cousin_it comments on Unsolved Problems in Philosophy Part 1: The Liar's Paradox - Less Wrong

4 Post author: Kevin 30 November 2010 08:56AM

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Comment author: cousin_it 30 November 2010 09:57:05AM 6 points [-]

I'm not sure if I like this paper (it seems to be trying to do too much), but it did contain something new to me - Yablo's non-self-referential version of the Liar Paradox: for every natural number n, let S(n) be the statement that for all m>n S(m) is false. Also there is a funny non-self-referential formulation by Quine: “Yields falsehood when preceded by its quotation” yields falsehood when preceded by its quotation.

Comment author: Thomas 30 November 2010 08:30:55PM 2 points [-]

Interestingly, the Yablo's paradox vanishes when there is no infinity. If the last statement of the Yablo's sequence exists, it is true. And all at the preceding positions are false. Everything is well. Another reason, I am an infinity atheist.

Comment author: nshepperd 30 November 2010 09:07:17PM 2 points [-]

The "last statement"? This would require that there exists a highest natural number. That seems like it would be a weirder occurrence than the mostly harmless Yablo's paradox.

Although I suppose we can always choose to work in "the natural numbers mod N", for some value of N, which is one way to banish "infinity".

Comment author: Thomas 30 November 2010 10:07:14PM -1 points [-]

There is no need for ridiculously large numbers. There is always the last statement in a row and this way and only this way, no Yablo paradox arises.

Comment author: nshepperd 02 December 2010 12:45:11AM -1 points [-]

I'm not sure what you mean by this. "There is no need"? So is there a highest natural number, or not? Because if not:

  • If S(N) is the last statement, N is a natural number.
  • Therefore N + 1 is a natural number and N + 1 > N.
  • Therefore the statement S(N + 1) exists.
  • Therefore S(N) is not the last statement. Contradiction.
Comment author: wedrifid 02 December 2010 12:58:15AM *  0 points [-]

So is there a highest natural number, or not?

If there is no infinity (the premise) then there must be.

Comment author: FAWS 02 December 2010 01:06:54AM *  -1 points [-]

If there is no infinity there must not be a highest natural number, but there could be if there is infinity?

Comment author: wedrifid 02 December 2010 01:19:00AM *  1 point [-]

s/not //

Edit: That looks bad. Let's see.

s/.ot /

That works.

Comment author: ShardPhoenix 30 November 2010 11:10:28AM *  1 point [-]

The second has an implied "This sentence ..." so I'd say it's still self-referential.

edit: actually I don't think that's required (the quote is the subject) so it does count I suppose.

Comment author: shokwave 30 November 2010 12:55:02PM 1 point [-]

If I remember rightly, the process is called "quining" and while it produces similar paradoxes and problems, it is distinct from self-reference. Linguistically, at least - logically one might be a form of the other.

Comment author: wedrifid 30 November 2010 11:18:40AM 0 points [-]

(Upvoted the edit!)

Comment author: red75 30 November 2010 12:06:12PM 0 points [-]

Yablo's version looks like unrolled infinite loop of function

s :: Bool
s=not s
Comment author: cousin_it 30 November 2010 12:19:50PM 1 point [-]

Not to me it doesn't. Yablo's version has a "forall" that your translation misses. So in Yablo's version there's no consistent way to assign truth values to S(n), but in your version we could make S(n) = "n is odd" or something.

Comment author: red75 30 November 2010 09:04:03PM *  0 points [-]

Not exactly. My version is incorrect, yes. But there is, uhm, controversial way of consistent assignment of truth values to Yablo's statements.

In my version n-th step of loop unrolling is

S'(n) = not not ... {n times} ... S

or

S'(n)=not S'(n+1)

Yablo's version

S(n)=not exists m>n such that S(m)=true

or

S(n)=(not S(n+1)) && (not exists m>n+1 such that S(m)=true)

If we extend set of natural numbers by element omega such that

forall n in N : (omega>n),
not exists n in N : (n+1=omega),
omega=omega+1

Than we can assign S(n)=false for all n in N, and S(omega)=true.

Edit: Oops, second version of Yablo's statement, which I included to demonstrate why I had an idea of loop unrolling, is not consistent when n equals omega. Original Yablo's statement is consistent although.

Edit: Meta. The thing I always hated about my mind is that it completely refuses to form intuitions about statements which aren't directly connected to object level (but then what is object level?).

Edit: Meta Meta. On introspection I don't feel anything about previous statement. Pretty damn consistent...

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