Re: Theories and (denumerable) models

On 12 May 2006 15:26:19 -0700, mordov <-knowledge-@xxxxxxxxxx> said:
So your argument seems to be this. Consider a predicate calculus
without equality. Now let's add the equality predicate and let K be
the first-order theory consisting of the usual axioms for equality
plus the axiom "(Ex)(Ey)(~x=y & (z)(z=x v z=y))", i.e., intuitively,
the axiom "there are exactly two things". K obviously has a model.
Hence, by 2.21, K has a denumerable model. But that's a
contradiction, as all of K's models have only two elements!

Well I was thinking of '=' being interpreted as equality, not merely
an equivalence relation ...

Right, but that's just what you don't get to stipulate in predicate
calculus without identity.

I had intended that '=' also be given the replacement property that
does not hold of equivalence relations in general.

What do you mean by the "replacement property"?

Only they don't. :-) The reason is that, since you are working in the
context of the pred calc without equality, if you add an equality
predicate to your language, you are not forced to interpret that
predicate as *true* equality in an interpretation of the language to get
a model; all you have to do is satisfy the equality axioms, and they are
not strong enough to rule out "unintended" interpretations.

I have another question then: Is there any object-language formula that
can be added as axiom to rule out these "unintended" interpretations?

Nope. If you could, identity would be definable and you'd be able to do
things that we know you can't do in pred calc without identity.

I am thinking that the principle of substitutivity, i.e. x=y -> (A(x)
<-> A(y)) for any wff A, does the trick.

No, I was assuming all instances of this principle (a.k.a. the
indiscernibility of identicals) were in the theory K in the earlier
post. It is an easy induction to show that every instance of it will be
true in the denumerable model M*. (I glibly suggested that it was easy
just to "see" this in the earlier post).

For clarification, by 'the equality axioms', you just mean the three
corresponding to reflexivity, symmetry, and transitivity, right?

As noted -- I also meant all instances of the schema above.

.

Relevant Pages

  • Re: a short article about equality in Lisp (for beginners)
    ... | equality predicate + hash function. ... | according to whether the equality function is eq, eql, equal or equalp. ... tests and hashing methods. ... hashing function for an arbitrary equality predicate. ...
    (comp.lang.lisp)
  • Re: Are you guys really such a bunch of wimps .....
    ... second order definition, and I have not systematically studied identity ... The definition is called "Leibniz law", ... given by Leibniz' definition of equality. ... For first order theories that have only finitely many predicate symbols ...
    (sci.math)
  • Re: E! quantification and equivalence
    ... equality in FOL. ... Suppose P is the only predicate symbol? ... equality axioms to the sentence doesn't ensure that either. ...
    (sci.logic)
  • Re: Theories and (denumerable) models
    ... Now let's add the equality predicate and let K be the ... all you have to do is satisfy the equality axioms, ... you can always satisfy them in a denumerable model, ...
    (sci.logic)
  • Re: Goedel - interesting problem?
    ... > These sentences are trivial to represent, predicate calculus to form the ... >> on what axioms you are allowed to use. ... >> Truth has to do with interpretations of sentences. ...
    (sci.logic)
Loading