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"All statements are false" is NOT false!?!

Pippen August 27, 2017 at 12:03 16575 views 65 comments
Most logicians and philosophers seems to agree that the statement (S) "All statements are false" is plain false, because if S is true it's a contradiction, so by RAA it follows ~S ("Not all statements are false") and since that statement is not inconsistent it must be true.

But I think this is wrong. S is logically & semantically equivalent to (S') "All statements, but this statement, are false and this statement is false". But S' is illogical since its part "this statement is false" can't be given a truth value. Since S = S' this result must also hold for S.

So in fact S is not false, but illogical!!! So who's right, me or the rest of the world?

Comments (65)

Michael Ossipoff August 27, 2017 at 18:49 #100544
If "All statements are false" is false, then either 1) all statements are true, or else 2)some statements are true and some statements are false. Of course, if the statement is false, then that rules out implication #1.

Then the falsity of "All statements are false" would just mean that some statements are true, and some statements are false.

So, for "All statements are false" to be false, doesn't mean that it, itself, can't be false. It only means that there are at least some true statements.

Michael Ossipoff
Michael Ossipoff August 27, 2017 at 18:52 #100546
Quoting Pippen
But I think this is wrong. S is logically & semantically equivalent to (S') "All statements, but this statement, are false and this statement is false". But S' is illogical since its part "this statement is false" can't be given a truth value.


But I remind you that no one's saying that that's so, because people are saying that the statement isn't true. Yes, if the statement is true, then it's false. But if it's false, that doesn't make it true. It just means that some startements are true.

Michael Ossipoff
Michael Ossipoff August 27, 2017 at 22:55 #100610

A.Typo. Here's what I meant to say:

Quoting Michael Ossipoff
Then the falsity of "All statements are false" would just mean that some statements are true, and some statements are false.


Michael Ossipoff
Pippen August 27, 2017 at 23:08 #100620
But if (S) "All statements are false" is illogical - like I want to prove - then it can't be false (and therefore it's negation be true) like the majority wants us to believe. And (S) "All statements are false" seems to have the exact same meaning like (S') "All statements, but this statement, are false and this statement is false", just that the last one explicitly shows what's implicit in the first one. If S' is an illogical statement - which it is - then since S' has the same content than S, S must also be illogical.

I hope it makes my point clearer.
Meta August 28, 2017 at 10:00 #100689
In formal logic directly self referential statements like the one you showed do not exist and can not be defined. So your problem is a problem of naive logic or just playing with words and is not exact at all.

We know other self referential sentences like:
A: A is false
B: If B is true then 1+1=1
C: C is true
D: All statements are false

A and B leads to contradiction directly.
C and D do not, however.

I think it is a matter of opinion whether we say the truth value of C and D can be defined or not. Again: in formal logic these sentences do not exist hecne the problem has no meaning.

My personal opinion is that if there is a truth value (true or false) for which a statement do not imply contradiction then we can potentially assign that truth value to the statement, so D is false.
(If we have 3 truth values: T,F and X then T or X is true imo. your logic can be different, but that has nothing to do with being right or wrong)
Michael Ossipoff August 28, 2017 at 20:46 #100733
The statement:

"All statements are false" is false...

means:

Some statements are true.

That's completely uncontroversial and unproblematic.

Saying that that statement is true would be meaningless, self-contradictory, without truth-value.

Michael Ossipoff
Pippen August 29, 2017 at 01:31 #100799
But I say that "All statements are false" has no truth value and therefore can't be false.

And my proof is simple: We know that (S') "All statements, but this one, are false and this statement is false" has no truth value because of the paradoxical sencond half sentence. But wouldn't we all agree that S' means the same like "All statements are false"? But then "All statements are false" must have the same fate: no truth value.

And that would be huge, e.g. truth skepticism "All statements are false" would be non-refutable instead of plain false.



TheMadFool August 29, 2017 at 10:29 #100881
Reply to Pippen

S = All statements are false
S' = All statements, but S, are false AND S is false

So you're saying S can't be false because S', the equivalent statement, can't be false because of the "S is false" part.

But there's a contradiction in your claim:

All statements, but S, are false is literally saying All statements, but S, are false AND S is true ("but S") and then you contradict this claim by saying, in the latter part, S is false.
Michael Ossipoff August 29, 2017 at 17:38 #100931
Quoting Pippen
But I say that "All statements are false" has no truth value and therefore can't be false.


Are you sure it doesn't have truth-value?

That means you're saying that it can't be true or false. But of course it obviously can be false, as we've discussed.

You seem to be saying that it doesn't have truth value because it can't be true.

The statement is false-if-true, but not true-if-false.

It can be false without any problem..

Michael Ossipoff
Pippen September 02, 2017 at 16:30 #101908
Quoting TheMadFool
So you're saying S can't be false because S', the equivalent statement, can't be false because of the "S is false" part.


Yes, but maybe this proof makes everything clearer:.

If "All statements are false" would have a truth value and therefore would be a statement, it'd follow by universal instantiation + introducing conjunction: "All statements are false and this statement is false". But that deduction has no truth value since it's a conjunction containing "this statement is false" which has no truth value which poisons the whole conjunction eventually. Therefore "All statements are false" cannot be a statement with a truth value as well.

Again: This is HUGE, basically all philosophers agree that truth skepticism (All statements are false) is refuteable, but that seems just false, because it's not even a statement. Is nobody here with real university logic knowledge?

Michael September 02, 2017 at 16:47 #101912
Reply to Pippen Actually, if you look at a truth-table for three-valued logic then a conjunction of a false statement ("all statements but this statement are false") and a statement that is neither true nor false ("this statement is false") is itself false.

So "all statements but this statement are false and this statement is false" is false. Therefore "all statements are false" is false.

Quoting Pippen
which poisons the whole conjunction eventually


Quoting Pippen
Is nobody here with real university logic knowledge?


Which university taught you about logic "poison"?
TheMadFool September 03, 2017 at 08:06 #102036
Reply to Pippen What I'm saying is that your interpretation of ''all statements are false'' is a contradiction from the get go. So, it can't be the correct interpretation.

The only way to make sense of ''P = all statements are false'' is to interpret it as ''all but P are false.'' and that we know is false because we can find at least one statement, other than P, that's true e.g. ''Trump is the president of USA''.
Meta September 03, 2017 at 08:31 #102037
Reply to Pippen
University logic says you did not prove anything. Your statements are meaningless. You're welcome.
edit: Who are the "most logicians and philosophers" you are referring to?
Pippen September 05, 2017 at 12:11 #102630
Quoting Meta
University logic says you did not prove anything. Your statements are meaningless.


Why?

1. We assume (A) "All statements are false" has a truth value.

2. Since we can always go from "All x are y" (x may stand for 1,2,3 here) to for instance "All x are y and 2 are y" without changing the truth value, it must hold that A is logically equivalent to (A') "All statements are false and this statement is false", so A <-> A'.

3. Now we prove that A' can't have a truth value because of its part "this statement is false". This is self referential and can't have a truth value (if it's true it would be false and vice versa), so it follows for the whole conjunction if A'.

4. That leads to a contradiction, because it's impossible that A <-> A' (2.) and A' doesn't have a truth value (3.). That means that the assumption, A has a truth value, leads to a contradiction and is false, A has no truth value.

That looks like a legit proofsketch to me.
Meta September 06, 2017 at 01:34 #102795
Reply to Pippen "Why?"
Because Your arguments can not be formalized. You can't speak about "all statements" formally.

Nvm I like your idea. I just don't see why we should choose your idea of giving truth value instead of the other one.
Nagase September 06, 2017 at 21:11 #103002
Quoting Meta
Because Your arguments can not be formalized. You can't speak about "all statements" formally.


That's incorrect. Assuming a modicum amount of arithmetic, it's possible to formalize the syntax of first-order logic inside a given mathematical theory (say, primitive recursive arithmetic). This technique is known as the arithmetization of syntax and is due to Gödel. Given one such formalization, we can define a predicate S(x) which is true of all and only the (codes of) sentences of the language. So a sentence such as "all statements" would be regimented as "for every x, if S(x), then ... ". Notice that this allows for self-reference, by employing the Carnap-Gödel diagonalization lemma, which is an element in the original proof of Gödel's incompleteness theorem (though it's not essential for proving the result).
Nagase September 06, 2017 at 21:13 #103004
Quoting Pippen
2. Since we can always go from "All x are y" (x may stand for 1,2,3 here) to for instance "All x are y and 2 are y" without changing the truth value, it must hold that A is logically equivalent to (A') "All statements are false and this statement is false", so A <-> A'.


As I mentioned in a private message, I'm not entirely sure this step holds. What is the reference of "this" above? If it is "this statement is false" (taking it to have small scope), then your proposed rule would take us from "All statements are false" to ""This statement is false" is false". But the latter one is false, not truth-valueless. So the whole conjunction is false.
Meta September 07, 2017 at 08:48 #103129
Reply to Nagase I have to correct myself. Can't speak about all statements in a definition of a statement.

edit: But.... If we think about statements more naively and generally (or philosophically) then my statement is correct, because your post is about formal statements and not all statements. I think OP counts every self-referential sentence as a statement. In this sense I was correct in my post.
andrewk September 07, 2017 at 11:11 #103152
Reply to Nagase Using the Godel function we can create a statement that is analogous in some sense to the unary predicate S with the properties you describe. But I don't think we would one formalise 'is false', could we? The Godel function arithmetises syntax, not semantics, and 'false' is concerned with semantics.

IIRC, in some interpretations of Godel, his diagonalised sentence is associated with the purely syntactic notion of 'is provable'.

I am rusty on these issues - It's been a few years since I was involved with them, so I'm happy to have my leaky memory corrected.
Nagase September 07, 2017 at 15:37 #103181
Reply to andrewk

You are right that we can't define a truth predicate for the language in question in the language itself---that's Tarski's theorem (though you can define it in a metalanguage, and one can then restrict the quantifier of the op to "All statements of the object language"; admittedly, this would make his argument evaporate for rather trivial reasons). But there are other options, namely to introduce a new predicate, say "Tr" to the language, try to fix its extension by introducing new axioms, say "Tr("Q") <--> Q" for every sentence Q. Of course, you need to be careful if you want both to preserve classical logic and avoid inconsistencies, and things can get complicated rather quickly here, especially considering iterations ("Tr("Tr("Q")")", for instance). But there are some reasonable ways of doing it.
Nagase September 07, 2017 at 15:49 #103185
Reply to Meta

I'm not sure I understand. What do you mean by "can't speak about all statements in a definition of a statement"? In any case, it seems to me false; in defining the predicate "S", we will presumably use universal quantifiers, e.g. "(x) (Sx <--> ...)", where "..." is the definition of "S". But then, this quantifier will range over all statements (and more besides), so we will be in a sense "speaking about" all statements. If you mean that we can't mention all statements in the definition of a particular statement, that seems false too. We could define a new statement, say Q, such that Q is "All statements imply themselves", which is presumably true.

As for your point regarding taking statements naively, I'm unconvinced. Formal linguistics try to capture what is meant by a "naive" statement, and it's not unreasonable to suppose that in such a theories we quantify over all such statements. There may be other problems lurking in the background, though: if you construe statement broadly enough (naively?), consider a statement about every statement which is not about itself. Is it about itself?

Meta September 07, 2017 at 17:35 #103208
Reply to Nagase Let me clarify what I want to say.
Let's take the Berry paradox. In the paradox there is a definition D which is talking about every (arithmetical) definition. D is a well formed definition in the natural language. However if we could formally model D in arithmetic that would mean arithmetic is inconsistent. When I call a concept "naive" I think of this. So there is a (naive) definition that can exist only outside of formal logic.

Same goes for sentences. We can't have a statement A where the definition of A mentions A. When you are talking about the S predicate you automatically restrict the universe to well-formed formulas and the sentence in the OP is no such statement. So there are statements outside of logic and outside of the S predicate. This is why the OP's sentence is outside of the scope of logic and his proof is not a formal proof.

We can't define Q formally (only in the informal metalanguage). There isn't such definition as
Q <-> (Q->Q and X)
Q is talking about all statements but Q itself is a statement and not part of the (subject) universe of statements.
andrewk September 07, 2017 at 22:41 #103242
Quoting Meta
D is a well formed definition in the natural language.

How can we say it is well-formed when natural language is informal, and hence does not have a notion of well-formed statement? Do you just mean it is grammatically correct? If so, that doesn't tell us much as the statement 'The cheese of five is sad' is also grammatically correct.Quoting Meta
Same goes for sentences. We can't have a statement A where the definition of A mentions A.

Depending on what language we are using, this may be possible. For instance, the sentence:

'This sentence contains the first letter of the alphabet'

is meaningful and true, and can be formalised without difficulty.

Meta September 08, 2017 at 07:48 #103316
Reply to andrewk If we have a first order language L we can define a relation symbol (with one variable) R by saying:
R(x) <-> Phi(x)
Where Phi(x) is a formula of L. R is not part of L so Phi does not contain R.
I call a statement a (naive) well formed definition if there is no such restriction.
In this sense 'This sentence contains the first letter of the alphabet' is not a formal definition.

You can say we can be more general and use higher order logic but higher order logic is also problematic.

edit: And yes i think being grammatically correct and being a well formed statement (or definition or whatever) are very close concepts in the natural language.
Nagase September 08, 2017 at 11:55 #103324
Reply to Meta

There is no need to use higher-order logic. Let T be a first-order theory containing enough arithmetic to capture the primitive recursive functions (you can let T be primitive recursive arithmetic, Robinson's Arithmetic, etc.). Then T has enough resources to code its own syntax: in particular, it has enough resources to express the following notions: , , , , , (where Q is any particular expression of the language: it can be part of the alphabet, a term, a formula, anything), etc. The basic idea here is that we can use numbers to stand as codes for things, for example people (ID cards), products (bar codes), strings of symbols (word processing), etc. Anyway, the important point is that using any reasonable coding scheme, T can prove the following lemma:

Carnap-Gödel Diagonalization Lemma: Let Q(x) be any formula of the language with one free variable. Then there is a sentence D such that T proves "D <-> Q("D")", where "D" is the number coding D.

Notice that the above implies that, for any property Q, there is a sentence D which says of itself that it has the property Q (notice also that this immediately implies that, if the theory is consistent, truth is not definable, otherwise we would have a version of the liar paradox---again, which is not to say we cannot introduce a predicate Tr by axiomatic stipulation). In particular, if we order the alphabet of the language and construct a predicate F(x) such that a (code of a) sentence satisfies F(x) iff it contains the first letter of the alphabet (a tedious, but entirely routine exercise, once you get the hang of the coding machinery), there will be a sentence D which says of itself that it contains the first letter of the alphabet. So that particular sentence can be given a formal definition.

Meta September 08, 2017 at 12:55 #103329
Reply to Nagase This argument fails when applied to a natural language, because there is no precise definition for . Only our intuition can tell us what do we consider a WFF. There isn't a fixed set of relation symbols either.

edit: Any anyways we are not talking about real self-reference just some kind of reflection. Let's modify the statement of the Berry paradox:
"The definition with the least Godel number not definable in fewer than 20 words."
This is also paradoxical.
Or "The statement with the least Godel number that does not contain the first letter of the alphabet."
This can be a paradox.

Nagase September 08, 2017 at 14:05 #103331
Quoting Meta
This argument fails when applied to a natural language, because there is no precise definition for . Only our intuition can tell us what do we consider a WFF. There isn't a fixed set of relation symbols either.


I don't agree with the premises here, but even granting them that's irrelevant: you originally claimed that the op's argument was not formalizable, since it contained self-referent statements. So the claim was about the expressive powers of formal languages, not natural languages. The Carnap-Gödel diagonal lemma literally disproves this assertion: if T is a nice arithmetic theory, for any one-place predicate of the language, we can form a sentence which says of itself that it has that predicate.

Quoting Meta
Any anyways we are not talking about real self-reference just some kind of reflection. Let's modify the statement of the Berry paradox:
"The definition with the least Godel number not definable in fewer than 20 words."
This is also paradoxical.


This just shows that "definable" is not, well, definable in a formal language (even this, strictly speaking, is not true: we can both define definability for a given language in a meta-language with richer expressive resources, or, some times, we can make do with "local" definability---e.g. the constructible sets in ZFC). The problem is with the concept of definability, not with self-reference.

Quoting Meta
Or "The statement with the least Godel number that does not contain the first letter of the alphabet."
This can be a paradox.


But it's not, and is perfectly definable in, say, Robinson's Arithmetic.
Meta September 09, 2017 at 00:03 #103416
Reply to Nagase We use the expression self-reference differently. When I say self-reference I mean (as I have explained earlier) that the definiens can not contain the definiendum. And I am not talking about Godel numbers and codings.

You restrict the universe of statements to a set of formulas of a given language. Even in this case the truth predicate is undefinable so the OP's sentence can't be formalized. Or if you formalize the truth predicate in a metalanguage then when talking about all statements you will only talk about all statements of the object language.

I think OP meant something much more general than statements of a fixed object language L when he was talking about all statements.

So either:
1) "All statements are false" is self-referential (as I mean it), or
2) if we assume Godel numberings and that the sentence talks about its Godel-number then
2a) we don't have a truth predicate defined or
2b) we have a truth predicate but then we have meta statements outside of the quantification range
of "All statements are false".

This is why it can't be formalized.

"But it's not, and is perfectly definable in, say, Robinson's Arithmetic."
But it is because "a" is the first letter of the alphabet and
"The statement with the least Godel number that does not contain the first letter of the alphabet." contains "a".

edit: "the definiens can not contain the definiendum" in an explicit definition.
Pippen September 12, 2017 at 17:56 #104238
Quoting Nagase
As I mentioned in a private message, I'm not entirely sure this step holds. What is the reference of "this" above? If it is "this statement is false" (taking it to have small scope), then your proposed rule would take us from "All statements are false" to ""This statement is false" is false". But the latter one is false, not truth-valueless. So the whole conjunction is false.


You are right with the reference, but if the part "...this statement is false" is false then it'd be true (and vice versa false if it's true), isn't it? So therefoe it can't have a truth value and so does the conjunction.

Just for reference the proof again:

1. We assume (A) "All statements are false" has a truth value.

2. Since we can always go from "All x are y" (x may stand for 1,2,3 here) to for instance "All x are y and 2 are y" without changing the truth value, it must hold that A is logically equivalent to (A') "All statements are false and this statement is false", so A <-> A'.

3. Now we prove that A' can't have a truth value because of its part "this statement is false". This is self referential and can't have a truth value (if it's true it would be false and vice versa), so it follows for the whole conjunction of A'.

4. That leads to a contradiction, because it's impossible that A <-> A' (2.) and A' doesn't have a truth value (3.). That means that the assumption, A has a truth value, leads to a contradiction and is false, A has no truth value.

Michael September 12, 2017 at 18:05 #104239
Quoting Pippen
Now we prove that A' can't have a truth value because of its part "this statement is false". This is self referential and can't have a truth value (if it's true it would be false and vice versa), so it follows for the whole conjunction of A'.


I addressed this mistake here.
Pippen September 12, 2017 at 18:13 #104240
@Michael: We are dealing just in the scope of classical (two-valued) logic and there a statement like "This statement is false" is just without a truth-value as far as I know.
Michael September 12, 2017 at 18:18 #104242
Quoting Pippen
We are dealing just in the scope of classical (two-valued) logic and there a statement like "This statement is false" is just without a truth-value as far as I know.


So not having a truth value is the third option. If we have the conjunction p ? q and if p is false and q doesn't have a truth value then the conjunction as a whole is false.
Michael September 12, 2017 at 18:31 #104244
Although actually there's also Bochvar's internal three-valued logic which has a different truth table to ?ukasiewicz's, and has a conjunction of this kind not having a truth value (e.g. propositions like "thiggledy piggledy and grass is green" are meaningless), which is consistent with your view.

I suppose which to choose is just a matter of preference.
Pippen September 12, 2017 at 21:09 #104281
Quoting Michael
So not having a truth value is the third option. If we have the conjunction p ? q and if p is false and q doesn't have a truth value then the conjunction as a whole is false.


Not in classical logic. Not having a truth value is not a third option there. I think 2. of my proof is dubious, maybe nagase will check that out.
Michael September 12, 2017 at 21:24 #104286
Quoting Pippen
Not in classical logic. Not having a truth value is not a third option there. I think 2. of my proof is dubious, maybe nagase will check that out.


But your argument rests on your own truth table in which there are three options. You're saying that if p is false (or true) and if q doesn't have a truth value then p ? q doesn't have a truth value.

If there wasn't a third option then q ("this sentence is false") must be either true or false.

If it helps, don't think of them as truth values but as predicates. You can have "true" as a predicate, "false" as a predicate, or "neither true nor false" as a predicate.

So the question is on what predicate a conjunction has if one of its operators has "false" as its predicate and the other has "neither true nor false" as its predicate. According to ?ukasiewicz such a conjunction has "false" as its predicate and according to Bochvar such a conjunction has "neither true nor false" as its predicate. Unfortunately I don't know enough to determine how one goes about choosing one truth table over another. Maybe it's simply axiomatic. But as you say, perhaps @Nagase has some insight into the matter.
Michael September 12, 2017 at 22:04 #104299
However, the above might not even be relevant to the particular issue at hand. It could be that your argument conflates. To explain this, consider the statement "this statement is false and grass is red". There are two different ways to interpret this, depending on what we consider to be the referent of "this statement":

1. a) "this statement is false" is false and b) grass is red
2. a) "this statement is false and grass is red" is false and b) grass is red.

We might say that if "this statement is false" is false then "this statement is false" is true, giving us our paradox . But can we say that if "this statement is false and grass is red" is false then "this statement is false and grass is red" is true, giving us another paradox? I don't think we can. So given that grass isn't red, both 2a and 2b are false, and so the conjunction of 2a and 2b is false.

I also think that 2 is the correct interpretation of the statement "this statement is false and grass is red".

Now replace "grass is red" with "every other statement is false".
Srap Tasmaner September 13, 2017 at 01:48 #104327
Reply to Pippen
Let S be the set of all statements.
Let z be the string "If x ? S, then x is false."
Assume z ? S.
If z is true, then z is false.
If z is false, then it is false that if x ? S, then x is false.
Therefore if z ? S, z is false, and it is false that if x ? S, then x is false.

Let z* be the string "z* is false and if x ? S/z*, then x is false."
Assume z* ? S.
Let S/z* be the complement of z* in S.
If z* is true, then z* is false.
If z* is false, then either z* is true or it is false that if x ? S/z*, then x is false.
Therefore if z* ? S, z* is false, and it is false that if x ? S/z*, then x is false.
Pegasys September 13, 2017 at 12:05 #104445
Reply to Pippen

It's true. If "all statements" implies that the statement "all statements are false" is a part of the set of all statements, then "false" is identical. But this does not mean that the statement is false, if the definition of identity is merely reflexive. If however, "identity" is transitive, then this would imply a higher resolution in the proof, implying that more statements are added to the block of proof. And since these statements are elements of the set of all statements, then it follows that some statements which are not on the right side of the equation, are not in the game, but they remain elements of the set of all statements. It follows that the "belief" you mention induces meta-perception, intelligent observation, but this does not imply omni. It only means that self-referentials implicitly define the truth of the statement if "limited truth" is observed by climbing the meta-ladder disregarding the left-hand side, if symmetry is not defined. but how-could it be symmetrical if "identity" is implicit and "false" is not? So it must be true since the interpretation of "all statements" is "false" and "false" is an non-valid intepretation on a lower level but on the level in question it is a "non-valid interpretation" reflecting some symmetrical property of belief dynamics, that is true, and hence a part of the negation of your statement "all statements are false". This implies the importance of the "some" quantifier.

Human nature has a tendency to contradict in order to control environment. Logic is the web in that its elements are of a dual nature, but they only are dual through the lens the spider uses. If an element becomes conscious of itself, then it can only do so by attaching meaning to the situation, to the set it belongs to, like using Gödel numbering for example. Then this element needs to evoke contradiction in order for its statements to make sense, but only to itself, the rest may challenge the element or not. It's a game for machines and rises the question if machines are conscious or not, amongst other things, dependend on the individual, or machine, or element, dependent on the sociology involved.
Owen September 18, 2017 at 19:20 #105937
Ep(p).Reply to Pippen

(All statements are false) is false, and it is equivalent to
(Some statements are true).

1. (All p)(~p) -> ~q.
2. (All p)(~p) -> ~(~q).
3. (All p)(~p) -> q.
4. (All p)(~p) -> (~q & q).
5. ~(All p)(~p).
6. (Some p)(p).

Also,
(All statement are true) is false, and it is equivalent to
(Some statements are false).
Pippen September 26, 2017 at 17:35 #108580
Here is what I found out so far:

The logicians formulate "All (S)tatements are (F)alse" as follows: All x: (Sx -> Fx). If you do that you can indeed prove that this is just plain false since if it's true it's a contradiction and so by RAA it's its negation that is consistent.

But why do I have to formulate the statement like above? Why can't I just formulate: All x: (Sx & Fx). This statement is not false, it is not well formed since it entails the liars paradox.

Both versions of the upper statement say roughly the same - that every x in the set of statements is false - but their form is different and so are their results. So who is right? Or why am I wrong?
Srap Tasmaner September 26, 2017 at 20:18 #108602
Quoting Pippen
The logicians formulate "All (S)tatements are (F)alse" as follows: All x: (Sx -> Fx). If you do that you can indeed prove that this is just plain false since if it's true it's a contradiction and so by RAA it's its negation that is consistent.

But why do I have to formulate the statement like above? Why can't I just formulate: All x: (Sx & Fx). This statement is not false, it is not well formed since it entails the liars paradox.

Both versions of the upper statement say roughly the same - that every x in the set of statements is false - but their form is different and so are their results. So who is right? Or why am I wrong?


Because [math]\small ?x(Sx?Fx)[/math] and [math]\small ?x(Sx ? Fx)[/math] don't say the same thing.
BlueBanana September 26, 2017 at 20:48 #108606
Quoting Owen
(All statements are false) is false, and it is equivalent to
(Some statements are true).


A small error there: ((All statements are false) is false) is equivalent to (some statements are true). (All statements are false) is equivalent to (there are no statements that are true).

And this is, of course, only considering the classical logic with only two truth values, which we aren't considering here: more correct statement would be ((All statements are false) is false) is equivalent to (some statements are either true or have no truth value). This actually proves OP wrong, because (this statement is false) implies (there is a statement with no truth value), which of course implies (there is a statement that is either true or has no truth value), which then implies ((All statements are false) is false).
Pippen September 29, 2017 at 19:36 #109580
Quoting Srap Tasmaner
Because ?x(Sx?Fx) and ?x(Sx?Fx) don't say the same thing.


Ok, but why is my formulation of "All statements are false" as ?x(Sx?Fx) impossible and only ?x(Sx?Fx) the correct formulation? Is there any reason? Because, again, both formulations lead to different results for the proposition. In case of ?x(Sx?Fx) it's false, in case of ?x(Sx?Fx) it's not well formed.
Srap Tasmaner September 29, 2017 at 20:16 #109610
Reply to Pippen
One says, if x is a statement, then it is false. There is a longish tradition of so interpreting universal statements (Russell and Ramsey both for slightly different reasons), and thus denying them existential import.

The other version says, everything is a statement and everything is false. Is that what you want to claim?
Pippen October 06, 2017 at 15:58 #111889
I assume a set of all statements as the domain. Then indeed ?x(Sx?Fx) means: everything is a statement and everything is false. But that's exactly what "all statements are false" says, isn't it? Why can't I formulate it that way? My guess is there is no real reason, it's just an agreement that we formalize universal statements as implications (and use universal sets as domains).

But the consequences are huge, because if my case is formally possible then ?x(Sx?Fx) fails to be true or false in the one and only case of applying it to itself. It'd be false because it says so, it'd be true because it's false and says so. Wouldn't that be enough to make ?x(Sx?Fx) a not-wff-formula?
Srap Tasmaner October 07, 2017 at 01:16 #112034
Reply to Pippen
Read this again. I did it that way specifically to leave room for the conclusion that the claim must not be a statement (i.e., not truth-apt), and that would be the result for the Liar. (You could think of that as concluding "If there's a set of all truth-apt strings, the Liar isn't in it.")

There was no reason to conclude either version of the claim is not truth-apt; both versions must be false.
Srap Tasmaner October 07, 2017 at 01:21 #112035
Reply to Pippen Besides which, there are loads of reasons to think this claim is simply false. You'd have to have a language without the logical constants.

Even if you changed it to "All atomic sentences of L are false", then the question is: under what interpretation? Systematic falsehood under one interpretation is systematic truth under another.
Michael October 07, 2017 at 10:51 #112143
Quoting Pippen
But the consequences are huge


They're not. Whether you want to say "all statements are false" is false or "all statements are false" is neither true nor false, it is still the case that "all statements are false" isn't true and "at least one statement is true" is true.
Pippen October 07, 2017 at 16:01 #112164
Quoting Michael
They're not. Whether you want to say "all statements are false" is false or "all statements are false" is neither true nor false, it is still the case that "all statements are false" isn't true and "at least one statement is true" is true.


Yeah, but that statement wouldn't be false either! And it's a huge difference if a statement is false or not false (no matter if it's true or not besides that).

In summary, my problem is why nobody interprets the statement "All sentences are false" as "All sentences are false and this very sentence is false", because in this version the whole thing wouldn't be true or false. I just don't see an error in infering one from the other.
Michael October 07, 2017 at 16:48 #112172
Quoting Pippen
And it's a huge difference if a statement is false or not false (no matter if it's true or not besides that).


Not really. What difference does it make if we consider the sentence "grass is red and wub a lub a dub dub" to be false or neither true nor false?

Either way, it isn't true. It doesn't describe some fact about the world.
Pippen October 10, 2017 at 15:22 #113430
@Michael: It may not be true, but it wouldn't be false either. It would hang in the middle between the two values.
Srap Tasmaner October 10, 2017 at 18:27 #113463
Quoting Pippen
In summary, my problem is why nobody interprets the statement "All sentences are false" as "All sentences are false and this very sentence is false", because in this version the whole thing wouldn't be true or false. I just don't see an error in infering one from the other.


You can make the inference if you like, but the second version is also false, as I showed a month ago. I'll do it again:

Let S be "S is false and all statements are false".

Now we try assigning truth values to S to see if it is possible for S to be true or false. We may find that it must be assigned the value "true" in all models of English (that it is a tautology), that it must be assigned the value "false" in all models (that it is a contradiction), or that it can be true in some models and false in others, like most statements, or that it cannot be assigned a truth value in any model (like the Liar).

1. Assume S is true.
2. If S is true, then both conjuncts are true.
3. If both conjuncts are true, then the first is true, so it is is true that S is false.

This is contradiction, because at the moment we are assuming S is true. So there are no models of English in which S is true. But perhaps S is not false either, as you claim.

4. Assume S is false.
5. If S is false, then at least one of the conjuncts is false.

We now try each conjunct in turn. First conjunct:

6. Assume it is false that S is false.
7. If it is false that S is false, then S is true.

But we assumed in (4) that S is false, so this is a contradiction; thus there are no models of English in which S is false and its first conjunct is false. Second conjunct:

7. Assume it is false that all statements are false.

Here at last we have a possible truth-value assignment that doesn't immediately produce a contradiction. Thus all models of English must assign truth values this way: S is false (our assumption 4) and it is the second conjunct that is false, so it is also false that all statements are false. This is the only possible way to assign truth values without the model contradicting itself.

Your view I think is something like this: the usual way of determining whether a conjunction is a tautology, a contradiction, contingently true or false, or just not truth-apt at all, is to assign all possible truth-value combinations to the conjuncts and use a little truth table to see how the conjunction comes out. An arbitrary P & Q is

T & T : T
T & F : F
F & T : F
F & F : F

so it could be either true or false. On the other hand, P & ~P goes like this:

T & F : F
F & T : F

It's always false. So you're thinking that since you have, in essence, "[the Liar] & P" as your conjunction, we'll be unable to construct a truth table because the first conjunct is not truth-apt. True.

But conjunction is a short circuit with respect to falsehood, as disjunction is with respect to truth. If we can establish that one of a pair of conjuncts is false, we are never forced to evaluate the other conjunct to know that the conjunction as a whole is false. In my version, I never do assign a truth value to the first conjunct, but it doesn't matter. In essence we end up treating it as a pseudo-conjunction: blah-blah-blah P, and P happens to be a contradiction.

So we could skip most of the steps above and go that way instead. We show that "All statements are false" can only be false, which you agree to, and then short-circuit any conjunction it appears in.

I claim this is reasonable because whether to count some string of English words as a statement at all, as truth-apt, is up to us. There is no formal system on offer here to tell us whether some string is or isn't well-formed. We can rule out anything we like, but we can only rule in strings we can successfully assign a truth value to. We can't rule in the Liar. But we can rule in "All statements are false" by calling it always false, and we can rule in "This statement is false and all statements are false" by calling it always false. We're not compelled to, but we can.

ADDED: I would claim further that this approach is reasonable precisely on the grounds that we would ordinarily expect "All statements are false" and "This statement is false and all (other) statements are false" to be equivalent.

Your version has a truth-apt statement being equivalent to the Liar. If we can avoid that, we ought.

AND STILL MORE: Note that short-circuiting is consistent: "Dinosaurs are extinct" is contingently true; there are possible models in which it is false. But "Dinosaurs are extinct and all statements are false" is always false. "It's raining or it's not, and all statements are false" is always false, even though "It's raining or it's not" is always true.
Jake Tarragon October 10, 2017 at 20:21 #113490
Quoting Meta
In formal logic directly self referential statements like the one you showed do not exist and can not be defined.


Surely there's a "New Logic" somewhere that incorporates them? An extension of formal logic ...
Meta October 10, 2017 at 21:22 #113506
Reply to Jake Tarragon
I think a better expression instead of self- referential would be self-containing maybe?
Like the symbol of the meta language denoting a sentence can not be a part of that sentence.

You must be right but these new logics must not have the same axioms as our classical one and must be weaker in at lest one aspect. (Could be stronger in other aspects.)
Pippen October 10, 2017 at 21:27 #113507
Quoting Srap Tasmaner
So you're thinking that since you have, in essence, "[the Liar] & P" as your conjunction, we'll be unable to construct a truth table because the first conjunct is not truth-apt. True.


Yes, that's my point.

Now, a conjunction of the form "[liar] & P" is neither a conjunction nor a proposition at all, because a conjunction is formally defined as "p & q" and p,q need to be propositions with truth values. That's just predetermined in the language of predicate and propositional logic. So "[liar] & p" can't be true and can't be false, it's just syntactically not a wff formula if I am right. I see your point and you are quite right from a practical point of view, but formally it's not possible I think.
Srap Tasmaner October 10, 2017 at 22:04 #113518
Reply to Pippen
If your point is only that this sentence can't be represented within classical logic, then duh.

I still say the English sentence in question can justifiably and consistently be considered false.

Logic is a tool, one of my faves, but it does not sit in judgment of natural language, which may very well be just as formal only vastly more complex and powerful.
Pippen October 10, 2017 at 22:21 #113523
Quoting Srap Tasmaner
If your point is only that this sentence can't be represented within classical logic, then duh.


That's my point. The problem is this: To prove that (S) "All sentences are false" is not a wff, I have to assume that S = S' with (S') "All sentences are false and this sentence is false". Then I can prove that S' is not a wff and since S = S' then S can't bei either, but how can a not well formed formula like S' be equal/equivalent to anything at all? It's not working. This is the reason why I wanted to formalize "All sentences are false" right away as a conjunction "All sentences are false and this sentence is false" and wondered if it would be possible and if not why.
Srap Tasmaner October 10, 2017 at 22:26 #113525
Reply to Pippen
You can't do this in classical logic because you would need the predicates "… is a statement" and "… is false". You can't have either of those. Classical logic is swell, but it gets some of its swellness from being carefully circumscribed.
Meta October 10, 2017 at 22:28 #113527
Reply to PippenSticking with naive logic. You say that the statement in the OP cant be false because of the liar paradox.
The liar paradox says that the statement A for which A <-> not A can't have a truth value.
Your statement has the form A <-> not A & B
This sentence is not contradictiory in itself because A can be false. There is no need to add a new truth value. However it is not prohibited. Just totally unnecessary.

Edit: Saying A is a paradox because it has similarities with the liar paradox is dogmatic and irrational imo and totally misses the point of what makes a paradox paradoxical.

Edit2: Let's say X is the statement: Every statement is either true false or a paradox.
Now X has similarities with the liar paradox and based on your logic it is a paradox. But I would say X is trivially true.
Pippen October 10, 2017 at 22:45 #113536
@srap: Why can't I have "is a statement" and "is false" in predicate logic?

@Meta: Your formulation in propositional logic is too simple, you can't express there what "All statements are false" want to say. In propostional logic, as a matter of fact, the liar paradox is just plain false since it says there: p <-> ~p.
Meta October 10, 2017 at 23:09 #113546
Reply to Pippen I was a bit sloppy there. When I said A <-> not A I meant that the equivalence is the defining equivalence of A and in order to be a valid definition the equivalence must be true. The liar paradox is there. In propositional calculus. Not A cant be the definition of A.

For me the fact that the negation of a sentence occurs at the right side of the definition of that sentence does not make it a paradox (in naive logic), see my previous post.

Question: do you think that "All statements are true, false or paradoxical." has no truth value?
Srap Tasmaner October 10, 2017 at 23:17 #113551
Quoting Pippen
Why can't I have "is a statement" and "is false" in predicate logic?


The short answer is: to avoid crap like this. The predicate logic we use was designed to formalize mathematics. It's supposed to help, not hinder, and there is no reason to think you can do everything in it.

The long answer is: in part Tarski, and in the other part some metalogic I don't know.

Partial explanation, ignoring the metalogic, which someone else would have to speak to: truth is the fundamental primitive in the system. It's already there, so there's no reason to introduce an "… is true" predicate. (This is easy to overlook because modern notation dropped Frege's early assertion stroke and judgment stroke. Start reading "p" as "It is true that p" and you'll see what I mean.)

If you had an "… is true" predicate, you'd want to apply it to propositions. But it's not perfectly clear that propositions are objects. What we've got is a way to say things like "If x is divisible by 5, then x is not prime." Does what's in between the quotation marks look like an object to you?

Keep in mind: you should not be able to get the result you want. Your argument, in brief, is that if A is false and A is equivalent to B which is neither true nor false-- wait, what? If it is equivalent, this won't happen. If they're not equivalent, then you've nothing to say. Pick your poison.

I think what you really have is an "apparent" paradox. That is, an ambiguity. You use the ordinary English word "equivalent" to mean something like, "Another way to say this is …" but then you take it to mean "must have the same truth value."
TheMadFool October 11, 2017 at 07:03 #113669
Quoting Pippen
So in fact S is not false, but illogical!!!


S = all statements are false
S' = all statements but S is false AND S is false
--------------------------------------------------------------------------------
Argument A
You say S can't be true because that would make S false: the contradiction S & ~S. Then you conclude, by RAA, that S is false.

--------------------------------------------------------------------------------
Argument B. You go on...
S = S'

You say S' can't be false because of the clause ''S is false'' which is the Liar statement and is neither true nor false. So S can't be false too.
--------------------------------------------------------------------------------

Your argument rests on the assumption that a conjunction of a proposition with a nonproposition can't have a truth value. Am I right?

The problem is such conjunctions, as you've used, are syntactically wrong. Conjunctions, or any logical operator, are restricted to propositions. Since ''S is false'' (the Liar statement) isn't a proposition, S' is not a proposition and so has no truth value. Do you mean this?

What's important is that S is interpreted as ''all statements but S are false''.

Meta October 11, 2017 at 07:50 #113674
In fact if you want to define S without having a truth value then you have to give up the X <-> X & (A or not A) tautology.
Because lets say A <-> X is true.
Based on your logic A <-> X & (A or not A) does not have a truth value. And therefore X and
X & (A or not A) are not equivalent.
Your logic is fatally broken I think. (Unless you specify it even more to get a standard naive multivalued logic)
Pippen October 19, 2017 at 02:39 #116412
First of all I disagree with Srap that you can't have a predicate "is true/false" in PL. I think you can have, but sometimes it leads to inconsistencies or violation of formal rules and only in that last case it is forbidden. That's why "This statement is false" is forbidden in PL, because you couldn't assign a truth value to it which is necessary to apply PL.

I disagree with Meta that one can construt the liar paradox in propositional calculus. It's too weak.

I think I found the solution (at least for myself). "All statements are false" can never be equivalent to "All statements are false AND this statement is false" because the last sentence would not be a wff in predicate logic. From "Alle statements are false" one cannot deduce "All statements are false AND this statement is false" because of the same formal reason. Therefore I cannot prove "All statements are false" to be illogical. I could only argue that when you say "All statements are false" you really mean "Alle statements are false AND this statement too" and that's indeed illogical. And of course that's what a truth skeptic would argue. But this is not logic anymore, but how you formalize everyday-language into PL.