Re: Finite schema?
- From: Aatu Koskensilta <aatu.koskensilta@xxxxxxxxx>
- Date: Tue, 22 May 2007 05:41:25 GMT
On 2007-05-21, in sci.math, zuhair wrote:
How can I know that a formula contain a quantifier over classes or
not. I want examples?
You look at it. Supposing you're working with a language with a different
kind of variable for classes, you inspect the formula to see whether any
quantifier is binding such a variable. If you're working in a single sorted
language, you determine whether every quantified variable is required to
fulfill the relevant condition in the theory of being a set, e.g. being a
member of something, satisfying a predicate Set, ...
what the word parameters exactly mean?
In this context it means that the formula is allowed to have free class
variables; these act, in effect, as parameters to the formula. So the scheme
allows you to conclude, for example, that given a class A there is a class
consisting of all singletons {a} with a in A.
What is the relationship between a formula not containing a quantifier
over classes and this axiom being a finite schema?
The comprehension schema restricted to formulas without bound class variables
can be replaced with a finite number of its instances. This does not hold if
we allow class variables. The schema itself consists of an infinite number of
axioms.
--
Aatu Koskensilta (aatu.koskensilta@xxxxxxxxx)
"Wovon man nicht sprechen kann, daruber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
.
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- Finite schema?
- From: zuhair
- Finite schema?
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