What if P implies Q is false when both P and Q are false?

Question

This is actually a problem that our professor gave us, and I'm clueless of how to answer this. I browsed through various sources, but none were helpful regarding this question.

The question is,

In the definition of semantics of logic, P implies Q is defined as true under the assignment of both P and Q are false. Although this is rather unusual at a glance, explain what would be the issue with logic, if the definition is differently.

Any helpful answer is highly appreciated.

Answer 1

This question is a soft question IMO, it assumes there's common ground about what "should be true" independent of definitions but it doesn't clarify what those assumptions are. If I were to break it down, I'd say it's looking for a particular bit of reasoning (e.g. a proof or an inference rule) to hold and then asking you to show that that reasoning would be invalid if we defined the semantics of implication differently. In essence it's asking if there exists a proof that, under this new model, would be invalid.

Consider $\neg (Q \vee \neg Q) \to \neg (Q \vee \neg Q)$ which is provable, and should be valid in all models (no matter what we assign to $Q$ that is). As a general rule of reasoning it should always hold that $P \to P$ no matter what. I posit this is sufficiently fundamental common ground of "what should be true". So if we find this reasoning invalid, we have our soft contradiction. We can construct instances of $P$ for which $P$ is certainly false as shown above. So despite the very reasonable proof of $\neg (Q \vee \neg Q) \to \neg (Q \vee \neg Q)$ that would actually be false under the model...in fact it's negation would be valid in the model!

Answer 2

In classical propositional logic, we define $P \rightarrow Q$ to be $\neg P \vee Q$. But if we redefined the semantics of implication as suggested above, this would no longer hold.