Naming Set Theory Interprets NFU

[zuhair]

Dear Sirs,

-The following is and exposition of Naming Set Theory "N.S.T"
-Below it are the Proofs of interpretation of NFU in this theory
-Below it is Ontology and general remarks to familiarize on with this
rather hard to work with theory.

EXPOSITION:
N.S.T is the set of all sentences entailed (from first order
logic with identity and epsilon membership "e" and the
primitive one place function symbol "N" standing for "name" and the
primitive two place function symbol "<,>" standing for ordered pair.)
by the non logical axioms outlined below the following definitions:

Define(Ur-element): x is a Ur-element iff Exist y ( x=N(y) )

Define(set): x is a set iff ~ Exist y ( x=N(y) )

Define(in): y in x iff N(y) e x

AXIOMS:

1) Extensionality 1:

for all sets x,y (for all z ( z e x <-> z e y) -> x=y)

2) Extensionality 2:

For all x,y ( N(x)=N(y) -> x=y )

3) Ordered pairs: for all a,b,c,d (<a,b>=<c,d> -> (a=c & b=d))

4) Sets: For all x,y ( y e x -> (y is a Ur-element & x is a set) )

5) Naming: For all x: ~ N(x) = x

6) Pair coding:

For all x,y ( <N(x),N(y)> = N(<x,y>) )

7) Comprehension: If Phi is a formula in which x is not free, and
which do not use the symbol "N", then all closures of

Exist a set x for all y ( y in x iff Phi )

are axioms.

Define: x={y|Phi} iff for all y ( y in x iff Phi )

Define: x=[N(y)|Phi(y)] iff for all y ( N(y) e x iff Phi(y) )

8) Back coding:

For all x
(For all y ( y e x -> Exist z ( y=N(z) & ~ z is a set))
->
Exist x' For all y ( y e x' iff Exist m ( m e x & m=N(y)))).

9) Union:

For all c Exist x for all y
(y in x <-> Exist z ( z in c & y in z ))

Theory definition finished/

INTERPRETING NFU IN NST.

First the membership relation "e" that is present in NFU is
interpreted by the defined membership relation "in" in this theory

So for example we take the first axiom in NFU which Extensionality
written in the finite axiomatization of NFU by Holmes in the
following
manner:

For all sets a,b ( For all x ( x e a iff x e b ) -> a=b )

Now this is interpreted in this theory as

For all sets a,b ( For all x ( x in a iff x in b ) -> a=b )

This is a theorem of this theory since this would be reduced to

For all sets a,b ( For all x ( N(x) e a iff N(x) e b ) -> a=b )

Now it is not difficult to prove that

For all sets a,b ( For all x ( N(x) e a iff N(x) e b ) ->
For all y ( y e a iff y e b ) ).

The rational beyond that is it is impossible to have an object y e a
and ~y e b and vice versa!

Now for a proof by negation Suppose we have the object y e a
and ~ y e b, now since y e a then y is a Ur-element (axiom 4) which
mean that there should exist an object z such that y=N(z) , but by
then we'll have y e b. A contradiction! thus we cannot have any
object
y such that y e a and ~ y e b, thus we have for all y: y e a -> y e
b ;same thing will be proved of the opposite direction and so we'll
have
for all y: y e b -> y e a, Theorem proved.

Now we come to the second axiom in NFU which is of Atoms, which are
interpretable as Ur-elements here, both are not sets and both are
empty objects, this is the consequence of definition of Ur-elements
and of axiom 4 in this theory.QED

Sets in NFU is interpretable to sets here.

The third axiom of NFU asserts the existence of the set of all
objects

Also here we do have the set of all objects in this theory, simply
take Phi to be y=y and substitute in comprehension and we'll have the
set of all objects.QED

so V={x|x=x} do exist in this theory.

The fourth axiom in NFU is about absolute complements, this is proved
using the formula ~y e A and substituting in comprehension and then
using axiom of back coding to derive the complementary set of A.QED

The fifth axiom in NFU is that of Boolean unions. This is also proved
simply by using the formula y e A or y e B, substituting in
comprehension and then back coding and we'll have the Boolean union
of
A and B.

The sixth axiom in NFU is that of Set union, which is interpreted to
the axiom of Union here.

The seventh axiom of NFU is axiom of singletons, which is proved by
simply taking Phi to be y=A and substituting in comprehension and
well
get {A}.QED.

The eighth axiom of that of ordered pair, which is interpretable to
the one present here (axiom 3)

The ninth axiom of NFU is that of Cartesian product of two sets.

This is proved here simply because any two sets are nothing but
epsilon collections of Ur-elements, so simply take Phi to be
Exist a,b (a e A & b e B & y=<a,b>) then substitute in comprehension
and then back code the resulting set and we'll get Cartesian product
AxB. QED

The tenth axiom in NFU is that of the Diagonal which asserts that the
set {<x,x>|x e V} exists.

which is interpreted here as the Diagonal {<x,x>| x in V} exists.

This is simply proved by taking the formula "Exist x (y=<x,x>)"
substitute in comprehension and you get the Diagonal set.QED

The eleventh axiom of NFU asserts that the projection sets {<x,y>,x>|
x,y e V} and {<x,y>,y>|x,y e V} exists.

Which is interpreted in this theory to:
{<x,y>,x>|x,y in V} and {<x,y>,y>|x,y in V} exists.

These are proved by simply taking Phi to be
Exist a,b ( y=<a,b>,a>)
Exist a,b ( y=<a,b>,b>)

and substitute in comprehension and well get the projection sets.QED

The twelve's axiom of NFU is the axiom of converses: which asserts
that for any relation R , its converse relation R^-1={<y,x>|xRy}
exist.

This is simply proved by taking Phi to be
Exist a,b (<a,b> e R & y=<b,a>)

Substitute in comprehension and back code and we get R^-1.

The thirteenth axiom of NFU is axiom of relative products written as:

IF R,S are relation the set

(R|S)={<x,y>|Exist z, x R z & z S y}

called the relative product of R and S exists.

This is also simply proved by taking Phi to be:

Exist a,b,c ( y=<a,c> & Exist b (<a,b> e R & <b,c> e S) )

substitute in comprehension and back code and we'll get (R|S).

The fourteenth axiom of NFU is the axiom of Domains
stated as: for some relation R the set

dom(R)={x| Exist y (<x,y> e R)}

called the domain of R exists.

also this is simply proved by taking Phi to be Exist z( <z,y> e R)
and substitute in comprehension back code and well get dom(R).

The fifteen axiom of is the axiom of Singleton images which is stated
as:

For any set R, the set

SI(R)={<{x},{y}>| <x,y> e R }

called the singleton image of R exist.

This is also simply proved by the formula
Exist a,b,m,n (<a,b> e R & m={a} & n={b} & y=<{a},{b}>).

substitute in comprehension and we get SI(R).

The last axiom of NFU is axiom of inclusion which is:

The set {<x,y>| x subset_of y} exists.

This also can be simply proved by taking Phi in comprehension to be
Exist a,b ( y=<a,b> & Az( z e a -> z e b))
and we get the above set.

Or further restriction to avoid Ur-elements would
be

Exist a,b( ~Exist k( a e k)& ~Exist k (b e k) & y=<a,b> & Az( z e a -
>
z e b))).

So all sixteen axioms of NFU are proved here.

Axiom of Choice can also be added to this theory, since it contain
Ur-elements, in exactly the same manner as how it is added to NFU.

So this theory clearly interprets NFU.

But is it consistent?

IF consistent, can we find an interpretation of this theory in NFU?

What would be the consistency strength of this theory?

All questions to be solved?


ONTOLOGY AND GENERAL REMARKS:

Objects are divided here into "Naming objects" called Ur-elements,
and
these are the objects that name some objects; and non naming objects
axiomatized to be collections of Ur-elements and these are called
sets, now sets do not name any object.

So for an object x to be called a Ur-element then there must exist an
object y such that x is the name of y, or simply N(y)=x, while on the
other hands for sets which are non naming objects there do not exist
an object that they name.

Sets are actually epsilon collections of Ur-elements, i.e. objects
that have Ur-elements related to them by membership relation "e"
which
is the primitive membership relation in this theory. While Ur-
elements
do not have any object related to them by the membership e, i.e they
are empty objects. However the empty set on the other hand has not
object related to it by membership relation e, but the difference
between it and the Ur-elements is that it is a non naming object,
while the Ur-elements are empty objects but they are naming objects.

Sets do not have Sets related to them by membership relation e, so in
this they you cannot say that x and y are sets and at the same time
have x e y or have y e x, Sets only have Ur-elements as their
members.

Ur-elements are non extensional objects, that's why axiom of
Extensionality 1 is only restricted to "sets", so Extensionality 1
has
nothing to do with Ur-elements, i.e two Ur-elements can be different
although they have the same membership which is the empty membership,
i.e. the identity of Ur-elements do not depend on their members,so
for
every two Ur-elements x and y we do have
for all z( z e x <-> z e y) but yet x can be not identical to y.

On the other hand "sets" are extensional objects, i.e their identity
can be determined from their membership, so any two sets that have
the
same members are identical.

Now we come to explain the defined membership relation "in".

Simply speaking when one have x in y then this simply mean
the name of x has epsilon membership relation to y, i.e.
N(x) e y

So: For all x: x in y <-> N(x) e y

Now the symbols used to differentiate between these two membership
relations are the solid and curved brackets [] and {}
so x=[y] means y e x and every object other than y is not e x

For all z ( z e x <-> z=y ) <-> x=[y]

In general we used [y|Q] to mean the set which contain the all
objects
that satisfy property Q, of course all y should be
Ur-elements.

While x={y} mean y in x, which is N(y) e x and every object other
than y then it is not in x (i.e. its name is not e x )

so we do have {y}=[N(y)]

In general {y|Q}=[N(y)|Q(y)]

i.e x={y|Q} is the set of all of the names of objects that satisfy Q
i.e all objects that are names of objects y that satisfy Q have
epsilon membership relation to x, while other objects that are not
names of objects y that satisfy Q then they do not have epsilon
membership relation to x.

Now an ordered pair of x and y would be a Ur-element if x and y are
Ur-
elements, while if x and y are sets they it would be a set, Ordered
pairs of Ur-elements and sets are ignored in this theory.

So the big picture is that we have non naming objects called Sets and
we have their names and the names of their names and the names of
these later names etc... infinitely, and Sets themselves are epsilon
collections of these naming objects.

The names are unique identifiers so there is one name for each object
and one object named by one name. And the name is always different
from the object it names.

One property of this theory is back coding, that is suppose you have
a
set of names of names, then you will have the set of the named names.

So suppose you have sets , x, y, z
you have their first names N(x), N(y), N(z)
and we have their second names N(N(x)), N(N(y)),N(N(z))
Now suppose we have the set K=[N(N(x)), N(N(y)),N(N(z))]
which is of course the set {N(x), N(y), N(z)}
So by back coding on K we should have the set [N(x), N(y), N(z)]
which is of course the set {x,y,z}.

so if we have a set x of the i-th names of some sets, then we have a
set of the i-1 names of them that are named by i-th names in(e) x.

The comprehension axiom in this theory is perhaps the nearest
comprehension axiom to Naive comprehension in form.

Applying these axioms we can interpret NFU easily actually.

I hope that this account would be helpful in familiarizing this
rather
hard to work with theory.

Regards

Zuhair