Re: what follows from denying an axiom

On Jan 23, 5:49 pm, stevendaryl3...@xxxxxxxxx (Daryl McCullough)
wrote:
It seems to me that the key feature that makes
a language second-order is a comprehension schema.

This is not MERELY mistaken. This is THE OPPOSITE of the truth.
Having to state a schema LIMITS what subsets are allowed.
What makes 2nd-order ANYthing 2nd-order is that there is SOME
component
of it that is somehow relating to THE CLASS *OF ALL* subsets of the
(first-order
universal) class. YOU DON'T use or need a SCHEMA for that! ALL just
MEANS
ALL!

A comprehension schema is going to tell you what sorts of
comprehensive
collections exist and what don't. ZFC's axiom schema of separation
is
a comprehension schema. The schematic variable there is instantiated
to FORMULAS OF THE LANGUAGE AND NOT subsets of the domain!

If you don't have some comprehension schema,
then I don't know in what sense it is
a second-order language. It's just a two-sorted language with
a binary relation epsilon.

Chris Menzel is the first person I have ever heard say "second-order
language".
EVEN IF YOU DO have a comprehension schema, IT IS STILL a two-sorted
first-
order language, ESPECIALLY if you are claiming the relevance of the
Henkin
semantics.

My point is simply that you CANNOT attach the order TO THE LANGUAGE in
ANY case!
The strings and symbols are going to be EXACTLY THE SAME, COMPLETELY
IRrespecitve of how you treat the predicate quantifiers!
You can, in the first-order case, allege, semantically, that the
"predicate variables" ,
when THEY are being quantified over, are ranging, IN THE SAME
canonical
FIRST-ORDER sense of quantifying over a domain of discourse,
over a DIFFERENT (or even disjoint) domain from the one that variables
of the lower sort are ranging over. You can say THAT EVEN if you DO
have
a comprehension schema!

What makes the enterprise second-order IS THE REQUIREMENT that
the domain over which the "predicate" or "higher" sort of variables
are
ranging MUST BE the class of ALL subsets of the lower-sort domain.
This requirement gets imposed WITHOUT CHANGING ONE JOT OR TITTLE
OF THE *LANGUAGE*!

So the phrase "second-order language" IS JUST STUPID as long as a two-
sorted
first-order language IS ALSO BEING CONTEMPLATED in the same lesson,
UNLESS, of course, you really do have Some ACTUALLY DIFFERENT
language in mind.

But the presence of a comprehension schema is going to cut the OTHER
way;
it is going to make the whole enterprise LESS 2nd-order.
.

Relevant Pages

  • Re: what follows from denying an axiom
    ... a language second-order is a comprehension schema. ... to FORMULAS OF THE LANGUAGE AND NOT subsets of the domain! ... of first-order logic by allowing quantification of predicate symbols ...
    (sci.logic)
  • Re: what follows from denying an axiom
    ... "second-order language" can be defined completely syntactically. ... Having a comprehension schema ... The mapping then takes an atomic formula Pt_1...t_n of L2 to the atomic ...
    (sci.logic)
  • Re: what follows from denying an axiom
    ... "second-order language" can be defined completely syntactically. ... a language second-order is a comprehension schema. ... predicate variables, for each n>0, whereas a first order language has ...
    (sci.logic)
  • Re: what follows from denying an axiom
    ... "second-order language" can be defined completely syntactically. ... a language second-order is a comprehension schema. ... predicate variables, for each n>0, whereas a first order language has ...
    (sci.logic)
  • Re: what follows from denying an axiom
    ... fact specify the intended model of, say, the language of PA or ZF all ... I know you know that first-order consequence is a purely model theoretic ... You aren't thinking of Quine's claim that second-order logic is set ... in set theory picks out exactly the infinite sets therein. ...
    (sci.logic)