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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?

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  • $\begingroup$ See truth table for $P \lor Q$: at least one of the two disjuncts must be true in order for the formula being true. And apply it to $P \lor \lnot P$: in whatever case, at least one of the two disjuncts is true. $\endgroup$ Feb 28 at 11:27
  • $\begingroup$ en.wikipedia.org/wiki/Law_of_excluded_middle $\endgroup$ Feb 28 at 11:29
  • $\begingroup$ And a disjunction is true if both its constituent propositions are true, as well. So is it right to say that $P \lor \neg P$ is a tautology because if both $P$ and $\neg P$ are true, then $P \lor \neg P$ must be true, even if in classical settings it is never the case that both $P$ and $\neg P$ are true? $\endgroup$ Feb 28 at 11:31
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    $\begingroup$ Yes, in the truth table for $P \lor Q$ you have four rows, with only one "flagged" False (that for both $P,Q$ false), while the t-t for $P \lor \lnot P$ bolis down to two rows only: that for $P$ true and that for $P$ false. There is no row where $P$ and $\lnot P$ are both true. $\endgroup$ Feb 28 at 11:35
  • $\begingroup$ Okay, thank you! $\endgroup$ Feb 28 at 11:37

4 Answers 4

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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.

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"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.

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    $\begingroup$ Thank you for the detailed answer. Was I right though in noting implicitly that this holds necessarily only in classical settings? If the truth valuation sends its inputs to the set {$0, 1$}, then the truth value of either $P$ or $\neg P$ is 'constrained' by the truth value of the other, so they cannot be assigned truth values arbitrarily. But in a non-classical setting where the codomain of the truth valuation was, e.g., $[0, 1]$, this wouldn't be true, right? $\endgroup$ Feb 28 at 13:29
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    $\begingroup$ Yes , this is true in most varieties of classical models. We have non-classical settings where this is not tautological. The most important such non-classical model is "Constructive Mathematics" where "Principle of Excluded Middle" is invalid. The other common non-classical model is fuzzy logic where the truth values are continuous. There are various other non-classical models like 3-valued logic & 4-valued logic. Various models will treat $P \lor \lnot P$ in various ways. $\endgroup$
    – Prem
    Feb 28 at 15:15
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$P \lor \neg P~$ 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 $A \to B$:

enter image description here

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:

enter image description here


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
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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.

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