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Topic: A logically motivated theory
Replies: 15   Last Post: May 21, 2013 8:22 AM

 Messages: [ Previous | Next ]
 fom Posts: 1,968 Registered: 12/4/12
Re: A logically motivated theory
Posted: May 18, 2013 1:58 PM

On 5/18/2013 10:40 AM, Zuhair wrote:
> In this theory Sets are nothing but object extensions of some
> predicate. This theory propose that for every first order predicate
> there is an object extending it defined after some extensional
> relation.
>
> This goes in the following manner:
>
> Define: E is extensional iff for all x,y: (for all z. z E x iff z E y)
> -> x=y
>
> where E is a primitive binary relation symbol.
>

So,

<X,E>

is a model of the axiom of extensionality.

> Now sets are defined as
>
> x is a set iff Exist E,P: E is extensional & for all y. y E x <-> P(y)
>

So,

xEX <-> ...

where

... is a statement quantifying over relations and predicates.

> Axioms:
>
> [1] If E, D are primitive binary relation symbols then:
>
> E,D are extensional -> (For all x,y: (for all z. z E x iff z D y) ->
> x=y).
>
> is an axiom.
>

So, for definiteness, let

E be membership

D be initial segment

in a theory for which every limit ordinal is a model.

> [2] If E, D are primitive binary relation symbols; P,Q are first order
> language formulas in which x do not occur free, then
>
> E,D are extensional ->
> for all x ((for all y. y E x iff Q) & (for all y. y D x iff P)) ->
> (for all y. P<->Q)
>
> is an axiom
>

You are referring to relations here using
free variables.

You probably mean 'axiom schema' here.

> [3] If P is first order predicate, then
>
> Exist E,x: E is extensional & for all y. y E x iff P(y)
>
> is an axiom.
>
> where E range over primitive binary relations only.
>

You are quantifying over relations here.

You probably mean 'axiom schema' here.

> /
>
> It is possible that this might interpret PA?
>
> The whole motivation beyond this theory is to extend any first order
> predicate by objects.

Could you please clarify this remark?

> It is a purely logical motivation.

Which logic? This is stronger than first-order
because of the quantifications.

> If this does
> interpret PA and no inconsistency is shown with it, then PA can in a
> sense be seen as a kind of a logical theory. IF we extend [3] to allow
> infinitely long formulas, then possibly second order arithmetic would
> be provable? if so then it would be a kind of a logical theory also.
>
> All of that motivates logicism.
>

Date Subject Author
5/18/13 Zaljohar@gmail.com
5/18/13 fom
5/18/13
5/18/13 Zaljohar@gmail.com
5/18/13 fom
5/18/13 Zaljohar@gmail.com
5/18/13 fom
5/19/13 Zaljohar@gmail.com
5/19/13 fom
5/19/13 ross.finlayson@gmail.com
5/20/13 Zaljohar@gmail.com
5/20/13 fom
5/20/13 fom
5/21/13 Zaljohar@gmail.com
5/18/13 Zaljohar@gmail.com
5/18/13 fom