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

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.

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!
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.
• This is completely accurate but feels like it is missing the point to me. It reduces the question to a different question: "Why is $\neg P \vee Q$ the correct way to think about the concept of implication in classical logic?" Mar 10 '20 at 19:42