The logical statement
$$(P \implies Q) \vee (Q \implies P)$$
is an example of a tautology. However, if I choose logical statements for $P$ and $Q$, it is not always true that either $Q \implies P$ or $P \implies Q$. For example, we may take
$$P(x) = x \text{ is odd}, \qquad Q(x) = x \text{ is prime}$$
Odd numbers are not necessarily prime (e.g., $9$ is odd but not prime), and prime numbers are not necessarily odd (e.g., $2$ is a prime which is not odd). Why is this?
On
You are misunderstanding the meaning of a tautology, especially when free variables are involved.
You have a propositional tautology, in terms of propositions $P$ and $Q$. You then substitute in propositions for $P$ and $Q$ which have a free variable $x$ in them. If we don’t know what $x$ is, we cannot determine whether $x$ is prime, nor whether $x$ is odd. Nevertheless, we do know that either ($x$ is prime implies $x$ is odd), or vice versa.
What you have demonstrated is that $\forall x (A(x) \lor B(x))$ is not equivalent to $\forall x (A(x)) \lor \forall x (B(x))$.
On
The open formula $$\big(P(x) \to Q(x)\big) \vee \big(Q(x) \to P(x)\big)\tag1$$ and the sentence $$∀x\,\Big(\big(P(x) \to Q(x)\big) \vee \big(Q(x) \to P(x)\big)\Big)\tag2$$ and the tautology $$(P \to Q) \vee (Q \to P)\tag3$$ are all logically valid. On the other hand, the sentence $$∀x\big(P(x) \to Q(x)\big) \;\vee\; ∀x\big(Q(x) \to P(x)\big)\tag4$$ is not logically valid and certainly not a tautology.
For example, we may take $$P(x) = x \text{ is odd}, \qquad Q(x) = x \text{ is prime}$$ Odd numbers are not necessarily prime (e.g., $9$ is odd but not prime), and prime numbers are not necessarily odd (e.g., $2$ is a prime which is not odd).
Here, you are demonstrating that sentence $(4)$ is not logically valid.
On
As others have pointed out, this English statement is not captured very well by the truth-functional logic statement of the form $(P \to Q) \lor (Q \to P)$, and it would be better captured by something like $\forall x (P(x) \to Q(x)) \lor \forall x (Q(x) \to P(x))$, which is not a necessarily true statement.
However, I don't think that really gets us to the core issue that you are having. Consider the following two statements: "Sylvester Stallone lives in Illinois" and "Sylvester Stallone lives in Ohio". Both sentences are very specific sentences, and there are no implicit universal quantifiers involved here, so it seems we can safely use $P$ and $Q$ here. Moreover, most people would say that "If Sylvester Stallone lives in Illinois, then Sylvester Stallone lives in Ohio" is false, and the same for "If Sylvester Stallone lives in Ohio, then Sylvester Stallone lives in Illinois". Hence, we would consider "If Sylvester Stallone lives in Illinois, then Sylvester Stallone lives in Ohio, or if Sylvester Stallone lives in Ohio, then Sylvester Stallone lives in Illinois" to be false, even though this time it does seem to neatly correspond with $(P \to Q) \lor (Q \to P)$. So what is going here? How to square our strong intuition that here we have a case where $(P \to Q) \lor (Q \to P)$ seems to be false, even though logical analysis shows it is a tautology? This time, we can't resolve this by appealing any implicit quantifiers ... it goes deeper than that.
What is really at the core of this issue is that there is often a mismatch between what we mean when we express an English 'if ... then ...' claim (and consequently, how we interpret English 'if ... then ...' claims) on the one hand, and the logical truth-functional operator $\to$ (called the 'material conditional') on the other hand.
That is, when you are given a logical expression like $(P \to Q) \lor (Q \to P)$, you are reading this to yourself as the English sentence 'If $P$ then $Q$, or if $Q$ then $P$', but in doing so you evoke the ways in which you normally interpret an English sentence like that, and find that that does not sound like a tautology at all, unlike what logicians claim $(P \to Q) \lor (Q \to P)$ to be.
Specifically, we say something stronger with our English "If Sylvester Stallone lives in Illinois, then Sylvester Stallone lives in Ohio" statement than what the logical $P \to Q$ captures. Remember from its truth-table definition that the logical $P \to Q$ is set to true as soon as $P$ is false, or $Q$ is true. Now that is a very weak statement: it represents a very low bar to clear for the statement to be true.
OK, but now notice that if we were to use this very low bar of what would make a material conditional true, and apply that to what would make English conditionals true, then the English statement "If bananas are yellow, then the sky is blue" would be true in our world simply because the sky is blue ... the bananas being yellow has nothing further to do with it. But that is of course completely counter-intuitive: when we say something like "If bananas are yellow, then the sky is blue", we mean to express some kind of connection or relationship between bananas being yellow and the sky is blue.
Indeed, often we express a causal connection, temporal connection, or logical connection when we make English 'if ... then ..' statements. And since we see no such connection between bananas being yellow and the sky being blue at all, most of us would deem the claim that "If bananas are yellow, then the sky is blue" to be false, because our thinking goes: "No, the sky isn't caused to be blue by bananas being yellow. No, it is not as if the sky turns blue after bananas turn yellow. No, the sky being blue isn't logically implied by bananas being yellow. So, this statement is false!"
So now we've reached what I said at the beginning: there is a mismatch between the logical material conditional, and English conditionals. As such, if you try to apply the truth-conditional analysis from truth-functional logic and apply it to English statements, very strange things can happen, in the same way that if you were to apply Newtonian physics to situations involving very large objects or very high speeds, you are going to get the wrong result, because Newtonian physics is only an approximation, a model. And here is a website that shows some other very unintuitive consequences of trying to make sense of English conditionals from a purely truth-functional point of view. Your $(P \to Q) \lor (Q \to P)$ is one example of such an intuitive consequence .... but note that an even 'worse' one is that $(P \to Q) \lor (Q \to R)$ is a tautology as well!
Finally, so what does all this mean? Is logic not a good way to analyze English conditionals then? No, that is not what I am saying. There are many situations in physics where a Newtonian analysis will do just fine, where it is a 'good enough' approximation. And the same is true for logic. At the same time, you will also need to recognize those times where the truth-functional analysis is really not appropriate. This is something that will take some experience. As a beginning student of logic, just hold your nose, and simply master truth-functional logic, its notation, its rules, and its methods. Same for any other logics. But once you have a firm grasp of that, once you're good at the purely mathematical 'plug-and-chug', then the kind of 'wisdom' that will tell you when and how (and when not and how not) to apply logic to the real world will gradually come as well.
Yes, it is always true. I guess you are thinking about real life examples and that is confusing you because of the time dimension added to the situation. For instance: Let $Q$ be "I eat" and $P$ be "I drink water". You might think "Oh well sometimes I eat and I don't drink water, and also some times I drink water and I don't eat". But here, the statements you are making (say, "I drink water") refer to different situations in the implications (different times). $P$ and $Q$ must refer to the SAME thing, so it would look more like:
"This time I was drinking water and NOT eating ($Q \nRightarrow P$) and, at the same, time, I was eating and NOT drinking water ($P \nRightarrow Q$)"
I'm pretty sure you see why that doesn't make any sense: for $Q \nRightarrow P$ to be false, $Q$ must be true and $P$ must be false, so for $Q \nRightarrow P$ and $P \nRightarrow Q$ to be false at the same time, $Q$ (and $P$) would be false and true at the same time. That's why one of the implications must always be true, therefore the disjunction is a truism.
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