Given a $P$ does this always imply either $Q$ or $\neg Q$? (not both at the same time)

Question

Given a $P$ does this always imply either $Q$ or $\neg Q$? I want to show that given a $P$, then either $Q$ holds or $\neg Q$ holds (that is, not both at the same time). My attempted proof is the following: since $P$ is given, then this implies that it is "true", then one could say that $P$ implies $Q$ or $\neg Q$, but these last two cases are exclusive , then $P$ implies only one of them and not both.

If that proof is valid, then it seems strange to me, since it is not necessary that $Q$ (or $\neg Q$) can be logically deduced from $P$, as an example: "If it rains tomorrow, then I either study or I don't study", the proposition is true, although I cannot deduce that actually raining implies that I study or not.

Edited:

I think it can get much weirder, for example if $P$ implies $Q$ or $\neg Q$, and $\neg Q$ implies $R$. Then $P$ implies $Q$ or $R$.

I have written this, assuming that the above written before editing this message is correct.

Answer 1

You need to be careful separating between the various ways in which we use ‘implies’.

If by ‘implies’ you mean ‘logically implies’, then please note that $P$ does indeed logically imply $Q \lor \neg Q$, but this does not mean that either $P$ logically implies $Q$ or that $P$ logically implies $\neg Q$. Indeed, we can use your very example as a counterexample.

If by ‘implies’ you mean the truth-functional operator or connective known as the material implication, then it is true that for any $P$ and $Q$ at least one of $P \to Q$ or $P \to \neg Q$ is true. But that is only weird if you treat the material implication as something more than it is. Indeed, if you treat it as a natural language conditional (‘if … then ...’). you will often get counterintuitive results, known as the Paradoxes of the Material Implication.

Answer 2

Implication does not imply causation. It merely implies that if the antecedent is true, then the consequent will also be true. In other words, the implicaiton states that the consequent is true whenever the antecedent is true. For instance, consider the statement

"if a person is pregnant, then that person has a uterus"

The above implication is obviously true, and we know that being pregnant does not cause one to have a uterus. You may find it helpful to understand the implication in terms of set theory. The above implication conveys that the set of instances in which a person is pregnant is a subset of the set of instances in which that person has a uterus. For this reason, if we are presented with an instance of a person being pregnant, then we must also have an instance in which that person has a uterus. There are numerous relationships like this that exist, and not all imply causation between the antecedent and consequent.

With that being said, lets take a closer look at your statement

"if it rains tomorrow, then either I study or I don't study"

The set of instances in which you either study or don't study is in fact the set off all possible instances. The set of instances in which rains tomorrow is a subset of this set. Hence, if tomorrow is an instance in which it rains, then tomorrow must be instance in which you either study or don't study.

In addition, I hesitate to call your proof valid because saying "then one could say that $P$ implies $Q$ or $\neg Q$" does not exactly prove $P \to (Q \vee \neg Q)$ on the basis of $P$ alone. The chain of reasoning you're using is obscure.

I think a poof of $P \to (Q \vee \neg Q)$ would be better stated as follows: Assume $P$ is true. Now consider that either $Q$ or $\neg Q$ is true. So, under the assumption that $P$ is true, we have shown that either $Q$ or $\neg Q$ is true. Thus, if $P$ is true, then either $Q$ or $\neg Q$ is true.