck inside (though in fact I know that I am stuck inside).

The logical connectives simply don’t match up perfectly with English usage of the words or, and, if ... then, etc., and the connective $\to$ is the worst of the bunch. It’s probably best to remember that when we write $p\to q$, what we’re really doing is denying the possibility that $p$ is true and $q$ is false; any other combination of truth values for $p$ and $q$ is consistent with $p\to q$. In your example, for instance, the main implication denies the possibility that on the one hand I am both stuck inside and reading, but on the other hand if I’m stuck inside, then I’m reading is a false statement. And if I’m stuck inside, then I’m reading is a false statement precisely when I’m stuck inside but not reading, so the whole thing actually amounts to saying this:

• It is not possible that on the one hand I am both stuck at home and reading, and on the other hand I am stuck at home and not reading.

In other words:

• It is not possible that on the one hand $p\land q$ is true, and on the other hand $p$ is true and $q$ is false.

And because $p\to q$ really just means that it’s impossible to have $p$ true and $q$ false, that last version is just another way to say:

• It is not possible that on the one hand $p\land q$ is true, and on the other hand $p\to q$ is false.

Answer 2

I'm reading because I'm stuck inside.

This is an ambiguous statement about causality. Taken literally, it would mean that being "stuck inside" would cause you to do nothing but read non-stop. Strictly speaking then, this statement is probably false (e.g. a humorous exaggeration) and not subject to logical analysis.

Classical propositional logic can only be applied to propositions that are unambiguously either true or false at the same instant in time (usually the present). My favourite example is the sentence: If it is raining, then it cloudy.

$Raining \to Cloudy$

This does not mean that rain causes cloudiness. It means only that, at present, it is not both raining and not cloudy.

$Raining \to Cloudy \space \space \equiv \space \space \neg (Raining \land \neg Cloudy)$

Answer 3

Implication ($\rightarrow$) cannot be read as " because".

The connective " P because Q " is not truth functional and , hence, does not belong to propositional logic. Implication ( material implication) is truth-functional.

It means that, as soon as you know the truth value of $P$ and of $Q$, you can determine the truth value of $(P\rightarrow Q)$ .

One cannot do this with " because".

So, you are right when you say that (S&R) does not imply ( R because R)

Now, supposse your stuck and you are reading.

In particular, you are reading.

So, since you are reading, a fortiori, it is not the case that : you are stuck & you are not reading. So ( S & not-R) is false .

Now if ( S & not-R) is false, $\neg$ ( S & not-R) is true. But this is the definition of material implication : a proposition implies another one iff it's not the case the first is true and the second is false.

So S materially implies R

since : it is not the case that S is true and R is false.

How could it be the case,since R is true!