I have a homework assignment which declares that following:

If $P \implies Q$ is valid, and $P$ is valid, then $Q$ must also be valid (prove true or false).

My interpretation of validity in logic is that it basically means that there is a tautology. Its hard for me to wrap my head around the fact that we can declare $P \implies Q$ valid for any propositional formulas $P$ and $Q$ since logically $P \implies Q$ is not a tautology.

Should I be interpreting $P \implies Q$ as being true and that if $P$ is true then $Q$ must be true? That seems to go against the textbook interpretation of validity, but I know how to prove that statement.

Is there something I'm missing about the concept of validity? If $P \implies Q$ is declared as valid does that mean I should only be looking at the "true" sections of the truth table and then proceed with my proof? And that if we declare $P \implies Q$ as satisfiable it means to look at all the entries of the truth table? Just getting confused on this new terminology I am being introduced to.


  • $\begingroup$ khanacademy.org/partner-content/wi-phi/wiphi-critical-thinking/… may help $\endgroup$ – user451844 Sep 15 '17 at 21:23
  • $\begingroup$ This is just another way of saying, "If P implies Q, and P, then Q." Which is the same as just defining what "P implies Q" means. $\endgroup$ – Wildcard Sep 15 '17 at 23:08
  • $\begingroup$ $P$ and $Q$ must be read as meta-variables, i.e. variables representing formulas, and not as propositional letters, i.e. variable representing single sentences. You are right: a sentence $P \to Q$ cannot be a tautology; but modus ponens must be expresses with variables for formulas: "if $P$ is a true formula and $P \to Q$ is a true formula, then also the formula $Q$ is true". $\endgroup$ – Mauro ALLEGRANZA Sep 16 '17 at 8:33

You are right that the terminology can be confusing. Validity is usually seen as a property of arguments ... but sometimes statements themselves are said to be valid when they are necessarily true (i.e are a tautology). But then, like you say, how can something like $P \rightarrow Q$ be valid? Well, maybe $P$ and $Q$ are used as statement variables, meaning that they can stand for complex statements such as $A \lor \neg A$ ... and as such they can be valid.

  • $\begingroup$ So if I am understanding correctly, P -> Q can be valid because if P is valid (a tautology) and Q is valid (a tautology) that would make P -> Q a tautology/valid? $\endgroup$ – chevybow Sep 15 '17 at 21:27
  • $\begingroup$ @AdamS That is correct ... but the question you have to answer is different: you have to show that if $P$ is valid and if $P \rightarrow Q$ is valid then $Q$ is valid. $\endgroup$ – Bram28 Sep 15 '17 at 23:34

$P$ and $Q$ are here placeholders (at the meta-level) for particular formulas.

One instance of the claim is therefore

If $A\to A$ is valid, and $A$ is valid, then $A$ must also be valid.

Another is

If $(A\to A)\to A$ is valid, and $A\to A$ is valid, then $A$ must also be valid.

A third is

If $(A\to(B\to C))\to(B\to (A\to C))$ is valid, and $A\to(B\to C)$ is valid, then $B\to(A\to C)$ must also be valid.

In each of these three examples, one but not both of $P\to Q$ and $P$ is in fact valid, but $Q$ is not valid. It may also be that all of them are valid, of course:

If $(A\to A)\to(B\to(A\to A))$ is valid, and $A\to A$ is valid, then $B\to(A\to A)$ must also be valid.

The homework asks you do decide whether the claim is always true, no matter which particular formulas you insert into the $P$ and $Q$ positions.


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