Re: Why Regularity?


Rupert wrote:
zuhair wrote:
Rupert wrote:
zuhair wrote:
Rupert wrote:
zuhair wrote:
Rupert wrote:
zuhair wrote:
Rupert wrote:
Jesse F. Hughes wrote:
"Rupert" <rupertmccallum@xxxxxxxxx> writes:

That implies that the set {m|mex or m=x'} exists, and since every
set is a member of itself it will be either x or x'.

Sorry, I don't follow.

Suppose that mex. That does not imply that m=x, does it? Every set
is self-membered, but not every set satisfies z={z}.

How does our resident philosopher define x'? I might have missed
that.


It was zuhair who confused me. First he said the successor of U was
{U,U'}, which as I proved can't exist, then elsewhere he said the
successor of x was {m|mex or m=x'}.

Suppose {m|mex or m=x'} exists. It's a member of itself, and it can't
be x. So it must be in x. So in any model of zuhair's axioms there are
infinite descending membership chains.

right, but all of these chains would make any difference. the axiom of
extentiality man,
Let me tell you what I mean.

for example 0={0}={0,{0}} = {0,{0},{0,{0}} } , etc... all of those are
nothing but zer0.

Now 1={0,0',1} were 0' is the complementary set of 0.

Now you can say that {0,0',1} = {0,0',1,{0,0',1}} =
{0,0',1{0,0',1},{0,0',1,{0,0',1}} } =.....

All of these cyclations should not bother you since by extensiality we
can reduce all of them
to {0,0',1}.

2= {0,0',1',2}

3= { 0,0',1',2',3}


You seem to keep changing your mind about how you define successor. I
thought the successor of x was {m|mex or m=x'}. That means 1=0 union
{0'}. It won't work if you start with 0 as the unique singleton.

So how is successor defined?

y=S(x) <-> y:(Amex -> mey) /\ x'ey

By axiom of Pairing if x is a set, y is a set, so is {x,y}
so {x,y} is the paired set of x and y.


Make up your mind. Do you want {x,y} or x union {y}?

The axiom of Pairing isn't consistent with your other axioms.

Let me use this definition.

Axiom of infinity:

EN:QeN /\ ( Ax:xeN -> (y:(Amex -> mey) /\ x'ey) eN )

so the essence of this axiom is:

EN:QeN/\(Ax:xeN -> S(x) e N ).

Q which is the unique singlton set in this theory, which I used to
symbolize it as U.


Say 0={0}. Then S(0)=0 too, regardless of which definition you use. So
this axiom is trivial and follows from the other axioms. If you drop
the requirement that the first element of the set be Q, then the axiom
says something interesting and is consistent (provided you use S(x)=x
union {x'} rather than S(x)={x,x'}, which doesn't work).

You are saying big statements without giving any proof for what you are
saying. you are simply speaking what you like to beleive.


My claims are very easy to prove - so easy that I didn't see the need
to give a proof. If you can't work out how to prove them, just ask me.

Say 0={0}. Then S(0)={0,0'}. So either S(0)=0 or S(0)=0',

wait a minute. from were you broght this. you have the following
implication;

S(0) = { 0,0' } -> S(0)=0 v S(0)=0'


Yes, that follows from your axiom that every set is a member of itself.

You forgot S(0)= { 0,0' } . or according to your terminology ,

you forgot S(0)= S(0)=0 v S(0)=0'


Sorry, I don't follow this. Does S(0)=S(0)=0 mean (S(0)=S(0) & S(0)=0)?

It is clear , how like sun in the ski is clear, that you are not
familiar with irregular sets.


These consequences of your axioms are trivial. I don't need to be
familiar with anything to derive them. You haven't found a problem with
my argument.


but since
0eS(0) the latter is clearly impossible.

what are you talking? this is not understood.


We can't have S(0)=0', because 0eS(0) so it would imply 0e0', which is
impossible.

you are saying 0eS(0) -> ~S(0)=0', why?????


Because (S(0)=0'&0eS(0))->0e0', and we have ~0e0'.

Now I know what you are trying to do. How many times I told you that
you should read my recent changes about my notation, I changed the
definition of complementay set. no Rupert we don't have ~0e0'. on the
contrary we have 0e0'.

You should know my definition of 0', you seem to think in the line that


AxEx'Ay yex'<->y!ex.

I admit that this was my definition of x' at the begining. BUT I
changed this definition into the following one.

AxEx'Ay:y=/=Q yex'<->y!ex.

this is different from the first one. According to this definition 0
which is Q, can be
in 0'. In reality according to my last set theory with four axioms, 0'
MUST have 0 as a member in it.

I hope that this resolve the confusion.

Zuhair

So S(0)=0. Actually this is
clearly impossible as well, so it is actually inconsistent that 0 has a
successor, I didn't realize this the first time around.

all of this reasoning is erronus.

Wrong.


What are you saying is not understandable at all.


Not to you, maybe. If you're having trouble seeing how to prove my
claims, just ask me.

0={0} then S(0)=0 too, this is one of your strange statements, what do
you mean by this ? it looks pritty okward to me.


Actually, if 0={0} then S(0) doesn't exist. See above.

I defined S(0) to be the set which has all the elements set 0 has and
in addition it contain 0', which is the complementary set of 0, so how
come S(0) =0 as you say, this is nonsense.
there is a great difference between { 0,0' ,1 } and 0, if you cannot
see that, then I don't think you understood anything from what I am
saying.

Oh, so you think S(0)={0,0',1}? What's 1?

1= S(0), it goes like that S(0)={0,0',S(0)}, but since we S(0)=1
then S(0)={0,0',1}, it seems that you are forgetting that here in this
theory Ax xex.

So you should really have said S(x) is the unique set such that
Ay(y!=S(x)->(yeS(x)<->(yex or y=x'))). All right, fine. Then your
axioms will work, I think.


If 0={0} and S(x)=x union {x'}, then S(0)={0,0'}. And this is
impossible because then S(0)=0 or S(0)=0', since every set is a member
of itself, but we can't have 0={0,0'} or 0'={0,0'}. So S(0) does not
have a successor.

If you want S(0)={0,0',1} you'll have to come up with another
definition of successor.

NO you are mistaken.

No, I'm not. The earlier definition you gave was wrong. I've fixed it
for you now.

.

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