Re: ZFC in another shape.

On 6 Feb., 04:24, "MoeBlee" <jazzm...@xxxxxxxxxxx> wrote:
On Feb 5, 6:17 pm, "zuhair" <zaljo...@xxxxxxxxx> wrote:

I don't know what you are hinting at. Perhaps you want me to remove =
as a primitive. since if I am already working in a first order logic
with identity, there is not need to mention = as a primitive. Is that
what is in your mind.

No, not at all. I'm not hinting, I'm asking clearly, whether your
theory is an extension of first order logic with identity. That is a
precise mathematical question. Put another way, you are adopting some
usual set of axioms and rules of inference for first order logic with
identity?

Frankly speaking, I can answer your question because I don't have the
sufficient information.
For instance I thought first order logic does't use identity. Anyhow.
I can answer this question.

There are at least three variations:

1. Pure first order logic without identity.
2. First order logic with identity axioms also.
3. First order logic with identity axioms and special identity
semantics too.

Unless you have something else special in mind, we might as well
regard you as taking option 3.

And I further think that this can be shortened to the following :

x is an ordinal <-> AyAz(yex&zey ->(zex & Aw(wez->wey))).

what do you think?

I don't know. It seems to say that x is transitive and if z is
obtained, by transitivity, as a member of x, then z is transitive.
Might work. Why don't you just go ahead to try to prove equivalence?

hmmm... , Or more elegantly.

"Axiom Schema of Ordinal succession".

You don't need 'schema'. It's a single formula.

It is not one axiom as you think, it has an axiom for every value of
d. since we have uncountable number of ordinals d. then we have
uncountable number of axioms.

That is a huge confusion on your part.

1. It is a universally quantified formula. The fact that we can
instantiate to many different values for 'd' does not make it a
schema.

2. It is not possible to have an uncountable number of axioms in a
countable language (and, I'm supposing that you have only a countable
language).

4. The reason we'd like only a countable language is that an
uncountable language does not permit your axiomatization to be
recursive. Or put another way, with an uncountable language, we are
not ensured that there is an algorithm to determine even whether a
given string is or is not a formula in the language of your theory.

Anyway, it IS a single formula. It's right in front of me. A single
formula.

hmmm...., I confess my massive ignorance.
Anyhow I think it is the strong replacement, which implies the
existance of the empty set, and separation. anyhow. I will leave this
to you to choose since you are more informed than me.

That's good enough. The strong version (as in Enderton and Suppes),
which does entail both the exixtence of an empty set and the theorem
schema of separation.

5) Infinity

You have not defined '0', 'u' and '{}' that are used in the axiom of
infinity. If you don't first prove the needed existence theorems, then
your axiom might not say what you want it to say. I already pointed
this out before.

Or, just state your axiom of infinity in primitive notation.

Yes, that is correct. But you forgot many things:

0 can be defined simply for the axiom schema of ordinal succession,
and from replacement.

'0' and '{}' can be defined, for sure, now that I know you've chosen
the strong version of replacement. But you haven't proven union.

Is that right? Can replacement produce an empty set {F(x): x in A} if
A is not empty?


Union and Pairing are theorums in this theory.

Pairing okay now that we know you've chosen the strong version of
replacement along with power set (maybe the weak version of
replacement proves it also along with power set, but I don't know).
But you haven't proven union.

so I can define {x} from pairing .
so {x} is the paired set of x and x.
AxAxEzAy ( yez<->(y=x or y=x))
so here z={x}.

Right. Now we have pairing from the strong version of replacement with
power set.

So all the elments for the axiom of infinity are their.

No need to state infinity in a primitive manner.

IF you're proven union.

6) Choice

You have not defined 'is a function' nor functional notation that are
usually used for this axiom. If you don't first prove the needed
existence theorems (for unordered pairs, upon which the definition of
'is a function' ultimately relies for 'is a function' to have any more
than a "hypothetical force") then your axiom might not say what you
want it to say. I already pointed this out before.

unordered pairing is a theorum in this theory.
so I can define function after it.

There is no problem here.

Now that it's clear that pairing is in place, okay. And functional
notation usually requires union, but we can use the Fregean method for
improper descriptions, thus bypassing union on this particular point.

Or, just state your axiom of choice in primitive notation.

Yea, why not. But still since I have unordered pairing as a theorum in
this theory then I can actually state the axiom in the functional
manner.

With pairing in place, we're okay here (per the remarks I just made).

7) Power.

It seems that the existence of the empty set will depend on your
particular formulation of the axiom schema of replacement.

The existance of the empty set is quarenteed from the Axiom of Ordinal
succession.

Meanwhile, it is entailed anyway by the strong version of replacement.

See that when d is not an ordinal, then by Ordinal succession axiom we
have the empty set existing.

Yes, that works too, since we can easily prove Ed ~ d is an ordinal.

One is tempted to try to prove the existence of the non-ordinal {{}}.
However that would in itself require {}.



.

Relevant Pages

  • Re: ZFC in another shape.
    ... First order logic with identity axioms also. ... countable language (and, I'm supposing that you have only a countable ... Anyhow I think it is the strong replacement, ... But you haven't proven union. ...
    (sci.math)
  • Re: truth DOES NOT MATTER for incompleteness
    ... non-schematic axioms are direct formalisations of statements that are ... combining the observation that properties expressible in the language ... we conclude that each instance of the replacement schema is true. ... (Incidentally, my justification for replacement comes from reflection, ...
    (sci.logic)
  • Re: primitive recursive: obsolete?
    ... we have to add more axioms. ... all true sentences of arithmetic in the language of PA. ... logic we define the set of sentences true in the standard model of PA ... theorems" or not has no bearing on the fact that what I said is ...
    (sci.logic)
  • Re: Cantors circular "proof" that evens = integers
    ... independently of any choice of axioms. ... side CALLS "a language" is completely IRrelevant to ... I KNOW what an interpretation is! ... predicates and term for term-functors) and -arity. ...
    (sci.logic)
  • Re: Undecidability of undecidability: some examples?
    ... > we presuppose an interpretation of the statements of the language. ... And it is THOSE axioms and THOSE models that are under consideration! ... commonly-used sets of FIRST-order axioms phrased in FIRST-order ... to accept it) is to stop blaming the bafflee for being baffled and ...
    (sci.logic)
Loading