Class Theory /\
Zalj...@gmail.com
Theory /\ is the set of all sentences entailed by (using FOL with
identity and the primitives "in" and "V" were V is a constant) the
following non logical axioms:
Definition: x is a set iff x in V
Definition: x is a class iff [ x in V or ~ x in V ]
So we have: for all x : x is a class.
1) Axiom of Extensionality:
forall z (z in A <-> z in B) -> A=B
2) Axiom of complementary class.
forall A, Exists X : forall y ( y in X <-> ~ y in A )
Define: X=A' iff forall y ( y in X <-> ~ y in A )
A' is read as 'the complementary class of A'
3) Axiom of Intersection:
forall A, Exists X : forall y ( y in X <-> forall z ( z in A -> y in
z ) )
Define: X= /\ A iff forall y ( y in X <-> forall z ( z in A -> y in
z ) )
/\ A is read as 'the intersectional class of A'.
4) All axioms of Z except regularity and sepration.
Define: X = { A,B } iff forall y ( y in X <-> (y=A or y=B) )
{A,B} is read as the unordered pair of A and B.
Define: X= Union A iff forall y ( y in X <-> Exists z ( z in A and
y
in z ) ).
Union A is read as "the class union of A".
Define: X= Power A iff forall y ( y in X <-> forall z ( z in y -> z
in A ) ).
Power A is read as "the power class of A".
5) Axiom of set existence: Exists X : X in V
6) Axiom of reflexiseness of V: forall X in V , forall y ( y in X ->
y
in V ).
i.e: any member of a set is a set.
or: any set is a subclass of V.
7.1) Axiom of pairing of sets: The unordered pair of two sets is a
set.
7.2) Axiom of set union: The class union of a set is a set.
7.3) Axiom of power set: The power class of a set is a set.
7.4) Axiom of Infinity:
Exists N in V ( 0 in N and forall y (y in N -> yU{y} in N) ).
7.5) Axiom schema of separation for sets: If P is a formula in which X
is not free, then all closures of
forall A in V, Exists X in V, forall y ( y in X <-> ( y in A and
P(y) ) )
are axioms.
7.6) Axiom schema of replacement for sets: If P is a formula in which
B is not free, then all closures of
[(forall x in V, Exists! y in V:P(x,y))
and (forall x in V ( P(x,y) -> y in V ))]
-> forall A in V ,Exists B in V, forall y ( y in B <-> Exists x in A:
P(x,y) ).
are axioms.
7.7) Axiom of Regularity for sets: For every non empty set there
should exist an element that is disjoint of it.
7.8) Axiom of choice for sets: V is well orderable.
8) Axiom schema of class comprehension over V: if P is a formula in
which X is not free, then all closure of
Exists X: forall y ( y in X <-> ( y in V and P(y) ) ).
are axioms.
9) Axiom schema of partial replacement for classes: If P is a formula
in which B is not free, then all closures of:
[forall x in V, Exists! y : P(x,y)] ->
forall A subset_of V, Exists B, forall y ( y in B <-> Exists x in
A:P(x,y) ).
are axioms.
/Theory definition finished
Perhaps, and I say perhaps Axiom 8 is redundant.
I think that we can of course reduce the number of axioms in this
theory, however I think the above presentation is a good one to begin
with.
This theory do not have Ur-elements
classes in this theory are divided into:
sets: which are classes that are members of V.
proper classes: which are subsets of V but not members of V.
ultraclasses: which are not subsets of V, but smaller of equal in size
to V.
universal classes: like U the class of all classes, and classes that
are near to U in size, like U except 0 for example,etc...
Now the question here is: Is this theory consistent?
Zuhair
Rupert
Yes, I believe that I can prove that this theory is consistent
provided it is consistent with ZFC that an inaccessible cardinal
exists and I may post a proof of this later.
Zalj...@gmail.com
Note: Subset in the above axiom schema means subclass.
>
> /Theory definition finished
>
> Perhaps, and I say perhaps Axiom 8 is redundant.
>
> I think that we can of course reduce the number of axioms in this
> theory, however I think the above presentation is a good one to begin
> with.
>
> This theory do not have Ur-elements
> classes in this theory are divided into:
> sets: which are classes that are members of V.
> proper classes: which are subsets of V but not members of V.
> ultraclasses: which are not subsets of V, but smaller of equal in size
> to V.
> universal classes: like U the class of all classes, and classes that
> are near to U in size, like U except 0 for example,etc...
>
> Now the question here is: Is this theory consistent?
>
> Zuhair
I am thinking of adding an axiom that is similar to inverse
intersection. An axiom that reveal the maximal class the intersection
of which is C for example.
Axiom of maximal inverse intersection.
forall C, Exists X : forall y ( y in X <-> C subclass of y ).
Zuhair
Zalj...@gmail.com
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -
Also I will add axiom schema of equivalence classes:
If R is equivlance relation, then all closures of
forall A, Exists X, forall y ( y in X <-> y R A )
is an axiom.
equivalence relation mean a relation that is reflexive,symmetric and
transitive.
By this schema, on can have the equivalence class of all classes
bijective to A.
also one can have equivalence class of all classes ordinally
isomorphic to A.
In this way one can define cardinal numbers as these equivalence
classes of bijective classes, and also define ordinal numbers as
equivalance classes of ordinally isomorphic classes. i.e. Frege's
method can be adopted here in this thoery.
This theory seem's to be nearer actually to Cantor's set theory, since
Cantor's original theory had a class of all classes in it (see
Cantor's paradox), and also Cantor defined cardinality and ordinality
in terms of equivalence classes.
So this theory is the nearest to Cantor's concepts of set theory, not
ZFC.
Zuhair
MoeBlee
> Also I will add axiom schema of equivalence classes:
>
> If R is equivlance relation, then all closures of
>
> forall A, Exists X, forall y ( y in X <-> y R A )
>
> is an axiom.
I don't think that works. You're in the meta-language talking about R
being an equivalence relation, which actually belongs in the object-
language. I don't see that it can't be a direct axiom rather than a
schema. So maybe this instead:
R is an equivalence relation -> AzExAy(yex <-> <y z>eR)
MoeBlee
Zalj...@gmail.com
a relation is not necessarily an object. so I think the way I put it
is the right one.
MoeBlee
> On Jul 25, 3:36 pm, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > On Jul 25, 3:04 pm, Zaljo...@gmail.com wrote:
>
> > > Also I will add axiom schema of equivalence classes:
>
> > > If R is equivlance relation, then all closures of
>
> > > forall A, Exists X, forall y ( y in X <-> y R A )
>
> > > is an axiom.
>
> > I don't think that works. You're in the meta-language talking about R
> > being an equivalence relation, which actually belongs in the object-
> > language. I don't see that it can't be a direct axiom rather than a
> > schema. So maybe this instead:
>
> > R is an equivalence relation -> AzExAy(yex <-> <y z>eR)
> a relation is not necessarily an object. so I think the way I put it
> is the right one.
It doesn't parse correctly. You use 'R' as a variable of the object
language, but mention R ITSELF in the meta-language. That doesn't make
sense, as well as there is no need for that kind of thing.
And what do you think is lacking in this much simpler formulation,
just a plain axiom:
R is an equivalence relation -> AzExAy(yex <-> <y z>eR)
Or maybe you mean this:
For all formulas P, all closures of
Axyz(P(x x) & (P(x y) -> P(y x)) & ((P(x y) & P(x z)) -> P(x z)))
->
AzExAy(yex <-> P(y z))
are theorems.
MoeBlee
Zalj...@gmail.com
You committed a typo twixe, at the transtivity of P and at saying
theorems.
I think you mean
Axyz(P(x x) & (P(x y) -> P(y x)) & ((P(x y) & P(y z)) -> P(x z)))
->
AzExAy(yex <-> P(y z))
are AXIOMS.
I agree with the above, that's what was in my mind.
But I prefer to state it in the following manner:
If R is a binary relation in which x is not free, then all closures of
[Ax(R(x,x)) & Axy(R(x,y)->R(y,x)) & Axyz((R(x,y)&R(y,z)) ->R(x,z))]
->
AzExAy(yex <-> R(y,z))
are axioms.
Zuhair
>
> MoeBlee- Hide quoted text -
Zalj...@gmail.com
> On Jul 25, 3:51 pm, Zaljo...@gmail.com wrote:
>
>
>
>
>
> > On Jul 25, 3:36 pm, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > > On Jul 25, 3:04 pm, Zaljo...@gmail.com wrote:
>
> > > > Also I will add axiom schema of equivalence classes:
>
> > > > If R is equivlance relation, then all closures of
>
> > > > forall A, Exists X, forall y ( y in X <-> y R A )
>
> > > > is an axiom.
>
> > > I don't think that works. You're in the meta-language talking about R
> > > being an equivalence relation, which actually belongs in the object-
> > > language. I don't see that it can't be a direct axiom rather than a
> > > schema. So maybe this instead:
>
> > > R is an equivalence relation -> AzExAy(yex <-> <y z>eR)
> > a relation is not necessarily an object. so I think the way I put it
> > is the right one.
>
> It doesn't parse correctly. You use 'R' as a variable of the object
> language, but mention R ITSELF in the meta-language. That doesn't make
> sense, as well as there is no need for that kind of thing.
>
> And what do you think is lacking in this much simpler formulation,
> just a plain axiom:
>
> R is an equivalence relation -> AzExAy(yex <-> <y z>eR)
What is lacking is that R here is an object. While I mean relations
that are not necessarily objects. And in FOL we cannot quantify over
such relations.
>
> Or maybe you mean this:
>
> For all formulas P, all closures of
>
> Axyz(P(x x) & (P(x y) -> P(y x)) & ((P(x y) & P(x z)) -> P(x z)))
> ->
> AzExAy(yex <-> P(y z))
>
> are theorems.
>
> MoeBlee- Hide quoted text -
Zalj...@gmail.com
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -
However Moe, I am coming to actually think of an axiom schema, that
states that for every two classes a relation that is a subclass of the
cartesian product between them should exist.
Axiom of relations: if P is a formula in which X is not free,
then all closures of
forall A, forall B, Exists X, forall y ( y in X <->
Exists z in A, Exists u in B ( y=<z,u> and P(y) )
are axioms
or according to your symbolisim
Axiom of relations: if P is a formula in which x is not free,
then all closures of
AabExAy(yex<->Ezu(zea,ueb,y=<z,u>,P(y) )
are axioms.
However I don't know why I have the vague sense that this axiom would
lead to a contradiction, but I am not sure.
Zuhair
Zalj...@gmail.com
>
> - Show quoted text -
Now I think the better way to wright this schema is
Axiom schema of relations: if P is a binary formula in which c is not
free, then all closures of
Axyz(P(x x) & (P(x y) -> P(y x)) & ((P(x y) & P(y z)) -> P(x z)))
->
AzEcAy(yec <-> P(y z))
are axioms.
MoeBlee
> I think you mean
>
> Axyz(P(x x) & (P(x y) -> P(y x)) & ((P(x y) & P(y z)) -> P(x z)))
> ->
> AzExAy(yex <-> P(y z))
>
> are AXIOMS.
Right, I meant 'axioms' not 'theorems'.
> I agree with the above, that's what was in my mind.
>
> But I prefer to state it in the following manner:
>
> If R is a binary relation in which x is not free, then all closures of
>
> [Ax(R(x,x)) & Axy(R(x,y)->R(y,x)) & Axyz((R(x,y)&R(y,z)) ->R(x,z))]
> ->
> AzExAy(yex <-> R(y,z))
> are axioms.
You may prefer it, but it's incorrrect. I explained why in the earlier
version. It's even more plain in your latest version: It makes sense
to talk about x not being free in a FORMULA R, but not in a binary
relation R. You conflate 'R' as an individual variable in the object
language with 'R' as a variable in the meta-language.
MoeBlee
MoeBlee
> Axiom schema of relations: if P is a binary formula
What is a "binary formula"?
> in which c is not
> free, then all closures of
>
> Axyz(P(x x) & (P(x y) -> P(y x)) & ((P(x y) & P(y z)) -> P(x z)))
> ->
> AzEcAy(yec <-> P(y z))
>
> are axioms.
MoeBlee
Zalj...@gmail.com
Sorry , the axiom schema is of equivalence classes and not of
relations.
Binary formula is a formula in two free variables.
x=y , Ez( z in x -> z in y ) , y in x , Ef( f:x->y, f is bijective) ,
etc....
all are binary formula's
Zuhair
MoeBlee
> Binary formula is a formula in two free variables.
>
> x=y , Ez( z in x -> z in y ) , y in x , Ef( f:x->y, f is bijective) ,
> etc....
>
> all are binary formula's
But I don't think you need to limit to only two variables, since
you're closing the whole thing with the clause 'all closures of'
anyway,
MoeBlee
Zalj...@gmail.com
do you mean that there is a contradiction when I say a "binary
formula"
and when I say "all closures of" . just a question.
Zuhair
MoeBlee
> On Jul 28, 3:05 pm, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > On Jul 28, 2:43 pm, Zaljo...@gmail.com wrote:
>
> > > Binary formula is a formula in two free variables.
>
> > > x=y , Ez( z in x -> z in y ) , y in x , Ef( f:x->y, f is bijective) ,
> > > etc....
>
> > > all are binary formula's
>
> > But I don't think you need to limit to only two variables, since
> > you're closing the whole thing with the clause 'all closures of'
> > anyway,
> do you mean that there is a contradiction when I say a "binary
> formula"
> and when I say "all closures of" . just a question.
No, not a contradiction, but then 'all closures of' is otiose since,
if there are no other free variables, then in your formula you've
already bound all the variables.
MoeBlee
Zalj...@gmail.com
>
> - Show quoted text -
So if I don't say all closures, then what I should say then.
Zuhair
Zalj...@gmail.com
>
> - Show quoted text -
Should I state it like that:
Axiom schema of equivalence classes: If P is a formula in two free
variables, then the closure of
Axyz(P(x x) & (P(x y) -> P(y x)) & ((P(x y) & P(y z)) -> P(x z)))
->
AzExAy(yex <-> P(y z))
is an axiom.
MoeBlee
In this particular case, since your formula has only two free
variables, and you quantify all substitutions of those variables, all
you have to say is
"If P is a formula with exactly two free variables, then all
instances of the following [put schema here] are axioms."
MoeBlee
MoeBlee
> Should I state it like that:
>
> Axiom schema of equivalence classes: If P is a formula in two free
> variables, then the closure of
>
> Axyz(P(x x) & (P(x y) -> P(y x)) & ((P(x y) & P(y z)) -> P(x z)))
> ->
> AzExAy(yex <-> P(y z))
>
> is an axiom.
You can, but the "closure" part is now unneeded.
All you need is:
If P is a formula with exactly two free variables, then
Axyz(P(x x) & (P(x y) -> P(y x)) & ((P(x y) & P(y z)) -> P(x z)))
->
AzExAy(yex <-> P(y z))
is an axiom.
MoeBlee