IMPROVED AXIOM OF REGULARITY... <X1 e X2 e X3 .. e Xn e X1>

[Graham Cooper]

**********
AXIOM OF REGULARITY
A(Z)
E(Y) YeZ -> E(Y)( YeZ ^ !E(X)( XeY ^ XeZ ) )
**********

AOR is poorly written if it intends to stop this cyclic membership
sequence from occurring in ZFC!

X1 e X2 e X3 .. e Xn e X1

It barely works by breaking ONE DIRECT CHAIN in the transitive network

X e X e X e X ...

and fails on the counterexample:

X e Y e X e Y ...

-------AXIOM OF REGULARITY II------

AXIOM OF TRANSITIVITY
A(X) A(Z) X t Z <-> (X e Z) v E(Y) (X e Y) ^ (Y t Z)

AXIOM OF REGULARITY
A(X) A(Z) (X t Z) -> ~(X e Z)

-----------------------------------

e.g.
IF (A e B) & (B e C) ^ (C e D)
THEN (A t D)
A is a Transitive Element of D

THEREFORE BY AOR-II
A e D would be disallowed!


Graham Cooper (BInfTech)
KINGS BEACH QUEENSLAND
--
1 X ^ NOT(X)
2 G = NOT(PRV(G))
3 S > INF
4 R = {X | NOT(X e X)}
5 IF HALT() GOTO 5
6 ALL(F) MAX(F)
=
THE 6 DEAD ENDS IN MATHEMATICS
but only 4 are recognised contradictions

[Graham Cooper]

Post by Graham Cooper
-------AXIOM OF REGULARITY II------
AXIOM OF TRANSITIVITY
A(X) A(Z) X t Z <-> (X e Z) v E(Y) (X e Y) ^ (Y t Z)
AXIOM OF REGULARITY
A(X) A(Z) (X t Z) -> ~(X e Z)
-----------------------------------

This suggests an:

AXIOM OF WEAK REGULARITY
A(X) A(Z) (X t Z) -> ~(Z e X)

which will allow a directed acyclic graph of set membership
this is a more general network than a hierachy of set ranks.

e.g.

1 e SINGLETONS
SINGLETONS e MATHS_TERMS
1 e MATHS_TERMS

A relative rank and order is apparent, without cycles,
though an absolute rank is lost since SINGLETONS is mid-way
between MATHS_TERMS and 1, hence if 1 e MATHS_TERMS was
defined 1st a difference in set rank of 1 would have to be
split to accommodate the set SINGLETONS.

Graham Cooper (BInfTech)
KINGS BEACH QUEENSLAND

[MoeBlee]

Theorem of Z set theory: There is no f such that f1 e f2 .. e fn e
f1.
Let '/\' stand for binary intersection.
axiom of regularity (which is included in Z set theory): Ax(~x=0 ->
Em(mex & x/\m = 0))
Now suppose f1 e f2 ... e fn e f1.
So ~Em(m e {f1 f2 ... fn} & m/\{f1 f2 ... fn} = 0), which contradicts
the axiom of regularity

it doesn't hold for any general membership cycle
it only holds for a simple membership cycle

You disputed that ZFC proves that there are no X and Y such that
XeYeX.
I corrected you by pointing that not only does ZFC prove that there
are no X and Y such that XeYeX but even more generally ZFC proves that
for any natural number n, there are no x1, x2, ..., xn such that x1 e
x2 ... e xn e x1. You asked me to provide a proof. And now I've given
you a proof.

Right

Right. And you should've stopped right there. You claimed that ZFC
does not prove that there are no X and Y such that XeYeX. But I showed
that ZFC does prove that there are no X and Y such that XeYeX.

You need POWERSET(V) to use AOR.

Like I said, you should have stopped at 'Right'.

My proof (including any lemma upon which the proof depends) makes no
use nor mention of powersets, the power set axiom, or V (I surmise you
mean 'V' as the class of all sets).

My proof depends only on the axiom schema of separation, the pairing
axiom, the union axiom, the axiom of regularity, and for ease, the
axiom of extensionality (I could have avoided even the axiom of
extensionality, but it's easier to present the proof using
extensionality).

Anyway, even if the proof did use the powerset axiom, then the proof
would still be within ZFC (the proof is actually within Z set theory
not including the power set axiom and the axiom of infinity).

MoeBlee

[MoeBlee]

Post by MoeBlee
My proof depends only on the axiom schema of separation, the pairing
axiom, the union axiom, the axiom of regularity, and for ease, the
axiom of extensionality

Actually, even the axiom of schema of separation is not needed, since
whenever intersection is mentioned, we could just as easily formulated
in a different way that would not require the "sethood" of
intersections. So, the proof really only relies on pairing, union, and
regularity.

MoeBlee

[Graham Cooper]

Post by MoeBlee

Post by MoeBlee
My proof depends only on the axiom schema of separation, the pairing
axiom, the union axiom, the axiom of regularity, and for ease, the
axiom of extensionality

Actually, even the axiom of schema of separation is not needed, since
whenever intersection is mentioned, we could just as easily formulated
in a different way that would not require the "sethood" of
intersections. So, the proof really only relies on pairing, union, and
regularity.
MoeBlee

Great debating points MoeBlee, now yell *STOP^ when you get to the
first sentence that you disagree with!

Subject:
*IMPROVED* AXIOM OF REGULARITY... <X1 e X2 e X3 .. e Xn e X1>

AOR Ax(~x=0 -> Em(mex & x/\m = 0))

The problem is X is not a formulated_set in ZFC like the other
constructive_axioms.
AOR applied to an infinite countable list of theorems would need |
P(N)| invocations!

A(X) A(Z) (X t Z) -> ~(Z e X)
is equivalent to AOR but is a programmatic method

Herc

[MoeBlee]

Post by Graham Cooper

Post by MoeBlee

Post by MoeBlee
My proof depends only on the axiom schema of separation, the pairing
axiom, the union axiom, the axiom of regularity, and for ease, the
axiom of extensionality

Actually, even the axiom of schema of separation is not needed, since
whenever intersection is mentioned, we could just as easily formulated
in a different way that would not require the "sethood" of
intersections. So, the proof really only relies on pairing, union, and
regularity.

Great debating points MoeBlee,

They're not debating points. I'm just telling you finitistic facts
about provablity in a system.

Post by Graham Cooper
now yell *STOP^ when you get to the
first sentence that you disagree with!
*IMPROVED* AXIOM OF REGULARITY... <X1 e X2 e X3 .. e Xn e X1>
AOR Ax(~x=0 -> Em(mex & x/\m = 0))
The problem is X is not a formulated_set in ZFC like the other
constructive_axioms.

Whether of not the axiom of regularity should be considered
constructive, I do not opine. However, the question was whether the
ZFC proves ~Exy xeyex. And you see now that ZFC does prove ~Exy
xeyex.

Post by Graham Cooper
AOR applied to an infinite countable list of theorems

I didn't aply the axiom of regularity to an infinite countable list of
theorems. I don't even know what it would mean to "apply an axiom to a
list of theorems".

The axiom of regularity is used exactly once in the proof. It is
exactly one line in the proof.

The proof uses only certain axioms of Z set theory (by the way, power
set axiom is not used in this proof) and first order logic. The proof
is in Z set theory, perforce in ZFC.

MoeBlee

[Graham Cooper]

Post by MoeBlee

Theorem of Z set theory: There is no f such that f1 e f2 .. e fn e
f1.
Let '/\' stand for binary intersection.
axiom of regularity (which is included in Z set theory): Ax(~x=0 ->
Em(mex & x/\m = 0))
Now suppose f1 e f2 ... e fn e f1.
So ~Em(m e {f1 f2 ... fn} & m/\{f1 f2 ... fn} = 0), which contradicts
the axiom of regularity

it doesn't hold for any general membership cycle
it only holds for a simple membership cycle

You disputed that ZFC proves that there are no X and Y such that
XeYeX.
I corrected you by pointing that not only does ZFC prove that there
are no X and Y such that XeYeX but even more generally ZFC proves that
for any natural number n, there are no x1, x2, ..., xn such that x1 e
x2 ... e xn e x1. You asked me to provide a proof. And now I've given
you a proof.

Right

Right. And you should've stopped right there. You claimed that ZFC
does not prove that there are no X and Y such that XeYeX. But I showed
that ZFC does prove that there are no X and Y such that XeYeX.

You need POWERSET(V) to use AOR.

Like I said, you should have stopped at 'Right'.
My proof (including any lemma upon which the proof depends) makes no
use nor mention of powersets, the power set axiom, or V (I surmise you
mean 'V' as the class of all sets).

Yes! x is not an element of V

x e P(V)

Herc

[MoeBlee]

Post by Graham Cooper

Post by MoeBlee
My proof (including any lemma upon which the proof depends) makes no
use nor mention of powersets, the power set axiom, or V (I surmise you
mean 'V' as the class of all sets).

Yes!

Yes, right. The proof makes no use of power sets, the power set axiom,
or of the class of all sets. (And even if the power set axiom were
used, still the proof would be in Z set theory, perforce in ZFC).

Post by Graham Cooper
x is not an element of V
x e P(V)

You're confused. V is not invoked in the proof. Really, I don't see
why you won't read an introductory textbook in set theory.

MoeBlee

[Graham Cooper]

Post by MoeBlee

Post by Graham Cooper

Post by MoeBlee
My proof (including any lemma upon which the proof depends) makes no
use nor mention of powersets, the power set axiom, or V (I surmise you
mean 'V' as the class of all sets).

Yes!

Yes, right. The proof makes no use of power sets, the power set axiom,
or of the class of all sets. (And even if the power set axiom were
used, still the proof would be in Z set theory, perforce in ZFC).

Post by Graham Cooper
x is not an element of V
x e P(V)

You're confused. V is not invoked in the proof. Really, I don't see
why you won't read an introductory textbook in set theory.
MoeBlee

You really do not see the difference between the equivalent formula

Ax(~x=0 -> Em(mex & x/\m = 0))
and
A(X) A(Z) (X t Z) -> ~(Z e X)

X is a theorem, x is a set of theorems.

The purpose of set theory axioms is not to prove that set theory
works, they are supposed to be utilised to instantiate new formula.

If you had 10^7 sets in your theory and added 1 more:

a) my set level AOR-ii would take O(n log(n)) checks
to instantiate 1 more formula.

b) the powerset level ZFC-AOR would take |P(10^7)| checks
to instantiate 1 more formula.


e.g.
V = { SET1, SET2, SET3 }
<=>
V = { {SET8}, {SET3,SET4}, {SET1,SET2,SET5} }

i.e.
SET1 = {8}
SET2 = {3,4}
SET3 = {1,2,5}

P(V) = {
{},
{{8}},
{{3,4}},
{{1,2,5}},
{{8},{3,4}},
{{8},{1,2,5}},
{{3,4},{1,2,5}}, **
{{8},(3,4),{1,2,5}}
}

NOW you can invoke ZFC-AOR on P(V)

Ax(~x=0 -> Em(mex & x/\m = 0))

WHEN
x = {{3,4},{1,2,5}}
ie. x = {{SET3, SET4}, {SET1, SET2, SET5}}
ie. x = {SET2, SET3}

m=SET2
x/\m = x/\SET2 = x/\{SET3,SET4} = {SET3,SET4} =/= 0

m=SET3
x/\m = x/\SET3 = x/\{SET1,SET2,SET5} = {SET1,SET2,SET5} =/= 0

AOR is not satisfied... eventually!

***********************************


Compare that to a simple TRANSITIVE RELATION and AOR-II

-------AXIOM OF REGULARITY II------
AXIOM OF TRANSITIVITY
A(X) A(Z) X t Z <-> (X e Z) v E(Y) (X e Y) ^ (Y t Z)

AXIOM OF REGULARITY
A(X) A(Z) (X t Z) -> ~(Z e X)
-----------------------------------


Herc

[Graham Cooper]

Post by Graham Cooper
e.g.
V = { SET1, SET2, SET3 }
<=>
V = { {SET8}, {SET3,SET4}, {SET1,SET2,SET5} }
i.e.
SET1 = {8}
SET2 = {3,4}
SET3 = {1,2,5}
P(V) = {
{},
{{8}},
{{3,4}},
{{1,2,5}},
{{8},{3,4}},
{{8},{1,2,5}},
{{3,4},{1,2,5}},   **
{{8},(3,4),{1,2,5}}
}
NOW you can invoke ZFC-AOR on P(V)
Ax(~x=0 -> Em(mex & x/\m = 0))
WHEN
x = {{3,4},{1,2,5}}
ie. x = {{SET3, SET4}, {SET1, SET2, SET5}}
ie. x = {SET2, SET3}
m=SET2
x/\m = x/\SET2 = x/\{SET3,SET4} = {SET3,SET4} =/= 0

Should read:

x/\m = x/\SET2 = x/\{SET3,SET4} = {SET3} =/= 0



Herc

--
PROGRAM UNIVERSE
BEGIN: NIL
WHILE TRUE# DO
FOR X,Y,Z INC BY PLANCH LENGTH
IF (SCHRODINGERS_CAT == ALIVE)^(SCHRODINGERS_CAT == DEAD)
REDUCE(STATEVECTOR(X,Y,Z))
END_IF
NEXT_BIT
END_WHILE
THE_END

#TRUE IS UNTIL THE END OF TIME

[MoeBlee]

Post by Graham Cooper
You really do not see

Enough.

You said that

"it is not the case that there exist X and Y such that XeYeX"

is not provable in ZFC.

But I proved it in ZFC.

The rest of your confusions are of no import in that matter.

I do suggest (though you reject such suggestions) that you get an
introductory textbook in set theory.

MoeBlee

[Graham Cooper]

Post by MoeBlee

Post by Graham Cooper
You really do not see

Enough.
You said that
"it is not the case that there exist X and Y such that XeYeX"
is not provable in ZFC.
But I proved it in ZFC.
The rest of your confusions are of no import in that matter.
I do suggest (though you reject such suggestions) that you get an
introductory textbook in set theory.
MoeBlee

I said RIGHT the 1st time.

You ignored that topic in the other thread then pounced on this thread
instead which is a different topic, it only mentioned XeYeXeY.. as an
adjunct to AOR being unworkable, just omit that 1 line from the OP.

Herc

[MoeBlee]

Post by Graham Cooper
I said RIGHT the 1st time.

Good.

But then you went on with confused clutter.

Post by Graham Cooper
You ignored that topic in the other thread

I don't promise to address each topic, even the main topic, of a
thread. What interested me in particular in that thread was your
incorrect claim that ZFC does not prove that there are no X and Y such
that X e Y e X.

Post by Graham Cooper
then pounced on this thread

And you repeated that incorrect claim in this thread, so I replied
here too.

More basically, you have too many self-imposed confusions about set
theory. You'd do yourself a great favor by reading an introductory
textbook in the subject.

MoeBlee

[Graham Cooper]

Post by MoeBlee

Post by Graham Cooper
I said RIGHT the 1st time.

Good.
But then you went on with confused clutter.

Post by Graham Cooper
You ignored that topic in the other thread

I don't promise to address each topic, even the main topic, of a
thread. What interested me in particular in that thread was your
incorrect claim that ZFC does not prove that there are no X and Y such
that X e Y e X.

Post by Graham Cooper
then pounced on this thread

And you repeated that incorrect claim in this thread, so I replied
here too.
More basically, you have too many self-imposed confusions about set
theory. You'd do yourself a great favor by reading an introductory
textbook in the subject.
MoeBlee

More good debating points!

MoeBlee, if O(P(N)) complexity issues don't interest you feel free to
allow others to contribute instead of trashing the topic.

If you can explain how to use Axiom Of Regulation on this example
without invoking

ALL SUBSETS OF ALL SETS IN V (i.e. the POWERSET)

V = { SET1, SET2, SET3 }
<=>
V = { {SET8}, {SET3,SET4}, {SET1,SET2,SET5} }

i.e.
SET1 = {8}
SET2 = {3,4}
SET3 = {1,2,5}

P(V) = {
{},
{{8}},
{{3,4}},
{{1,2,5}},
{{8},{3,4}},
{{8},{1,2,5}},
{{3,4},{1,2,5}}, **
{{8},(3,4),{1,2,5}}
}

NOW you can invoke ZFC-AOR on P(V)
Ax(~x=0 -> Em(mex & x/\m = 0)

*************

then do so!

Herc

[MoeBlee]

Post by Graham Cooper
feel free to
allow others to contribute

I'm not stopping anybody from posting whatever they want.

MoeBlee

[Graham Cooper]

Post by MoeBlee

Post by Graham Cooper
feel free to
allow others to contribute

I'm not stopping anybody from posting whatever they want.
MoeBlee

You trashed all the content in the thread insisting AOR has nothing to
do with powersets or V.

Can you actually use AOR on a trivial example MoeBlee?

Assume V = {SET1, SET2}

SET1 = {SET8}
SET2 = {SET3, SET4}
SET3 = {SET1, SET2, SET5}

Will SET3 be an allowed set using Axioms Of ZFC?

HINT: SET2 e SET3 and SET3 e SET2

Herc