Re: Honorary truth?

"Dunk" <pdunkel@xxxxxxxxxxxxxxx> wrote in message
news:42db9313.1777921@xxxxxxxxxxxxxxxxxxxxxxxxxx
>
> Thanks to all who answered 'What isn't a tautology?'
> That's all cleared up. A slightly different question, the status of
> definitions, was only partly clarified. I suggested that they are
> rules, hence normative rather than true-false. It is sometimes said
> that they are true by definition, but the special definition that
> would make other definitions true has not been produced. Truth tables
> don't help. If in an informal way they are viewed as "Left hand side
> if and only if right hand side" the table is the usual T - F - F - T
> in vertical form. Perhaps they are regarded as true by convention - a
> sort of gentleman's agreement. But why? To avoid contemplating a
> different sort of sentence?
>
> Pete

I can give my own answer for first order languages. Some context might help.
This is the way I think of this subject, and I don't intend to speak for how
others make their own formulations.


Let's say we use Polish notation, and we have a language whose symbols are
pared down to:



* a binary sentential connective



* a quantifier



* an infinite set of variables



* a set Fsymbolset of sets of function symbols, each of the members of
Fsymbolset being a set of function symbols, all of the same arity



I.e., each member of Fsymbolset is, for some n>=0, a set of n-place function
symbols. A member of Fsymbolset can be of any cardinality, including 0.
I.e., for any n, there may be zero n-place function symbols, or many,
possibly infinitely many, n-place function symbols.



* a set Psymbolset of sets of predicate symbols, each of the members of
Psymbolset being a set of predicate symbols, all of the same arity, and the
union of Psymbolset is not empty (so that there is at least one predicate
symbol).



I.e., each member of Psymbolset is, for some n>=0, a set of n-place function
symbols. And there is at least one non-empty member of Psymbolset. I.e.,
there is at least one predicate symbol. All other members of Psymbolset may
be of any cardinality, including 0. I.e., if one member of Psymbolset is a
non-empty set of n-place predicates, then for all m not=n, there may be zero
n-place predicate symbols, or many, possibly infinitely many, n-place
predicate symbols.



Sidenotes:



(1) If I'm not mistaken (please correct me if I am), if we have a fixed
binary predicate for identity, then we can dispense with Fysmbolset. I.e.,
if we have the equality predicate, then we can do with just predicate
symbols and we don't have to have function symbols.



(2) If I'm not mistaken (please correct me if I am), if we have binary
predicate symbols, then we can do without predicate symbols of all other
arity.



The symbols for a language are thus specified. Each time we add a function
symbol or a predicate symbol to a language, we've specified a new language.
And, for any theory using the language, each time we add a function symbol
or predicate symbol, we do so with a definitional axiom.



However, it is not clear to me whether adding a sentential connective or
quantifier specifies a new language. I'll assume not.



In what follows, I may use parentheses, and loosely, but, it should be
understood that officially the formulas are in Polish notation. Also, I
won't always bother with single quote marks to distinguish use-mention,
since this will be clear by context.



* So I'll say that we add a sentential connective as a notational
convention, not with a definitional axiom. For example, if we've started
with the Sheffer stroke, '|', then we agree that ~Phi is an abbreviation for
Phi|Phi.



* And I'll say that we add a quantifier as a notational convention, not with
a definitional axiom. For example, we agree that ExPhi is an abbreviation
for ~Ax~Phi.



* We don't add variables, so we don't define variables.



** When we add a function or predicate symbol for use in a theory, we give a
definitional axiom. However, these axioms must be of special forms that
ensure that two criteria are met: eliminability and non-creativity.



The criterion of eliminability is that for any formula that can be formed
with the new symbol, there must be an equivalent formula that can be formed
without the new symbol.



The criterion of non-creativity is that any theorem derived after the
definitional axiom must have an equivalent formula that is derivable as a
theorem in the original language.



I won't delve into the forms themselves. See Patrick Suppes's
[i]Introduction To Logic[/i] for a nice, leisurely discussion of the
subject. But for each of these definitional forms, we should prove that they
ensure the two criteria. (An interesting complication arises with
conditional defintions, some of which don't satisfy the criteria, but are
used frequently nonetheless, such as with the famous problem of division by
zero or the problem of taking the intersection of the empty set.)



The question was whether these definitional axioms are true. If we apply the
notion of true in a model, then, since these definitional axioms are
non-logical axioms (if I'm not mistaken, a definition that was not
non-logical wouldn't work to define anything), I believe the answer would be
that they are true in some models and not true in others.



I very much welcome any corrections or suggestions that might be made and I
thank anyone in advance for their help.


.