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

From: Andrew Boucher (Helene.Boucher_at_wanadoo.fr)
Date: 06/25/04 

Date: 24 Jun 2004 22:24:27 -0700

"Jeffrey Ketland" <ketland@ketland.fsnet.co.uk> wrote in message news:<cb9pni$qjp$1@news5.svr.pol.co.uk>...
> Paul Holbach 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.)
> 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. This is no surprise, since its instances refer to syntactical items.

Just to clarify (?) this point.

The T-schema is like a universal generalization: you
can substitute any proposition you want in for p, but you still have to
have one to substitute. It therefore does not imply the existence of any
p, nor for that matter the existence of any two p. The existence of p (and
the existence of "not p" given the existence of p) must come elsewhere,
from e.g. assumptions about what propositions can be formulated. The
argument shows only (as Jeffrey says) that, if p and "not p"
exist, then the T-schema implies that they cannot be identical; it does not
show (which was perhaps less clear) that, from the T-Schema, one can
deduce that there exists any p or any "not p" given p.

In brief, the content of the T-schema implies only that a proposition
cannot be identical to its negation. The T-schema implies the existence of
a proposition only if you're taking the Cartesian route and are considering
the statement itself and not the content of the statement. As in: I've
used a proposition to state the T-schema - therefore at least one
proposition exists! Still this doesn't imply the existence of the negation
of the proposition - that too still has to come from elsewhere.