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The Liar Paradox Revisited... Again!
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Dan Christensen  
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 More options Aug 21 2012, 12:36 pm
Newsgroups: sci.logic
From: Dan Christensen <Dan_Christen...@sympatico.ca>
Date: Tue, 21 Aug 2012 09:36:01 -0700 (PDT)
Local: Tues, Aug 21 2012 12:36 pm
Subject: The Liar Paradox Revisited... Again!
Following is my attempt, using predicate calculus, to resolve the so-called Liar Paradox. Informally:

A man says, "Everything I say is a lie."

If the sentence, "Everything I say is a lie," is true, then the man does say at least one thing that is true, namely that sentence itself. This a contradiction, therefore the sentence must be false.

The formal argument:

We have 3 predicates:

S(x) means x is sentence
T(x) means x is true
M(x) means man says x

Using indirect proof...

1. S(x) /\ M(x) & Ay(S(y) -> (M(y) -> ~T(y)))  Premise

2. S(x)  Splitting premise, 1

3. M(x)  

4. Ay(S(y) -> (M(y) -> ~T(y)))

5. T(x)  Premise

6. S(x) -> (M(x) -> ~T(x))  Universal Specification, 4

7. M(x) -> ~T(x)  Detachment, 2, 6

8. ~T(x)   Detachment, 3, 7

9. T(x) & ~T(x)  Contradiction

10. ~T(x)   Conclusion, 5

11. ~Ax((S(x) /\ M(x) & Ay(S(y) -> (M(y) -> ~T(y))) -> ~T(x))  Conclusion, 1

Comments? Does this resolve the paradox?

Dan
Download my DC Proof 2.0 software at http://www.dcproof.com


 
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Dan Christensen  
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 More options Aug 21 2012, 12:42 pm
Newsgroups: sci.logic
From: Dan Christensen <Dan_Christen...@sympatico.ca>
Date: Tue, 21 Aug 2012 09:42:39 -0700 (PDT)
Local: Tues, Aug 21 2012 12:42 pm
Subject: Re: The Liar Paradox Revisited... Again!
(Correcting typos)

Following is my attempt, using predicate calculus, to resolve the so-called Liar Paradox. Informally:

A man says, "Everything I say is a lie."

If the sentence, "Everything I say is a lie," is true, then the man does say at least one thing that is true, namely that sentence itself. This a contradiction, therefore the sentence must be false.

The formal argument:

We have 3 predicates:

S(x) means x is sentence
T(x) means x is true
M(x) means man says x

Using indirect proof...

1. S(x) /\ M(x) /\ Ay(S(y) -> (M(y) -> ~T(y)))  Premise

2. S(x)  Splitting premise, 1

3. M(x)  

4. Ay(S(y) -> (M(y) -> ~T(y)))

5. T(x)  Premise

6. S(x) -> (M(x) -> ~T(x))  Universal Specification, 4

7. M(x) -> ~T(x)  Detachment, 2, 6

8. ~T(x)   Detachment, 3, 7

9. T(x) & ~T(x)  Contradiction

10. ~T(x)   Conclusion, 5

11. Ax((S(x) /\ M(x) /\ Ay(S(y) -> (M(y) -> ~T(y))) -> ~T(x))  Conclusion, 1

Comments? Does this resolve the paradox?

Dan
Download my DC Proof 2.0 software at http://www.dcproof.com


 
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Frederick Williams  
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 More options Aug 21 2012, 1:00 pm
Newsgroups: sci.logic
From: Frederick Williams <freddywilli...@btinternet.com>
Date: Tue, 21 Aug 2012 18:00:45 +0100
Local: Tues, Aug 21 2012 1:00 pm
Subject: Re: The Liar Paradox Revisited... Again!

Dan Christensen wrote:

> Following is my attempt, using predicate calculus, to resolve the so-called Liar Paradox. Informally:

> A man says, "Everything I say is a lie."

That isn't the liar paradox because you can resolve it by saying that
the man lies on that occasion.

The liar paradox is

  This sentence is false.

which isn't so readily dealt with.  Hence "paradox".

> If the sentence, "Everything I say is a lie," is true, then the man does say at least one thing that is true, namely that sentence itself. This a contradiction, therefore the sentence must be false.

Is it really plausible that a stupid cunt like you can resolve the
problem so easily?

--
     The animated figures stand
     Adorning every public street
     And seem to breathe in stone, or
     Move their marble feet.


 
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Dan Christensen  
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 More options Aug 21 2012, 1:42 pm
Newsgroups: sci.logic
From: Dan Christensen <Dan_Christen...@sympatico.ca>
Date: Tue, 21 Aug 2012 10:42:12 -0700 (PDT)
Local: Tues, Aug 21 2012 1:42 pm
Subject: Re: The Liar Paradox Revisited... Again!

My resolution, if indeed it is one, goes beyond that. It says, in effect, that EVERY such sentence that a man says must be false.

> The liar paradox is

>   This sentence is false.

Historically, that is only one form. And I'm not sure that it captures of essence of the original from antiquity, which talks about a potentially infinite number of sentences. The paradox in the original can also be removed by restating the liar's claim: "But for this sentence which is true, everything I say is a lie." I don't think that any such patch can repair "This sentence is false."

> which isn't so readily dealt with.  Hence "paradox".

> > If the sentence, "Everything I say is a lie," is true, then the man does say at least one thing that is true, namely that sentence itself. This a contradiction, therefore the sentence must be false.

> Is it really plausible that a stupid cunt like you can resolve the

> problem so easily?

Geez, Willy, it seems you have given up already and are starting right in with the personal insults! Get your butt kicked one too many times?

Dan
Download my DC Proof 2.0 software at http://www.dcproof.com


 
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Dan Christensen  
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 More options Aug 21 2012, 2:15 pm
Newsgroups: sci.logic
From: Dan Christensen <Dan_Christen...@sympatico.ca>
Date: Tue, 21 Aug 2012 11:15:38 -0700 (PDT)
Local: Tues, Aug 21 2012 2:15 pm
Subject: Re: The Liar Paradox Revisited... Again!

Using the same predicates, "This sentence is false," we obtain:

S(x) /\ (T(x) <-> ~T(x))

where x = "This sentence is false."

This nothing more than a simple contradiction. It's like saying the sky is blue and the sky is not blue. Not very interesting. Have I missed something?

Dan
Download my DC Proof 2.0 software at http://www.dcproof.com


 
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Dan Christensen  
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 More options Aug 21 2012, 11:51 pm
Newsgroups: sci.logic
From: Dan Christensen <Dan_Christen...@sympatico.ca>
Date: Tue, 21 Aug 2012 20:51:57 -0700 (PDT)
Local: Tues, Aug 21 2012 11:51 pm
Subject: Re: The Liar Paradox Revisited... Again!
(Important corrections)

5. S(x) -> (M(x) -> ~T(x))  Universal Specification, 4

6. M(x) -> ~T(x)  Detachment, 2, 5

7. ~T(x)   Detachment, 3, 6

8. Ax((S(x) /\ M(x) /\ Ay(S(y) -> (M(y) -> ~T(y))) -> ~T(x))  Conclusion, 1

There is nothing intrinsically paradoxical about someone always lying. As far as I can tell, a contradiction arises only from the fact that, as in my scenario, our constant liar admits to being one.

Dan

Download my DC Proof 2.0 software at http://www.dcproof.com


 
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William Elliot  
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 More options Aug 22 2012, 1:47 am
Newsgroups: sci.logic
From: William Elliot <ma...@panix.com>
Date: Tue, 21 Aug 2012 22:47:10 -0700
Local: Wed, Aug 22 2012 1:47 am
Subject: Re: The Liar Paradox Revisited... Again!

5, 9 and 10 aren't needed to conclude 8.

> 11. ~Ax((S(x) /\ M(x) & Ay(S(y) -> (M(y) -> ~T(y))) -> ~T(x))  
> Conclusion, 1

How did you conclude that?  Clearly

S(x) & M(x) & Ay(S(y) -> (M(y) -> ~T(y))) -> ~T(x)
Ax(S(x) & M(x) & Ay(S(y) -> (M(y) -> ~T(y))) -> ~T(x))

in contradiction to "conclusion" 1.

> Comments? Does this resolve the paradox?

No.  That's not the pardox which requires 2nd order logic.

 
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William Elliot  
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 More options Aug 22 2012, 3:58 am
Newsgroups: sci.logic
From: William Elliot <ma...@panix.com>
Date: Wed, 22 Aug 2012 00:58:28 -0700
Local: Wed, Aug 22 2012 3:58 am
Subject: Re: The Liar Paradox Revisited... Again!

On Tue, 21 Aug 2012, Dan Christensen wrote:
> Following is my attempt, using predicate calculus, to resolve the
> so-called Liar Paradox. Informally:
> A man says, "Everything I say is a lie."

> If the sentence, "Everything I say is a lie," is true, then the man does
> say at least one thing that is true, namely that sentence itself. This a
> contradiction, therefore the sentence must be false.

> The formal argument: We have 3 predicates:

> S(x) means x is sentence
> T(x) means x is true
> M(x) means man says x

The liar's paradox
S(p) & M(p) & (p <-> Ay(S(y) & M(y) -> ~T(y))

Theorem.  p -> ~T(p).

Assume ~p.  Then Ey(S(y) & M(y) & T(y))
        Denote by y such a proposition.  Thus T(y) & ~T(y).
Consequently p.  Thusly ~T(p).  
Is ~T(p) -> ~p a theorem?  If so, p & ~p.

Resolution of this paradox is that p -> T(p) is not a theorem.


 
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Rupert  
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 More options Aug 22 2012, 9:45 am
Newsgroups: sci.logic
From: Rupert <rupertmccal...@yahoo.com>
Date: Wed, 22 Aug 2012 06:45:00 -0700 (PDT)
Subject: Re: The Liar Paradox Revisited... Again!
On Aug 21, 7:42 pm, Dan Christensen <Dan_Christen...@sympatico.ca>
wrote:

> On Tuesday, August 21, 2012 1:00:45 PM UTC-4, Frederick Williams wrote:
> > Dan Christensen wrote:

> > > Following is my attempt, using predicate calculus, to resolve the so-called Liar Paradox. Informally:

> > > A man says, "Everything I say is a lie."

> > That isn't the liar paradox because you can resolve it by saying that

> > the man lies on that occasion.

> My resolution, if indeed it is one, goes beyond that. It says, in effect, that EVERY such sentence that a man says must be false.

What happens if all the other statements the man makes throughout his
life are statements we know to be false?

 
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LudovicoVan  
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 More options Aug 22 2012, 10:54 am
Newsgroups: sci.logic
From: "LudovicoVan" <ju...@diegidio.name>
Date: Wed, 22 Aug 2012 15:54:58 +0100
Local: Wed, Aug 22 2012 10:54 am
Subject: Re: The Liar Paradox Revisited... Again!
"Dan Christensen" <Dan_Christen...@sympatico.ca> wrote in message

news:962c3acf-aa2d-4dfc-92b0-bfeff0c796ff@googlegroups.com...

> There is nothing intrinsically paradoxical about someone always
> lying.  As far as I can tell, a contradiction arises only from the fact
> that, as in my scenario, our constant liar admits to being one.

Paraphrasing something from Smullyan's GIT:

Imagine a land where people are either Athenians, or Cretan, or neither.

You are given that every inhabitant of the land either always lies or never
does so, and that Athenians always tell the truth, while Cretans always lye.

What would be a statement from an inhabitant of the land that would convince
you that he (always) tells the truth?

What would be a statement from an inhabitant of the land that would convince
you that he (always) lies?

-LV


 
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Frederick Williams  
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 More options Aug 22 2012, 11:43 am
Newsgroups: sci.logic
From: Frederick Williams <freddywilli...@btinternet.com>
Date: Wed, 22 Aug 2012 16:43:53 +0100
Local: Wed, Aug 22 2012 11:43 am
Subject: Re: The Liar Paradox Revisited... Again!

Dan Christensen wrote:

> Following is my attempt, using predicate calculus, to resolve the so-called Liar Paradox. Informally:

> A man says, "Everything I say is a lie."

> If the sentence, "Everything I say is a lie," is true, then the man does say at least one thing that is true, namely that sentence itself. This a contradiction, therefore the sentence must be false.

> The formal argument:

> We have 3 predicates:

> S(x) means x is sentence
> T(x) means x is true
> M(x) means man says x

It is clear that those predicates alone cannot be used to formalize the
liar because it is about a sentence (or a supposed sentence, if you
prefer) saying something about itself (or supposedly saying something
about itself, if you prefer).  The self-referential nature of the liar
is an important aspect of it, this you ignore completely.  It was not
long ago that you became an instant expert in group theory, and thus
managed to spout bollocks about it.  Have you now become an instant
expert in the liar with a similar outcome?

--
     The animated figures stand
     Adorning every public street
     And seem to breathe in stone, or
     Move their marble feet.


 
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Dan Christensen  
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 More options Aug 22 2012, 12:59 pm
Newsgroups: sci.logic
From: Dan Christensen <Dan_Christen...@sympatico.ca>
Date: Wed, 22 Aug 2012 09:59:18 -0700 (PDT)
Local: Wed, Aug 22 2012 12:59 pm
Subject: Re: The Liar Paradox Revisited... Again!

A glutton for punishment, Willy?

In hindsight, my "proof" did not contribute much to the discussion (just "brainstorming"), but what does self-reference have to do with it? The classical paradox arises from two contradictory assumptions:

1. Everything the liar says is false. (Not always made explicit by writers, including me.)

2. The claim by the liar that everything he says is false.

The negation of the liar's claim is, of course, that something he says is true.

Without either of these two assumptions, there is no paradoxical contradiction. If the liar simply refrained from making such a claim, there would be no contradiction. And self-reference doesn't seem to be the culprit. If the liar had said, "Everything I say is true" (a self-reference), there would also be no contradiction.

Dan
Download my DC Proof 2.0 software at http://www.dcproof.com


 
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Frederick Williams  
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 More options Aug 22 2012, 1:37 pm
Newsgroups: sci.logic
From: Frederick Williams <freddywilli...@btinternet.com>
Date: Wed, 22 Aug 2012 18:37:15 +0100
Local: Wed, Aug 22 2012 1:37 pm
Subject: Re: The Liar Paradox Revisited... Again!

Nor did I say it was.  What I did say was "The self-referential nature
of the liar is an important aspect of it, this you ignore completely."

> If the liar had said, "Everything I say is true" (a self-reference), there would also be no contradiction.

--
     The animated figures stand
     Adorning every public street
     And seem to breathe in stone, or
     Move their marble feet.

 
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Dan Christensen  
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 More options Aug 22 2012, 3:40 pm
Newsgroups: sci.logic
From: Dan Christensen <Dan_Christen...@sympatico.ca>
Date: Wed, 22 Aug 2012 12:40:57 -0700 (PDT)
Local: Wed, Aug 22 2012 3:40 pm
Subject: Re: The Liar Paradox Revisited... Again!

On Wednesday, August 22, 2012 1:37:15 PM UTC-4, Frederick Williams wrote:

[snip]

How is self-reference important? As I see, it is just a matter of having made two contradictory assumptions (see above). Remove one of them and the paradoxical aspect of this scenario vanishes.

I know that self-reference is often cited in the literature as part of the problem -- as if it was something to be avoided -- but I just don't see it. As I pointed out above, you can still have self-reference without any contradictions arising, with only a slightly modified scenario (where the liar says everything he says is TRUE).

Dan
Download my DC Proof 2.0 software at http://www.dcproof.com


 
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Frederick Williams  
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 More options Aug 22 2012, 3:50 pm
Newsgroups: sci.logic
From: Frederick Williams <freddywilli...@btinternet.com>
Date: Wed, 22 Aug 2012 20:50:49 +0100
Local: Wed, Aug 22 2012 3:50 pm
Subject: Re: The Liar Paradox Revisited... Again!

The claim that self-reference is a problem in some particular case
should not be read as a claim that it is always a problem.  Only a
witless cunt like you would think otherwise.

--
     The animated figures stand
     Adorning every public street
     And seem to breathe in stone, or
     Move their marble feet.


 
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Dan Christensen  
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 More options Aug 22 2012, 4:02 pm
Newsgroups: sci.logic
From: Dan Christensen <Dan_Christen...@sympatico.ca>
Date: Wed, 22 Aug 2012 13:02:45 -0700 (PDT)
Local: Wed, Aug 22 2012 4:02 pm
Subject: Re: The Liar Paradox Revisited... Again!

[snipping abuse]

So, self-reference is NOT important in this case after all?

Dan
Download my DC Proof 2.0 software at http://www.dcproof.com


 
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Frederick Williams  
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 More options Aug 22 2012, 4:56 pm
Newsgroups: sci.logic
From: Frederick Williams <freddywilli...@btinternet.com>
Date: Wed, 22 Aug 2012 21:56:19 +0100
Local: Wed, Aug 22 2012 4:56 pm
Subject: Re: The Liar Paradox Revisited... Again!

It is relevant to the liar.

--
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     Adorning every public street
     And seem to breathe in stone, or
     Move their marble feet.


 
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Dan Christensen  
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 More options Aug 22 2012, 6:08 pm
Newsgroups: sci.logic
From: Dan Christensen <Dan_Christen...@sympatico.ca>
Date: Wed, 22 Aug 2012 15:08:43 -0700 (PDT)
Local: Wed, Aug 22 2012 6:08 pm
Subject: Re: The Liar Paradox Revisited... Again!

Ummmm.... OK.

Dan
Download my DC Proof 2.0 software at http://www.dcproof.com


 
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Graham Cooper  
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 More options Aug 22 2012, 7:40 pm
Newsgroups: sci.logic, sci.math
From: Graham Cooper <grahamcoop...@gmail.com>
Date: Wed, 22 Aug 2012 16:40:00 -0700 (PDT)
Subject: Re: The Liar Paradox Revisited... Again!
On Aug 23, 8:08 am, Dan Christensen <Dan_Christen...@sympatico.ca>
wrote

> > > So, self-reference is NOT important in this case after all?

> > It is relevant to the liar.

> Ummmm.... OK.

> Dan

S:  IF HALT(S) GOTO S

This is informally,

"If this program itself always stops, then keep running forever and
never stop".

THEREFORE:  OMEGA ~e COMPUTABLE-REALS
THEREFORE:  |R| > |N|

Correct?

Herc


 
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Daryl McCullough  
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 More options Aug 23 2012, 10:35 am
Newsgroups: sci.logic
From: Daryl McCullough <stevendaryl3...@yahoo.com>
Date: Thu, 23 Aug 2012 07:35:50 -0700 (PDT)
Local: Thurs, Aug 23 2012 10:35 am
Subject: Re: The Liar Paradox Revisited... Again!

Dan Christensen wrote:
> Following is my attempt, using predicate calculus, to resolve the
> so-called Liar Paradox. Informally:

> A man says, "Everything I say is a lie."

> If the sentence, "Everything I say is a lie," is true, then the
> man does say at least one thing that is true, namely that sentence
> itself. This a contradiction, therefore the sentence must be false.

Suppose that is the only statement that the man has ever made.

 
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Dan Christensen  
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 More options Aug 23 2012, 10:47 am
Newsgroups: sci.logic
From: Dan Christensen <Dan_Christen...@sympatico.ca>
Date: Thu, 23 Aug 2012 07:47:37 -0700 (PDT)
Local: Thurs, Aug 23 2012 10:47 am
Subject: Re: The Liar Paradox Revisited... Again!

What does this have to do with the topic at hand?

Dan
Download my DC Proof 2.0 software at http://www.dcproof.com


 
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Dan Christensen  
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 More options Aug 23 2012, 11:40 am
Newsgroups: sci.logic
From: Dan Christensen <Dan_Christen...@sympatico.ca>
Date: Thu, 23 Aug 2012 08:40:21 -0700 (PDT)
Local: Thurs, Aug 23 2012 11:40 am
Subject: Re: The Liar Paradox Revisited... Again!
(Correction)

I think my analysis still holds up in this case. By assumption, his statement is false. The negation of his statement, of course, is that he must say something that is true -- a contradiction.

Again, the paradox arises from making two contradictory assumptions:

1. Everything the liar says is false.

2. The liar himself claims that everything he says is false.

Eliminate either one of these assumptions and the paradoxical nature of this scenario vanishes.

Dan
Download my DC Proof 2.0 software at http://www.dcproof.com


 
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Frederick Williams  
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 More options Aug 23 2012, 12:42 pm
Newsgroups: sci.logic
From: Frederick Williams <freddywilli...@btinternet.com>
Date: Thu, 23 Aug 2012 17:42:18 +0100
Local: Thurs, Aug 23 2012 12:42 pm
Subject: Re: The Liar Paradox Revisited... Again!

Why are you assuming that?  Look, if the man has only ever said
"Everything I say is a lie" then that is equivalent to (him saying)

  This statement is false.  (*)

Which _is_ the liar (unlike your garbled version).  Now, _if_ that is
false then it is true; and _if_ it is true then it is false.  That, in
the light of two assumptions, is paradoxical.  The two assumptions are:
1: (*) really is a statement, and 2: statements are always either true
or false and never both.

> The negation of his statement, of course, is that he must say something that is true -- a contradiction.

> Again, the paradox arises from making two contradictory assumptions:

> 1. Everything the liar says is false.

> 2. The liar himself claims that everything he says is false.

That is not an assumption.  If you think in terms of a man actually
making an utterance you might call it an empirical fact.

> Eliminate either one of these assumptions and the paradoxical nature of this scenario vanishes.

No, forget 2 because it is a fact over which you have no control.

--
     The animated figures stand
     Adorning every public street
     And seem to breathe in stone, or
     Move their marble feet.


 
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Dan Christensen  
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 More options Aug 23 2012, 2:00 pm
Newsgroups: sci.logic
From: Dan Christensen <Dan_Christen...@sympatico.ca>
Date: Thu, 23 Aug 2012 11:00:29 -0700 (PDT)
Local: Thurs, Aug 23 2012 2:00 pm
Subject: Re: The Liar Paradox Revisited... Again!
(Corrections)

On Thursday, August 23, 2012 12:42:18 PM UTC-4, Frederick Williams wrote:

[snip]

I neglected to mention the assumption that everything the liar says is indeed a lie. Without that assumption, there is no contradiction and no paradox. As one reader pointed out, his statement could then be a false and other of his statements could be true.

>   This statement is false.  (*)

> Which _is_ the liar (unlike your garbled version).  

Mine more closely resembles the original Liar Paradox of antiquity. Much more interesting.

We are talking about a hypothetical situation here. As such, your distinction is moot.

> > Eliminate either one of these assumptions and the paradoxical nature of this scenario vanishes.

> No, forget 2 because it is a fact over which you have no control.

We cannot forget 2. The "fact" is, if the liar never made this famous claim, there would be no contradiction and no paradox. If he had even said, "Everything I say is TRUE," there would be no contradiction and no paradox.

Dan
Download my DC Proof 2.0 software at http://www.dcproof.com


 
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Frederick Williams  
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 More options Aug 23 2012, 3:10 pm
Newsgroups: sci.logic
From: Frederick Williams <freddywilli...@btinternet.com>
Date: Thu, 23 Aug 2012 20:10:39 +0100
Local: Thurs, Aug 23 2012 3:10 pm
Subject: Re: The Liar Paradox Revisited... Again!

Dan Christensen wrote:

> (Corrections)

> On Thursday, August 23, 2012 12:42:18 PM UTC-4, Frederick Williams wrote:

> > That is not an assumption.  If you think in terms of a man actually

> > making an utterance you might call it an empirical fact.

> We are talking about a hypothetical situation here.

No we're not. At about 19:44 BST I uttered these words: "This statement
is false."  That's a fact.  In bivalent propositional calculus these are
laws:

   (P -> ~P) -> ~P              (1)
   (~P -> P) -> P               (2)

for any statement P.  Furthermore, the rule,

   from P and P -> Q, Q follows (3)

for any statements P and Q applies in bivalent propositional calculus.
If my utterance was true, then it would be false.  Letting P in (1) be
my utterance, and applying (3) we find that

   not-(this statement is false).

On the other hand, if my utterance was false, then it would be true.
Letting P in (2) be my utterance, and applying (3) we find that

   this statement is false.

There is a contradiction.  What have we assumed?  (a) that my utterance
is a statement, and (b) that certain laws and rules of bivalent
propositional calculus apply.

> As such, your distinction is moot.

> > > Eliminate either one of these assumptions and the paradoxical nature of this scenario vanishes.

> > No, forget 2 because it is a fact over which you have no control.

> We cannot forget 2. The "fact" is, if the liar never made this famous claim, there would be no contradiction and no paradox.

There is no need for the liar to do anything.  I acted in the necessary
way at about 19:44 BST.

>                If he had even said, "Everything I say is TRUE," there would be no contradiction and no paradox.

An irrelevance.  Clearly you're struggling to understand what the liar
is about; introducing irrelevancies won't help you.

--
     The animated figures stand
     Adorning every public street
     And seem to breathe in stone, or
     Move their marble feet.


 
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