Tautologies in classical logic

Question

According to its truth table, $P \lor \neg P$ is a tautology, i.e. it is true for all truth values of its constituent propositions. But how come that is true in classical logic? $\lor$ is the inclusive 'or', so $P \lor \neg P$ means 'either $P$, or $\neg P$, or both of them'. But in classical logic it is never true that both $P$ and $\neg P$, i.e. $P \land \neg P$ is a contradiction (false for all truth values of its constituent propositions).

So why is $P \lor \neg P$ always tautologically true?

Answer 1

In Classical Logic it is always the case that either $P$, $\neg P$, or both. You will always be dealing with one of those three options, even if the third never happens.

It is also always true for Classical logic that at least one of the following holds: $P$, $\neg P$, pigs fly, grass is pink.

Answer 2

"According to its truth table, $P \lor \lnot P$ is a tautology, i.e. it is true for all truth values of its constituent propositions."

OP is missing the Point that the constituent propositions should be independent to be assigned truth values arbitrarily & independently.
We can not assign truth values to derived terms : We have to evaluate the derived terms to get the truth values using the independent variables.

"$X \lor Y$ means 'either (A1) $X$ , or (A2) $Y$ , or (A3) both of them'"
Here , $X$ & $Y$ are independent and we can assign the truth values independently & arbitrarily to check whether it is a tautology.
When we assign $X=Y=0$ , we get $X \lor Y = 0$. Very other Case [ A1 , A2 , A3 ] we get $X \lor Y = 1$
Hence it is not a tautology.

"$P \lor \lnot P$ means 'either (B1) $P$ , or (B2) $\lnot P$ , or (B3) both of them'"
Here , $P$ & $\lnot P$ are not independent and we can not assign the truth values independently & arbitrarily , to check whether it is a tautology.
When we assign $P = 0$ , we automatically get $\lnot P = 1$ , then $P \lor \lnot P = 1$ via Case B2.
When we assign $P = 1$ , we automatically get $\lnot P = 0$ , then $P \lor \lnot P = 1$ via Case B1.
Hence it is a tautology.

What OP observed here is that we never get ( & never use ) Case B3 here. That is ok. It is the outcome of logical consistency & model necessity where we can not assign $P = 1$ & $\lnot P = 1$ at the same time , because these two terms are not independent.
We do not use B3 with $P \lor \lnot P$ , because B1 & B2 are enough to show that it is a tautology.

EDITORIAL COMMENT :
This answer is too elaborate , to assist the OP at the very low foundational level.
Advanced users may find it too wordy , though that is the unfortunate outcome of catering to elementary nature of the confusion here.

Answer 3

$P \lor \neg P~$ is an axiom. It can be seen as the theoretical basis for the construction of truth tables, which are simply a convenient visualization of a proof by cases.

Example

Consider the truth table for $A \to B$:

There are 4 cases to consider. It translates:

$~~~~~~[(A \land B) \to (A\to B)]~~~~~~~~~~~~~$ (Case 1)

$~~\land [(A \land \neg B) \to \neg (A\to B)]~~~~~~~~$ (Case 2)

$~~\land [(\neg A \land \ B) \to (A\to B)]~~~~~~~~~~$ (Case 3)

$~~\land [(\neg A \land \neg B) \to (A\to B)]~~~~~~~~$ (Case 4)

Formal Proof of Case $1$ using a form of natural deduction:

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Plain text version:

Case 1:
1   A & B
    Premise
2   A
    Premise

3   B
    Split, 1


Discharge premise on line 2
4   A => B
    Conclusion, 2
Discharge premise on line 1

As Required:

5   A & B => [A => B]
    Conclusion, 1

Answer 4

As mentioned or hinted at in some of the comments and answers, $P \lor \neg P$, or the law of the excluded middle is axiomatically true in classical logic. So it is true because we assume it to be true without proof.

One alternative axiom is the double negation rule: $\neg \neg P \Rightarrow P$. From double negation, you can prove the law of the excluded middle as a tautology. An example of such a proof can be seen here. The basic idea of the proof is to assume $\neg (P \lor \neg P)$ and show that it leads to a contradiction, therefore $\neg \neg (P \lor \neg P)$ is tautologically true. Then apply the double negation rule.