Re: Exception to the rule? (Tarski´s T-scheme)

From: Jeffrey Ketland (ketland_at_ketland.fsnet.co.uk)
Date: 06/22/04 

Date: Wed, 23 Jun 2004 00:27:24 +0100

Paul Holbach wrote in message
<881c8779.0406221452.42ac0931@posting.google.com>...
>> "Jeffrey Ketland" <ketland@ketland.fsnet.co.uk> wrote in message news:<
>> cb9pni$qjp$1@news5.svr.pol.co.uk>...
>> > Paul Holbach (paulholbachSPAMBAN@freenet.de) wrote in message
>> > <881c8779.0406211820.43c78402@posting.google.com>...
>
>
>> >Let´s consider Tarski´s famous T-scheme:
>> >
>> >True("p") <-> p
>> >
>> >Now what about the statement "Nothing exists"?
>> >
>> >True("Nothing exists") <-> Nothing exists
>> >
>> >Truth is a property of statements, and if nothing exists, there aren´t
>> >any statements either. The point is that nonexistent statements are
>> >neither true nor false, and so it is not the case that "Nothing
>> >exists" is true iff nothing exists.
>
>
>> The T-scheme implies the existence of at least two things. In particular,
a
>> syntactic item A must be distinct from its negation ~A. (For each item,
the
>> T-scheme implies "~A is true if and only if A is not true", so "A = ~A"
>> would be inconsistent with the T-scheme.)
>
>
>(You mean "A <=> ~A", don´t you?)

No. I mean the equation "A is identical to ~A".

>> If one considers a model with at least two objects---and preferably one
>> where all syntactic items are elements of the domain---then the relevant
>> restricted T-scheme can be made *true* (including the instance using
>> "~Ex(x=x)").
>>
>> But this merely tells us that the T-scheme itself implies that something
>> exists.
>
>
>So we must write:
>
>[T("p") <-> p] -> Ex(x = x)

You cannot compress the T-scheme to a single sentence, since it is a scheme,
and thus has infinitely many instances. However, let DT be the theory in the
language of arithmetic L_{Tr} extended with a primitive predicate Tr(x),
containing the restricted T-scheme

         Tr("phi")<->phi,

for all sentences phi not containing Tr. Then

         DT |- ExEy(x =/= y).
(for proof, see below)

>or even
>
>[T("p") <-> p] <-> Ex(x = x)
>
>~Ex(x = x) <-> ~[T("p") <-> p]
>
>For p = ~Ex(x = x) we get:
>
>~Ex(x = x) <-> ~[T("~Ex(x = x)") <-> ~Ex(x = x)]
>
>- If it is the case that nothing exists:
>
>1 <-> ~(0 <-> 1)
>1 <-> ~0
>1 <-> 1
>1
>
>- If it is not the case that nothing exists:
>
>0 <-> ~(0 <-> 0)
>0 <-> ~1
>0 <-> 0
>1
>
>Ok, works fine.
>
>
>> This is no surprise, since its instances refer to syntactical items.
>> What is wrong with that? No one ever said the T-scheme was a tautology,
and
>> it isn't a tautology.
>> So, this is not an "exception to the rule". It merely notes that the
>> T-scheme has some existential content.
>
>
>You are right--iff the T-scheme in fact implies the existence of something.

The T-scheme is actually inconsistent (with syntax), as Tarski pointed out
in 1933.
If you restrict it, so that in
        Tr("phi") <-> phi,
the formula phi does not contain the symbol Tr, then it is easy to show that
the scheme implies ExEy(x =/= y). That is, the restricted T-scheme implies
the existence of at least two objects.

Indeed, take any formula you like. Say 0=0.
Then

   {Tr("0=0") <-> 0=0, Tr("~(0=0)") <-> ~(0=0)} |- ExEy(x =/= y)

Proof:

    1. Tr("0=0") <-> 0=0,
    2. Tr("~(0=0)") <-> ~(0=0)
    3. ~ExEy(x =/= y)
                    |
    4. AxAy(x = y)
                    |
    5. Tr("~(0=0)") <-> ~Tr("0=0")
    6. "~(0=0)" = "0=0"
    7. Tr("0=0") <-> ~Tr("0=0")
                 contradiction

>> Actually, the unrestricted T-scheme ...
>
>
>The restriction you´re referring to is Ex(x = x) / ExE!x, isn´t it?

The restriction I mean is the restriction needed to avoid the Liar paradox,
that in instances of Tr("phi") <-> phi, the formula phi does not contain the
truth predicate symbol Tr.

It is easy to show that the theory PA + T-scheme is inconsistent.
Proof: By the Diagonal Lemma, construct a LIar sentence L such that PA |- L
<-> ~Tr("L"). This contradicts the instance L<-> Tr("L"). So, PA +
(unrestricted) T-scheme is inconsistent.

Roughly, the restricted T-scheme is not self-applicative, and only applies
to the formulas of the _object-language_, and intuitively the formula Tr(x)
means "x is a true sentence of the object language".

It can be shown that PA + restricted T-scheme is a conservative extension of
PA. So, if you can prove an arithmetic formula in the theory PA + restricted
T-scheme, then it is provable in PA. In fact, this holds for a very wide
class of theories (certainly for any reasonable theory of syntax). In this
sense, the (restricted) T-scheme represents a very weak notion of truth, and
I have argued in print that this is why it can be considered "deflationary".
(See my paper in "Deflationism and Tarski's Paradise", Mind 108 (1999), pp.
69-94.)

But you are correct in noting that the T-scheme (even the consistent,
restricted T-scheme) has non-trivial ontological commitment: it implies the
existence of at least two things (roughly, some sentence and its negation).
So, the T-scheme implies "Something exists".

However, also note that so long as your background theory is not
ridiculously weak, the instance of the T-scheme with the sentence "Nothing
exists", i.e., the formula

         Tr("~Ex(x =x)") <-> ~Ex(x = x)

will be a perfectly acceptable theorem. It's just that both sides of the
biconditional are false.
When people first come across the T-scheme, they often forget to consider
that it also applies to instances where the sentence of the object language
used is obviously untrue. For example,

       "The moon is made of cheese" is true iff the moon is made of cheese.
       "0 =1" is true iff 0 = 1
       etc.

--- Jeff