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In propositional logic, a proposition is a statement that is either true or false, but not both. In a text I am reading and in many others, "this statement is false" is not considered a proposition. But is this because it is both true and false or because it is neither true or false, i.e., doesn't have a truth value?

EDIT 1:
The text I am reading says that the truth or falsity of a proposition may be clearly understood or arbitrarily assigned, which I interpret as meaning that what is important is that a proposition must be able to hold a single "stable" truth value. When we attempt to assign a truth value to "this statement is false" what is the problem?

EDIT 2: I understand that in order for an assertion to be considered a proposition, we must be able to associate a truth value to it. I have seen the following terse reasoning about why the assertion "this sentence is false" is not a proposition: If it is true, then it is false, and if it is false, then it is true. Does this mean that assigning it a truth value leads to it being both true and false, or that assigning it a truth value leads to a contradiction? Please explain.

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    $\begingroup$ If "true" is the same as "not false" and "false" is the same as "not true", then "both true and false" is the same as "neither true nor false". $\endgroup$
    – Alex Provost
    Commented Mar 14, 2016 at 21:16
  • $\begingroup$ @AlexProvost I meant it in the sense of not having a truth value. $\endgroup$
    – user185744
    Commented Mar 14, 2016 at 21:19
  • $\begingroup$ @AsafKaragila The text also says that the form of this sentence makes it impossible to designate it as true or as false. Is this what you were referring to? $\endgroup$
    – user185744
    Commented Mar 15, 2016 at 0:39

3 Answers 3

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It's a paradox, meaning any interpretation of it's truth value results in a contradiction. The statement is neither true nor false. If "this statement is false" were false, then it would be true, contradicting the assumption that it's false. Similarly if it were true. It can be shown using the rules of logic that a contradiction $A \wedge \neg A$ implies $B$, for any proposition $B$. Hence when constructing a formal logical system we want to guarantee that contradictions are excluded, and therefore define propositions in ways that do not allow for the kind of paradox you mentioned.

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  • $\begingroup$ So it's not considered a proposition because it can't be assigned a single truth value? $\endgroup$
    – user185744
    Commented Mar 14, 2016 at 21:11
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    $\begingroup$ Yes see my edits I think I explained a bit better $\endgroup$
    – Exit path
    Commented Mar 14, 2016 at 21:14
  • $\begingroup$ Essentially, this isn't a proposition because assigning it any truth value isn't possible? $\endgroup$
    – user185744
    Commented Mar 14, 2016 at 21:33
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    $\begingroup$ We could imagine a system in which we can assign truth values to any statement. But if this were true, then our system would contain a contradiction, and therefore imply everything, which is undesirable. Hence the definition of a proposition as a statement which is either true or false but not both $\endgroup$
    – Exit path
    Commented Mar 14, 2016 at 21:44
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Statements are very specific type of strings which are constructed recursively:

  1. If $p$ is an atomic proposition, then $p$ is a statement.
  2. If $p,q$ are statements, then: $p\land q$, $p\lor q$, $p\to q$ and $\lnot p$ are also statements.

If you try to construct $p$ which states "$p$ is true" or "$p$ is false" then you quickly realize this is not a well-formed formula.

And just like in natural language, not every string of characters has any actual meaning. The vile bile danced for a while, as the liver was guile.

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  • $\begingroup$ What is a well-formed formula and why are "p is true" and "p is false" not well-formed? $\endgroup$
    – user185744
    Commented Mar 14, 2016 at 23:18
  • $\begingroup$ The text also says that the form of this sentence makes it impossible to designate it as true or as false. Is this what you were referring to? $\endgroup$
    – user185744
    Commented Mar 15, 2016 at 0:41
  • $\begingroup$ A well-formed formula is essentially what I called "a statement" in my answer. It is constructed inductively from atomic formulas. While you might be able to write it "$p$ states that $p$ is [true/false]" this is not a well-formed formula. Since we only assign truth value to well-formed statements, we don't assign a truth value to such $p$. We could assign it a truth value, but then we would lose all sort of properties that we're interested regarding the properties of truth (e.g. $p$ is true if and only if $\lnot p$ is false). $\endgroup$
    – Asaf Karagila ♦
    Commented Mar 15, 2016 at 15:30
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You can use propositional logic to prove this statement is not a proposition:

Suppose the statement, $S$, "This statement is false", is a proposition

That is $S\iff\lnot S$

Then $S$ is true or false, but both of those lead to a contradiction, so our supposition is false:

"This statement is false" is not a proposition.

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    $\begingroup$ $S\iff\lnot S$ is a contradiction for any proposition S. $\endgroup$
    – user185744
    Commented Mar 15, 2016 at 2:09
  • $\begingroup$ @K.Hotz No it's not, it's a false proposition (a proposition that you can reduce to false). $\endgroup$
    – Graffitics
    Commented Mar 15, 2016 at 15:23
  • $\begingroup$ @Graffitics The text I am reading says that a compound proposition that is always false is a contradiction. $\endgroup$
    – user185744
    Commented Mar 15, 2016 at 18:57
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    $\begingroup$ @K.Hotz Oh, you're right, I think I am participating in the confusion. Let's clarify things. When he wrote "$S \Leftrightarrow \neg S$", he meant that at the meta-level, which is the paradox; if you write "$S \leftrightarrow \neg S$", it is a 'contradiction', i.e. a well-formed formula that is always false, but there is no contradiction at the meta-level (no paradox). $\endgroup$
    – Graffitics
    Commented Mar 15, 2016 at 19:06
  • $\begingroup$ @Graffitics Ok. However, I am wondering what it means to say that S (where S is "this statement is false") is true if and only if it is false. When we assign true to S does it "become" false and when we assign it false does it then "become" true, sort of like it's switching between truth values? $\endgroup$
    – user185744
    Commented Mar 15, 2016 at 19:46