FOM: Re: Re: Ontology in Logic and Mathematics

[Jeffrey Ketland]

Dear Roger

>If you leave open the semantics of the language, then what you assert is
>also left open.

If I assert something (say, "Cheese is edible"), then have I *not* asserted
something, until I have previously specified the semantics of English? In
that case, how come I can understand the assertions of my 2 year-old niece?

The semantics of English cannot be expressed in English (Tarski's theorem).

English talks about *everything*. That's why we have a word "everything", so
we can refer to everything. (I think that Harvey Friedman has worked out a
more precise version of this!).

The semantics of English can (almost) trivially be expressed in meta-English
using *disquotational* axioms, such as

    "Cheese" designates cheese
    "Edible" is true of x iff x is edible

(NB1., the first and second sentences in (3) above are meta-English, not
base-English).
(NB2., on the disquotational approach, you have to understand base-English
before you can understand meta-English. But that's not a problem, since
base-English is meaningful and the disquotation sentences are true sentences
of meta-English. In fact, the disquotation sentences are not only true, but
they even form a conservative extension, so they're almost "logically
true").

Following a Tarskian definition, here are the semantics for the language of
arithmetic:

    "0" designates the number 0
    "s" designates the successor operation
    "+" designates the addition operation
    "x" designates the multiplication operation
    "N" designates the set of whole numbers (the closure of {0} under
successor}

Are you saying that the sentences of arithmetic do *not* have these
meanings? What meanings do they have? Is the language of arithmetic
*meaningless*? If so, the sentence "The number of people on Earth (today) is
larger than 6,000,000,000" would be meaningless, rather than true (which it
is).

What exactly is wrong with Quine's analysis of ontological commitment. I.e.,

    "There is something x such that x is an F" is true if and only if there
is an F?

I.e., Quine's analysis of ontological commitment = the disquotation clause
for existential quantifiers (in a Tarskian truth definition in the
metalanguage).

I do not understand your use of the notion "ontological commitment". And I
do not understand your use of the word "absolute". On standard usage, a
sentence that starts "There is a ..." is true only if there is such a thing.
I don't see what's wrong with that?

Regards - Jeff

~~~~~~~~~~~ Jeffrey Ketland ~~~~~~~~~
Dept of Philosophy, University of Nottingham
Nottingham NG7 2RD United Kingdom
Tel: 0115 951 5843
Home: 0115 922 3978
E-mail: jeffrey.ketland at nottingham.ac.uk
Home: ketland at ketland.fsnet.co.uk
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~