Modular Arithmetic Can't Be Well Founded
RussellE
standard axioms of set theory requires the existence
of sets which are elements of themselves.
http://en.wikipedia.org/wiki/Axiom_of_regularity
The axioms of MA:
Ex (S(x)=0)
Ax ((S(x)=S(y)) -> (x=y))
Ax (x+0=x)
AxAy (x+S(y)=S(x+y))
Ax (x*0=0)
AxAy (x*S(y)=x*y+x)
First Order Induction
Ay (P(0,y) and Ax (P(x,y) -> P(S(x),y))) -> Ax P(x)
where y is a finite number of free variables
The axioms of General Set Theory are
Extensionality, Specification schema,
and Adjunction.
http://en.wikipedia.org/wiki/General_set_theory
Adjunction says if x and y are sets then
x union {y} is a set. I will use the symbol
"J" to mean adjunction.
This is my definition of ordinal.
The empty set is an ordinal.
If x is an ordinal then x J x is
an ordinal.
It is standard to define the successor
function as:
S(x) = x J x
Letting the empty set represent 0 and
assuming x is an ordinal and substituting
into Ex (S(x)=0) we get:
Ex (x J x = {})
All non-empty ordinals have {} as an
element. xJx must be non-empty.
MA + GST proves there exists an
ordinal, k, such that all ordinals
greater than or equal to k must
contain themselves as an element.
Russell
- Integers are an illusion
Frederick Williams
>
> Combining the axioms of Modular Arithmetic with
> standard axioms of set theory requires the existence
> of sets which are elements of themselves.
> http://en.wikipedia.org/wiki/Axiom_of_regularity
>
> The axioms of MA:
>
> Ex (S(x)=0)
> Ax ((S(x)=S(y)) -> (x=y))
> Ax (x+0=x)
> AxAy (x+S(y)=S(x+y))
> Ax (x*0=0)
> AxAy (x*S(y)=x*y+x)
>
> First Order Induction
> Ay (P(0,y) and Ax (P(x,y) -> P(S(x),y))) -> Ax P(x)
> where y is a finite number of free variables
A remark that proves that you don't know what a free variable is.
> The axioms of General Set Theory are
> Extensionality, Specification schema,
> and Adjunction.
> http://en.wikipedia.org/wiki/General_set_theory
>
> Adjunction says if x and y are sets then
> x union {y} is a set. I will use the symbol
> "J" to mean adjunction.
>
> This is my definition of ordinal.
>
> The empty set is an ordinal.
> If x is an ordinal then x J x is
> an ordinal.
>
> It is standard to define the successor
> function as:
>
> S(x) = x J x
You have some axioms governing the behaviour of S, and GST's axioms
govern the behaviour of J. Can you prove that those axioms and your
definition are consistent?
> Letting the empty set represent 0 and
> assuming x is an ordinal and substituting
> into Ex (S(x)=0) we get:
>
> Ex (x J x = {})
>
> All non-empty ordinals have {} as an
> element. xJx must be non-empty.
>
> MA + GST proves there exists an
> ordinal, k, such that all ordinals
> greater than or equal to k must
> contain themselves as an element.
>
> Russell
> - Integers are an illusion
--
When a true genius appears in the world, you may know him by
this sign, that the dunces are all in confederacy against him.
Jonathan Swift: Thoughts on Various Subjects, Moral and Diverting
Dan
> Combining the axioms of Modular Arithmetic with
> standard axioms of set theory requires the existence
>
> The axioms of MA:
>
> Ex (S(x)=0)
> Ax ((S(x)=S(y)) -> (x=y))
> Ax (x+0=x)
> AxAy (x+S(y)=S(x+y))
> Ax (x*0=0)
> AxAy (x*S(y)=x*y+x)
>
> First Order Induction
> Ay (P(0,y) and Ax (P(x,y) -> P(S(x),y))) -> Ax P(x)
> where y is a finite number of free variables
>
> The axioms of General Set Theory are
> Extensionality, Specification schema,
>
> Adjunction says if x and y are sets then
> x union {y} is a set. I will use the symbol
> "J" to mean adjunction.
>
> This is my definition of ordinal.
>
> The empty set is an ordinal.
> If x is an ordinal then x J x is
> an ordinal.
>
> It is standard to define the successor
> function as:
>
> S(x) = x J x
>
> Letting the empty set represent 0 and
> assuming x is an ordinal and substituting
> into Ex (S(x)=0) we get:
>
> Ex (x J x = {})
>
> All non-empty ordinals have {} as an
> element. xJx must be non-empty.
>
> MA + GST proves there exists an
> ordinal, k, such that all ordinals
> greater than or equal to k must
> contain themselves as an element.
>
> Russell
> - Integers are an illusion
The successor function from ordinals doesn't need to be the same as S
from modular arithmetic . And in fact isn't . Names used in different
contexts/theories mean different things .
For your effort , Bugs Bunny will have your palm red for free
http://www.youtube.com/watch?v=CeUeN2sQeDo&t=2m36s.
Frederick Williams
In
> Ay (P(0,y) and Ax (P(x,y) -> P(S(x),y))) -> Ax P(x)
should the P(x) at the end by P(x,y)? And is the scope of the first
quantifier this:
(P(0,y) and Ax (P(x,y) -> P(S(x),y)))
or this:
(P(0,y) and Ax (P(x,y) -> P(S(x),y))) -> Ax P(x)
?
RussellE
wrote:
> RussellE wrote:
>
>
> > It is standard to define the successor
> > function as:
>
> > S(x) = x J x
>
> You have some axioms governing the behaviour of S, and GST's axioms
> govern the behaviour of J. Can you prove that those axioms and your
> definition are consistent?
FOL with equality can prove the axioms of MA
have a model. Since FOL with = is complete and
consistent, this proves MA is consistent with
respect to FOL with =.
I don't know if GST is consistent or if
MA+GST is consistent. Do you have
some reason to think MA+GST is inconsistent?
I am using a standard definition of successor.
Everything is a set in set theory. Again, it is
standard to associate ordinals with elements
in the universe. How else can we represent
numbers in set theory other than associating
each number with some set?
Russell
- Zeno was right. Motion is impossible.
David Libert
> On Jul 23, 1:21=A0pm, Frederick Williams <freddywilli...@btinternet.com>
> wrote:
>> RussellE wrote:
>>
>>
>> > It is standard to define the successor
>> > function as:
>>
>> > S(x) = x J x
>>
>> You have some axioms governing the behaviour of S, and GST's axioms
>> govern the behaviour of J. Can you prove that those axioms and your
>> definition are consistent?
>
> FOL with equality can prove the axioms of MA
> consistent, this proves MA is consistent with
>
> I don't know if GST is consistent or if
> MA+GST is consistent. Do you have
> some reason to think MA+GST is inconsistent?
>
> I am using a standard definition of successor.
> Everything is a set in set theory. Again, it is
> standard to associate ordinals with elements
> in the universe. How else can we represent
> numbers in set theory other than associating
> each number with some set?
>
>
> Russell
> - Zeno was right. Motion is impossible.
There is the literal question as asked above, and variant questions,
according the various sensible ways to consider combining GST and MA
into one theory.
The literal question posed above, is to consider numbers as in MA
intepreted into sets in the usual way, as for example the usual
interpretation of PA into ZF, so interpreting numbers as von Neumann
ordinals, and intepreting the sucessor function on numbers S as the
set theoretic adjunction function J as you write above.
So you could consider the theory, GST about sets, with the MA axioms
intepreted into set language added.
That theory is inconsistent.
There are other variations, similar to Dan's post above. Just add S
as a new primitive, with no axioms relating to sets, and put in the
MA axioms about this and the other primitives from the MA language.
The most straightforward way to do this is just copy all axioms of
the two theories GST and MA into the same combined language. So in
each case quantifiers just copy as simple quantifiers. Ie meaning that
the universe of the resulting theory is both a GST model and an MA model,
but same underling universe for both models.
In this case, if we just smash together the two theories GST and MA
as they literally read originally, the resulting theory is consistent.
It only has infinite models though, no finite ones. Since GST only as
infinite models, as you were saying recently Russell in another thread.
There are a couple more sensble variations to take on defining the
new theory combining GST and MA. You could also consider it to be a
theory of a universe combining a set theory part with a modular arithmetic
part. You could do this as a two sorted logic, or equiavelently put in
one or 2 new primitives as unuary predicates to distinguish these 2 parts
of the universe, and relativise the quantigfiers from the respective GST and
MA subtheoreis to the respective predicates.
Then the next question is what sort of crosstalk you allow between the two
parts.
The most literal version would be the variables for members of set in the
GST axioms only range over the set theory part. Also all other quantifiers
in each subtheory just stay in their own predicate.
Then any GST model and any MA can be stuck together to make a model of this
theory. So this theory has infinite models.
A more seneisible theory would be to let the set part have members from the
MA side. So the variables rnging over set members would be over the combined
universe.
The most conservative theory along these lines would only make those variables
for set members be on the full universe. Maybe the other quantifiers in induction
axioms in the MA part are realtivized just to the MA predicate and the
quatifiers in separation in GST are just realtized to the GST part.
That theory is consistent, with infinite models only. Namely the set part of
the models is always infinite. There are models with the MA part of the model
being finite, and other models with it being infinite.
We can even make a more powerful version of the last theory. This version
would be allow the quantifiers in the separtion axioms in the GST part to
range over the combined GST and MA predicates.
This theory is similar to the last case. Consistent, all models infinite
set part. MA part of models can be finite for some models and infinite for
others.
Next consuder variations. Allow quantifited in the induction axiom on
the MA part to range over both sets and MA numbers.
There could be 2 subcases of this: do quantifiers for separation axioms
on sets range only over sets or over numbers and sets.
In both these subcases the resulting theories are inconsistent.
--
David Libert ah...@FreeNet.Carleton.CA
Frederick Williams
> FOL with equality can prove the axioms of MA
> have a model.
You can't even express "the axioms of MA have a model" in the language
of FOL with equality.
> I don't know if GST is consistent or if
> MA+GST is consistent. Do you have
> some reason to think MA+GST is inconsistent?
You also have S(x) = x J x, where x J y = x union {y}.
RussellE
> RussellE (reaste...@gmail.com) writes:
>
> The literal question posed above, is to consider numbers as in MA
> intepreted into sets in the usual way, as for example the usual
> interpretation of PA into ZF, so interpreting numbers as von Neumann
> ordinals, and intepreting the sucessor function on numbers S as the
> set theoretic adjunction function J as you write above.
The usual interpretation of PA into ZF is to interpret numbers as
von Neuman ordinals and to interpret successor using adjunction.
> So you could consider the theory, GST about sets, with the MA axioms
> intepreted into set language added.
>
> That theory is inconsistent.
So, if we negate one axiom of PA+ZF, Ax (S(x) !=0), we get
an inconsistent theory? Doesn't this mean Ax (S(x) != 0) is
NOT independent of the other axioms of PA+ZF and this
axiom can be derived from the other axioms of PA+ZF?
I don't know why you think MA+GST is inconsistent.
GST does not have an axiom of regularity.
> There are other variations, similar to Dan's post above. Just add S
> as a new primitive, with no axioms relating to sets, and put in the
> MA axioms about this and the other primitives from the MA language.
Why add axioms for set theory then?
I did not include axioms of set theory in MA because
I know some axioms of set theory require the existence
of arbitrarily large sets.
How can I represent the elements of the universe
in set theory without using sets? How can I define
a successor function in set theory without using sets?
Russell
- 2 many 2 count
RussellE
wrote:
> RussellE wrote:
> > FOL with equality can prove the axioms of MA
> > have a model.
>
> You can't even express "the axioms of MA have a model" in the language
> of FOL with equality.
FOL with equality can prove there exists a structure that satisfies
the axioms of MA.
> > I don't know if GST is consistent or if
> > MA+GST is consistent. Do you have
> > some reason to think MA+GST is inconsistent?
>
> You also have S(x) = x J x, where x J y = x union {y}.
This is a standard definition of successor in set theory.
Can you suggest a different definition for the
successor of a set?
David Libert
In this thread I posted
[1] David Libert "Re: Modular Arithmetic Can't Be Well Founded"
sci.logic, sci.math July 24, 2011
http://groups.google.com/group/sci.logic/msg/6cda4ec6c1c15e99
Russell posted a followup to [1]:
[2] RussellE "Re: Modular Arithmetic Can't Be Well Founded"
sci.logic, sci.math July 24, 2011
http://groups.google.com/group/sci.logic/msg/53db90930d403381
This present article is a followup to [2]. This article will
comment on points from [2], but also return back for comments on
[1] directly.
In [1] I was commenting on an earlier question raised by
Russell previously in this thread. Since positng [1] I noticed
some errors of mine in [1]. I will be commenting on those
below. Also in general I will give a clearer statement of things
from [1] apart from those errors.
I will also respond to some questions and comments from [2],
apart from those issue above regarding [1].
I will first respond to [2].
In [2], Russell wrote:
>On Jul 23, 10:53 pm, ah...@FreeNet.Carleton.CA (David Libert) wrote:
>> RussellE (reaste...@gmail.com) writes:
>> The literal question posed above, is to consider numbers as in MA
>> intepreted into sets in the usual way, as for example the usual
>> interpretation of PA into ZF, so interpreting numbers as von Neumann
>> ordinals, and intepreting the sucessor function on numbers S as the
>> set theoretic adjunction function J as you write above.
>
>
>The usual interpretation of PA into ZF is to interpret numbers as
>von Neuman ordinals and to interpret successor using adjunction.
>> So you could consider the theory, GST about sets, with the MA axioms
>> intepreted into set language added.
>> That theory is inconsistent.
>
>
>So, if we negate one axiom of PA+ZF, Ax (S(x) !=0), we get
>an inconsistent theory? Doesn't this mean Ax (S(x) != 0) is
>NOT independent of the other axioms of PA+ZF and this
>axiom can be derived from the other axioms of PA+ZF?
>I don't know why you think MA+GST is inconsistent.
>GST does not have an axiom of regularity.
Yes, I think negating that one axiom Ax (S(x) !=0)
makes an inconsistent theory. Here is why.
GST includes in its axioms the schematum of separation. Some
set exists by pure FOL, and applying separation on formula ~x=x
to such a set gives emptyset, as an existing set in GST.
The interpretation above of numbers into sets interprets number
0 as set emptyset.
The MA axiom in question says ~Ax (S(x) !=0). So working in
this combined MA+GST theory under discussion, let x be such an
an interpreted number from MA s.t. S(x) = 0.
Our interpretation under discussion interprets S(x) as x J x,
so this last says x J x = 0.
x J x = x union {x}, so x is a member of x J x.
So this x would be member of 0.
By our interpetation under discussion, 0 is another name for
emptyset.
So x is a member of emptyset. Contrary to what the set theory
part GST of MA+GST proves about emptyset.
That concludes my claim about inconsistency of this formulation
of MA+GST under this version based on the usual interpretation of
the arithmetical 0 and S into sets.
You comment above that GST does not have an axiom of
regularity. That is true. And I even assert the stronger claim
that GST does not prove regularity as a theorem.
But my inconsistency claim was based on the proof above, which
doesn't use regularity.
[2] continues:
>> There are other variations, similar to Dan's post above. Just add S
>> as a new primitive, with no axioms relating to sets, and put in the
>> MA axioms about this and the other primitives from the MA language.
>
>Why add axioms for set theory then?
>I did not include axioms of set theory in MA because
>I know some axioms of set theory require the existence
>of arbitrarily large sets.
>How can I represent the elements of the universe
>in set theory without using sets? How can I define
>a successor function in set theory without using sets?
[2] was quoting [1], where [1] had just noted the inconsistency
of MA+GST based on the usual interpretation of 0, S, +, * into
sets as discussed above.
[1] turned to consider alternative formalizations of MA+GST.
At this point in [1], [1] was introducing formulations of
MA+GST which would be in a larger language with symbols =,
epsilon, 0, S, +, *. This version would not try to define
0, S, +, * in terms of =, epsilon, but instead just use those
symbols directly, and copy the MA axioms or variations on them to
those available symbols.
The [1] discussion of this began with the section that [2]
quoted above:
>> There are other variations, similar to Dan's post above. Just add S
>> as a new primitive, with no axioms relating to sets, and put in the
>> MA axioms about this and the other primitives from the MA language.
When that [2] quote said there were no axioms relationg to
sets, it meant there were now no axioms directly axiomatizing S
involving the set part of the MA+GST theory. There is no axiom
defining S in terms of =, epsilon.
It did not mean that the entire MA+GST theory dropped all set
related axioms from the entire theory. Just those axioms that
defined S (and 0, +, *) in terms of = , epsilon.
So regarding the questions which followed in [2]:
>Why add axioms for set theory then?
>How can I represent the elements of the universe
>in set theory without using sets? How can I define
>a successor function in set theory without using sets?
The set theory axioms of MA+GST provide set theory for its
theory, but they don't do the job of defining 0, S, +, *.
We had already tried that in the preceding version and got the
problem of inconsistency. So this was considering some
alternatives.
This alternative is not using the set theory part to do those
jobs of the last questions. This version of MA+GST doesn't
involve such a definability reduction.
[1] had its section introducing this definition of MA+GST. [2]
above quoted from that section of [1], but [2] did not quote
the full [2] section.
The full [2] section was:
> There are other variations, similar to Dan's post above. Just add S
>as a new primitive, with no axioms relating to sets, and put in the
>MA axioms about this and the other primitives from the MA
>language.
>
> The most straightforward way to do this is just copy all axioms of
>the two theories GST and MA into the same combined language. So in
>each case quantifiers just copy as simple quantifiers. Ie meaning that
>the universe of the resulting theory is both a GST model and an MA model,
>but same underling universe for both models.
[2] above only made quoted this part from that of [1]:
>> There are other variations, similar to Dan's post above. Just add S
>> as a new primitive, with no axioms relating to sets, and put in the
>> MA axioms about this and the other primitives from the MA language.
The full [1] quote shows more clearly how this version of MA+GST
involves set theory, but just not in the connection to the MA
part.
[2] also icluded some comments I quoted earlier but not did not
repeat in this last section. I will quote those again and
comment:
>I did not include axioms of set theory in MA because
>I know some axioms of set theory require the existence
>of arbitrarily large sets.
Yes, in the other thread introducing FS you commented that you
would maybe go further to ultra finitism, with skepticism
about unboundedly large finite sets or numers (I don't remember
which now).
My theories FS and SS were not trying to formalize this and
they don't. That is also an interesting exercise, but is a
separate topic from these.
I was led to the original FS and SS by your comments, how
raising questions about my analysis of statements discussing MA
if mine rested on assumptions that infiinte sets existed.
My own working framework for mathematics is readily formaalized
in set theory, so I was trying to come up with an alternative
formalization similar to the usual one (allowing abribtraily
large finite sets, to be closer to how we usually do math),
without including the assumption you were then questioning
(infinite sets exist).
At that earlier time ultrfinitism had not yet been raised.
I was also inspired by WM, and he writes about potential
infinity with no actual infinity. To get something like
potential infinity I was also allowing arbitrarily large finite
sets.
That concludes my comments on [2].
I will return to some points about [1] that I had realized on
my own, apart from [2].
In [1] I wrote about the issue of separation axioms for the GST
part of MA+GST allowing quantification over the MA part of the
universe. And similarly the induction axioms from the MA part
allowing quantification over the GST part of the universe.
I should have added a stronger condition, which I had forgotten
to mention in [1] in that part. The separation axioms should not
merely allow quantification into MA, but should also allow using
0, S, +, *. And the induction axioms of MA should not merely
allow quqantification into the GST part of the universe, but
should also allow using epsilon.
My claims in [1] about those cases should have included those
conditions in the cases.
Also, one case from [1] was combined MA+GST language with no
quqntitifer relativization. For this case the condition above
about allowing full quantification becomes vacuous. But it still
an issue for axioms of one side accessing symbols 0, S, +, *
or epsilon from the other side.
At the end of [1] I wrote:
> Next consuder variations. Allow quantifited in the induction axiom on
>the MA part to range over both sets and MA numbers.
>
> There could be 2 subcases of this: do quantifiers for separation axioms
>on sets range only over sets or over numbers and sets.
>
> In both these subcases the resulting theories are inconsistent.
This was a mistake. I was thinking of my proof for the case of
a common universe with no realtization to MA part or GST part.
As I wrote that I forgot that this was claiming for more cases
than the proof applied to.
There is an additonal point to extend the old discussion from
[1], which I am only writing about for the first time now.
I had posted about my theories FS and SS. On my own, I had
thought about a continuing topic, of writing about a combined
theory of MA+FS.
I had never gotten to writing such articles. Before I did,
you posted about MA+GST. Seeing that, I thought it was more or
less as I had been thinking, so I started writing about your
topic, including some that I had intended to add later on my own
topic.
In my mind I was thinking of my own case MA+FS. As I thought
about it, I overlooked that there were two distinct parallel
cases: over GST or FS.
So this gives more subcases to consider. But it turns out to
not matter so much in the end. It was not obvious to me at the
outset, but as I work through all the proofs or example I know, I
realize the GST and FS cases work out the same way anyway.
But just to note we can state examples and proved claims for
these more cases.
I think I can give the general outline of what [1] was saying
with better overall organization.
So below I will give a new restatement of [1] along the lines
above. Correcting the errors noted above, and noting the cases
GST vs FS, and organziing it as just noted.
So this restatement of [1] will be various things about
theories of style MA+GST or MA+FS. There are different ways
to formulate such theories, so the discussion will be about
different such variants.
The discussion will be about which cases are inconsistent, and
for the consistent cases the possble sizes of models both
regarding the MA part and the theory part.
The new organization of the overall statement will be like a
(if then elseif then elseif then elseif ... elseif then )
construct.
So I will present a chart of test conditions, if tests. Given
a case, you read down the chart to find the first line where
the condition tests as true. The the final anwser for that case
is on that line to the right of the condition.
Each answer will just be read from one line. The test
conditions say which line.
At the bottom of the chart is an answer applicable if all tests
on all previous lines failed.
The first line of the chart tests if we are defining 0, S from
MA as the usual set theoretic defintion of 0 as emtptyset and
S as self adjunction, as discussed above. If so, the answer
is all such theories are inconsistent.
Second line is applicable if the 1st line didn't test
positive. This tests are we in the case of common universe. no
realtization of quantifiers to MA part and set theory part, and
are we also in the case that MA induction axioms can mention
epsilon. If so, then the theory is inconsistent. Both GST and FS
cases.
Next case. Common universe again, but not in previous cases,
in other words MA induction axioms can't mention epsilon. The
the theory is consistent, but all models are infinite. Both GST
and FS versions
Last case. No new test condition, just if all previous tests
failed. In other words, MA induction can't mention epsilon. The
thoery is consistent. All models have infinite set theory part.
Some models have MA art finite, others infinite.
To sumarize:
0,S defined by emptyset, J inconsistent
common universe, induction has epsilon inconsistent
common universe, no epsilon in induction consistent
infinite MA
otherwise consistent
MA can be finite
or infinite
--
David Libert ah...@FreeNet.Carleton.CA
Frederick Williams
But it is not "standard" to add MA to set theory and then to define the
S(x) in MA (previously an undefined notion) to be x union {x}.
There is no reason to suppose that because two theories are each
consistent, the theory that arises by taking the axioms of both of them,
and defining a function symbol in one by using function symbols in the
other, will be consistent. David Libert has explained why, in the
particular case of MA + GST + S(x) = x J x, inconsistency results.
If you are interested in joint consistency and definability you should
read Beth and Robinson on these issues.
Meanwhile, I feel no need to suggest a different definition for the
successor of a set. You will be familiar with the saying
If you're in a hole, stop digging.
I've just thought up an addition to it:
and don't get others to do any digging in your place.
Frederick Williams
>
> On Jul 23, 10:53 pm, ah...@FreeNet.Carleton.CA (David Libert) wrote:
> > RussellE (reaste...@gmail.com) writes:
> >
> > The literal question posed above, is to consider numbers as in MA
> > intepreted into sets in the usual way, as for example the usual
> > interpretation of PA into ZF, so interpreting numbers as von Neumann
> > ordinals, and intepreting the sucessor function on numbers S as the
> > set theoretic adjunction function J as you write above.
>
> The usual interpretation of PA into ZF is to interpret numbers as
> von Neuman ordinals and to interpret successor using adjunction.
Not really. Natural numbers may be defined to be finite von Neumann
ordinals. They could be defined in Zermelo's way. In the von Neumann
case successor of x is defined as x union {x}, which is a particular
case of adjunction. But not only can we interpret the symbols of PA in
ZF, we can also ...
> > So you could consider the theory, GST about sets, with the MA axioms
> > intepreted into set language added.
> >
> > That theory is inconsistent.
>
> So, if we negate one axiom of PA+ZF, Ax (S(x) !=0), we get
> an inconsistent theory?
... prove the axioms of PA as theorems of ZF. So negating an axiom of
PA means (in effect) negating a theorem of ZF (and, a fortiori, ZF +
anything).
> Doesn't this mean Ax (S(x) != 0) is
> NOT independent of the other axioms of PA+ZF and this
> axiom can be derived from the other axioms of PA+ZF?
With the appropriate definitions, ZF proves PA.
>
> I don't know why you think MA+GST is inconsistent.
I think this is the fourth time I've said it, it's not MA + GST, it's MA
plus GST plus S(x) =df x union {x}. MA and GST are about different
things (integers modulo an integer vs sets), they have disjoint
vocabularies, so MA + GST is pointless _until_ a definition is made
which links the two. Your definition: S(x) =df x union {x} results in
inconsistency.
> [...]
>
> How can I represent the elements of the universe
> in set theory without using sets? How can I define
> a successor function in set theory without using sets?
The elements of the (I'd rather say 'a') universe in the set theory ZF
are sets. They may not be "real" sets, but they are the things that set
theory are about. Everything in ZF is about sets.
Frederick Williams
> >
> > How can I represent the elements of the universe
> > in set theory without using sets? How can I define
> > a successor function in set theory without using sets?
>
> The elements of the (I'd rather say 'a') universe in the set theory ZF
> are sets. They may not be "real" sets, but they are the things that set
> theory are about. Everything in ZF is about sets.
Bletch. "universe in the set theory ZF" should be "universe of a model
of the set theory ZF".
I should also have written "is about" not "are about".
MoeBlee
> This is my definition of ordinal.
>
> The empty set is an ordinal.
> If x is an ordinal then x J x is
> an ordinal.
That's not a definition in the limited context you gave.
(1) You left off the exclusion clause.
(2) To allow such an inductive definition requires proving that there
is a least set closed under the operation J. You haven't done that in
your GST.
(3) By the way, you've allowed only for finite ordinals.
(4) An actual definition:
x is an ordinal <-> (x is well ordered by membership on x & x is
membership transitive)
MoeBlee
RussellE
wrote:
> RussellE wrote:
>
> > On Jul 23, 10:53 pm, ah...@FreeNet.Carleton.CA (David Libert) wrote:
> > > RussellE (reaste...@gmail.com) writes:
>
> > > The literal question posed above, is to consider numbers as in MA
> > > intepreted into sets in the usual way, as for example the usual
> > > interpretation of PA into ZF, so interpreting numbers as von Neumann
> > > ordinals, and intepreting the sucessor function on numbers S as the
> > > set theoretic adjunction function J as you write above.
>
> > The usual interpretation of PA into ZF is to interpret numbers as
> > von Neuman ordinals and to interpret successor using adjunction.
>
> Not really. Natural numbers may be defined to be finite von Neumann
> ordinals. They could be defined in Zermelo's way. In the von Neumann
> case successor of x is defined as x union {x}, which is a particular
> case of adjunction. But not only can we interpret the symbols of PA in
> ZF, we can also ...
By "interpret the symbols", I assume you mean assign
some set to represent 0, define some operation on sets
to represent successor, addition, etc.
> > > So you could consider the theory, GST about sets, with the MA axioms
> > > intepreted into set language added.
>
> > > That theory is inconsistent.
>
> > So, if we negate one axiom of PA+ZF, Ax (S(x) !=0), we get
> > an inconsistent theory?
>
> ... prove the axioms of PA as theorems of ZF. So negating an axiom of
> PA means (in effect) negating a theorem of ZF (and, a fortiori, ZF +
> anything).
The axioms of PA are theorems of ZF?
I assume you mean PA interpreted as operations on sets.
> > Doesn't this mean Ax (S(x) != 0) is
> > NOT independent of the other axioms of PA+ZF and this
> > axiom can be derived from the other axioms of PA+ZF?
>
> With the appropriate definitions, ZF proves PA.
We can derive Ax (x union {x} != {}) in ZF?
Which axioms of ZF do we need to do this?
I don't think we need the axiom of infinity.
I think we might need the axiom of foundation.
>
> > I don't know why you think MA+GST is inconsistent.
>
> I think this is the fourth time I've said it, it's not MA + GST, it's MA
> plus GST plus S(x) =df x union {x}. MA and GST are about different
> things (integers modulo an integer vs sets), they have disjoint
> vocabularies, so MA + GST is pointless _until_ a definition is made
> which links the two. Your definition: S(x) =df x union {x} results in
> inconsistency.
The axioms of PA are not theorems of ZF.
The axioms of PA interpreted as sets may be theorems of ZF.
Axiom of PA:
Ax (S(x) != 0)
Theorem of ZF
Ax (x union {x} != {})
Using a standard interpretation of natural numbers
and successor, I show GST proves MA must have
sets that are not well founded.
GST has no axiom of foundation and I don't
see why MA as interpreted in GST is inconsistent.
Dave Libert's example uses separation to
"prove" x is a member of the empty set.
Let x={}
x union {x} = {{}}
{{}} is not empty. In some models
of MA we will have {{}}={}, but the
set {{}} will still exist in that model.
Separation just says a set exists.
I see no contradiction.
My primary interest in MA is as a
tool to prove PA is inconsistent.
I already have a finitistic proof that
MA is consistent. This means I
have a finitistic proof of all but
one of the axioms of PA.
If Ax (x union {x} != {}) is a theorem
of ZF, maybe there is a finitistic proof.
Russell
- The universe is one dimensional
Marshall
>
> My primary interest in MA is as a
> tool to prove PA is inconsistent.
Why do you want to try to prove a falsehood?
Marshall
David Libert
>We can derive Ax (x union {x} != {}) in ZF?
Yes. X is a member of the LS. x is not a member of the RS.
If LS = RS then LS and RS would agree on membership fact about
x. This is basically what I posted before.
>Which axioms of ZF do we need to do this?
If you define the terms above in the obvious way, x union {x}
is the set of those y s.t. y member of x or y = x, and
{} is the set with no members, those definitions would seem to
support the above. Those terms are not really in the language of
ZF, so we define them as in Betrand Russell's theory of definite
descriptions. If we take the definition using existential
quantifier, then using the term includes assuming it denotes.
This is the more common way. If so we should also have enough
axioms to prove such setrs exist. Union and singleon and
separation suffice. Or having separation or empty set, singelton
can be replaced by adjunction.
>I don't think we need the axiom of infinity.
>I think we might need the axiom of foundation.
The above doesn't need either.
>Theorem of ZF
>Ax (x union {x} != {})
Yes, this is what I have been saying.
>Using a standard interpretation of natural numbers
>and successor, I show GST proves MA must have
>sets that are not well founded.
I will comment more related to the above below.
>GST has no axiom of foundation and I don't
>see why MA as interpreted in GST is inconsistent.
>Dave Libert's example uses separation to
>"prove" x is a member of the empty set.
>Let x={}
>x union {x} = {{}}
>{{}} is not empty. In some models
>of MA we will have {{}}={}, but the
>set {{}} will still exist in that model.
>Separation just says a set exists.
>I see no contradiction.
This was my proof from previous post, similar to the
above.
Given we are interpeting MA's 0 as set {}, the proof
doesn't involve the successor of 0, it involves the predecessor
of 0.
Above given x = {}, you consider x union {x}.
In other words the sucessor of 0 by the interpretation.
But that was not the proof. Instead it considered the
PREDECESSOR of 0, not SUCCESSOR.
So instead it finds x s.t. x union {x} = 0.
The backward direction from what you were just talking about.
And from there notes as above, these are not =, because x is
a member of the LS and not a member of the RS.
>My primary interest in MA is as a
>tool to prove PA is inconsistent.
>I already have a finitistic proof that
>MA is consistent. This means I
>have a finitistic proof of all but
>one of the axioms of PA.
>
>If Ax (x union {x} != {}) is a theorem
>of ZF, maybe there is a finitistic proof.
Ax (x union {x} != {}) is a theorem of ZF.
Here are some new points I just thopught of, maybe getting
closer to what you had in mind.
Suppose we drop the requirement that MA 's 0 interpet as {}.
Then MA + GST + S interprets a J is consistent.
But all models of that theory are non-wellfounded, similar to
what you said above.
Further stronger similar results. Suppose M is a model of
MA and ZF is consistent. Then there is a ZF-regularity model
containing as member a set M' which is set interpretation of the
MA language and is ismorphic to M, where S's interetation
in M is isorphic to the adjunction operation in the ZF model on M.
Furthermore, if the starting M was a member of a V=L ZF
model, then we can arrange that the final ZF-regularity model N
with M' has M' definable N in the language of =, epsilon.
All such GST and ZF - regularity models interepresting MA with
MA 's S interepted as adjunction must be non-well-founded.
--
David Libert ah...@FreeNet.Carleton.CA
MoeBlee
> We can derive Ax (x union {x} != {}) in ZF?
You don't know how to prove
~ xu{x}=0
in Z set theory?
> I think we might need the axiom of foundation.
You think that because you haven't studied page one of a book.
MoeBlee
MoeBlee
> x union {x}
> is the set of those y s.t. y member of x or y = x, and
> {} is the set with no members, those definitions would seem to
> support the above. Those terms are not really in the language of
> ZF, so we define them as in Betrand Russell's theory of definite
> descriptions.
We don't need definite descriptions for this. Plain definitions
suffice.
MoeBlee
RussellE
I am glad you have an open mind about the problem.
Actually, I am trying to prove PA is consistent.
I can already prove MA is consistent using just
FOL with equality. If I can prove PA is consistent
using just FOL and the axioms of PA, I will be done.
RussellE
> >We can derive Ax (x union {x} != {}) in ZF?
>
> Yes. X is a member of the LS. x is not a member of the RS.
> If LS = RS then LS and RS would agree on membership fact about
> x. This is basically what I posted before.
>
> >Which axioms of ZF do we need to do this?
>
> If you define the terms above in the obvious way, x union {x}
> is the set of those y s.t. y member of x or y = x, and
> {} is the set with no members, those definitions would seem to
> support the above. Those terms are not really in the language of
> ZF, so we define them as in Betrand Russell's theory of definite
> descriptions. If we take the definition using existential
> quantifier, then using the term includes assuming it denotes.
> This is the more common way. If so we should also have enough
> axioms to prove such setrs exist. Union and singleon and
> separation suffice. Or having separation or empty set, singelton
> can be replaced by adjunction.
>
> >I don't think we need the axiom of infinity.
> >I think we might need the axiom of foundation.
>
> The above doesn't need either.
You need foundation because we can have {{}}={}
without foundation. Then Ax (x union {x} != {}) would
not be true for x={}.
> >Theorem of ZF
> >Ax (x union {x} != {})
>
> Yes, this is what I have been saying.
ZF has the axiom of regularity.
> >Using a standard interpretation of natural numbers
> >and successor, I show GST proves MA must have
> >sets that are not well founded.
>
> I will comment more related to the above below.
>
> >GST has no axiom of foundation and I don't
> >see why MA as interpreted in GST is inconsistent.
> >Dave Libert's example uses separation to
> >"prove" x is a member of the empty set.
> >Let x={}
> >x union {x} = {{}}
> >{{}} is not empty. In some models
> >of MA we will have {{}}={}, but the
> >set {{}} will still exist in that model.
> >Separation just says a set exists.
> >I see no contradiction.
>
> This was my proof from previous post, similar to the
> above.
>
> Given we are interpeting MA's 0 as set {}, the proof
> doesn't involve the successor of 0, it involves the predecessor
> of 0.
>
> Above given x = {}, you consider x union {x}.
>
> In other words the sucessor of 0 by the interpretation.
>
> But that was not the proof. Instead it considered the
> PREDECESSOR of 0, not SUCCESSOR.
This has nothing to do with successor or predecessor.
Separation says x union {x} defines a set.
If x={} then x union {x} = {{}} and {{}} is a set in any
interpretation of MA in GST. This satisfies separation.
> So instead it finds x s.t. x union {x} = 0.
>
> The backward direction from what you were just talking about.
>
> And from there notes as above, these are not =, because x is
> a member of the LS and not a member of the RS.
There is a model of MA where {{}} = {}.
{{}} is still a set in this model.
Thank you. This is what I think, too.
I am not really sure where you stand on the
consistency of MA. Would you please clarify some questions?
Is MA consistent with respect to FOL with equality?
By this I mean we can't derive a contradiction from
the axioms of MA using just FOL with equality.
Can PA prove MA has a 0-,model "with semantics"?
Can ZFC prove MA has a 0-model with semantics?
Can MA+"MA has an infinite model" prove MA has
a 0-model with semantics?
Jesse F. Hughes
> You need foundation because we can have {{}}={}
> without foundation. Then Ax (x union {x} != {}) would
> not be true for x={}.
You really are betraying your ignorance.
By definition of singleton, we know that {} in {{}}. By definition of
empty set, we know that {} not in {}. Therefore, {{}} != {}.
--
Jesse F. Hughes
"Now 'pure math' makes sense as well as clearly it's a peacock game,
where some of you see it as a way to show you as being highly
intelligent and thus more desirable to women." -- James S. Harris
Frederick Williams
> Actually, I am trying to prove PA is consistent.
> I can already prove MA is consistent using just
> FOL with equality.
You must show us the proof some day. I'd be delighted just to see
0 =/= 0 is not provable in MA
expressed in FOL=. Or
M is a model of MA
ditto.
> If I can prove PA is consistent
> using just FOL and the axioms of PA, I will be done.
You will have proved PA inconsistent in that case.
Frederick Williams
> Separation says x union {x} defines a set.
Pairing and union say that if x is a set, x union {x} is a set.
If you are working in a pure set theory, every term "defines" a set. No
axioms required.
MoeBlee
> > >We can derive Ax (x union {x} != {}) in ZF?
> You need foundation because we can have {{}}={}
> without foundation. Then Ax (x union {x} != {}) would
> not be true for x={}.
No, the absence of the axiom of regularity does NOT allow that "0 =
{0}" is consistent with ZF\{regularity}.
Yes, "Ex x = {x}" is consistent with ZF\{regularity}. But that does
not entail that "0 = {0}" is consistent with ZF\{regularity}.
By definition:
0 = y <-> Ax ~xey
By definition:
0e{0}.
So ~0={0}.
So 0={0} is NOT consistent with even a lot less than full ZF\
{regularity}.
You're confused as a result of your ignorance as a result of your
refusal to pick up a basic textbook.
> Separation says x union {x} defines a set.
No, the union, pairing, and extensionality axioms provide that there
is a unique set that has as members all and only those y such that y
in x or y=x. (Pairing is derivable, though, from the schema of
replacement and power).
Definition: {x} = {x x}
Definition: xuz = U{x z}
So xu{x} = U{x {x x}}
The axiom schema of separation is not used for this.
You're confused as a result of your ignorance as a result of your
refusal to pick up a basic textbook.
MoeBlee
David Libert
RussellE (reas...@gmail.com) writes:
> On Jul 26, 1:56=A0am, ah...@FreeNet.Carleton.CA (David Libert) wrote:
>> >We can derive Ax (x union {x} != {}) in ZF?
>>
>> Yes. X is a member of the LS. x is not a member of the RS.
>> If LS = RS then LS and RS would agree on membership fact about
>> x. This is basically what I posted before.
>>
>> >Which axioms of ZF do we need to do this?
>>
>> If you define the terms above in the obvious way, x union {x}
>> is the set of those y s.t. y member of x or y = x, and
>> {} is the set with no members, those definitions would seem to
>> support the above. Those terms are not really in the language of
>> ZF, so we define them as in Betrand Russell's theory of definite
>> descriptions. If we take the definition using existential
>> quantifier, then using the term includes assuming it denotes.
>> This is the more common way. If so we should also have enough
>> axioms to prove such setrs exist. Union and singleon and
>> separation suffice. Or having separation or empty set, singelton
>> can be replaced by adjunction.
>>
>> >I don't think we need the axiom of infinity.
>> >I think we might need the axiom of foundation.
>>
>> The above doesn't need either.
>
>
>You need foundation because we can have {{}}={}
>without foundation. Then Ax (x union {x} != {}) would
>not be true for x={}.
You are saying in ZF - Foundation we can have {{}} = {}.
I say ZF - Foundation proves {{}} ~= {}.
Jessie and MoeBlee already pointed this out in this thread.
{{}} is defined to be the set with only member {}.
So {} is a member of {{}}. Not using foundation, just using
the definition of {{}}.
And if you want to know what axioms support that there is a set
with only member {}, it is separation to get {} and then
pairing. MoeBlee already pointed this out.
So {} and {{}} being proper terms, that the underlying
definitions really are definitions (existence and unqiueness)
doesn't need foundation, and {} member of {{}} doesn't need
foundation.
As noted {} is a proper term wotithout foundation. And {} is
defined to be the set with no members. So ZF - Foundation proves
the set defined to be the set with no members has no members.
So {} is a member of {{}} and {} is not a member of {}.
That proof not assuming foundation.
So {{}} and {} are not =, since they have different members.
This by the axioms of = for FOL.
This proof is very short and simple.
I posted accounts of it in two previous articles, that should
have been enough for ordinary mathematical readers to read and
understand. Then just before this article in this thread Jesse
and MoeBlee each posted complete proofs.
Now I have just done it again.
This short simple little proof at the baby level of set theory
has now been posted 5 times in 5 separate articles by 3
different authors.
>> >Theorem of ZF
>> >Ax (x union {x} != {})
>> Yes, this is what I have been saying.
>
>
>ZF has the axiom of regularity.
Right. Ok I will add the proof I claim in ZF for this does not
use regularity. The same proof works in ZF - regularity, or
indeed in GST.
Later on, after some deletions:
>> Here are some new points I just thopught of, maybe getting
>> closer to what you had in mind.
>> Suppose we drop the requirement that MA 's 0 interpet as {}.
>>
>> Then MA + GST + S interprets a J is consistent.
>>
>> But all models of that theory are non-wellfounded, similar to
>> what you said above.
>>
>> Further stronger similar results. Suppose M is a model of
>> MA and ZF is consistent. Then there is a ZF-regularity model
>> containing as member a set M' which is set interpretation of the
>> MA language and is ismorphic to M, where S's interetation
>> in M is isorphic to the adjunction operation in the ZF model on M.
>>
>> Furthermore, if the starting M was a member of a V=L ZF
>> model, then we can arrange that the final ZF-regularity model N
>> with M' has M' definable N in the language of =, epsilon.
>>
>> All such GST and ZF - regularity models interepresting MA with
>> MA 's S interepted as adjunction must be non-well-founded.
>
>
>Thank you. This is what I think, too.
That >> is me, two articles back. I have some more to add
about that topic.
I will start by introducing terminolgy to help in stating
things more clearly.
The MA language includes individual constant 0 and unary
function symbol S.
As you point out, there is a common cenvention to interpret PA
into ZF, namely to interpret constant 0 as emptyset, and
PA's successor function symbol S as the set theoretic
operation commonly called successor x J x in your notation.
So you have raised the issue of using the corresponding
interpretation of MA's 0 and S into those definitions in GST
or ZF.
We were also talking about making a combined theory MA+GST,
and in there the corresponding convetnions were raised, how to
interpret the symbols 0 and S in MA+GST that were ingerited
into MA+GST from the MA part.
I will be discussing cases below, of which parts of these
interpretations are to be done of not.
So the nw terminolgy will be when I want to refer to the
constant 0 or the function symbol S from theory MA, I will
write those as 0^MA and S^MA respectively.
Also, when I am writing about theories as MA+GST (we
discussed different variatiosn of these) and I want to refer to
symbols 0 and S in these theories that were inherited from MA, I
will still refer to these as 0^MA and S^MA, according to them
coming from the MA part of theory MA+GST.
If I am discussing interpreting MA into GST, and as doing so I
want to talk about the term for emptyset in GST, I will write
that term as emptyset^GST.
The set theoretic operation called successor in set theory,
is the operation taking any set x to x union {x), or also as
our notation to x J x.
In set theory this operation is usually also called S.
So when I want to write about this operation in GST I will
to it as S^GST. The GST superscript will avoid confuction
between S^MA and S^GST.
If I am writing about the common extension theory MA+GST but I
want to talk about that same theoretic operation, defined in
terms of the epsilon realtion in the MA+GST theory inherited from
the GST part, I will refer to that also as S^GST.
So theory MA+GST will have primitive unaryt function symbol
S^MA, and will have defined unary function symbol S^GST, and
these are distinct symbols in the language.
Similarly when I discuss interepreting MA into ZF or
ZF-regularity, I will write the defined constant for the empty
set as emptyset^ZF or emptyset^ZF-regulkarity, and I will
wrtie the defined unary function for the set theoretic operation
above as S^ZF or S^ZF-regularity.
That concludes the new terminology.
So one result I have been claiming, and was in the proof I gave
above was there is no interpetation of MA into GST based on
interpeting 0^MA as emptyset^GST and interpeting S^MA as
S^GST.
Similarly, if we make a version of MA+GST and include the
axioms which assert 0^MA interprets as empptyset^GST and
S^MA interprets as S^GST, then this theory is inconsistent.
Similarly, there is no interpetation of MA into ZF with
0^MA interpeting as emptyset^ZF and S^MA interpeting as
S^ZF.
And there is no intepretation of MA into ZF-regularity, with
0^MA interpeting as emptyset^ZF-regularity and S^MA
interpeting as S^Zf-regularity.
This along the lines of the proof posted above, posted 5 times
counting also the earlier articles.
I will comment on the further points I had made earlier, in
some cases just to repat in this nw language, in other cases
amending the previous beyond just this change of terminolgy.
I had written in the previous article, as quoted above:
>> Suppose we drop the requirement that MA 's 0 interpet as {}.
>>
>> Then MA + GST + S interprets a J is consistent.
>>
>> But all models of that theory are non-wellfounded, similar to
>> what you said above.
So this would be the same but rephreased as
S^MA interprets as S^GST.
I then wrote:
>> Furthermore, if the starting M was a member of a V=L ZF
>> model, then we can arrange that the final ZF-regularity model N
>> with M' has M' definable N in the language of =, epsilon.
>>
>> All such GST and ZF - regularity models interepresting MA with
>> MA 's S interepted as adjunction must be non-well-founded.
I have realized a better argument since the above. I don't
need to assume the starting MA model M comes from a V=L ZF model.
I don't need anything about V=L now.
I will make a better statement of the result. Namely, there is
a ZF definiton of a proper class N, a definition in free
paramater M, and a defintion of a proper class of ordered pairs
E, this definition in parameter M, so I will write these
definitions as N(M) and E(M) to show the dependency on M,
so that ZF proves the scematum of theorems: for each phi an
axiom of ZF other than regularity,
ZF proves <N(M), E(M)> |= phi, ie with
E(M) inrerpreting the epsilon symbol. And also ZF proves there
is M' in N(M) so <(N(M), E(M)> |= M' is an MA model,
and recovering M' back in our ZF metalanuggae, ie taking
underlying set to be all N(M) members m which are m E(M) M',
and taking the E(M) "members" of the M' interpretations of
0^MA, S^MA, + and * from MA, to make ZF metalanguage
interpetations on thjat structure, in ZF meta lauguage the
original M structure is isomrorphic tot he extracted version of
M' from the <N(M), E(M)> structure. And finanlly, in
<N(M), E(M)>, the MA model M' has S^MA interpeting as
S^ZF-regularity. Also, M' is definable in <N(M), E(M)>
in the lanuage of =, epsilon.
Basically, that we can interpret an arbitrary MA model into
ZF-regularity with S^MA interpeting as S^ZF-regularity. But
complications as the ZF-regularity "model" is only constructed as
a proper class.
Drop the V=L premise from the previous article. And more
carefully state the construction as only producing definable
(in parameter M) proper classes for the unoverse of the model
and for membership.
These constructions are of interest becasue they are similar to
the issue of interpreting theories, in the sense of Tarski.
For example, in this sense PA interprets into ZF.
Since the previous article, I realized another point.
Suppose we are looking for interpretations of MA into various
set theories, where we require S^MA interpret as
S^(the set thoery).
As above, we can't do this into ZF.
But here is a new point I realized after the previous article.
We can't interpret MA into GST that way, and we can't
interpret MA into ZF-regularity that way.
Above I claimed each model of MA could be ismorphoric in an
interpretation in some GST or ZF-regularity model.
Interpreting theory MA into theory GST or ZF-regularity
requires more than that.
It would require a single defintion in set theory model <N, E>
of intepreted model M', a single definition which works in all
<N, E> models.
The above result didn't get that. I only claimed some
<N(M), E(M)> models had an M'. I didn't claim every <N, E>
model has an M'. And among those <N, E> models I calimed to have
an M'.
And it turns out, not merely do my claims above not prove that
MA interprets as a theory into GST or ZF-regularity.
In fact it is actually refutable for intepretations of
S^MA to S^(set theory). ZF proves there is no interpretation into
GST and ZF + con(ZF) proves there is no interpretation into
ZF.
On the other hand if oyou drop the requirement that S^MA
interpret as S^(set theory), then everything interprets.
Not only GST and ZF-regularity just discussed, but also even
ZF.
That is everything I wanted to add about my article.
I will contonue with some further questions of yours:
>I am not really sure where you stand on the
>consistency of MA. Would you please clarify some questions?
>
>Is MA consistent with respect to FOL with equality?
>By this I mean we can't derive a contradiction from
>the axioms of MA using just FOL with equality.
MA is consistent with respect to FOL with equality.
We can't derive a contradiction of MA using just FOL with
equality.
>Can PA prove MA has a 0-,model "with semantics"?
Yes.
>Can ZFC prove MA has a 0-model with semantics?
Yes.
>Can MA+"MA has an infinite model" prove MA has
>a 0-model with semantics?
I don't know what theory MA+"MA has an infinite model"
is. MA can't readily talk about models, so I don't know how to
formalize "MA has an infinite model" in MA. I don't know how
to make this a theory in the language of MA.
Something closely related you may hjave been thinking of.
Add to MA the schematum of axioms, the nth one (n>0) saying
the universe has at least n members (as a sentence in FOL:
exists x1 ... xn [conjunction saying n xi 's are pairwise ~=]).
Normal model theory takes a model to be a set, which don't have
in the langyuuage of MA.
But it is reasonable to consider a related notion of model as a
definable relatization in the working universe. This is like
Godel's L inside ZF. Or like my construction above.
We can get a version like that started in MA or the infinite
universe version of i just above.
But we still can't do Tarski semantics for it. By the well
definedness clauses messing up in the inductive definition, as I
wrote before.
So we do sort of have a 0 model (depending how you want to
define wat that means, not in literal sense of usual model theory
but fairly close. But not with semantics.
This is possible. In the same loose talk sense. you could say
Godel's L as a definable class in ZF is like a model. But we
don't have semantics for it in ZF.
>Russell
>- Integers are an illusion
--
David Libert ah...@FreeNet.Carleton.CA
RussellE
> On Jul 26, 7:54 pm, RussellE <reaste...@gmail.com> wrote:
>
> > > >We can derive Ax (x union {x} != {}) in ZF?
> > You need foundation because we can have {{}}={}
> > without foundation. Then Ax (x union {x} != {}) would
> > not be true for x={}.
>
> No, the absence of the axiom of regularity does NOT allow that "0 =
> {0}" is consistent with ZF\{regularity}.
I think it does in GST.
ZF has all those other axioms.
> Yes, "Ex x = {x}" is consistent with ZF\{regularity}. But that does
> not entail that "0 = {0}" is consistent with ZF\{regularity}.
Why not?
> By definition:
>
> 0 = y <-> Ax ~xey
Definition is not an axiom.
> By definition:
>
> 0e{0}.
>
> So ~0={0}.
Which axiom do you use for the "so"?
There is no axiom that prevents {} = {{}}.
> So 0={0} is NOT consistent with even a lot less than full ZF\
> {regularity}.
>
> You're confused as a result of your ignorance as a result of your
> refusal to pick up a basic textbook.
>
> > Separation says x union {x} defines a set.
>
> No, the union, pairing, and extensionality axioms provide that there
> is a unique set that has as members all and only those y such that y
> in x or y=x. (Pairing is derivable, though, from the schema of
> replacement and power).
GST has extensionality, separation, and adjunction.
No union or paring or regularity.
I don't see how union, paring, or extensionality
prevent two sets from being equal.
Extensionality says two sets with exactly the
say elements are equal. It says nothing about
two sets with different elements.
Neither union nor pairing say anything equality.
> Definition: {x} = {x x}
>
> Definition: xuz = U{x z}
>
> So xu{x} = U{x {x x}}
>
> The axiom schema of separation is not used for this.
I am not sure what you are proving here.
If your argument applies to any set then
how can ZF\foundation be consistent?
Frederick Williams
>
> [...]
>
> This short simple little proof at the baby level of set theory
> has now been posted 5 times in 5 separate articles by 3
> different authors.
I, too, have pointed out that MA + a weak theory of sets + "S(x) = x J
x" is inconsistent.
> [...] MA can't readily talk about models, so I don't know how to
> formalize "MA has an infinite model" in MA.
I have made this point repeatedly (talking perhaps of "structures" or
"models" rather than "infinite models", but so what?) but Russell
refuses to accept the point.
MoeBlee
> On Jul 27, 8:13 am, MoeBlee <modem...@gmail.com> wrote:
>
> > On Jul 26, 7:54 pm, RussellE <reaste...@gmail.com> wrote:
>
> > > > >We can derive Ax (x union {x} != {}) in ZF?
> > > You need foundation because we can have {{}}={}
> > > without foundation. Then Ax (x union {x} != {}) would
> > > not be true for x={}.
That is INCORRECT.
> > No, the absence of the axiom of regularity does NOT allow that "0 =
> > {0}" is consistent with ZF\{regularity}.
>
> I think it does in GST.
WHAT does in GST?
Are you saying that "0 = {0}" is consistent with GST?
That makes no sense, since "{ _ }" is not even defined in GST.
GST doesn't have { _ }. Rather, GST has the 2-place 'J' (or whatever
symbol you want to use) defined by
Jxy = b <-> Az(zeb <-> (zex or z=y)).
'Jxy' is notated as 'xu{y}'.
In GST, neither 'u' nor '{ _ }' are defined standalone.
And your assertion was as to ZF anyway. And your assertion is
INCORRECT. Whether for ZF or for GST. It's incorrect as a function of
your abysmal ignorance which is a function of your abysmal arrogance
that you can understand set theory and logic without reading even page
one of a textbook.
> ZF has all those other axioms.
>
> > Yes, "Ex x = {x}" is consistent with ZF\{regularity}. But that does
> > not entail that "0 = {0}" is consistent with ZF\{regularity}.
>
> Why not?
READ the rest of the post:
> > By definition:
>
> > 0 = y <-> Ax ~xey
>
> Definition is not an axiom.
Whether the definition is called and axiom or not, in this context '0'
is the empty set.
And, by the way, in certain ordinary treatments, definitions ARE a
certain special kind of axiom called 'definitional axioms'.
> > By definition:
>
> > 0e{0}.
>
> > So ~0={0}.
>
> Which axiom do you use for the "so"?
Those of first order logic with identity.
> There is no axiom that prevents {} = {{}}.
If you're not using the notation '{}' to indicate the unique set that
has no members, then it's a different discussion. But as long as we
take '{}' to stand for the unqiue set that has no members, then ~ {}
={{}} and we don't need regularity to prove that.
And the question was as to ZF, and in ordinary notation regarding ZF,
we have {} = 0 = the unique set having no members.
> > So 0={0} is NOT consistent with even a lot less than full ZF\
> > {regularity}.
>
> > You're confused as a result of your ignorance as a result of your
> > refusal to pick up a basic textbook.
>
> > > Separation says x union {x} defines a set.
>
> > No, the union, pairing, and extensionality axioms provide that there
> > is a unique set that has as members all and only those y such that y
> > in x or y=x. (Pairing is derivable, though, from the schema of
> > replacement and power).
>
> GST has extensionality, separation, and adjunction.
> No union or paring or regularity.
(1) So what? The question was as to ZF.
(2) If we're in GST, as you formulated, we have, quoting you:
"Adjunction says if x and y are sets then
x union {y} is a set. I will use the symbol
"J" to mean adjunction."
That is to say:
EbAz(zeb <-> (zex or z=y))
then define:
Jxy = b <-> Az(zeb <-> (zex or z=y))
In that case, separation is still not involved.
You don't know how how formal languages and theories work.
> I don't see how union, paring, or extensionality
> prevent two sets from being equal.
They DON'T.
IDENTITY (first order logic with IDENTITY) provides: for any property
P that can be formulated in the language, if P(x) and ~P(y) then ~x=y.
So let P be the property of having a member (Ez zex).
Ez ze{0}
and
~Ez ze0
So ~ {0}=0.
What could be more clear?
But if you're in GST, then such things as {0} alone are not even
defined.
Rather there is the 2-place function symbol 'J'.
> Extensionality says two sets with exactly the
> say elements are equal. It says nothing about
> two sets with different elements.
> Neither union nor pairing say anything equality.
So what? First order logic with IDENTITY says that if x and y differ
in some property then x is not y.
If you're not using first order logic with identity, then you better
say what axioms you ARE using for the 2-place predicate symbol '='.
But the question was as to ZF, and in ZF (and in GST, for that
matter), we have use of all of first order logic with identity.
> > Definition: {x} = {x x}
>
> > Definition: xuz = U{x z}
>
> > So xu{x} = U{x {x x}}
>
> > The axiom schema of separation is not used for this.
>
> I am not sure what you are proving here.
I'm showing just what I said: In ZF, we don't need the axiom schema of
separation for xu{x}.
Meanwhile, also in GST, the axiom schema of separation is not needed
to have
xu{x}
which is just a way of notating
Jxx
> If your argument applies to any set then
> how can ZF\foundation be consistent?
I don't know why you think anything I've said indicates that ZF\
{regularity} is inconsistent.
Damn, just get an introductory textbook already!
MoeBlee
RussellE
> RussellE (reaste...@gmail.com) writes:
> > On Jul 26, 1:56=A0am, ah...@FreeNet.Carleton.CA (David Libert) wrote:
>
> >You need foundation because we can have {{}}={}
> >without foundation. Then Ax (x union {x} != {}) would
> >not be true for x={}.
>
> You are saying in ZF - Foundation we can have {{}} = {}.
>
> I say ZF - Foundation proves {{}} ~= {}.
>
> Jessie and MoeBlee already pointed this out in this thread.
>
> {{}} is defined to be the set with only member {}.
OK
> So {} is a member of {{}}. Not using foundation, just using
> the definition of {{}}.
{} is a member of {{}}.
In ZF-Foundation we can have a set that is a member of itself.
In a model of MA we can have S(0)=0.
Using the standard interpretation of
number as ordinal, we get {{}} = {}.
{{}} contains itself as a member.
How are you proving {{}} ~= {}?
> And if you want to know what axioms support that there is a set
> with only member {}, it is separation to get {} and then
> pairing. MoeBlee already pointed this out.
{} must be well founded. {{}} may not be.
> So {} and {{}} being proper terms, that the underlying
> definitions really are definitions (existence and unqiueness)
ZF has an axiom of uniqueness?
Union and paring only say certain sets exist.
> doesn't need foundation, and {} member of {{}} doesn't need
> foundation.
>
> As noted {} is a proper term wotithout foundation. And {} is
> defined to be the set with no members.
{{}} is defined to be the ordinal with exactly one member.
If you are claiming every set is unique because of it members
then we can't have sets that contain themselves as members.
It sounds like you are claiming you can derive
foundation from union and paring.
> So ZF - Foundation proves
> the set defined to be the set with no members has no members.
>
> So {} is a member of {{}} and {} is not a member of {}.
>
> That proof not assuming foundation.
>
> So {{}} and {} are not =, since they have different members.
> This by the axioms of = for FOL.
Foundation says a set can't be a member of itself.
Separation, union, and paring say certain sets exist
and say nothing about equality.
The only axioms that say anything about equality
are the axioms of equality and extensionality.
Extensionality says two sets are equal if they
have exactly the same elements.
It says nothing about sets with different elements.
We haven't specified which axioms of equality
we are talking about.
http://www.quickmind.net/mathpathways/algebra1/content/algebra1/topics/axiomequ.htm
Let's work in the 0-model of MA+GST with 0 == {} and S(x) == x J x.
(== means defined as.)
U={0}, S(0)=0, 0+0=0, 0*0=0
In this model every ordinal equals {}.
Types of equality are:
Reflexive axiom a = a
Symmetric axiom If a = b, then b = a.
Transitive axiom If a = b and b = c, then a = c.
Additive axiom If a = b and c = d, then a + c = b + d.
Multiplicative axiom If a = b and c = d, then ac = bd.
If MA+GST is inconsistent we can derive P&~P.
I don't see how you can prove the negation of any of these axioms.
Frederick Williams
> ZF has an axiom of uniqueness?
Extensionality is used to prove uniqueness.
> We haven't specified which axioms of equality
> we are talking about.
Surely "ZF" without qualification, means the first order theory with
equality of ZF. So you get all the logical truths "for free". There is
no more need to specify the axioms of equality than there is to specify
the axioms of propositional calculus.
RussellE
>
> > > >We can derive Ax (x union {x} != {}) in ZF?
> > You need foundation because we can have {{}}={}
> > without foundation. Then Ax (x union {x} != {}) would
> > not be true for x={}.
>
> No, the absence of the axiom of regularity does NOT allow that "0 =
> {0}" is consistent with ZF\{regularity}.
>
> Yes, "Ex x = {x}" is consistent with ZF\{regularity}. But that does
> not entail that "0 = {0}" is consistent with ZF\{regularity}.
This is interesting.
I want to consider a weaker theory than
ZF or GST. Replace the axiom schema of
specification in GST with:
Ax (x={} OR Ey (yJy=x))
where J is adjunction.
I think every non-empty set in this weak
version of GST has {} as an element.
Ex (xJx = {}) -> ~Foundation
Ax (xJx ~= {}) OR ~Foundation
Foundation -> Ax (xJx ~= {})
We know Ax(xJx ~= {}) is a theorem of ZF.
ZF has an axiom of Foundation.
The other axioms of PA are also theorems of ZF.
We also know Ax(S(x) ~= 0) is independent
of the other axioms of PA.
If we can prove Ax(xJx ~= {}) in ZF-Foundation,
how can Ax(S(x) ~= 0) be independent of the other
axioms of PA? What other axioms of ZF do we need
to remove to prove the independence of
Ax (S(x) ~= 0)?
Frederick Williams
>
> [...]
>
> If we can prove Ax(xJx ~= {}) in ZF-Foundation,
> how can Ax(S(x) ~= 0) be independent of the other
> axioms of PA? What other axioms of ZF do we need
> to remove to prove the independence of
> Ax (S(x) ~= 0)?
What is the relation between {} and 0 in your theory?
David Libert
I had 3 recent posts to this thread:
[1] David Libert "Re: Modular Arithmetic Can't Be Well Founded"
sci.logic, sci.math July 25, 2011
http://groups.google.com/group/sci.logic/msg/0205ccd93f816444
[2] David Libert "Re: Modular Arithmetic Can't Be Well Founded"
sci.logic, sci.math July 26, 2011
http://groups.google.com/group/sci.logic/msg/35747418d6ed4099
[3] David Libert "Re: Modular Arithmetic Can't Be Well Founded"
sci.logic, sci.math July 27, 2011
http://groups.google.com/group/sci.logic/msg/0d140a1ff9a082fd
This present article is a followup to [3].
[1]-[3] were about various versions of relations of MA to set
theories, both in terms of theories and interpretations and in
terms of models.
[1] was about various forms of the combined theory MA+GST,
regarding consistency and existence of models with certain
properties.
[2]-[3] were about relating MA to ZF, both in terms of
realizing a given MA model as isomorphic to an MA model sitting
as a set in a ZF model, and in terms of interpreting Theory MA
into theory ZF.
I have a correction I realized since [3] to add to [2]-[3].
Also I have further comments to add about all topics [1]-[3]
beyond what I wrote before.
In [2]-[3] I claimed given M an MA model, there could be in a
suitable sense (possibly only as a definable class in the
universe) a ZF model N with an isomorphic copy M' of M in it.
Furthermore I claimed we could even do the above for M'
interpreting S^MA as N 's S^ZF-regularity, if we drop from
N satisfying ZF to ZF-regularity and allow non-wellfrounded N.
I still claim the original just mentioned for getting M into
N satisfying ZF, where we don't require S^MA = S^ZF.
For the second result, requiring S^MA = S^ZF-regularity, I
only claim this now for M either a 1 element model, or M an
infinite model.
And for M finite but size > 1, I claim instead the negation of
that claim. It is impossible to have a finite loop of S^ZF of
size > 1. This provable. So it is provable there are no such ZF
models with such M' for M finiste of size > 1.
That completes the correction of [2]-[3].
Regarding the new comments.
[1], writing about MA+GST considered an issue, whether the
induction axioms of the MA part of MA+GST allow quantification
into the GST part in the induction formula, and whether they
allow the epsilon relation.
In [2]-[3] I never considered the corresponding issue for
relationg MA to ZF, given MA model M, if we seek ZF or
Zf-regularity model N containing M' isomorphic to M, what about
the issue of whether that M' satisfies induction axioms for the N
universe's set language (quantification over entire N universe,
and mention epsilon in the induction).
Regarding that, copying MA model M to M' in ZF model N, with no
requirement S^MA in M' is N 's S^ZF. For any finite M (size 1
or size > 1 and finite), this is still possible. In fact any M'
copy being finite will satisfy induction.
I don't know for infinite M.
On the other hand, if we also require M' has S^MA =
S^Zf-regularity into N satisfying ZF- regularity.
For M size 1, this is still possible.
For all M of size > 1, it is provably impossible.
Above I already ruled out finite M of size > 1 copying into
such M' in N. So those cases are still ruled out.
But above I claimed infinite M could copy to M' in N modelling
ZF-regularity.
But those examples will all have M' not satisfying full N
induction.
And it is provably impossible. If we require
S^MA = S^ZF-regularity, it is impossible to have an infinite
such MA model M' in N satisfying N set language induction.
--
David Libert ah...@FreeNet.Carleton.CA
MoeBlee
> On Jul 27, 8:13 am, MoeBlee <modem...@gmail.com> wrote:
>
> > On Jul 26, 7:54 pm, RussellE <reaste...@gmail.com> wrote:
>
> > > > >We can derive Ax (x union {x} != {}) in ZF?
> > > You need foundation because we can have {{}}={}
> > > without foundation. Then Ax (x union {x} != {}) would
> > > not be true for x={}.
>
> > No, the absence of the axiom of regularity does NOT allow that "0 =
> > {0}" is consistent with ZF\{regularity}.
>
> > Yes, "Ex x = {x}" is consistent with ZF\{regularity}. But that does
> > not entail that "0 = {0}" is consistent with ZF\{regularity}.
>
> This is interesting.
And it's correct. You understand it now?
> I want to consider a weaker theory than
> ZF or GST. Replace the axiom schema of
> specification in GST with:
>
> Ax (x={} OR Ey (yJy=x))
Wrong already. Without the axiom schema of separation, you have not
derived an existence/uniqueness theorem by which to define '{}'.
On the other hand, if '{}' is primitive, then you need to say so.
Also, just as a matter of style, I'd write 'Jyy' in ordinary style for
a 2-place function symbol, instead of 'yJy', so in general 'Jxy'
instead of 'xJy'.
MoeBlee
RussellE
wrote:
> RussellE wrote:
>
> > [...]
>
> > If we can prove Ax(xJx ~= {}) in ZF-Foundation,
> > how can Ax(S(x) ~= 0) be independent of the other
> > axioms of PA? What other axioms of ZF do we need
> > to remove to prove the independence of
> > Ax (S(x) ~= 0)?
>
> What is the relation between {} and 0 in your theory?
Most people interpret PA in ZF by representing 0 with {}.
What does it mean to interpret a theory?
Consider the theory of Peano Arithmetic (PA).
I can interpret PA as apples.
I can say no apples interprets as 0 and
one apple interprets as S(0), etc.
I can interpret successor as adding
another apple.
What does it mean to interpret PA in ZF?
A standard method is to interpret empty
set as 0 and x union {x} as successor.
I define the theory of Modular Arithmetic
by taking the axioms of PA and replacing
Ax (S(x) ~= 0) with Ex (S(x) = 0).
Again, I can interpret MA as apples.
Again, I can define successor as adding
another apple. The difference in MA is
that at some point I will add an apple
and get so hungry I eat all of them.
Call this the apple interpretation of MA.
We can easily prove MA has a model using
the apple interpretation. Assume the
universe has no apples.
Successor(no apples) = no apples
no apples + no apples = no apples
no apples * no apples = no apples
What happens when we try to interpret MA in
ZF using the same interpretation we use for PA?
We immediately get a contradiction.
We are forced to have some non-empty set
equal to the empty set.
It is not surprising interpreting MA in
ZF leads to contradiction. The axioms
of PA are theorems of ZF and the axioms
of MA and PA are mutually exclusive.
Which axioms of ZF do we have to get
rid of before we can interpret MA?
Certainly Foundation must go, but,
this is not enough.
The PA axiom Ax(s(x) ~= 0) becomes
Ax (x union {x} ~= {}) in ZF using
the standard interpretation.
We can prove Ax (x union {x} ~= {})
using just GST. The only axioms of GST
are Extensionality, Separation, and
Adjunction.
http://en.wikipedia.org/wiki/General_set_theory
One of these axioms has to go if we want
to interpret MA. As well as a lot of others.
I can easily transform the apple
interpretation of MA into an orange
interpretation. Why can't I interpret
MA using sets in set theory?
Several people have pointed out we
can use different interpretations
of sets as numbers. In ZF we could
represent 0 with omega and define
some set operation on omega as
successor. No one has actually
given such an interpretation for MA.
And it is not obvious why we need
another interpretation.
Why would it matter if we use oranges
instead of apples?
Supposedly, Ax(S(x) ~= 0) is independent
of the other axioms of PA. Obviously,
it is not independent of many of the
axioms of set theory.
If we can't consistently interpret MA in
set theory then we have to ask where
the inconsistency comes from. We
know the axioms of MA are consistent.
Russell
- Zeno was right. Motion is impossible.
Frederick Williams
>
> On Jul 31, 3:26 am, Frederick Williams <freddywilli...@btinternet.com>
> wrote:
> >
> > What is the relation between {} and 0 in your theory?
>
> Most people interpret PA in ZF by representing 0 with {}.
Your MA has 0 and GST has {}. You also have that S(x) = x union {x}.
Are all these things mutually consistent?
David Libert
I had 4 recent posts to this thread:
[1] David Libert "Re: Modular Arithmetic Can't Be Well Founded"
sci.logic, sci.math July 25, 2011
http://groups.google.com/group/sci.logic/msg/0205ccd93f816444
[2] David Libert "Re: Modular Arithmetic Can't Be Well Founded"
sci.logic, sci.math July 26, 2011
http://groups.google.com/group/sci.logic/msg/35747418d6ed4099
[3] David Libert "Re: Modular Arithmetic Can't Be Well Founded"
sci.logic, sci.math July 27, 201
http://groups.google.com/group/sci.logic/msg/0d140a1ff9a082fd
[4] David Libert "Re: Modular Arithmetic Can't Be Well Founded"
sci.logic, sci.math Aug 1, 2011
http://groups.google.com/group/sci.logic/msg/579fb053271acb52
[4] summarised the main topics of [1]-[3].
This present article is a followup to [4].
In [4] I wrote
> [1], writing about MA+GST considered an issue, whether the
>induction axioms of the MA part of MA+GST allow quantification
>into the GST part in the induction formula, and whether they
>allow the epsilon relation.
>
> In [2]-[3] I never considered the corresponding issue for
>relationg MA to ZF, given MA model M, if we seek ZF or
>Zf-regularity model N containing M' isomorphic to M, what about
>the issue of whether that M' satisfies induction axioms for the N
>universe's set language (quantification over entire N universe,
>and mention epsilon in the induction).
>
> Regarding that, copying MA model M to M' in ZF model N, with no
>requirement S^MA in M' is N 's S^ZF. For any finite M (size 1
>or size > 1 and finite), this is still possible. In fact any M'
>copy being finite will satisfy induction.
>
> I don't know for infinite M.
Since [4] I have found the following about that last question.
The issue is, given M an infinite model of MA, what can be said
about the existence of N modelling ZF with N containing a
MA model M' isomorphic to M, and satisfying full set theoretic N
induction.
First: ZF proves there exist M an infinite model of MA s.t.
for all N model of ZF N does not have as memeber any inductive
MA model M' isomorphic to M.
So for some M the answer is no.
On the other hand, for some M the answer can be yes, in the
following sense:
ZF + Con(ZF) proves exist M infinite model of MA and
N model of ZF such that N contains as member a inductive
MA model M' isomorphic to MA.
Another interesting question related this is whether such M'
can be definable in N in the language of set theory, ie a
definition in symbols =, epsilon in FOL in N's universe.
This relates to interpretations between theories, which is turn
can give us realtive consistency results. Often when we
conisider interpretations like this, we want to work with them
explicitly and it is interesting to know beyond the existential
claims above if we can have a definition for M'.
As above, some M have no N and M', so for these there is
obviously no definable M'.
So let us conisider those infinite MA models M for which
there exist ZF model N constaining as memeber inductive
MA model M' isomorphic to M.
Work in ZF + Con(ZF) as metatheory to discuss these cases.
All the MA models below mentioned inside ZF models will be
understood to be inductive.
Then for every infinite MA model M such that there exist
N modelling ZF and M' in N with M' isomorphic to M,
there exists N1 modeling ZF with M1' a member of N1 and
M1' isomorphic to M and M1' definable in N1 in the
language of set theory.
So by possibly changing N,M' to N1,M1' you can always have
the M1' definable in N1'.
Again, in metatheory ZF + Con(ZF), it is provable that the
following case each arise, and these are all cases.
There are infinite MA nmodels M2 such that
for every ZF model N2 with M2' memeber of N2
and M2' ismorphic to M2,
then there is M3' contained in N2 s.t. M3' is
isomorphic to M and M3' is definable in N2.
Ie, given M2 and N2, M2' without changing N2 but changing
M2' tp a possibly different M3', you can have M3' definable in
N2.
Again rephrasing: no matter what model N2 you pick for M, your
choice of N2 doesn't stop you from getting rthe final M3' to be
definable in set theory.
That is the first of the 2 cases.
The other case that provably in ZF + Con(ZF) can arise is:
There is M4 an infinite MA model s.t. there exist N4 model
of ZF and M4' contained in N4 isomorphic to M4, yet for every
M5' in N4 isomorphic to M4, M5' is not definable in N4 in the
language of set theory.
These last examples M4, by the earlier result, must have other ZF
models N6 and M6' is isomorphic to M4 and M6' is definable in
N6.
So for these M4 cases, both cases can arise: N4 which can
make isomorphic copies but none definable, and N6 which can
still make an isomorphic copy.
So the previous case, M2, you can pick N2 anything that could
have worked (ie that has any ismorphic copies) and staying
inside N2 you can always find a definable copy.
For the last case of M4, you must pick ZF models carefully as
N6, to get to definability, avoiding N4 models of ZF which are
doomed to find defibale M% copies.
To summarize all of the above:
M into N copy ?
Sometimes yes, sometimes no (depending on which M).
Among the yes M above: definable copies?
Depending on M: sometimes every N model making copies also
includes a definable copy.
For othjer M: sometimes the question to go from a copy to a
definable copy depends on which N.
--
David Libert ah...@FreeNet.Carleton.CA
David Libert
I had 5 recent posts to this thread:
[1] David Libert "Re: Modular Arithmetic Can't Be Well Founded"
sci.logic, sci.math July 25, 2011
http://groups.google.com/group/sci.logic/msg/0205ccd93f816444
[2] David Libert "Re: Modular Arithmetic Can't Be Well Founded"
sci.logic, sci.math July 26, 2011
http://groups.google.com/group/sci.logic/msg/35747418d6ed4099
[3] David Libert "Re: Modular Arithmetic Can't Be Well Founded"
sci.logic, sci.math July 27, 201
http://groups.google.com/group/sci.logic/msg/0d140a1ff9a082fd
[4] David Libert "Re: Modular Arithmetic Can't Be Well Founded"
sci.logic, sci.math Aug 1, 2011
http://groups.google.com/group/sci.logic/msg/579fb053271acb52
[5] David Libert "Re: Modular Arithmetic Can't Be Well Founded"
sci.logic, sci.math Aug 3, 2011
http://www.mathforum.org/kb/message.jspa?messageID=7511752&tstart=0
Google has been slow lately, and it didn't yet get [5].
[1]-[5] were followups to each other in that order.
This present article is going back to a topic from [3]-[4].
(I listed all [1]-[5] as I thought it would be less confusing to
have a complete list.)
This article is mainly a response to [3], and I am posting this
as a followup to [3].
[3] made a claim I will be discussiong below. [4] already made
a correction to that claim from [3].
Now in this article I will note a second distinct correction
from that of [4], to this claim from [3].
The claim from [3]:
> I will make a better statement of the result. Namely, there is
>a ZF definiton of a proper class N, a definition in free
>paramater M, and a defintion of a proper class of ordered pairs
>E, this definition in parameter M, so I will write these
>definitions as N(M) and E(M) to show the dependency on M,
>so that ZF proves the scematum of theorems: for each phi an
>axiom of ZF other than regularity,
>ZF proves <N(M), E(M)> |= phi, ie with
>E(M) inrerpreting the epsilon symbol. And also ZF proves there
>is M' in N(M) so <(N(M), E(M)> |= M' is an MA model,
>and recovering M' back in our ZF metalanuggae, ie taking
>underlying set to be all N(M) members m which are m E(M) M',
>and taking the E(M) "members" of the M' interpretations of
>0^MA, S^MA, + and * from MA, to make ZF metalanguage
>interpetations on thjat structure, in ZF meta lauguage the
>original M structure is isomrorphic tot he extracted version of
>M' from the <N(M), E(M)> structure. And finanlly, in
><N(M), E(M)>, the MA model M' has S^MA interpeting as
>S^ZF-regularity. Also, M' is definable in <N(M), E(M)>
>in the lanuage of =, epsilon.
[4] made the following correction to that claim just quoted
from [3]:
> For the second result, requiring S^MA = S^ZF-regularity, I
>only claim this now for M either a 1 element model, or M an
>infinite model.
>
> And for M finite but size > 1, I claim instead the negation of
>that claim. It is impossible to have a finite loop of S^ZF of
>size > 1. This provable. So it is provable there are no such ZF
>models with such M' for M finiste of size > 1.
That correction from [4] still stands.
My new correction below to the claim above from [3] is about
another issue than [4]'s correction.
For M a one element model, the original [3] claim above still
stands and is not being corrected here.
The correction concerns the [3] claim above as applied to case
of infinite M.
Everything of the [3] claim above for this case remains except
the closing sentence:
> Also, M' is definable in <N(M), E(M)>
>in the lanuage of =, epsilon.
My argument for that was mistaken.
And in fact by a new argument, I now think the opposite is
provable. For every such M' as above a memeber of the
<N(M), E(M)> model, such M' are NOT definable in
<N(M), E(M)> in the language of =, epsilon. This in case as I
said, M is infinite.
That is for the ZF case, as the claim above specified.
A similar proof shows M' is not definable for that claim above
with FS replacing ZF, FS the fibnitary set theory from the recent
thread. And the rest of the claim above still stands for FS
replacing ZF. So the original claim above except the closing
sentecne about M' being definable, and replace that by its
negation as just noted. All that also provable for the FS case
replacing ZF.
For case GST, I think the original [3] claim above can stand.
All the other clauses, and we can even arrange M' be definable in
<N(M), E(M)> modelling GST (instead of ZF). So for the GST
case the original [3] claim, not the negated version as from here
for FS and ZF.
The previous articles [3]-[4] only mentioned the ZF case.
That is new to this article to mention GST and FS about the claim
above from [3].
--
David Libert ah...@FreeNet.Carleton.CA
Frederick Williams
>
> RussellE wrote:
> >
> > On Jul 31, 3:26 am, Frederick Williams <freddywilli...@btinternet.com>
> > wrote:
> > >
> > > What is the relation between {} and 0 in your theory?
> >
> > Most people interpret PA in ZF by representing 0 with {}.
>
> Your MA has 0 and GST has {}. You also have that S(x) = x union {x}.
> Are all these things mutually consistent?
And answer came there none.