1

In a previous question I relied on the notion of an "illogical statement" which led to some debate and I ended up making its definition an addendum to the question. I'd like to ask whether the notion of an "illogical statement" as defined here is a worthwhile concept, is it a concept that exists already, is it logical or not, and is it fair to apply the notion to Gödel's modified liar paradox?

It would seem to me to have merit to define an illogical statement as follows: An illogical statement is a statement whose truth value is the inverse of its own truth value.

1: (Sentence $s_1$ is illogical if $s_1\implies\neg s_1$ or $\neg s_1\implies s_1$)

2: Let $s_1$ be the Liar Paradox "This sentence is false.".

3: Rewriting 2: $s_1\implies \neg s_1$

4: From 3,1: $\implies s_1$ is illogical.

And for the modified Liar: Modified Liar: This sentence is unprovable:

5: Let $s_2$ be "This sentence is unprovable."

6: $(s_2\implies s_2)\implies\neg s_2$

7: $s_2\implies\neg s_2$

8: From 7,1: $\implies s_2$ is illogical.

It would seem to me that we should only permit true and not false statements to be accepted as true. This rule would appear to be the implicit basis of proof by contradiction. In a proof by contradiction, we suppose a proposition to be true, show that this supposition leads us to the conclusion that a statement is "both true and false" and therefore it is not accepted to be true on the grounds that it isn't "true and not false".

And it would seem that as a basic rule, illogical statements should not be permitted by the metatheory since they introduce a contradiction to the logical system. The modified liar, the moment we say it, is both true and false and therefore is not "true and not false" so it would seem to me that by the same rule as any proof by contradiction, the moment we suppose the modified liar, we can immediately reject it by contradiction.

As surely it's rejected by contradiction regardless of the system into which it is introduced. So the conclusion that the decidability of the modified liar is independent of some set of axioms does not imply that it is undecidable in the combination of those axioms and the metatheory, since the metatheory in isolation already provides for decidability that the modified liar must be rejected.

The problem seems to arise for most logicians, when they extend this conclusion by saying that since the modified liar is rejected, it is not true, so it must be false. But this step is only valid for sentences whose truth and falsehood are mutually exclusive. The modified liar does not possess that property. Were it true, it would not be "true and not false" and were it false, it would not be "false and not true". Therefore with statements like this it is incorrect to make the leap that if it is not false it must be true.

I'm no maths professor (far from it) and I accept I'm naively proposing a challenge to a deeply-founded concept here so in anticipation of the inevitable cascade of downvotes please accept my sincere apologies for any misconception on my part.

UPDATE (Post accepting the answer): My current thinking is that it would be best to define an illogical statement as:

An illogical statement is a statement which is incapable of being true without being contradictory in any system in which it is provable and in which $s\implies s$

  • Very quick question: Under your characterisation the truth-teller comes out as not illogical (unless maybe your characterisation in (1.) was meant to be non-exhaustive), I'm not clear on this. – Neil Barton Aug 10 '16 at 11:25
  • Sorry @NeilBarton could you elucidate? I don't understand your question. I don't know if this answers your question but if you mean that the law 1. Sentence $s_1$ is illogical if... is not necessarily entirely equivalent to "statements which are not true and not false or false and not true"; then yes, it is a special case. It was perhaps a bit lazy of me not to make them exactly equivalent. – it's a hire car baby Aug 10 '16 at 12:55
  • 3
    "This sentence is unprovable" (when made rigorous by referring to a particular notion of formal proof) does not imply that itself is false. It only implies that itself is unprovable. Indeed, there are models where the quoted sentence is true and not false, and other models where it is false and not true. So line 6 in the question does not seem to follow from line 5. – Carl Mummert Aug 10 '16 at 13:27
  • I don't know who you are referring to by "most logicians", but when you bring your question into the scope of mathematical logic, then any reasonable mathematical formalisation of your modified liar paradox is true unless it the notion of provability that has been formalised is for a system that is either uselessly weak or inconsistent. – Rob Arthan Aug 10 '16 at 13:32
  • @CarlMummert My reasoning at that point was that if we adopt $s_2$ as true then it becomes trivially provable by virtue of $s_2\implies s_2$ and therefore it is can never be "true and not false". Would it overcome your objection therefore if I redefined an illogical statement as one which can never be assigned "true and not false" or "false and not true"? – it's a hire car baby Aug 10 '16 at 13:45
  • 2
    Well, the statement $1 = 1$ is certainly true and not false, and that statement should not be illogical. It is the case that, if we adopt $s_2$ as an axiom then it becomes trivially provable, but $s_2$ does not say it is unprovable from "any means whatsoever" - inside, $s_2$ has to specify a particular formal proof system in order to be a precise statement. If we considered another sentence $t$ that said "I am not provable even if you assume me as an axiom" then $t$ would simply be false. The phrase "I am not provable" leaves out the key point: the specific definition of "provable". – Carl Mummert Aug 10 '16 at 13:50
  • @RobArthan I think my question is asking then, can and should those reasonable systems be modified to only accept as true, statements which are both true and not false, rather than only requiring truth. Would such a formulation necessarily be uselessly weak or inconsistent? – it's a hire car baby Aug 10 '16 at 14:04
  • 2
    In the formal systems that are used in mathematics, in any particular model, it is impossible for a statement to be true and false simultaneously, so saying that a statement is true is the same as saying it is true and not false. So it seems that the modification you are suggesting is already part of the usual systems. Can you name any formal system in which a statement can be both true and false? The liar sentence is not expressible in the usual formal systems exactly because it can't have a particular truth value. – Carl Mummert Aug 10 '16 at 14:08
  • Do I understand correctly that the liar can't be expressed due to Tarsi's indefinability of truth, but the modified liar can? If so, I think I'm arguing the same should go for the modified liar. Is there any reasonable formal truth system in which $s_2$ being true, would not imply $s_2\implies s_2$, implying that $s_2$ would be provable, implying that $s_2$ is false and is therefore $s_2$ is not capable of ever being "true and not false"? So its unprovability in some language is a trivial matter. – it's a hire car baby Aug 10 '16 at 14:14
  • The modified liar can be expressed because provability is arithmetically definable. It's unclear what you are asking for in your sentence beginning "Is there any reasonable formal system". – Rob Arthan Aug 10 '16 at 14:21
  • I guess to find any system of logic would be a good start, in which the following argument is contradicted: $s_2$ being true by any means whatsoever - proof, being self-evidently true, assumption, or adoption as an axiom - would imply $s_2\implies s_2$, implying that $s_2$ would be provable, implying that $s_2$ is false. Therefore that $s_2$ is not capable of ever being true and not false, by any means whatsoever. – it's a hire car baby Aug 10 '16 at 14:39
  • @CarlMummert: Right, and I hope my answer can make the meaning of your two comments clear. Robert, you could indeed say that Tarski's undefinability theorem does not allow you to construct the liar paradox, but the modified one can't be constructed either because provability is relative to the formal system. – user21820 Aug 10 '16 at 17:30
  • @CarlMummert Please correct me if I'm mistaken but I think you've missed the point, namely that in any reasonable system, i.e. in which $s\implies s\forall s$, if the modified liar were true, it would be trivially provable by virtue of $s\implies s\forall s$ and therefore it is not capable of ever being true and not false or false and not true in any reasonable system. It is therefore just as illogical as the original liar. I think my grasping of your argument requires that I better understand the systems in which the modified liar is true and not false or false and not true – it's a hire car baby Aug 11 '16 at 07:54
  • 2
    It is not the case that a statement that happens to be true in a given model is provable. The usual systems do prove $s \to s$ for all statements $s$ but that does not mean that any statement true in any model is provable. You can't use $s \to s$ to prove $s$ by modus ponens unless you also assume $s$ as an axiom; just because a statement is true does not make it an axiom. I would recommend working carefully through an intro logic textbook, such as the one by Enderton, to learn more about the distinction between truth in a model versus probability. – Carl Mummert Aug 11 '16 at 11:18
  • Thanks for accepting my answer. =) And feel free to come to the logic chat-room if you wish to clarify any point, as I'd be glad to explain in detail if you like. If I'm not there just ping me and I'll respond the next time I'm on Math SE. – user21820 Mar 15 '17 at 05:51
  • @CarlMummert I have grasped your argument here about the possibility that some statement might be true but unprovable in a model and therefore the modified liar has the capacity to be non-contradictory. Therefore I would like to propose the following theorem: An illogical statement is a statement which is incapable of being true without also being false, in any system in which it is provable. Have I now reached an appropriate definition? – it's a hire car baby Nov 08 '17 at 11:05

1 Answers1

11

Your question has two main facets. The first is that you did not grasp the way logic does not fall to the liar paradoxes. The second is that there are deeper reasons as to why we have such apparently innocuous sentences in natural language that seem to defy assimilation into formal logic systems. $ \def\nn{\mathbb{N}} \def\prov{\square} \def\t#1{\text{#1}} \def\pa{\t{PA}} \def\eq{\leftrightarrow} $

Why the original liar paradox fails

In classical logic one is only allowed to refer to objects that exist. Because of that, it is impossible to construct the liar sentence, because it is equivalent to:

??? Let $P$ be a sentence such that $P$ is equivalent to $\neg P$.

Which is clearly invalid because we have not proven that such a sentence exists. Indeed, we can show that no such sentence exists. Similarly the barber paradox fails for the same reason, namely that no such barber with the specified property exists.

Why the modified liar paradox fails

For the modified liar paradox, you have made the common error in assuming that provability is a well-defined absolute notion. It is not. Provable in what formal system? So even if we ignore the problem with "this", the string "This sentence is unprovable." is in fact meaningless. It can easily be that "S is unprovable in PA" is true while "S is unprovable in ACA" is false! (Here PA and ACA are two actual formal systems but it is irrelevant here.)

It turns out that given any formal system $S$ with some nice properties, and any sentence $P$ over $S$, there is in fact a sentence denoted by "$\prov_S P$" in the language of arithmetic such that $S$ proves $P$ iff $\nn \vDash \prov_S P$. In English terms, the provability of a sentence over a sufficiently nice formal system is equivalent to whether the collection of natural numbers satisfies some arithmetical sentence. (It is important to note that we are working in some meta-system that has a notion of the collection of natural numbers and understands string manipulation, so that this all makes sense.)

Now this means that a sentence $P$ is unprovable over $S$ iff $\nn \vdash \neg \prov_S P$. Let us see what happens when we try to set up the modified liar paradox in $S$:

Let $G$ be a sentence over $S$ such that $S \vdash G \eq \neg \prov_S G$.

Guess what? Such a $G$ actually exists, which is now called the Godel sentence for $S$. The proof of this fact is the crucial core of Godel's incompleteness theorem.

Yet the paradox vanishes! Let us see what we might try to do within $S$.

Within $S$:

  $G \eq \neg \prov_S G$.

  If $G$:

    $\neg \prov_S G$.

    $\color{red}{\prov_S G}$.   (WRONG!!!) [Even if $G$ is true, $S$ may be unable to prove "$\prov_S G$".]

    [So no contradiction.]

  If $\neg G$:

    $\prov_S G$.

    $\color{red}G$.   (ALSO WRONG!!!) [Even if $\prov_S G$ is true, $G$ may not be.]

    [Again no contradiction.]

In fact, strangely but truly, given any sentence $φ$ over $S$, if $S \vdash \prov_S φ \to φ$, then $S \vdash φ$. This fact is known as Lob's theorem.

However, $S$ (if sufficiently nice) satisfies the Hilbert Bernay's provability conditions, so the following argument is valid:

$S \vdash G \eq \neg \prov_S G$.

If $S \vdash G$:

  $S \vdash \prov_S G$.   [by (D1)]

  $S \vdash \neg G$.

  $S$ is inconsistent.

If $S \vdash \neg G$:

  $S \vdash \prov_S G$.

  If every arithmetical sentence that $S$ proves is satisfied by $\nn$:

    $\nn \vDash \prov_S G$.

    $S \vdash G$.

    $S$ is inconsistent.

Technical details aside, this is essentially Godel's proof of his first incompleteness theorem, and shows that if $S$ is consistent and is arithmetically sound (namely that every arithmetical sentence proven by it is satisfied by $\nn$), then $S$ neither proves nor disproves $G$. Rosser later proved that the incompleteness theorem holds even if we drop the condition of arithmetical soundness completely, and Kleene found a very elegant computation-theory argument for the strengthened theorem that I detailed in this post.

Why Curry's paradox fails

The above is what is meant when some people say that a correct analysis of the liar paradox yields Godel's incompleteness theorem. Similar analysis of Curry's paradox yields Lob's theorem. I leave this as an exercise for the reader.

Why Quine's paradox fails

People always ask about the liar paradox, but it uses self-reference and so some might say that the problem lies with self-reference. That is not true. There is a deeper paradox in natural language that does not rely on self-reference at all, called Quine's paradox. I shall give the clearest version I can construct here.

It seems that we can always interpret "true sentence" to mean "true statement about reality" unless otherwise specified, and then every sentence is clearly either a true sentence or not a true sentence. Now consider the following sentence $Q$:

" preceded by the quotation of itself is not a true sentence." preceded by the quotation of itself is not a true sentence.

Flawed argument: Notice that $Q$ is a grammatically well-formed sentence of the form X is not a Y., so it ought to be true or false. If $Q$ is a true sentence, then by what it claims, $Q$ is not a true sentence. Therefore $Q$ is not a true sentence. But then by what it claims, $Q$ is a true sentence. Thus we get a contradiction.

Where is the error? Think for a while before continuing!

The answer is that classical logic is essentially based on the fact that there is only one reality, and so when we consider totally precise and unambiguous statements about this reality, it is necessarily the case that every such statement is either true or false about this reality, meaning that either it correctly describes reality, or it does not correctly describe reality. We do not care about statements that are ambiguous or not well-defined, just like we do not care about meaningless nonsense like ß\EÂ{8ÄäÉ5¨5;-c1÷ÌOm¶ÑzYè:ÏÁôví2QêIxú·9Ñ5u¤­åÉ¡nçßów⧸}tì-Ì«ÞB8r%sHÛæW¯*".vD.

The key is that we will never be able to justify that $Q$ is a statement about reality, and so we cannot apply the law of excluded middle (LEM) to it. In other words, $Q$ is neither a true sentence about reality nor a false sentence about reality, since it is not even a sentence about reality! (And we can actually justify this claim, as I shall do so below!)

Every sentence can be considered as a string of symbols in reality, so we can rightly say that $Q$ is a sentence, and that "$Q$ is a sentence" is a sentence, but once we want to say something about truth, it may not be a statement about reality anymore. Let us see what we can and cannot say regarding $Q$.

First note that $Q$ is the exactly the same string as " preceded by the quotation of itself is not a true sentence." preceded by the quotation of itself.

Thus $Q$ is equivalent to asserting that $Q$ is not a true sentence.   [(*)]

If $Q$ is a true sentence:

  $Q$.   [We can state $Q$ since it is a true sentence.]

  $Q$ is not a true sentence.   [By (*).]

  Contradiction.

If $Q$ is not a true sentence:

  $Q$.   [By (*).]

  $Q$ is a true sentence.   [What we can validly state must be a true sentence.]

  Contradiction.

But we do not have LEM for "$Q$ is a true sentence", since we did not prove that it is a statement about reality!

Thus all we can say is that "$Q$ is a true sentence" is not a statement about reality, and likewise $Q$ is not a statement about reality.

Why Berry's paradox and the Surprise test paradox fail

In general the above approach can be used to resolve all logic paradoxes, including Berry's paradox and the Surprise test paradox, which I shall leave as more exercises for the reader. They can also be translated to be in terms of provability, which is instructive to analyze (see this question and the answer).

user21820
  • 57,693
  • 9
  • 98
  • 256
  • Thanks. It's going to take me a little while to digest that. – it's a hire car baby Aug 10 '16 at 21:17
  • 3
    @user21820: so please tell us, in less than 100 words, how your essay answers the questions in the first paragraph of the question? – Rob Arthan Aug 10 '16 at 21:36
  • I haven't digested properly yet but... Do you understand my argument correctly; that even if we append to the end of the modified liar the phrase 'in X' where X is any reasonable system of logic in which $s\implies s$ for all $s$, there are no means by which it can ever be true and not false or false and not true and that fact is independent of X. Or are you disagreeing with this? – it's a hire car baby Aug 10 '16 at 21:38
  • I don't necessarily think that the existence of illogical statements means they defy assimilation. They just need to be categorised as illogical and handled accordingly. – it's a hire car baby Aug 11 '16 at 06:55
  • 1
    @RobertFrost: Let me answer your last comment first. I said "seem to defy" and anyway it's a natural language statement with all its ambiguity. Different logicians and philosophers have different ways of handling them, and they often disagree too. I've presented what I feel is the most satisfactory, because just saying it is "illogical" is useless if you cannot give an unambiguous explanation of how to classify which are "illogical" and a meaningful reason. "Assuming it leads to a contradiction." is not a meaningful reason for rejecting it! – user21820 Aug 11 '16 at 11:00
  • @RobertFrost: Now for your second-to-last comment. Please read my answer first, because it already addresses explicitly the version with any arbitrary sufficiently nice formal system $X$. By the way, "true and not false" is meaningless in classical logic, so your question makes no sense. – user21820 Aug 11 '16 at 11:14
  • @RobArthan: The first paragraph of the question refers to a question that was not linked and I'm not going to waste my time hunting for it. But if you're asking about the answer to I'd like to ask whether the notion of an "illogical statement" as defined here is a worthwhile concept, is it a concept that exists already, is it logical or not, and is it fair to apply the notion to Godel's modified liar paradox? the answers are "You didn't define it." and "I don't know if it exists because you've not defined it." and "Without defining it you can't apply it." – user21820 Aug 11 '16 at 11:20
  • @user21820 in reply to your last comment, to Rob Arthan... I've attempted to define an "illogical statement" in the question, and by later refinement and clarification in the comments, as follows: A statement $s$ is illogical in some formal system $X$ if, in the system $X$, $s$ cannot be true and not false, nor can it be false and not true, independently of whether it $s$ provable in $X$. If I understand correctly your last comment is saying that this definition fails to define an "illogical statement", for some reason which I will understand better when I fully understand your answer. – it's a hire car baby Aug 11 '16 at 11:56
  • @user21820 p.s. I'm not criticising, I'm just trying to confirm back to you what I think you mean to make sure I've understood you correctly. – it's a hire car baby Aug 11 '16 at 11:57
  • @RobertFrost: Well as I said in my comment, "true and not false" is meaningless in classical logic because "not false" is exactly the same as "true". So if you disagree on that, you'll not only have to provide your own definitions of "true" and "false" but you also must provide justification for the meaningfulness of your definitions. That is why your definition of "illogical" does not work. If you find any specific part of my answer unclear, feel free to ask about it. But I suggest you first read proper introductory texts on logic (see http://math.stackexchange.com/a/1684208). – user21820 Aug 11 '16 at 12:10
  • @RobertFrost: By the way I know you're not criticizing, and indeed it is true that you will completely understand my answer once you get a firm grasp of real logic. The first two references I suggest in the linked post should complement each other and provide both the intuition and the formal analysis of logic. – user21820 Aug 11 '16 at 12:20
  • There is no way that all possible first-order S aren't able to recognize their own constructable G. – Joshua Mar 06 '17 at 19:24
  • @Joshua: I guess you do not grasp the incompleteness theorems... I said "given any formal system $S$ with some nice properties"; if you want a full definition, see this post. And in case you think that PA is too special, consider that an alternative 'base' system of TC is also essentially undecidable despite 'not involving arithmetic'. Note that my linked post applies to TC easily. – user21820 Mar 07 '17 at 02:44
  • @Joshua: Also, please read my posts carefully. Notice that "first-order" occurs in your comment but not my post. It is trivially obvious that the first-order theory of any structure is complete, and the theory of $(\mathbb{Q},<)$ is also recursive, so the incompleteness theorems clearly apply only to formal systems with certain nice properties. Many textbooks restrict to systems built on first-order logic, and hence can get away with defining "nice" to be "with recursive axiom set", but my linked post generalizes it, such that "nice" includes every conceivable useful formal system! – user21820 Mar 07 '17 at 03:27
  • @Joshua: Wait a minute; I did link to that post in my above answer. So you should have read it first before commenting... – user21820 Mar 07 '17 at 03:31
  • @user21820: Since the Godel construction form is reasonably simple, a system could run it over its own input and add the resulting Gs as axioms until it matches or the length of G (it increases each time) exceeds the length of its input. Only the constructive form of Godel fails here; G-like statements still exist because it hasn't the wit to find them, but neither can Godel's proof sequence generate them. – Joshua Mar 07 '17 at 04:24
  • @Joshua: Did you not read my posts? From your responses, I am very certain that you do not understand the incompleteness theorems at all. You should start with some introductory textbooks to logic, as I had recommended to Robert as well. Come back to this only after you've learnt the basics properly. – user21820 Mar 07 '17 at 05:16
  • @user21820: Or is it you who does not understand? – Joshua Mar 07 '17 at 16:11
  • @Joshua: That's enough. I have a complete and rigorous grasp of the incompleteness theorems and their proof, as you can easily see from my linked posts. If you don't understand them, feel free to inquire, but if you wish to question my understanding then do not respond to me anymore. – user21820 Mar 07 '17 at 16:17
  • @user21820 I know i have accepted your answer but having re-read and reconsidered the various arguments I would like to propose the following improvement to my proposition: An illogical statement is a statement which is incapable of being true without being contradictory, in any reasonable system in which it is provable. Have I now reached an appropriate and worthwhile definition? I define a reasonable system is one in which $s\implies s$. – it's a hire car baby Nov 08 '17 at 11:07
  • @RobertFrost: No you have not reached an appropriate definition. First, you've not even defined "reasonable system" in any reasonable way. Unfortunately, it is impossible to understand what is reasonable with no foundation in logic. If you define "reasonable" as "first-order theory", you are trivially saying that a statement is illogical iff it is a contradiction. That does not help you at all. Please start a proper study of logic; first read forallx and then Hannes' notes (linked from here, and ask questions about them in the Logic room. =) – user21820 Nov 08 '17 at 12:51
  • @user21820 thanks. i'm not intending to annoy you and I appreciate where you're coming from in terms of "STUDY LOGIC!" but I wanted to give you a heads up I've been working for a little while on another question which is probably going to make you say the same thing again! – it's a hire car baby Nov 08 '17 at 14:16
  • @RobertFrost: Haha I wasn't annoyed just now. But believe me, if you had taken my advice last time, you would have learnt enough by now to have a much better grasp of all mathematics, not just logic, than you can ever have without logic. A few who frequent the Logic room have indeed greatly benefited from learning logic from scratch, within the time you have spent on your other 'pursuits', and I see no reason you can't do as well as them; the only problem you have is your choice not to start learning... – user21820 Nov 08 '17 at 14:49
  • @user21820 I will get there. Rome wasn't built in a day! – it's a hire car baby Nov 08 '17 at 14:58