When FOL Fails
RussellE
First Order Logic. FOL can produce
inconsistent results when certain
assumptions are made.
I am going to start with the axioms
of Peano Arithmetic:
Ax (S(x) =/= "0")
AxAy (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)
A model is a set of definitions and functions.
I will describe a model of the axioms
above. I am going to cheat, of course.
PA requires the definition of a constant, "0".
"0" is natural number in the domain of universal
quantification.
PA is the theory of natural numbers and there
are some things that we can't put in the domain
of universal quantification. For example, the set of
all natural numbers isn't a member of the domain
for universal quantification.
I am going to add a new constant, "1", to PA.
I will add it as an axiom.
Limit Axiom
Ax (x < "1")
"1" is not in the domain for universal
quantification, but it can be defined
in functions.
A model of PA has a set of natural numbers
and functions for successor, addition, and
multiplication.
This is my model of PA+Limit
"0" = 0
"1" = 1
N: {0} = domain of universal quantification
S: S(0)=1, S(1)=0
+: 0+0=0, 0+1=1, 1+0=1, 1+1=0
*: 0*0=0, 0*1=0, 1*0=0, 1*1=1
It is simple to prove this is a model of PA+Limit.
Consider the axiom Ax (S(x) =/= "0").
Since x can only range over {0}, there
is only one case: S(0) =/= 0, which is true.
Similarly, 0 * S(0) = (0*0) + 0,
0+S(0) = S(0+0), etc.
Since this is a model of PA+Limit,
it is also a model of PA. Except its not.
Why does FOL tells us these axioms
have a model?
As soon as I remove the Limit axiom,
the model has undefined functions.
Allowing objects not in the domain
of universal quantification fools
FOL into "proving" there is model.
Russell
- Zeno was right. Motion is impossible.
Rupert
You're not allowed to have a constant symbol referring to an object
not in the domain of universal quantification.
RussellE
How else would I define a limit?
Aren't real numbers definable in PA?
How can I define a real number without a limit?
Russell
- 2 many 2 count
hagman
First, you can defein "<" as
x < y :<=> Ez ( x + Sz = y)
OK so far
In order to define a constant "1" with property
Ax (x < "1"),
you first need to prove a theorem
E!e ( Ax (x < e) )
hagman
hagman
According to thread topic, however, you really want to ask:
How can I define a real number with only FOL from PA?
I.e. by only talking about natural numbers, not dets/sequences/... of
natural numbers?
RussellE
OK. Real numbers are a bad example for PA+FOL.
What if I want to define induction on the natural numbers?
Isn't all natural numbers part of the definition of induction?
Is "all natural numbers" part of the domain of universal
quantification?
How would I define induction as a function?
Russell
- logic must be idiot proof
Tim Little
> How would I define induction as a function?
A function from what to what?
--
Tim
Gus Gassmann
> What if I want to define induction on the natural numbers?
> Isn't all natural numbers part of the definition of induction?
Don't know if this is already a conceptual error on your part.
However, this sentence does not make sense. Once you fix the grammar,
one can debate the corresponding statement "All natural numbers is
part of the definition of induction".
MoeBlee
>
> What if I want to define induction on the natural numbers?
What do you MEAN "define induction"?
Wait, nevermind, you're in virtually complete confusion about the most
basic aspects of this material.
Just get a damned book already!
MoeBlee
MoeBlee
> Few people question the consistency of
> First Order Logic.
The consistency of first order logic has a basic proof.
> FOL can produce
> inconsistent results when certain
> assumptions are made.
Of course one can ADD axioms or assumptions to get an inconsistency.
That doesn't make first order logic itself inconsistent.
> I am going to start with the axioms
> of Peano Arithmetic:
>
> Ax (S(x) =/= "0")
I don't know why you put quote marks around '0'.
> AxAy (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)
>
> A model is a set of definitions and functions.
No, it's not.
> I will describe a model of the axioms
> above. I am going to cheat, of course.
>
> PA requires the definition of a constant, "0".
No, it doesn't.
> "0" is natural number in the domain of universal
> quantification.
For SOME models (including the standard model) the SYMBOL '0' is
mapped to the natural number zero.
> PA is the theory of natural numbers
PA is ONE theory among others whose intended model involves natural
numbers.
> and there
> are some things that we can't put in the domain
> of universal quantification.
No, that's completely wrong.
For ANY infinite set, there exists a model of PA with said set as the
universe.
> For example, the set of
> all natural numbers isn't a member of the domain
> for universal quantification.
The set of natural numbers could very well be a member of the universe
of a model of PA. Of course, the set of natural numbers is not a
member of the STANDARD model for the language of PA.
> I am going to add a new constant, "1", to PA.
The quote marks there make sense, but not below:
> I will add it as an axiom.
>
> Limit Axiom
> Ax (x < "1")
There is no purpose in using quotes there.
> "1" is not in the domain for universal
> quantification,
You're totally confused.
'1' is a SYMBOL of the language. For a given model, the SYMBOL '1'
maps to some object in the universe for the model. And it is not
precluded that a symbol map to ITSELF for some model.
> but it can be defined
> in functions.
Whatever you mean by that.
> A model of PA has a set of natural numbers
CERTAIN models of PA (including the standard model) have the set of
natural numbers as the universe.
> and functions for successor, addition, and
> multiplication.
Correct, in the sense that the symbols 'S', '+', and '*' are mapped,
per the model, to, respectively, a 1-place function, a 2-place
function, and a 2-place function.
> This is my model of PA+Limit
>
> "0" = 0
Don't use '='. '0' might not EQUAL zero.
Also, there is a tricky informality that we use that should be made
more rigorous if we are at this level of exactness. That is, '0' is a
symbol in the language. So, to be precise, we should use something
other than '0' for the natural number zero. So, use 'zero'.
So, for certain models we have:
'0' maps to (the natural number) zero.
> "1" = 1
'1' maps to (the natural number) one.
> N: {0} = domain of universal quantification
That is impossible unless zero = one.
Already, what you're giving is NOT a model.
I'll stop here.
You are HORRIBLY ignorant and confused about this material. Why do you
even waste time typing nonsense about it before you've even taken
minimal time to understand it?
It's like if I knew nothing about chemistry but still posted a bunch
of nonsense using the symbols and abbreviations of chemistry.
Really, it's idiotic what you're posting.
MoeBlee
MoeBlee
Your are terribly ignorant and confused about the basics of this
subject. And due to that fact, your arguments, questions, and claims
are nonsense. You don't even know what a model IS. You don't know how
to state a model. What you state is nonsense.
You really need to get a book on the subject and read it.
I really don't understand why you think you can accomplish anything,
make any correct point, or even ask any coherent question when you
haven't learned the most basic things about the subject.
MoeBlee
Frederick Williams
> >
> > Ax (S(x) =/= "0")
>
> I don't know why you put quote marks around '0'.
I has supposed it was to distinguish between numeral and number.
Similar with 1 elsewhere. Some authors might use emboldening or
underlining to make the distinction, but Russell, unable to do either of
those, settled for quote marks. Maybe.
Despite which, I agree with your:
> You are HORRIBLY ignorant and confused about this material. Why do you
> even waste time typing nonsense about it before you've even taken
> minimal time to understand it?
--
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
Frederick Williams
The first order form in which phi (say) is not quantified is a function
from the set of formulae (or perhaps formulae with one free variable) to
the set of truth values. If we're in luck it's the constant function
true.
The second order form in which phi is universally quantified is... um...
what? A function {true} -> {true, false}? Frege would know.
Btw, in notation such as f:A -> B is there a name for the A -> B bit? I
might call it the type or signature of f, but I don't know if there is a
standard term.
MoeBlee
<freddywilli...@btinternet.com> wrote:
> MoeBlee wrote:
>
> > On May 26, 10:53 pm, RussellE <reaste...@gmail.com> wrote:
>
> > > Ax (S(x) =/= "0")
>
> > I don't know why you put quote marks around '0'.
>
> I has supposed it was to distinguish between numeral and number.
> Similar with 1 elsewhere. Some authors might use emboldening or
> underlining to make the distinction,
Sure, but Russell uses it in the WRONG place.
The axiom is:
Ax ~Sx=0
That is an axiom in the language of PA.
The axiom is not:
Ax ~Sx='0'
That is not even an expression in the language of PA.
MoeBlee
RussellE
wrote:
> MoeBlee wrote:
> > On May 26, 10:53 pm, RussellE <reaste...@gmail.com> wrote:
> > > Ax (S(x) =/= "0")
> > I don't know why you put quote marks around '0'.
> I has supposed it was to distinguish between numeral and number.
> Similar with 1 elsewhere. Some authors might use emboldening or
> underlining to make the distinction, but Russell, unable to do either of
> those, settled for quote marks. Maybe.
I use "0" to mean the symbol that is part of
the theory. I should write:
Ax (S(x) =/= A("0")) where A(x) means the
natural number assigned to the symbol "0".
Ax (S(x) =/= 0) uses a natural number in
an axiom used to define natural numbers.
Just looks circular to me.
> Despite which, I agree with your:
>
> > You are HORRIBLY ignorant and confused about this material. Why do you
> > even waste time typing nonsense about it before you've even taken
> > minimal time to understand it?
Thanks for the encouragement.
Of course, this is no worse than things being
said about Voevodsky on fom archive.
And he is a Field's Medalist.
I make no secret of the fact I am trying
to prove PA is inconsistent. This does
not mean I am ignorant.
I admit I am an amatuer.
It would be hard to make a carrer of
proving what you do is inconsistent.
I agree with Voevodsky that the
consistency of PA is an important question.
We know from Godel we can't prove PA is consistent.
All we are left with is proving PA is inconsistent.
Yes, there are many "consistency" proofs.
All these proofs require PA or induction of some sort.
We can't assume the set of natural numbers
exists and then prove it exists.
We can't prove PA is consistent and we know it.
Why isn't there a journal dedicated to
proving PA is inconsistent?
This thread is about the limits of FOL.
I start with the axioms of Peano Arithmetic.
I added one axiom, the Limit axiom.
Ax ( x < 1 )
I assume "1" is a symbol for a
mathematical object not in the
domain of universal quantification.
I assume "1" is not a natural number.
This is not an unreasonable
assumption. Lots of mathematical
objects are not natural numbers.
Sets of natural numbers are not
natural numbers.
I give functions for 0,1,S,+,* and
show my defintions satisfy the
axioms of PA+Limit.
My model is so simple we can't
even derive a contradiction like 0=1
because "1" is not in the domain of
universal quantifaction.
Obviously, my model is inconsistent
with the "intended" model, ie, the
natural numbers.
So, what went wrong?
Allowing functions to use mathematical
objects not in the universal domain seems
like a bad idea. I guess our functions
better not return sets of natural numbers.
My Limit axiom is similar to non-standard models.
http://en.wikipedia.org/wiki/Non-standard_model_of_arithmetic
In a non-standard model, "a" is a non-standard
natural number, so, "a" is in the domain of
universal quantification.
We add an infinite number of axioms of the form:
0 < a
1 < a
2 < a
...
If PA has a model then PA+"a" also has a model.
As usual, I am going to cheat.
I am going to assume "0" is the only
standard natural number and there exists
an infinite number of non-standard
natural numbers: {"0", "a", "b", ... }
I will add an infinite number of axioms:
0 < a
0 < b
0 < c
,,,
If PA has an "intended" model then so does my theory.
I simply assign 1 to "a", 2 to "b", etc.
Of course, I will have lots of constants
left over. I wonder what I can do with those?
The added axioms basically say any
natural number not equal to 0 is
greater than 0. This is a non-standard
model where 0 is the only standard
natural number.
BTW, I must like set like set theory
or I wouldn't waste so much time on it.
Calculus works. We can add infinite sums.
Just becuase it works doesn't mean its true.
There are lots of things we don't know
about large natural numbers.
Russell
- The universe is one dimensional
Jesse F. Hughes
> On May 27, 9:55 am, Frederick Williams <freddywilli...@btinternet.com>
> wrote:
>> MoeBlee wrote:
>
>> > On May 26, 10:53 pm, RussellE <reaste...@gmail.com> wrote:
>
>> > > Ax (S(x) =/= "0")
>
>> > I don't know why you put quote marks around '0'.
>
>> I has supposed it was to distinguish between numeral and number.
>> Similar with 1 elsewhere. Some authors might use emboldening or
>> underlining to make the distinction, but Russell, unable to do either of
>> those, settled for quote marks. Maybe.
>
> I use "0" to mean the symbol that is part of
> the theory. I should write:
>
> Ax (S(x) =/= A("0")) where A(x) means the
> natural number assigned to the symbol "0".
Not if you're writing axioms in FOL, you shouldn't.
Really, Russell, perhaps you *should* learn the very basics of FOL
before trying to prove that it is inconsistent or broken (and that you
alone have the insight to discover this fact!).
>
> Ax (S(x) =/= 0) uses a natural number in
> an axiom used to define natural numbers.
> Just looks circular to me.
>
>> Despite which, I agree with your:
>>
>> > You are HORRIBLY ignorant and confused about this material. Why do you
>> > even waste time typing nonsense about it before you've even taken
>> > minimal time to understand it?
>
> Thanks for the encouragement.
> Of course, this is no worse than things being
> said about Voevodsky on fom archive.
> And he is a Field's Medalist.
Well, it stands to reason, then, that you are just as respectable as
Voevodsky and are probably right on the money with your criticism.
Congrats!
> I make no secret of the fact I am trying
> to prove PA is inconsistent. This does
> not mean I am ignorant.
No, but the fact that you display ignorance is a pretty good sign you're
ignorant.
[...]
> This thread is about the limits of FOL.
> I start with the axioms of Peano Arithmetic.
> I added one axiom, the Limit axiom.
>
> Ax ( x < 1 )
>
> I assume "1" is a symbol for a
> mathematical object not in the
> domain of universal quantification.
> I assume "1" is not a natural number.
Then you're not doing FOL.
But it's not at all clear why you are so keen on assuming something as
odd as this. Why not just relativize the relevant axioms like so?
(Ax)( x != 1 -> x < 1 )
--
Jesse F. Hughes
"Why do the dirty villains always have to tie your hands *behind* ya?"
"That's what makes them villains."
--Adventures by Morse (old radio show)
Frederick Williams
>
> On May 27, 9:55 am, Frederick Williams <freddywilli...@btinternet.com>
> wrote:
> > MoeBlee wrote:
>
> > > On May 26, 10:53 pm, RussellE <reaste...@gmail.com> wrote:
>
> > > > Ax (S(x) =/= "0")
>
> > > I don't know why you put quote marks around '0'.
>
> > I has supposed it was to distinguish between numeral and number.
> > Similar with 1 elsewhere. Some authors might use emboldening or
> > underlining to make the distinction, but Russell, unable to do either of
> > those, settled for quote marks. Maybe.
>
> I use "0" to mean the symbol that is part of
> the theory.
That is what I thought.
> I should write:
>
> Ax (S(x) =/= A("0")) where A(x) means the
> natural number assigned to the symbol "0".
Why? Having decided that "0" is the symbol, why not use 0 for the
number?
> Ax (S(x) =/= 0) uses a natural number in
> an axiom
You have to use symbols in axioms. (Some people will tell you that
_anything_ can be a symbol, including numbers. And, though there is
some truth in that, I don't think we should pursue the idea here. It
will just cause confusion.)
> [...]
> We know from Godel we can't prove PA is consistent.
I don't think we do. We know that if PA is consistent it can't prove
that PA is consistent.
> All we are left with is proving PA is inconsistent.
>
> Yes, there are many "consistency" proofs.
> All these proofs require PA or induction of some sort.
PA isn't required.
> I assume "1" is a symbol for a
> mathematical object not in the
> domain of universal quantification.
That's bizarre. In First order theories names always name things in the
domain quantification.
> [...]
>
> In a non-standard model, "a" is a non-standard
> natural number, so, "a" is in the domain of
> universal quantification.
> We add an infinite number of axioms of the form:
>
> 0 < a
> 1 < a
> 2 < a
> ...
>
> If PA has a model then PA+"a" also has a model.
Don't you mean PA + (Ex)(x = "a")?
>
> [...]
> Calculus works. We can add infinite sums.
Yes, we add them in the sense that we can define them. How they behave
is determined by the definition and (for example) the axioms for the
real numbers.
> Just becuase it works doesn't mean its true.
Just because it's true doesn't mean it works. How many physical
phenomena do you know of that calculus is true of? I mean really true,
not just approximately.
Frederick Williams
>
> [...]
> --Adventures by Morse (old radio show)
He was a dotty fellow: always dashing about.
Parkitesta Arctonium
Jesse F. Hughes
> "Jesse F. Hughes" wrote:
>>
>> [...]
>> --Adventures by Morse (old radio show)
>
> He was a dotty fellow: always dashing about.
Adventures by Morse totally kicks ass. Don't you poke fun at it.
Available at archive.org in the old radio section.
--
Jesse F. Hughes
"[I]t's the damndest thing. There's something wrong with every last
one of you, and I *never* thought that was a possibility. But now I
feel it's the only reasonable conclusion." --JSH sees some sorta light
MoeBlee
> I use "0" to mean the symbol that is part of
> the theory. I should write:
>
> Ax (S(x) =/= A("0")) where A(x) means the
> natural number assigned to the symbol "0".
>
> Ax (S(x) =/= 0) uses a natural number in
> an axiom used to define natural numbers.
> Just looks circular to me.
You're completely confused. And you ignore the explanations given you.
It's amazing that you think you can talk about a subject of which you
refuse to its basics.
> this is no worse than things being
> said about Voevodsky on fom archive.
It's VASTLY worse.
> I make no secret of the fact I am trying
> to prove PA is inconsistent. This does
> not mean I am ignorant.
I didn't say you are ignorant because you are trying to prove PA
inconsistent. Rather, you are ignorant since you don't even know what
PA IS; you don't know what formal theory is, what a model is, or the
most basic things about using SYMBOLS.
> I agree with Voevodsky that the
> consistency of PA is an important question.
> We know from Godel we can't prove PA is consistent.
No, we know that we can't prove PA consistent in PA. That does not
preclude that we can prove PA consistent by other means.
> All these proofs require PA or induction of some sort.
Formally, there are many theories that prove Con(PA) without using
induction. However, yes, some of the most famous proofs (in which
proofs are not as trivial as taking Con(PA) as an axiom, etc.) do use
induction.
> We can't assume the set of natural numbers
> exists and then prove it exists.
That's not what we do.
> We can't prove PA is consistent and we know it.
> Why isn't there a journal dedicated to
> proving PA is inconsistent?
Whatever journals exist or not, you still need to learn the basics of
the subject before you can comment on it intelligently and not in the
OFFENSIVELY and PATHETICALLY ignorant way you do.
> I start with the axioms of Peano Arithmetic.
No, YOUR rendering of the axioms is INCOHERENT.
> I added one axiom, the Limit axiom.
>
> Ax ( x < 1 )
>
> I assume "1" is a symbol for a
> mathematical object not in the
> domain of universal quantification.
> I assume "1" is not a natural number.
That is INCOHERENT.
> This is not an unreasonable
> assumption. Lots of mathematical
> objects are not natural numbers.
> Sets of natural numbers are not
> natural numbers.
This has been explained to you already.
> I give functions for 0,1,S,+,* and
> show my defintions satisfy the
> axioms of PA+Limit.
> My model is so simple we can't
> even derive a contradiction like 0=1
> because "1" is not in the domain of
> universal quantifaction.
That is HOPELESSLY confused.
> Obviously, my model is inconsistent
> with the "intended" model, ie, the
> natural numbers.
Models are not consistent or inconsistent with other models.
> So, what went wrong?
What went wrong is that you're posting confused and ignorant garbage.
I might as well post a bunch of nonsense using symbols of chemistry
and get some bizarre conclusion and ask, "So, what went wrong?"
MoeBlee
MoeBlee
wrote:
> Thanks for the encouragement.
I DO encourage you to study the subject. I encourage you to bring all
of your skepticism and critical thinking to a study of the subject.
But you must at least UNDERSTAND and read CORRECTLY the material, and
then you can scrutinize it to whatever extent you like.
It is typical narcissism of the crank when he thinks that HE is the
critical thinker but that the many actual mathematicians haven't
themselves brought critical thinking to their reading and study of the
material. Yes, the lone crank is the rare human being blessed with a
critical mind, but all the mathematicians who have actually STUDIED
the subject are just missing that critical ability.
It's funny, because I have brought agonizing skepticism to the study
of mathematical logic, to the extent that I VERIFIED for myself that
it (that portion I've been able to work through) does check out, at
least in the sense that it can be stated and proven formally in
certain ways. And with that I also have an appreciation of finitism,
constructivism, and all kinds of alternatives, even paraconsistency. I
understand cerrtain things about how those alternative criteria can be
applied too. But that doesn't confuse me and cause me to misunderstand
and misstate classical mathematics meanwhile.
I first learned the predicate calculus. I didn't know about such
things as proofs of the soundness and completeness of the predicate
calculus. When I came across soundness and completeness proofs I was
astonished that there are people who have critical capacity even to
take on the task of figuring out how to prove things about logic
itself! And, naturally, critically, I questioned: "But what is the
basis and logic used to prove things ABOUT predicate logic? What is
the basis for accepting induction on formulas and induction on natural
numbers, etc?" Then I looked into it more and found out about
formalization of meta-theory, about finitistic reasoning, and more of
this rich subject matter. All the while, I've applied as much
skepticism as I could reasonably bring. I have demanded of myself that
I verify that the proofs in mathematical logic can be formalized and
that the steps all do logically "lock". And meanwhile, I admit that
there are certain very fundamental related philosophical questions to
which I don't have adequate answers.
Because I don't have a great amount of time each day for math, I am
still at pretty much a beginner's level. But I don't allow myself to
go off in confusion and ignorance to blather a bunch of incoherent
garbage as cranks do.
MoeBlee
Rotwang
>
> [...]
>
> I first learned the predicate calculus. I didn't know about such
> things as proofs of the soundness and completeness of the predicate
> calculus. When I came across soundness and completeness proofs I was
> astonished that there are people who have critical capacity even to
> take on the task of figuring out how to prove things about logic
> itself! And, naturally, critically, I questioned: "But what is the
> basis and logic used to prove things ABOUT predicate logic? What is
> the basis for accepting induction on formulas and induction on natural
> numbers, etc?"
This was something that also troubled me when I learnt what little I
know about FOL some years back, though I never pursued it (apart from
some fruitless conversations with mathmo friends).
> Then I looked into it more and found out about
> formalization of meta-theory, about finitistic reasoning, and more of
> this rich subject matter.
Can you recommend a good book to get me started on this kind of thing?
RussellE
> RussellE <reaste...@gmail.com> writes:
> > On May 27, 9:55 am, Frederick Williams <freddywilli...@btinternet.com>
> > wrote:
> >> MoeBlee wrote:
>
> >> > On May 26, 10:53 pm, RussellE <reaste...@gmail.com> wrote:
>
> >> > > Ax (S(x) =/= "0")
>
> >> > I don't know why you put quote marks around '0'.
>
> >> I has supposed it was to distinguish between numeral and number.
> >> Similar with 1 elsewhere. Some authors might use emboldening or
> >> underlining to make the distinction, but Russell, unable to do either of
> >> those, settled for quote marks. Maybe.
> > I use "0" to mean the symbol that is part of
> > the theory. I should write:
> > Ax (S(x) =/= A("0")) where A(x) means the
> > natural number assigned to the symbol "0".
> Not if you're writing axioms in FOL, you shouldn't.
> Really, Russell, perhaps you *should* learn the very basics of FOL
> before trying to prove that it is inconsistent or broken (and that you
> alone have the insight to discover this fact!).
I routinely find bugs in large, complex systems
I know very little about. You don't have to be an
expert to find a mistake.
> > Ax (S(x) =/= 0) uses a natural number in
> > an axiom used to define natural numbers.
> > Just looks circular to me.
>
> >> Despite which, I agree with your:
>
> >> > You are HORRIBLY ignorant and confused about this material. Why do you
> >> > even waste time typing nonsense about it before you've even taken
> >> > minimal time to understand it?
I have never found reading math books very helpful.
If I can't figure it out myself then I don't understand it.
Limit ordinals are a good example.
Ordinals are hereditarily transitive sets.
In ZFC-Infinity, we can prove every ordinal
has a largest element. Yes, we can define sets
that might be ordinals and might not have a
largest element. So what?. We can define sets
that might be unicorns. Are these "infinite" sets
part of the domain of universal quantification in
ZFC-Infinity? Every ordinal in ZFC-Infinity has
a largest element.
As near as I can tell, ZFC-Infinity proves there
are no limit ordinals.
> > Thanks for the encouragement.
> > Of course, this is no worse than things being
> > said about Voevodsky on fom archive.
> > And he is a Field's Medalist.
> Well, it stands to reason, then, that you are just as respectable as
> Voevodsky and are probably right on the money with your criticism.
> Congrats!
Thank you.
> > I make no secret of the fact I am trying
> > to prove PA is inconsistent. This does
> > not mean I am ignorant.
>
> No, but the fact that you display ignorance is a pretty good sign you're
> ignorant.
>
> [...]
>
> > This thread is about the limits of FOL.
> > I start with the axioms of Peano Arithmetic.
> > I added one axiom, the Limit axiom.
>
> > Ax ( x < 1 )
>
> > I assume "1" is a symbol for a
> > mathematical object not in the
> > domain of universal quantification.
> > I assume "1" is not a natural number.
>
> Then you're not doing FOL.
Why not? Why can't I assume there
are objects not in the universal domain?
> But it's not at all clear why you are so keen on assuming something as
> odd as this. Why not just relativize the relevant axioms like so?
The first thing I do when I see a physics equation
is try to break it. Let's say I have a formula for excape velocity.
What happens if the escape velocity is equal to the speed of light?
> (Ax)( x != 1 -> x < 1 )
Isn't this a theorem of the system I describe?
I don't need or want the x != 1 part.
If I assume 1 is in the universal domain,
I can't define S,+, and * in a way that
satisfies the axioms.
The model I give is a model of PA
with one of the axioms reversed.
Replace Ax (S(x) =/=0) with Ex (S(x)=0).
Ax (x != 1 -> x < 1) would be true in
the model I give. I am not really sure
how "<" is defined in PA where we
have Ex (S(x)=0) instead of Ax (S(x) != 0).
If I can add an axiom to PA and prove the
resulting theory is consistent then PA can
prove its own consistency, proving PA is
inconsistent.
I describe a model of PA+Limit.
All I have to do is assume there is
an object not in the domain of
universal quantification.
RussellE
> On May 28, 12:17 am, RussellE <reaste...@gmail.com> wrote:
> > I agree with Voevodsky that the
> > consistency of PA is an important question.
> > We know from Godel we can't prove PA is consistent.
> No, we know that we can't prove PA consistent in PA. That does not
> preclude that we can prove PA consistent by other means.
Yes, it pretty much does.
> > All these proofs require PA or induction of some sort.
>
> Formally, there are many theories that prove Con(PA) without using
> induction. However, yes, some of the most famous proofs (in which
> proofs are not as trivial as taking Con(PA) as an axiom, etc.) do use
> induction.
Assume we prove PA is consistent in a system
weaker than PA. Can't PA prove anything the
weaker system proves?
Russell
- Never never means never in set theory
RussellE
> On May 28, 12:17 am, RussellE <reaste...@gmail.com> sarcastically
> wrote:
> It's funny, because I have brought agonizing skepticism to the study
> of mathematical logic, to the extent that I VERIFIED for myself that
> it (that portion I've been able to work through) does check out, at
> least in the sense that it can be stated and proven formally in
> certain ways. And with that I also have an appreciation of finitism,
> constructivism, and all kinds of alternatives, even paraconsistency. I
> understand cerrtain things about how those alternative criteria can be
> applied too.
I know you study these things and I respect your knowledge.
That is why it is so much fun to annoy you.
I am just trying to find more things for you to be skeptical about.
MoeBlee
> Can you recommend a good book to get me started on this kind of thing?
What books have you already studied?
MoeBlee
MoeBlee
> I have never found reading math books very helpful.
My guess is that you don't have the attention span to actually read
one.
> If I can't figure it out myself then I don't understand it.
Right. If it's not something from your own brain then you can't be
bothered trying to understand it. Chronic narcissism.
> In ZFC-Infinity, we can prove every ordinal
> has a largest element.
WRONG. In ZFC-I+~I we prove every ordinal has a greatest element.
> Yes, we can define sets
> that might be ordinals and might not have a
> largest element. So what?. We can define sets
> that might be unicorns. Are these "infinite" sets
> part of the domain of universal quantification in
> ZFC-Infinity?
They may be, depending on the model.
> Every ordinal in ZFC-Infinity has
> a largest element.
WRONG.
> As near as I can tell, ZFC-Infinity proves there
> are no limit ordinals.
WRONG.
> If I can add an axiom to PA and prove the
> resulting theory is consistent then PA can
> prove its own consistency,
No, that's stupid.
You're a bane.
MoeBlee
MoeBlee
> On May 28, 8:27 am, MoeBlee <modem...@gmail.com> wrote:
> > Formally, there are many theories that prove Con(PA) without using
> > induction. However, yes, some of the most famous proofs (in which
> > proofs are not as trivial as taking Con(PA) as an axiom, etc.) do use
> > induction.
>
> Assume we prove PA is consistent in a system
> weaker than PA.
Then PA is inconsistent.
> Can't PA prove anything the
> weaker system proves?
Of course.
MoeBlee
MoeBlee
>
> I know you study these things and I respect your knowledge.
> That is why it is so much fun to annoy you.
> I am just trying to find more things for you to be skeptical about.
What you're doing is being an ignorant fool.
MoeBlee
Rotwang
On logic, only Hamilton's "Logic for Mathematicians". Though I have
picked up a few bits and pieces elsewhere, for example I learnt a tiny
bit about intuitionistic logic from reading about topoi, and a tiny (and
rapidly forgotten) bit about modal logic from reading about
non-wellfounded sets.
Jesse F. Hughes
> I have never found reading math books very helpful.
> If I can't figure it out myself then I don't understand it.
> Limit ordinals are a good example.
> Ordinals are hereditarily transitive sets.
>
> In ZFC-Infinity, we can prove every ordinal
> has a largest element.
Well, don't be coy. Show us that proof.
Perhaps you mean that it's provable in ZFC - Infinity + ~Infinity?
_
> Yes, we can define sets
> that might be ordinals and might not have a
> largest element. So what?. We can define sets
> that might be unicorns. Are these "infinite" sets
> part of the domain of universal quantification in
> ZFC-Infinity? Every ordinal in ZFC-Infinity has
> a largest element.
Er, no, since any model of ZFC is also a model of ZFC - Infinity.
>
> As near as I can tell, ZFC-Infinity proves there
> are no limit ordinals.
As near as I can tell, you need to learn the subject matter.
Or continue to pretend that you know more than all them folk who
actually study the subject. Whichever.
>> > This thread is about the limits of FOL.
>> > I start with the axioms of Peano Arithmetic.
>> > I added one axiom, the Limit axiom.
>>
>> > Ax ( x < 1 )
>>
>> > I assume "1" is a symbol for a
>> > mathematical object not in the
>> > domain of universal quantification.
>> > I assume "1" is not a natural number.
>>
>> Then you're not doing FOL.
>
> Why not? Why can't I assume there
> are objects not in the universal domain?
Because that's not what the universal quantifier *means*.
>> But it's not at all clear why you are so keen on assuming something as
>> odd as this. Why not just relativize the relevant axioms like so?
>
> The first thing I do when I see a physics equation
> is try to break it. Let's say I have a formula for excape velocity.
> What happens if the escape velocity is equal to the speed of light?
Er, huh?
>
>> (Ax)( x != 1 -> x < 1 )
>
> Isn't this a theorem of the system I describe?
> I don't need or want the x != 1 part.
> If I assume 1 is in the universal domain,
> I can't define S,+, and * in a way that
> satisfies the axioms.
Again, not part of FOL.
--
"Sexual love makes of the loved person an Object of appetite; as soon
as that appetite has been stilled, the person is cast aside as one
casts away a lemon which has been sucked dry." -- Immanuel Kant
"Squeeze my lemon til the juice runs down my leg." -- Robert Johnson
RussellE
> RussellE <reaste...@gmail.com> writes:
> > I have never found reading math books very helpful.
> > If I can't figure it out myself then I don't understand it.
> > Limit ordinals are a good example.
> > Ordinals are hereditarily transitive sets.
>
> > In ZFC-Infinity, we can prove every ordinal
> > has a largest element.
>
> Well, don't be coy. Show us that proof.
>
> Perhaps you mean that it's provable in ZFC - Infinity + ~Infinity?
No, I mean ZFC-Infinity.
I can be more precise. To prove a set is transitive
and well ordered in ZFC-Infinity requires proving
the set has a largest element.
> > Yes, we can define sets
> > that might be ordinals and might not have a
> > largest element. So what?. We can define sets
> > that might be unicorns. Are these "infinite" sets
> > part of the domain of universal quantification in
> > ZFC-Infinity? Every ordinal in ZFC-Infinity has
> > a largest element.
>
> Er, no, since any model of ZFC is also a model of ZFC - Infinity.
My point exactly.
> > As near as I can tell, ZFC-Infinity proves there
> > are no limit ordinals.
>
> As near as I can tell, you need to learn the subject matter.
>
> Or continue to pretend that you know more than all them folk who
> actually study the subject. Whichever.
>
> >> > This thread is about the limits of FOL.
> >> > I start with the axioms of Peano Arithmetic.
> >> > I added one axiom, the Limit axiom.
>
> >> > Ax ( x < 1 )
> >> > I assume "1" is a symbol for a
> >> > mathematical object not in the
> >> > domain of universal quantification.
> >> > I assume "1" is not a natural number.
> >> Then you're not doing FOL.
Yes I am.
> > Why not? Why can't I assume there
> > are objects not in the universal domain?
> Because that's not what the universal quantifier *means*.
No it doesn't. The domain of universal quantification
is a set. I define four set: N, S, +, and *.
N is the set for universal quantification.
S is a set of ordered pairs. + and * are sets
of ordered triplets.
It is quite decidable if there exists some
element, x, of an ordered pair in S, or an
ordered triplet in + or * such that x is not a
member of N.
This question is independent of the
other axioms of PA. I have added an
axiom to say there is such a element.
Russell
- Integers are an illusion
Marshall
>
> I routinely find bugs in large, complex systems
> I know very little about. You don't have to be an
> expert to find a mistake.
As a professional programmer, I find it dismaying that
so many of the cranks on these newsgroups are programmers.
Marshall
RussellE
the successor function has
a non-number.
Revised Limit Axiom:
ExAy (S(x) != y)
Axioms of PA:
Ax (S(x) =/= 0)
AxAy (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)
This theory is provably
consistent because it has
a model. A finite model.
N: {0}
S: S(0)=z, S(z)=0
+: 0+0=0, 0+z=z, z+0=z, z+z=0
*: 0*0=0, 0*z=0, z*0=0, z*z=z
We can easily check if all
of the axioms are satisfied.
The model is so simple we can't
really prove much of anything,
let alone derive a contradiction.
Jesse F. Hughes
> On May 28, 6:16 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> RussellE <reaste...@gmail.com> writes:
>> > I have never found reading math books very helpful.
>> > If I can't figure it out myself then I don't understand it.
>> > Limit ordinals are a good example.
>> > Ordinals are hereditarily transitive sets.
>>
>> > In ZFC-Infinity, we can prove every ordinal
>> > has a largest element.
>>
>> Well, don't be coy. Show us that proof.
>>
>> Perhaps you mean that it's provable in ZFC - Infinity + ~Infinity?
>
> No, I mean ZFC-Infinity.
> I can be more precise. To prove a set is transitive
> and well ordered in ZFC-Infinity requires proving
> the set has a largest element.
I reckon I'm thick. Show me.
Show me a proof of the following (in ZFC - Infinity)
(Ax)( x is transitive & x is well-ordered ->
(E y)( y in x & (A z)( z in x -> z <= y ))).
I'm very interested in this startling proof. After all, it would be a
proof that ZFC is inconsistent. What could be more interesting than
that?
>> > Yes, we can define sets
>> > that might be ordinals and might not have a
>> > largest element. So what?. We can define sets
>> > that might be unicorns. Are these "infinite" sets
>> > part of the domain of universal quantification in
>> > ZFC-Infinity? Every ordinal in ZFC-Infinity has
>> > a largest element.
>>
>> Er, no, since any model of ZFC is also a model of ZFC - Infinity.
>
> My point exactly.
Great! Give the proof.
[...]
>> > Why not? Why can't I assume there
>> > are objects not in the universal domain?
>
>> Because that's not what the universal quantifier *means*.
>
> No it doesn't. The domain of universal quantification
> is a set. I define four set: N, S, +, and *.
> N is the set for universal quantification.
> S is a set of ordered pairs. + and * are sets
> of ordered triplets.
>
> It is quite decidable if there exists some
> element, x, of an ordered pair in S, or an
> ordered triplet in + or * such that x is not a
> member of N.
>
> This question is independent of the
> other axioms of PA. I have added an
> axiom to say there is such a element.
Once again, learn FOL before you declare that you've proved it's "bad".
--
Jesse F. Hughes
"It is not as satisfying to disagree with a book."
-- Russell Easterly, on why he argues against set theory without
reading a book on set theory.
Jesse F. Hughes
> I really only need to assume
> the successor function has
> a non-number.
>
> Revised Limit Axiom:
> ExAy (S(x) != y)
That is inconsistent with the usual axioms of equality.
Let x be given such that (Ay)(S(x) != y). It follows that S(x) != S(x).
Nice try, Russell.
--
Scientists have calculated that the chance of anything so patently
absurd actually existing are millions to one. But magicians have
calculated that million-to-one chances crop up nine times out of ten.
-- Terry Pratchett on Intelligent Design. Or something.
RussellE
> RussellE <reaste...@gmail.com> writes:
> > I really only need to assume
> > the successor function has
> > a non-number.
> > Revised Limit Axiom:
> > ExAy (S(x) != y)
> That is inconsistent with the usual axioms of equality.
Which axiom are you talking about?
These are the axioms I am using:
Ax (S(x) =/= 0)
AxAy (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)
ExAy (S(x) != y)
And S(0) = S(0) in my model.
Frederick Williams
>
> [...]
>
> And S(0) = S(0) in my model.
S(0) = S(0) is a theorem of logic with =, S, 0; whatever function symbol
S is, so long as it's unary, and whatever constant 0 is. No need for
any non-logical axioms.
Jesse F. Hughes
Why don't you read what I wrote?
Let x be given such that (Ay)(S(x) != y). It follows that S(x) != S(x).
Thus,
(Ex)(Ay)( S(x) != y ) -> (Ex)( x != x )
Do you really not understand this utterly trivial argument?
--
"Being in the ring of algebraic integers is just kind of being in a
weird place, but it's no different than if you are in an Elk's Lodge
with weird made up rules versus just being out in regular society."
-- James S. Harris, teacher
Mike Terry
news:6fe72301-06fa-423b...@34g2000pru.googlegroups.com...
> On May 29, 6:24 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> > RussellE <reaste...@gmail.com> writes:
>
> > > I really only need to assume
> > > the successor function has
> > > a non-number.
>
> > > Revised Limit Axiom:
> > > ExAy (S(x) != y)
>
> > That is inconsistent with the usual axioms of equality.
>
> Which axiom are you talking about?
It is an axiom of equality that S(x) = S(x), and so Ay (S(x) != y) is false.
So provably there cannot be an x that makes your axiom true...
Mike.
RussellE
Your argument S(x) != S(x) is provably false in my model.
I am not sure what you are complaining about.
The axiom ExAy (S(x) != y) is satisfied by x=0
and S(0) = S(0) in my model.
Russell
- 2 mnay 2 count
RussellE
<news.dead.person.sto...@darjeeling.plus.com> wrote:
> "RussellE" <reaste...@gmail.com> wrote in message
I gave the axioms I am using. Which one of those axioms
is the "axiom of equality"?
In my model ExAy (S(x) != y) is satisfied by x=0
and S(0) = S(0).
RussellE
There is no axiom saying S(x) has to be a member of N.
Ay(S(x)!=y) can be true if S(x) is not a natural number.
Jesse F. Hughes
You are *still* confused.
From (Ay)(S(x) != y), it is utterly trivial to conclude S(x) != S(x).
There is, after all, a reason that we call it the *universal*
quantifier.
Once again, you are pretending that certain terms are "outside" the
domain of quantification, but that's not how FOL works.
--
Jesse F. Hughes
"And I'm one of my own biggest skeptics as I had *YEARS* of wrong
ideas, and attempts that failed. Worse, for some of them it took
*MONTHS* before I figured out where I screwed up." -- James Harris
Jesse F. Hughes
This makes no sense.
The axiom says there is an x such that S(x) is not equal to y for *any*
y. The element 0 does not witness this claim.
But, in any case, as I've said before, this axiom is inconsistent with
the reflexivity of equality.
--
Jesse F. Hughes
Baba: Spell checkers are bad.
Quincy (age 7): C-H-E-K-E-R-S A-R-E B-A-D.
Daryl McCullough
>There is no axiom saying S(x) has to be a member of N.
>Ay(S(x)!=y) can be true if S(x) is not a natural number.
In ordinary, untyped first order logic, every closed
term is an element of the domain, the same domain
that quantifiers range over. It's an axiom of
first order logic that
(Ay Phi(y)) -> Phi(t)
for every formula Phi(y) and for every
closed term t.
So in particular, if we let
Phi(y) be S(x) != y, and we
let t be S(x), then we have:
(Ay S(x) != y) -> S(x) != S(x)
So your axiom Ex Ay S(x) != y
is a contradiction, according to
the rules of first order logic.
Now, you could go to many-sorted first-order logic,
where it is not assumed that every element is the
same type, and quantifiers are restricted by type.
--
Daryl McCullough
Ithaca, NY
Aatu Koskensilta
> But, in any case, as I've said before, this axiom is inconsistent with
> the reflexivity of equality.
A highly dubious principle, I've always thought.
--
Aatu Koskensilta (aatu.kos...@uta.fi)
"Wovon man nicht sprechen kann, darüber muss man schweigen."
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
Aatu Koskensilta
> Now, you could go to many-sorted first-order logic, where it is not
> assumed that every element is the same type, and quantifiers are
> restricted by type.
He could also blather incoherently about logical matters some more.
RussellE
Limit Axiom: ExAy (S(x != y)
> The axiom says there is an x such that S(x) is not equal to y for *any*
> y.
Where "any y" means any natural number.
S(x) does not have to be a natural number.
In my model, S(0)=z and z is not a natural number.
The point of the Limit axiom is to guarantee
the successor function returns something
that is not a natural number.
> The element 0 does not witness this claim.
Yes it does. x=0 satisfies ExAy (S(x) != y).
0 is the only natural number in my model.
> But, in any case, as I've said before, this axiom is inconsistent with
> the reflexivity of equality.
Only if we require S(x) to be a natural number.
For this we would need an axiom like:
AxEy (S(x) = y)
This statement is the inverse of my
Limit axiom and is false in my model.
Russell
RussellE
wrote:
> RussellE says...
> >There is no axiom saying S(x) has to be a member of N.
> >Ay(S(x)!=y) can be true if S(x) is not a natural number.
> In ordinary, untyped first order logic, every closed
> term is an element of the domain,
By closed term I assume you mean things
like S(x), x+y, etc. So, all of these things
are in the domain for functions.
> the same domain
> that quantifiers range over.
Uh, why? In PA I thought the
quantifiers range over the
natural numbers.
> It's an axiom of
> first order logic that
>
> (Ay Phi(y)) -> Phi(t)
OK. Would you have a reference
for these axioms?
> for every formula Phi(y) and for every
> closed term t.
> So in particular, if we let
> Phi(y) be S(x) != y, and we
> let t be S(x), then we have:
> (Ay S(x) != y) -> S(x) != S(x)
I don't really see how this follows.
> So your axiom Ex Ay S(x) != y
> is a contradiction, according to
> the rules of first order logic.
That would be bad news for FOL,
because it is easy to see S(0)=S(0)
in my model.
Aatu Koskensilta
> By closed term I assume you mean things like S(x), x+y, etc. So, all
> of these things are in the domain for functions.
Yes, naturally when Daryl speaks of closed terms he has in mind terms
with free variables.
> OK. Would you have a reference for these axioms?
The axioms for first-order logic are the following:
(x)Px --> Qx
(Ex)(Px & Fx \/ Qx)
(y)(Ex)(Rxy & ~Qx <--> Wxy)
You will find all this spelled out in detail, in somewhat more
cumbersome syntax of Hilbert's epsilon-calculus, in Kreisel's
autobiography (_Go Away -- how I was very rude to everyone I ever met
and made a name for myself as a very insightful logician and philosopher
by uttering random observations and insights at a time everyone else
thought bizarrely implausible gratuitously formal going on about
abstract philosophical theories was the thing_) p. 345 - 4324325.
> That would be bad news for FOL, because it is easy to see S(0)=S(0) in
> my model.
Oh no!
Marshall
> stevendaryl3...@yahoo.com (Daryl McCullough) writes:
> > Now, you could go to many-sorted first-order logic, where it is not
> > assumed that every element is the same type, and quantifiers are
> > restricted by type.
>
> He could also blather incoherently about logical matters some more.
Five bucks says he does what Aatu said.
Marshall
Marshall
> RussellE <reaste...@gmail.com> writes:
> > By closed term I assume you mean things like S(x), x+y, etc. So, all
> > of these things are in the domain for functions.
>
> Yes, naturally when Daryl speaks of closed terms he has in mind terms
> with free variables.
Exactly. "Closed" terms as in "closed for the winter" so the variables
are all "free" to go skiing.
> > OK. Would you have a reference for these axioms?
>
> The axioms for first-order logic are the following:
>
> (x)Px --> Qx
I love this axiom so much!
I chucked for a full minute at your post.
Marshall
kluto
>
> I chucked for a full minute at your post.
>
=kluto
Rupert
> On May 26, 9:05 pm, Rupert <rupertmccal...@yahoo.com> wrote:
>
>
>
>
>
>
>
>
>
> > On May 27, 1:53 pm, RussellE <reaste...@gmail.com> wrote:
>
> > > Few people question the consistency of
> > > First Order Logic. FOL can produce
> > > inconsistent results when certain
> > > assumptions are made.
>
> > > I am going to start with the axioms
> > > of Peano Arithmetic:
>
> > > Ax (S(x) =/= "0")
> > > AxAy (S(x) = S(y) -> x = y)
> > > Ax (x + "0" = x)
> > > Ax (x * "0" = "0")
> > > AxAy (x*S(y) = (x*y)+x)
>
> > > I will describe a model of the axioms
> > > above. I am going to cheat, of course.
>
> > > PA requires the definition of a constant, "0".
> > > "0" is natural number in the domain of universal
> > > quantification.
>
> > > PA is the theory of natural numbers and there
> > > are some things that we can't put in the domain
> > > of universal quantification. For example, the set of
> > > all natural numbers isn't a member of the domain
> > > for universal quantification.
>
> > > I am going to add a new constant, "1", to PA.
> > > I will add it as an axiom.
>
> > > Limit Axiom
> > > Ax (x < "1")
>
> > > "1" is not in the domain for universal
> > > quantification, but it can be defined
> > > in functions.
> > You're not allowed to have a constant symbol referring to an object
> > not in the domain of universal quantification
>
> How else would I define a limit?
> Aren't real numbers definable in PA?
No.
> How can I define a real number without a limit?
Rupert
> On May 26, 11:31 pm, hagman <goo...@von-eitzen.de> wrote:
>
> > According to thread topic, however, you really want to ask:
> > How can I define a real number with only FOL from PA?
> > I.e. by only talking about natural numbers, not dets/sequences/... of
> > natural numbers?- Hide quoted text -
>
> OK. Real numbers are a bad example for PA+FOL.
>
> What if I want to define induction on the natural numbers?
> Isn't all natural numbers part of the definition of induction?
> Is "all natural numbers" part of the domain of universal
> quantification?
>
> How would I define induction as a function?
>
> Russell
> - logic must be idiot proof
Induction is an axiom schema. The definition of the axiom schema is
done in the metatheory.
Daryl McCullough
>
>RussellE <reas...@gmail.com> writes:
>
>> By closed term I assume you mean things like S(x), x+y, etc. So, all
>> of these things are in the domain for functions.
>
> Yes, naturally when Daryl speaks of closed terms he has in mind terms
>with free variables.
Uh, yes. In other words, when I said "closed term" I meant
"not necessarily closed term".
Daryl McCullough
>
>On May 29, 9:26=A0pm, stevendaryl3...@yahoo.com (Daryl McCullough)
>wrote:
>> In ordinary, untyped first order logic, every closed
>> term is an element of the domain,
>
>By closed term I assume you mean things
>like S(x), x+y, etc. So, all of these things
>are in the domain for functions.
I should have just said "term" rather than "closed term".
>> the same domain
>> that quantifiers range over.
>
>Uh, why? In PA I thought the
>quantifiers range over the
>natural numbers.
Yes, and every term is a natural number.
>> So your axiom Ex Ay S(x) != y
>> is a contradiction, according to
>> the rules of first order logic.
>
>That would be bad news for FOL,
>because it is easy to see S(0)=S(0)
>in my model.
Why is it bad news for FOL? Your model
is not a model, in the sense of first-order
logic.
Frederick Williams
>
> On May 29, 8:14 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>
> Limit Axiom: ExAy (S(x != y)
>
> > The axiom says there is an x such that S(x) is not equal to y for *any*
> > y.
>
> Where "any y" means any natural number.
> S(x) does not have to be a natural number.
> In my model, S(0)=z and z is not a natural number.
Whatever the interpretation of z is, that thing must be in the domain of
quantification. If it isn't then you are not discussing what is usually
called a first order theory. That's all right, maybe you're discussing
something of your own devising. But if so the term "FOL" does not
belong in the title of your posts.
Jesse F. Hughes
> On May 29, 8:14 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>
> Limit Axiom: ExAy (S(x != y)
>
>> The axiom says there is an x such that S(x) is not equal to y for *any*
>> y.
>
> Where "any y" means any natural number.
That is not what ExAy(S(x) != y) means, Russell.
Learn some basic logic. If you meant something like
(Ex)(Ay)( y in N -> S(x) != y ),
then that's what you need to write. What you wrote was that there's an
x such that S(x) is not equal to *anything at all*.
> S(x) does not have to be a natural number.
> In my model, S(0)=z and z is not a natural number.
>
> The point of the Limit axiom is to guarantee
> the successor function returns something
> that is not a natural number.
Whatever your aim is, you've written an inconsistent axiom.
>> The element 0 does not witness this claim.
>
> Yes it does. x=0 satisfies ExAy (S(x) != y).
> 0 is the only natural number in my model.
>
>> But, in any case, as I've said before, this axiom is inconsistent with
>> the reflexivity of equality.
>
> Only if we require S(x) to be a natural number.
> For this we would need an axiom like:
>
> AxEy (S(x) = y)
>
> This statement is the inverse of my
> Limit axiom and is false in my model.
Still clueless.
--
Sam Vimes could parallel process. Most husbands can. They learn to
follow their own line of thought while /at the same time/ listening to
what their wives say.... At any time they could be challenged and
must be ready to quote the last sentence in full. -- Terry Pratchett
RussellE
> RussellE <reaste...@gmail.com> writes:
> (Ex)(Ay)( y in N -> S(x) != y ),
> then that's what you need to write. What you wrote was that there's an
> x such that S(x) is not equal to *anything at all*.
This sounds reasonable.
Why don't we do this in all of the axioms of PA?
Ax (x in N -> S(x) =/= "0")
AxAy ( x,y in N -> S(x) = S(y) -> x = y)
Ax (x in N -> x + "0" = x)
AxAy (x,y in N -> x+S(y) = S(x+y)
Ax (x in N -> x * "0" = "0")
AxAy (x,y in N -> x*S(y) = (x*y)+x)
My definitions of S,+, and * are consistent
with these axioms.
If you are going to assume everything
is a natural number why isn't there an
axiom stating this assumption?
Something like:
AxEy (S(x)=y)
You would need similar axioms for
addition and multiplication.
Frederick Williams
>
> If you are going to assume everything
> is a natural number
I'm not sure what you mean by that, but PA does have non-standard
models, and in them not everything is a natural number. This has been
known since the mid thirties.
> why isn't there an
> axiom stating this assumption?
To exclude non-standard models you need second order quantifiers.
Aatu Koskensilta
> I'm not sure what you mean by that, but PA does have non-standard
> models, and in them not everything is a natural number.
There is no necessity to everything being a natural number in a
standard model of PA. The set of hereditarily ordinal definable reals, a
model of constructive type theory and the first omega measurable
cardinals, for example, together make up a perfectly fine standard model
of PA when equipped with suitable addition and multiplication.
> To exclude non-standard models you need second order quantifiers.
To exclude non-standard models by means of a formal sentence, in a
well known logical language, required to hold in the model, yes. To
exclude non-standard models by a device of a less technical nature we
need do nothing but say, Here we will only consider models in which
every element of the domain is named by a numeral, or, Here we will only
consider models that have no substructures that are models of PA, or,
...
Marshall
> Frederick Williams <freddywilli...@btinternet.com> writes:
> > I'm not sure what you mean by that, but PA does have non-standard
> > models, and in them not everything is a natural number.
>
> There is no necessity to everything being a natural number in a
> standard model of PA.
I don't really understand why no one ever comes right out
and says what's really going on as far as the members of
a structure are concerned: that there is no necessity for
anything to be anything.
WHAT the members are is irrelevant. Completely irrelevant.
(As far as their role in the structure.) All that matters is their
identity. In fact, we can entirely "throw away" any information
about what the elements are and just abstract out the
identity. Because of this, some subset of the natural numbers
is sufficient to act as the members for ANY structure of
finite or countably infinite cardinality.
Come to think of it, maybe George said something like that
once upon a time.
Marshall
Jesse F. Hughes
> On May 30, 5:50 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> RussellE <reaste...@gmail.com> writes:
>
>> (Ex)(Ay)( y in N -> S(x) != y ),
>
>> then that's what you need to write. What you wrote was that there's an
>> x such that S(x) is not equal to *anything at all*.
>
> This sounds reasonable.
> Why don't we do this in all of the axioms of PA?
>
> Ax (x in N -> S(x) =/= "0")
> AxAy ( x,y in N -> S(x) = S(y) -> x = y)
> Ax (x in N -> x + "0" = x)
> AxAy (x,y in N -> x+S(y) = S(x+y)
> Ax (x in N -> x * "0" = "0")
> AxAy (x,y in N -> x*S(y) = (x*y)+x)
For one thing, "in" is not a relational symbol in PA. There are other
ways to work around it, but the reason this isn't used in PA is that the
intended model of PA consists only of natural numbers.
> My definitions of S,+, and * are consistent
> with these axioms.
>
> If you are going to assume everything
> is a natural number why isn't there an
> axiom stating this assumption?
> Something like:
>
> AxEy (S(x)=y)
You *really* need to study some FOL.
In standard FOL, all function symbols are interpreted as total
functions.
The above "assumption" is trivially proved as follows:
S(x) = S(x) (Reflexivity of =)
(Ey)(S(x) = y) (E-intro)
(Ax)(Ey)(S(x) = y) (A-intro)
> You would need similar axioms for
> addition and multiplication.
These axioms are similarly unnecessary.
--
Jesse F. Hughes
"I'm losing you and you are all I've got.
Thanks a lot. Thanks a lot." --Johnny Cash
RussellE
OK, in standard FOL we assume
everything is a number.
> In standard FOL, all function symbols are interpreted as total
> functions.
I am not having problems with
the domain for functions. I want
to limit the domain of universal
quantification.
Standard FOL just assumes
both domains are the same
and everything is a number.
> The above "assumption" is trivially proved as follows:
>
> S(x) = S(x) (Reflexivity of =)
> (Ey)(S(x) = y) (E-intro)
> (Ax)(Ey)(S(x) = y) (A-intro)
Yes, based on the assumption everything
is a number. The second two statements
aren't true if S(x) isn't a natural number.
Barb Knox
Frederick Williams <freddyw...@btinternet.com> wrote:
> RussellE wrote:
>
> >
> > If you are going to assume everything
> > is a natural number
>
> I'm not sure what you mean by that, but PA does have non-standard
> models, and in them not everything is a natural number. This has been
> known since the mid thirties.
>
> > why isn't there an
> > axiom stating this assumption?
>
> To exclude non-standard models you need second order quantifiers.
Plus you need to specify that you are using the full 2nd-order semantics
(as opposed to, for example, Henkin's general semantics).
--
---------------------------
| BBB b \ Barbara at LivingHistory stop co stop uk
| B B aa rrr b |
| BBB a a r bbb | Quidquid latine dictum sit,
| B B a a r b b | altum videtur.
| BBB aa a r bbb |
-----------------------------
Jesse F. Hughes
No.
>> In standard FOL, all function symbols are interpreted as total
>> functions.
>
> I am not having problems with
> the domain for functions. I want
> to limit the domain of universal
> quantification.
>
> Standard FOL just assumes
> both domains are the same
> and everything is a number.
No. What FOL *does* assume is that the universal quantifier really is
universal. That is, when you write "for all x", you mean for all x,
without exception.
>
>> The above "assumption" is trivially proved as follows:
>>
>> S(x) = S(x) (Reflexivity of =)
>> (Ey)(S(x) = y) (E-intro)
>> (Ax)(Ey)(S(x) = y) (A-intro)
>
> Yes, based on the assumption everything
> is a number. The second two statements
> aren't true if S(x) isn't a natural number.
This has nothing to do with whether the objects of our consideration are
numbers or not. It simply has to do with the meaning of the
quantifiers.
--
Jesse F. Hughes
"When my brain begins to reel from my literary labors, I make an
occasional cheese dip." -- Ignatius J. Reilly
RussellE
> > Standard FOL just assumes
> > both domains are the same
> > and everything is a number.
>
> No. What FOL *does* assume is that the universal quantifier really is
> universal. That is, when you write "for all x", you mean for all x,
> without exception.
Thanks for the explanations.
I want to assume the universal quantifier
isn't really universal.
I might be able to do this by conditioning
all of my axioms.
ExAy (y in N -> (S(x) != y))
Marshall
> On May 30, 8:10 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>
> > > Standard FOL just assumes
> > > both domains are the same
> > > and everything is a number.
>
> > No. What FOL *does* assume is that the universal quantifier really is
> > universal. That is, when you write "for all x", you mean for all x,
> > without exception.
>
> Thanks for the explanations.
> I want to assume the universal quantifier
> isn't really universal.
That means you're not using FOL any more.
Marshall
Frederick Williams
>
> OK, in standard FOL we assume
> everything is a number.
I'm not sure "we" do, but according to Thoralf Löwenheim, or was it
Leopold Skolem?, "you" may.
> [...] I want
> to limit the domain of universal
> quantification.
I find that genuinely funny.
MoeBlee
> Hamilton's "Logic for Mathematicians".
That's one I haven't seen, so it's hard for me to say what would go
after it.
Perhaps I should say what I did myself:
First, 'Logic: Techniques Of Formal Reasoning' by Kalish, Montague,
and Mar.
This gave me a solid grasp of working in the first order predicate
calculus itself.
'Elements Of Set Theory' by Enderton
'Axiomatic Set Theory' by Suppes
I used those two books together to learn enough basic set theory in
which to carry out mathematical logic.
'A Mathematical Introduction To Logic' by Enderton
This is the text I used as a base for a rigorous approach to
mathematical logic.
A few other books to help out (I have not worked through all of the
material in these books):
Hinman - Fundamentals Of Mathematical Logic
Monk - Mathematical Logic
Those give additional technical details. They're two of the most
detailed books I've found on mathematical logic.
Church - Introduction To Mathematical Logic
I find the introductory chapter to this book to be essential. It gives
the best overview of the general context I've found.
Chang & Keisler - Model Theory
The "bible" of model theory.
Davis - Computability And Unsolvability
I like the detail in this one, for the subject of computability.
Jech - Set Theory
Kunen - Set Theory
Those are the two "go to" texts for set theory that's more advanced
than introductory texts like Enderton and Suppes.
Smullyan - Godel's Incompleteness Theorems
Elegant and beautiful.
Frazen - Godel's Theorem
Beautiful book for for its informal explanations.
MoeBlee
MoeBlee
> To prove a set is transitive
> and well ordered in ZFC-Infinity requires proving
> the set has a largest element.
Plainly FALSE.
You're claiming that ZFC-I proves:
Ax((x is transitive & x is well ordered) -> x has an epsilon-greatest
member)
But ZFC-I does not prove that.
> > > Every ordinal in ZFC-Infinity has
> > > a largest element.
>
> > Er, no, since any model of ZFC is also a model of ZFC - Infinity.
>
> My point exactly.
No, your claim CONTRADICTS that it is NOT the case that ZFC-I proves
"every ordinal has an epsilon-greatest member".
> > >> > I assume "1" is a symbol for a
> > >> > mathematical object not in the
> > >> > domain of universal quantification.
> > >> > I assume "1" is not a natural number.
> > >> Then you're not doing FOL.
>
> Yes I am.
You're not using the ordinary semantics for first order logic. If the
symbol '1' (as a constant) is in the language, then, per a given model
for that language, the symbol '1' stands for some member of the
universe of the model.
MoeBlee
MoeBlee
> Revised Limit Axiom:
> ExAy (S(x) != y)
That axiom contradicts the identity theory that is presupposed for PA.
From identity theory, we have:
Ax x=x.
So Ax Sx=Sx
So ~ExAy ~Sx=y
You really don't know jack about any of this.
MoeBlee
MoeBlee
> On May 29, 6:24 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>
> > RussellE <reaste...@gmail.com> writes:
> > > I really only need to assume
> > > the successor function has
> > > a non-number.
> > > Revised Limit Axiom:
> > > ExAy (S(x) != y)
>
> Which axiom are you talking about?
> These are the axioms I am using:
>
> Ax (S(x) =/= 0)
> AxAy (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)
>
We also have the axioms of identity theory. If you're not adopting
identity theory (or some other axioms regarding '='), then you don't
have a way to make inferences regarding '='.
MoeBlee
MoeBlee
> The axiom ExAy (S(x) != y) is satisfied by x=0
No, with the standard semantics for '=', we have that ExAy ~Sx=y is
FALSE in EVERY model for a language that has 'S' as a 1-place function
symbol.
You're hopeless, as you won't bother to learn even the basics of this
subject.
MoeBlee
MoeBlee
> I gave the axioms I am using. Which one of those axioms
> is the "axiom of equality"?
We presuppose the axioms of equality in the LOGIC that includes
identity theory.
Otherwise, you have to give your own axioms regarding '=' and your own
semantics for '='.
Any INFORMED reader is going to presuppose that you adopt identity
theory (which is part of first order logic with identity) unless you
specify otherwise.
You're an ignoramus.
MoeBlee
MoeBlee
> I don't really see how this follows.
Because you don't know anything about this! Because you won't read a
damn book about it. Because you're so infatuated with yourself that
you think you're above reading what other human beings have done
already. You're sad and offensive.
MoeBlee
MoeBlee
> By closed term I assume you mean things
> like S(x), x+y, etc.
No, you fool, he means NOT such things. Those have free variables in
the term so they are not closed terms.
Damn, how do you expect to carry on a conversation (let alone be in a
position to find "faults") when you don't know the most basic things
about the subject?
What would you say if I started talking about chemistry and starting
talking about "Au" as if it meant "chain reaction" or some other
foolishness? What you're doing is not a bit less foolish.
MoeBlee
MoeBlee
> in standard FOL we assume
> everything is a number.
What?
Okay, in standard chemistry we assume everything is a rye biscuit.
My chemistry is just as good as your mathematical logic.
MoeBlee
Daryl McCullough
>
>On May 30, 12:20=A0am, RussellE <reaste...@gmail.com> wrote:
>
>> By closed term I assume you mean things
>> like S(x), x+y, etc.
>
>No, you fool, he means NOT such things. Those have free variables in
>the term so they are not closed terms.
Don't blame Russell for this one. I said "closed term" when I
just meant "term".
My point was that in ordinary first-order logic if t is any
term whatsoever, and you have an axiom
Ax Phi(x)
then you can conclude
Phi(t)
In some logics (constructive type theory), t has to be
closed (which is why I said that), but in ordinary
first-order logic there is no such restriction, which
allows us to prove
(Ax Phi(x)) -> (Ex Phi(x))
Rotwang
> On May 28, 5:41 pm, Rotwang<sg...@hotmail.co.uk> wrote:
>> Hamilton's "Logic for Mathematicians".
>
> That's one I haven't seen, so it's hard for me to say what would go
> after it.
>
> Perhaps I should say what I did myself:
>
Thanks.
Marshall
wrote:
> MoeBlee says...
> >On May 30, 12:20=A0am, RussellE <reaste...@gmail.com> wrote:
>
> >> By closed term I assume you mean things
> >> like S(x), x+y, etc.
>
> >No, you fool, he means NOT such things. Those have free variables in
> >the term so they are not closed terms.
>
> Don't blame Russell for this one. I said "closed term" when I
> just meant "term".
I hate to disagree with you, but in this case I believe you are
wrong when you say you were wrong earlier. Your original
quote:
"In ordinary, untyped first order logic, every closed
term is an element of the domain, the same domain
that quantifiers range over."
This is precisely correct, is it not? An open term like
S(x) will not resolve to "*an* element of the domain"
but is rather more like a function whose domain and
codomain each are the domain of discourse.
> My point was that in ordinary first-order logic if t is any
> term whatsoever, and you have an axiom
>
> Ax Phi(x)
>
> then you can conclude
>
> Phi(t)
This is certainly true, not merely of closed terms,
but it's not quite the same statement as you said
originally.
Marshall
MoeBlee
wrote:
> Don't blame Russell for this one. I said "closed term" when I
> just meant "term".
Okay.
But still the general point stands. He doesn't know anything about
this subject, including its basic terminology, yet he thinks he knows
how to find what's so very wrong in it.
> if t is any
> term whatsoever, and you have an axiom
>
> Ax Phi(x)
>
> then you can conclude
>
> Phi(t)
Maybe I'm not too pedantic to say that, as you know, we require: t is
free for x in Phi.
> (Ax Phi(x)) -> (Ex Phi(x))
Indeed, for any formula P whatsoever, we have:
AxP -> ExP
MoeBlee
Jesse F. Hughes
Yes, you can do that. Alternatively (since this particular
relativization seems to require bringing in set-theoretic notions into a
theory for natural numbers and such), just add a predicate N(x) that
means "x is a natural number) and relativize axioms like so:
ExAy ( N(y) -> (S(x) != y) )
--
"Start obeying math rules, or if you prefer I can proceed to tear away
your illusions of being rational and logical piece by piece over time,
as I reduce your society using advanced psychological tools that rival
the best psychological warfare techniques of world governments." --JSH