Can ZFC prove Addition is Associative?
Charlie-Boo
about proving that addition is associative, you know, A+(B+C) = (A+B)+C
for all numbers A,B,C? It is a simple statement so it shouldn't take
more than a few steps. Could someone show it (the formal derivation)
here?
C-B
DmitryKovtun
"Charlie-Boo" <shyma...@gmail.com> wrote in message
news:1166192828.2...@f1g2000cwa.googlegroups.com...
Yes, Zentuckey Frid Chicken theorm indicates that A+B = D - C and that B +
C = D - A
therfore A + B + B + C = D + D - C - A
Simplyfying A + C + 2*B = 2*D - ( A + C)
or B = D
Chip Eastham
What "numbers" do you have in mind, Charlie? Formal proof
depends on formal definitions. Are we talking natural numbers?
Ordinals? Cardinals? Integers? Reals? Complex?
Your assertion that it is "a simple statement so it
shouldn't take more than a few steps" is argumentative.
I know of no reason to suspect this is true. We could
sketch the development of ordinal arithmetic here, with
the natural numbers developed as an initial segment,
in a post of reasonable length.
--c
Bob Kolker
Showing that addition is associative using just PA takes more than a few
steps.
Bob Kolker
Charlie-Boo
> Charlie-Boo wrote:
> > Many people say that all of Mathematics can be done using ZFC. Ok, how
> > about proving that addition is associative, you know, A+(B+C) = (A+B)+C
> > for all numbers A,B,C? It is a simple statement so it shouldn't take
> > more than a few steps. Could someone show it (the formal derivation)
> > here?
>
> What "numbers" do you have in mind, Charlie? Formal proof
> depends on formal definitions. Are we talking natural numbers?
Let's just say natural numbers to keep it simple (and still preserve
the point.)
> Ordinals? Cardinals? Integers? Reals? Complex?
>
> Your assertion that it is "a simple statement so it
> shouldn't take more than a few steps" is argumentative.
> I know of no reason to suspect this is true. We could
> sketch the development of ordinal arithmetic here, with
> the natural numbers developed as an initial segment,
> in a post of reasonable length.
Ok. Or if you (anyone) would like to simply present the proof and we
can work backwards as needed (back to definitions and other theorems
utilized), that might also work.
C-B
> --c
Charlie-Boo
LOL : )
Yeh, it taste good - but is it nutritional? Did we follow the Rules of
the Colonel (Secret Recipe)? Or do we make up rules as we go along -
uh oh!
Don't you just love asking for a simple little example and people flip
out? Makes me think they do know the truth - and sense we are getting
a bit closer to it!
C-B
Charlie-Boo
PA I believe in. It describes a specific domain (recursively
enumerable universal set) and has rules particular to that. In fact,
it is an example of what I maintain must be done: rules for each
domain.
But how would ZFC do it? It seems to only describe what we should
allow as sets - nothing about Arithmetic. We can use sets in place of
the natural numbers, but that is true of a subset of the functions, of
directed graphs, of any infinite set for which we have a formal
(finite) representation of each element. But you need special rules to
go beyond that. ZFC is supposed to be only the rules it already set
out.
C-B
> Bob Kolker
Dave Seaman
If A and B are cardinal numbers, then A+B is defined to be the
cardinality of the disjoint union. That is,
A+B = | Ax{0} U Bx{1} |,
where "x" denotes Cartesian product. Notice that if A and B are any
sets, then |Ax{0}| = |A|, |Bx{0}| = |B|, and the sets Ax{0} and Bx{1} are
necessarily disjoint, so that the sum of the cardinalities is the
cardinality of the union.
From the definition it follows that
A+(B+C) = A + | Bx{0} U Cx{1} |
= | Ax{0} U (Bx{0} U Cx{1})x{1} |
= | Ax{0} U (Bx{1} U Cx{2}) |
= | (Ax{0} U Bx{1}) U Cx{2} | (using associativity of union)
= | (Ax{0} U Bx{1})x{0} U Cx{1} |
= | Ax{0} U Bx{1} | + C
= (A+B) + C.
--
Dave Seaman
U.S. Court of Appeals to review three issues
concerning case of Mumia Abu-Jamal.
<http://www.mumia2000.org/>
Charlie-Boo
Dave Seaman wrote:
> On 15 Dec 2006 06:27:08 -0800, Charlie-Boo wrote:
> > Many people say that all of Mathematics can be done using ZFC. Ok, how
> > about proving that addition is associative, you know, A+(B+C) = (A+B)+C
> > for all numbers A,B,C? It is a simple statement so it shouldn't take
> > more than a few steps. Could someone show it (the formal derivation)
> > here?
>
> If A and B are cardinal numbers, then A+B is defined to be the
> cardinality of the disjoint union. That is,
>
> A+B = | Ax{0} U Bx{1} |,
>
> where "x" denotes Cartesian product. Notice that if A and B are any
> sets, then |Ax{0}| = |A|, |Bx{0}| = |B|, and the sets Ax{0} and Bx{1} are
> necessarily disjoint, so that the sum of the cardinalities is the
> cardinality of the union.
>
> From the definition it follows that
>
> A+(B+C) = A + | Bx{0} U Cx{1} |
> = | Ax{0} U (Bx{0} U Cx{1})x{1} |
> = | Ax{0} U (Bx{1} U Cx{2}) |
> = | (Ax{0} U Bx{1}) U Cx{2} | (using associativity of union)
Thanks. However, here you do as I predicted: you add the rule for the
Associativity of Union (which in this context means Addition!) So ZFC
(alone) did not prove it. As I said, you have to add new rules for
each branch of Mathematics e.g. Arithmetic.
Is there a real proof (using only ZFC)?
C-B
Dave Seaman
>> = | (Ax{0} U Bx{1})x{0} U Cx{1} |
>> = | Ax{0} U Bx{1} | + C
>> = (A+B) + C.
Let I be the set {A,B,C}. Then (AUB)UC = AU(BUC) = UI, where the existence of
UI is guaranteed by the union axiom of ZFC.
Charlie-Boo
LOL : )
Herman Rubin
>Bob Kolker
This is NOT the case. The customary treatment from Peano
postulates, after addition has been characterized, makes
this the first theorem, easily proved by induction on C.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
Dave L. Renfro
I don't know why you think simple statments must
have simple proofs. Probably the best known "hard"
mathematical achievement of the past 20 years is
a counterexample to this (Fermat's Last Theorem).
Anway ...
See: [1] Chapter 4, Theorem 4K(a), p. 81 in
Enderton's book; [2] Chapter 5, Section 2, Theorem 30,
pp. 146-147 in Suppes' book. Or, see any number
of other elementary set theory texts.
Incidentally, the relevant pages of Patrick
Suppes' book are available at google-books.
When you get to his book there, type "by
Theorem 28" (including quotes) into the window
'Search in this book'.
Herbert B. Enderton, ELEMENTS OF SET THEORY,
Academic Press, 1977, xiv + 279 pages.
Patrick Suppes, AXIOMATIC SET THEORY, Dover
Publications, 1960/1972, xii + 267 pages.
http://books.google.com/books?vid=ISBN0486616304
Dave L. Renfro
Herman Rubin
Charlie-Boo <shyma...@gmail.com> wrote:
>Dave Seaman wrote:
>> On 15 Dec 2006 06:27:08 -0800, Charlie-Boo wrote:
>> > Many people say that all of Mathematics can be done using ZFC. Ok, how
>> > about proving that addition is associative, you know, A+(B+C) = (A+B)+C
>> > for all numbers A,B,C? It is a simple statement so it shouldn't take
>> > more than a few steps. Could someone show it (the formal derivation)
>> > here?
>> If A and B are cardinal numbers, then A+B is defined to be the
>> cardinality of the disjoint union. That is,
>> A+B = | Ax{0} U Bx{1} |,
>> where "x" denotes Cartesian product. Notice that if A and B are any
>> sets, then |Ax{0}| = |A|, |Bx{0}| = |B|, and the sets Ax{0} and Bx{1} are
>> necessarily disjoint, so that the sum of the cardinalities is the
>> cardinality of the union.
>> From the definition it follows that
>> A+(B+C) = A + | Bx{0} U Cx{1} |
>> = | Ax{0} U (Bx{0} U Cx{1})x{1} |
>> = | Ax{0} U (Bx{1} U Cx{2}) |
>> = | (Ax{0} U Bx{1}) U Cx{2} | (using associativity of union)
>Thanks. However, here you do as I predicted: you add the rule for the
>Associativity of Union (which in this context means Addition!) So ZFC
>(alone) did not prove it. As I said, you have to add new rules for
>each branch of Mathematics e.g. Arithmetic.
Associativity of union of disjoint sets is proved
from the theorem of logic that disjunction is associative.
Mathematics does use the first-order predicate calculus
together with its axioms to prove theorems.
>Is there a real proof (using only ZFC)?
>C-B
>> = | (Ax{0} U Bx{1})x{0} U Cx{1} |
>> = | Ax{0} U Bx{1} | + C
>> = (A+B) + C.
>> --
>> Dave Seaman
>> U.S. Court of Appeals to review three issues
>> concerning case of Mumia Abu-Jamal.
>> <http://www.mumia2000.org/>
--
Charlie-Boo
LOL : )
CBT...@gmail.com
Yes. Again, ZFC says which sets exist, but where does it say that
(AUB)UC = AU(BUC) - how do we establish that?
(Answer: With additional axioms relating to union, intersection,
subsets et. al., which parallel Arithmetic.)
C-B
Charlie-Boo
Dave Seaman wrote:
> On 15 Dec 2006 06:27:08 -0800, Charlie-Boo wrote:
> > Many people say that all of Mathematics can be done using ZFC. Ok, how
> > about proving that addition is associative, you know, A+(B+C) = (A+B)+C
> > for all numbers A,B,C? It is a simple statement so it shouldn't take
> > more than a few steps. Could someone show it (the formal derivation)
> > here?
>
> If A and B are cardinal numbers, then A+B is defined to be the
> cardinality of the disjoint union. That is,
>
> A+B = | Ax{0} U Bx{1} |,
>
> where "x" denotes Cartesian product. Notice that if A and B are any
> sets, then |Ax{0}| = |A|, |Bx{0}| = |B|, and the sets Ax{0} and Bx{1} are
> necessarily disjoint, so that the sum of the cardinalities is the
> cardinality of the union.
>
> From the definition it follows that
>
> A+(B+C) = A + | Bx{0} U Cx{1} |
> = | Ax{0} U (Bx{0} U Cx{1})x{1} |
> = | Ax{0} U (Bx{1} U Cx{2}) |
> = | (Ax{0} U Bx{1}) U Cx{2} | (using associativity of union)
Thanks. However, here you do as I predicted: you add the rule for the
Associativity of Union (which in this context means Addition!) So ZFC
(alone) did not prove it. As I said, you have to add new rules for
each branch of Mathematics e.g. Arithmetic.
Is there a real proof (using only ZFC)?
C-B
Dave L. Renfro
See Chapter 4, Theorem 4K(a), p. 81 in Enderton's book.
See Chapter 5, Section 2, Theorem 30, pp. 146-147 in Suppes' book.
Or any number of other elementary set theory texts.
Dave L. Renfro
See Chapter 4, Theorem 4K(a), p. 81 in Enderton's book.
See Chapter 5, Section 2, Theorem 30, pp. 146-147 in Suppes' book.
Or any number of other elementary set theory texts.
Incidentally, the relevant pages of Patrick Suppes' book
are digitized at google's book project. When you get to
Dave L. Renfro
See Chapter 4, Theorem 4K(a), p. 81 in Enderton's book.
See Chapter 5, Section 2, Theorem 30, pp. 146-147 in Suppes' book.
Or any number of other elementary set theory texts.
Incidentally, the relevant pages of Patrick Suppes' book
are available at google-books. When you get to his book
Arturo Magidin
Charlie-Boo <shyma...@gmail.com> wrote:
>Chip Eastham wrote:
>> Charlie-Boo wrote:
>> > Many people say that all of Mathematics can be done using ZFC. Ok, how
>> > about proving that addition is associative, you know, A+(B+C) = (A+B)+C
>> > for all numbers A,B,C? It is a simple statement so it shouldn't take
>> > more than a few steps. Could someone show it (the formal derivation)
>> > here?
>>
>> What "numbers" do you have in mind, Charlie? Formal proof
>> depends on formal definitions. Are we talking natural numbers?
>
>Let's just say natural numbers to keep it simple (and still preserve
>the point.)
And what properties are you allowing as given?
Anyway... We define the function +_n :N --> N using
induction/recursion, by
+_n(0) = n
+_n(s(m)) = s(+_n(m))
where s is the successor function of the natural numbers. This is
interpreted as saying that
0 + n = n
s(m) + n = s(m+n)
and this tells you how to "add n" to any natural number. Using Peano's
fifth postulate, you can verify that this defines a function from N to
N.
Induction is of course an application of Peano's 5th postulate, which
can be justified in ZFC since N is proven to be the smallest inductive
set.
Then we define addition +: N x N --> N by currying:
+(m,n) = +_n(m)
and interpret +(m,n) as "m+n".
First:
LEMMA: For all natural number n, n+0 = n.
Proof: Induction on n. 0+0 = +_0(0) = 0 by definition of +_0,
If k+0 = k, then s(k)+0 = +_0(s(k))
= s(+_0(k)) by definition of +_0
= s(k) (by the induction hypothesis).
QED.
THEOREM: For all natural number m and n, m+n = n+m.
Proof: induction on n. If n =0, then
n+m = 0+m = +_m(0) = m by definition.
m+n = m+0 = m by Lemma.
QED
LEMMA 2: For all natural numbers m and n, m+s(n) = s(m)+n.
Proof: m+s(n) = s(n)+m by Theorem
= s(n+m) by definition
= s(m+n) by Theorem
= s(m)+n by definition of +_n.
QED
THEOREM: For all natural number m, n, k,
(m+n)+k = m+(n+k).
Proof: Induction on k.
(m+n)+0 = +_0(m+n) (by definition)
= m+n (by Lemma)
= +_n(m) by definition.
m+(n+0) = +_(n+0)(m) by definition
= +_n(m) by Lemma (since n+0 = n).
So the result holds for k=0.
Assume the result is true for k and any natural numbers m and n. Then
(m+n)+s(k) = s(m+n)+k (by Lemma 2)
= (s(m)+n) + k (by definition)
= s(m) + (n+k) (by the induction hypothesis)
= m + s(n+k) (by Lemma 2)
= m + (s(n)+k) (by definition)
= m + (n+s(k)) (by Lemma 2).
so the results holds for s(k) if it holds of k.
QED
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================
Arturo Magidin
magidin-at-member-ams-org
Dave Seaman
By using basic properties of logical primitives such as "and" and "or".
For all x, if x in U{A,B,C} <-> (x in U{A,B} or x in C)
<-> (x in A or x in U{B,C}).
> (Answer: With additional axioms relating to union, intersection,
> subsets et. al., which parallel Arithmetic.)
No, I haven't used any axioms relating to union, intersection, or subsets
other than the union axiom of ZFC.
CBT...@gmail.com
Dave L. Renfro wrote:
> Charlie-Boo wrote:
>
> > Many people say that all of Mathematics can be
> > done using ZFC. Ok, how about proving that addition
> > is associative, you know, A+(B+C) = (A+B)+C for
> > all numbers A,B,C? It is a simple statement so
> > it shouldn't take more than a few steps. Could
> > someone show it (the formal derivation) here?
>
> I don't know why you think simple statments must
> have simple proofs.
Not any one, just this one. Because it asks for a formal proof within
a formal system.
> Probably the best known "hard"
> mathematical achievement of the past 20 years is
> a counterexample to this (Fermat's Last Theorem).
That's not a formal proof within a formal system.
The problem is people will make claims and point to publications (and
refer to terms like "elementary" and "introductory") when the
publication is BS or doesn't include what is claimed, then refuse to
simply present their supposed substantiation here so that it can be
critiqued. I don't want any excuses as to why the proof can't be
shown here. (I know the real reason it will never be.)
> Anway ...
>
> See: [1] Chapter 4, Theorem 4K(a), p. 81 in
> Enderton's book; [2] Chapter 5, Section 2, Theorem 30,
> pp. 146-147 in Suppes' book. Or, see any number
> of other elementary set theory texts.
Suppes borrows from PA. Now if you want to claim that ZFC includes PA
in its definition, then I will have to propose a theorem outside of
both Number Theory and the question of which sets exist.
C-B
Arturo Magidin
That may be "an" answer, but it is not the correct answer.
ZFC includes the basic propositional logic, in which the associativity
of the disjunctive "or" holds.
The fact that (AUB)UC = AU(BUC) follows as a theorem from this. From
Union and Separation, and Extension, we get that
X U Y = { x | x in X or x in Y}.
So
(AUB)UC = {x | x in (AUB) or (x in C) }
= {x | ((x in A) or (x in B)) or (x in C)}
= {x | x in A or ((x in B) or (x in C)) } (Extension)
= {x | x in A or (x in BUC) }
= A U (BUC).
CBT...@gmail.com
You have to specify those, otherwise it isn't a formal proof (the set
of theorems isn't r.e.) and you haven't stayed within ZFC. Each time
you try, you wil find yourself needing principles (read: axioms) not
included in ZFC, as we have seen by your last several attempts.
> For all x, if x in U{A,B,C} <-> (x in U{A,B} or x in C)
> <-> (x in A or x in U{B,C}).
(The above is one possible axiom to add to ZFC to be able to prove
Associativity of Addition. You have not shown it to be a formal
theorem.)
Thanks,
C-B
Charlie-Boo
Dave Seaman wrote:
> On 15 Dec 2006 06:27:08 -0800, Charlie-Boo wrote:
> > Many people say that all of Mathematics can be done using ZFC. Ok, how
> > about proving that addition is associative, you know, A+(B+C) = (A+B)+C
> > for all numbers A,B,C? It is a simple statement so it shouldn't take
> > more than a few steps. Could someone show it (the formal derivation)
> > here?
>
> If A and B are cardinal numbers, then A+B is defined to be the
> cardinality of the disjoint union. That is,
>
> A+B = | Ax{0} U Bx{1} |,
>
> where "x" denotes Cartesian product. Notice that if A and B are any
> sets, then |Ax{0}| = |A|, |Bx{0}| = |B|, and the sets Ax{0} and Bx{1} are
> necessarily disjoint, so that the sum of the cardinalities is the
> cardinality of the union.
>
> From the definition it follows that
>
> A+(B+C) = A + | Bx{0} U Cx{1} |
> = | Ax{0} U (Bx{0} U Cx{1})x{1} |
> = | Ax{0} U (Bx{1} U Cx{2}) |
> = | (Ax{0} U Bx{1}) U Cx{2} | (using associativity of union)
Thanks. However, here you do as I predicted: you add the rule for the
Associativity of Union (which in this context means Addition!) So ZFC
(alone) did not prove it. As I said, you have to add new rules for
each branch of Mathematics e.g. Arithmetic.
Is there a real proof (using only ZFC)?
C-B
CBT...@gmail.com
That's right. My question is whether you can do it using ZFC (axioms),
as some claim implicitly.
C-B
CBT...@gmail.com
If you're going to claim that PA is part of ZFC, then I will have to
propose a theorem outside of Number Theory.
C-B
Dave Seaman
See the two lines below. Which part do you not consider as "specified"?
>> For all x, if x in U{A,B,C} <-> (x in U{A,B} or x in C)
>> <-> (x in A or x in U{B,C}).
> (The above is one possible axiom to add to ZFC to be able to prove
> Associativity of Addition. You have not shown it to be a formal
> theorem.)
It's not an axiom of ZFC. It's a principle of logic.
MoeBlee
> Bob Kolker wrote:
> > Charlie-Boo wrote:
> >
> > > Many people say that all of Mathematics can be done using ZFC. Ok, how
> > > about proving that addition is associative, you know, A+(B+C) = (A+B)+C
> > > for all numbers A,B,C? It is a simple statement so it shouldn't take
> > > more than a few steps. Could someone show it (the formal derivation)
> > > here?
> >
> > Showing that addition is associative using just PA takes more than a few
> > steps.
>
> PA I believe in. It describes a specific domain (recursively
> enumerable universal set) and has rules particular to that. In fact,
> it is an example of what I maintain must be done: rules for each
> domain.
Every axiom of PA is a theorem of Z set theory.
Associativity of addition of natural numbers is a theorem of Z set
theory.
Another poster has since mentioned such textbooks as Suppes's and
Enderton's. They both have proofs in Z set theory of the associativity
of addition.
> But how would ZFC do it? It seems to only describe what we should
> allow as sets - nothing about Arithmetic.
Wrong. In Z set theory we can define the basic operations on natural
numbers such as addition and prove basic theorems about addition such
as associativity.
> We can use sets in place of
> the natural numbers, but that is true of a subset of the functions, of
> directed graphs, of any infinite set for which we have a formal
> (finite) representation of each element. But you need special rules to
> go beyond that. ZFC is supposed to be only the rules it already set
> out.
Using only the axioms of Z set theory (which, of course, subsumes first
order logic with identity) we prove the associativity of addition. Of
course, along the way toward that proof, we will have made defintions,
but only as supported by proving, from the axioms of Z set theory
alone, the existence of the defined objects. Anyway, if we didn't mind
longer formulas, we could give the proof of the associativity of
addition of natural numbers strictly in primitive set theoretic
notation.
It's simply a plain stone cold fact that we do prove the associativity
of addition of natural numbers from the axioms of Z set theory alone.
You claim to know the basics of set theory, yet you don't know this
very basic first semester material.
MoeBlee
MoeBlee
> Every axiom of PA is a theorem of Z set theory.
>
> Associativity of addition of natural numbers is a theorem of Z set
> theory.
Just to be clear, I am not saying that associativity of addition is an
axiom of PA.
MoeBlee
MoeBlee
> Thanks. However, here you do as I predicted: you add the rule for the
> Associativity of Union (which in this context means Addition!) So ZFC
> (alone) did not prove it. As I said, you have to add new rules for
> each branch of Mathematics e.g. Arithmetic.
Associativity of union is a theorem of Z set theory. That is covered in
chapter one Set Theory 101. Yet you claim to be familiar with basic set
theory.
MoeBlee
MoeBlee
> Yes. Again, ZFC says which sets exist, but where does it say that
> (AUB)UC = AU(BUC) - how do we establish that?
Easily. Just look at chapter one of just about any basic set theory
textbook. Enderton or Suppes are both fine.
I could type out the proof for you in a blink, but you really do need
to learn chapter one set theory for yourself anyway.
MoeBlee
Mitch
MoeBlee wrote:
> It's simply a plain stone cold fact that we do prove the associativity
> of addition of natural numbers from the axioms of Z set theory alone.
>
So associativity of addition (of suitably defined numbers in ZFC)
depends on the associativity of conjunction in FOL (which is
understood as part of the specification of ZFC). Maybe CB doesn't
think of ZFC as the content full ZFC plus the logic FOL?
Mitch
Michael Stemper
>Many people say that all of Mathematics can be done using ZFC. Ok, how
>about proving that addition is associative, you know, A+(B+C) = (A+B)+C
>for all numbers A,B,C?
I don't know it off the top of my head, but _Axiomatic Set Theory_ by
Suppes has a proof of this -- for naturals. Once you've got that, it's
easy enough to use it to prove associativity for the integers (defined
as an equivalence class of ordered pairs of naturals), and then the
rationals (defined as an equivalence class of ordered pairs of integers),
and then the reals (defined as an equivalence class of Cauchy sequences
of rationals), and finally the complex numbers (defined as an equivalence
class of ordered pairs of reals).
> It is a simple statement so it shouldn't take
>more than a few steps.
As I recall (I'm at work and don't have Suppes with me), it takes about
a page and a half. Of course, FLT and 4CT have simple statements, as
well.
--
Michael F. Stemper
#include <Standard_Disclaimer>
No animals were harmed in the composition of this message.
MoeBlee
> Suppes borrows from PA.
Not in proving associativity of addition of natural numbers. Rather, he
proves that all the axioms of PA are theorems of Z set theory. And he
gives a proof of associativity straight from Z set theory anyway.
> Now if you want to claim that ZFC includes PA
> in its definition, then I will have to propose a theorem outside of
> both Number Theory and the question of which sets exist.
No, in Z we prove the existence of a unique least successor inductive
set, which we define as 'omega'; we prove the existence of the addition
function on omegaXomega; and we prove the associativity of that
function.
That is basic first semester set theory 101. And you don't even know
about it. Wow.
MoeBlee
MoeBlee
> MoeBlee wrote:
> > It's simply a plain stone cold fact that we do prove the associativity
> > of addition of natural numbers from the axioms of Z set theory alone.
> >
>
> So associativity of addition (of suitably defined numbers in ZFC)
> depends on the associativity of conjunction in FOL (which is
> understood as part of the specification of ZFC).
Sentential DISjunction (not conjunction, though it is also a sentential
theorem), I think you meant.
And, as you and I understand, of course Z set theory subsumes
sentential logic. But even then, I don't recall (I'd have to look back
at all the steps) that the other proof that comes from the recursive
definition of addition (as opposed to the definition based on unions of
disjunct sets) uses associativity of disjunction (though it might, as I
mentioned, if we looked back at all steps).
> Maybe CB doesn't
> think of ZFC as the content full ZFC plus the logic FOL?
Maybe. Clearly, he's unfamiliar with even the basic notions.
MoeBlee
Alan Smaill
> Many people say that all of Mathematics can be done using ZFC. Ok, how
> about proving that addition is associative, you know, A+(B+C) = (A+B)+C
> for all numbers A,B,C? It is a simple statement so it shouldn't take
> more than a few steps. Could someone show it (the formal derivation)
> here?
CB says that CBL is a universal system.
How does CBL show that A+(B+C)=(A+B)+C, over the natural numbers?
Be careful not to add any additional ad hoc axioms to the 8
universal rules of inference.
> C-B
>
--
Alan Smaill
Arturo Magidin
I'm not "claiming" something. It is a fact that the "axioms of PA" are
in fact ->theorems<- of ZF.
The axiom of infinity guarantees the existence of a set that contains
{}, and contains the successor of each of its elements.
From such a set A, we can define the set N to be the intersection of
all subsets of A which contain {} and contain the successor of each of
its elements.
It is then a theorem that if B is ANY set that contains {} and
contains the successor of each of its elements, then the intersection
of all subsets of B which contain {} and contain the successor of each
of its elements is in fact equal to N, so that N is well-defined and
does not depend on the choice of A.
Then the "Peano axioms" are theorems of ZF:
1. {} is in N follows by construction;
2. If n is in N, then s(n) is in N follows by construction;
5. If S is a subset of N such that {} is in S, and for every n in N,
if n is in S then s(n) is in S, then S=N follows by construction of N;
3. If n is in N, then s(n) is not equal to {} follows by definition of
s(X) for any set X, since S(X) = X \/ {X}, hence is not empty.
4. If n and m are in N, and s(n)=s(m), then n=m is the only nontrivial
one, but it is still a theorem of Z.
First you prove that for every n in N, if x is in n, then n is not a
subset of x. This can be done without invoking regularity, given the
construction of N, using induction:
Let S be the set of all n in N such that n is not a subset of any of
its elements. Then 0 is in S, since 0 satisfies the
property. Suppose that n is in S; since n is a subset of n, it
follows that n is not an element of n, so s(n)=n \/ {n} is not a
subset of n (since n is an element of s(n) but not of n). If s(n)
were a subset of an element of s(n), then it would therefore be a
subset of an x in n. But n is a subset of s(n), hence n would be a
subset of s(n), which would be a subset of x, which is an element of
n; this contradicts the assumption that n is in S, hence s(n) is not
a subset of any element of x, and is not a subset of n, hence s(n)
is not a subset of any element of s(n). Thus, if n is in S, then
s(n) is in S; By 5, S=N.
Then you prove that elements of N are transitive; that is, if n is in
N, and m is an element of n, then m is a subset of m.
This is also done using 5. Let S be the set of all n in N which are
transitive. Then 0 is in S vacuously. Assume that n is in S. If x is
an element of s(n) = n\/{n}, then either x is in n, or x=n. If x=n,
then x is a subset of s(n), since n is a subset of s(n) by
construction. If x is an element of n, then x is a subset of n since
n is in S, hence x is a subset of n, n is a subset of s(n), and so x
is a subset of s(n). Therefore, s(n) is in S. By 5, S=N, so every
natural number is transitive.
Now we can prove the 4th "Peano Axiom". Suppose that n,m are in N, and
s(n)=s(m). Since n is an element of s(n), then n is an element of
s(m) = m \/ {m}. So either n=m, and we are done, or else n is an
element of m. Symmetrically, since m is in s(m), it is in s(n)=n\/{n},
hence m = n, or else m is an element of n. If n is an element of m
and m is an element of n, then n would be a subset of an element of n,
which we proved was impossible. Therefore, we cannot have both "m is
an element of n" and "n is an element of m". Therefore, we must have
n=m, proving the "4th Peano Axiom" as a theorem of ZF.
So the 5 Peano Axioms are ->theorems<- of ZF.
>then I will have to
>propose a theorem outside of Number Theory.
In other words, you are just looking for a question which is
sufficiently hard to explain easily so you can claim some kind of
pyrrhic victory, rather than trying to actually learn something? I
should have guessed.
Charlie-Boo
You're doing what everyone else is doing: using an axiom from another
domain that parallels Arithmetic. Addition, Union, and Disjunction are
"recursive functions away from each other" (my term.) You have not
shown a formal derivation. (ZFC states what sets exist, not which set
definitions define the same sets!)
Now an aside: How do we abstract all of these into a higher level
(LOA)?
Answer: I call a function a "set function" if the same result is
obtained when it is applied to all elements of a (generally finite) set
in any order. This is easily proved to be associative, commutative,
etc. from this one axiom.
So we need to relate these 3 functions to this new concept.
> So
>
> (AUB)UC = {x | x in (AUB) or (x in C) }
> = {x | ((x in A) or (x in B)) or (x in C)}
> = {x | x in A or ((x in B) or (x in C)) } (Extension)
You're using an Axiom of The Associativity of Disjunction, not ZFC.
(Just like I described at the beginning of this thread.)
C-B
Charlie-Boo
Proof? Just think of models of ZFC in which addition (defined as it is
in ZFC) is NOT associative.
C-B
Arturo Magidin
Oh, well. In that case, you are "an idiot" (my term).
>You have not
>shown a formal derivation. (ZFC states what sets exist, not which set
>definitions define the same sets!)
ZFC not only states what sets exists. It also states a condition under
which sets are "the same." It's called the Axiom of Extension.
ZF states that two sets are the same if they have the same
elements. It also states that given a set, one can use certain types
of expressions to define another set (separation).
Those two already imply that if two expressions which are distinct are
equivalent in first order logic, then the two expressions will yield
the same set (by extension).
But you aren't really asking questions. You are pontificating; and you
are making false claims. Good for you. As Darrow said to Bryan:
--
======================================================================
Bryan: "When you display my ignorance, could you not give me the
facts so I would not be ignorant any longer? [...]"
Darrow: "You know, some of us might get the facts and still be
ignorant."
-- The examination of William Jennings Bryan by Clarence Darrow,
during the Scopes Trial. (From _Attorney for the Damned_,
edited by
Arthur Weinberg)
======================================================================
Arturo Magidin
magidin-at-member-ams-org
Arturo Magidin
Given a set Z, the Axiom of Union guarantees that
{x | exists y (x in y and y in Z)}
is a set.
Given sets X and Y, the Axiom of Pairing guarantees that {X,Y} is a
set. (Other axiomatizations only guarantee the existence of a set A
such that X in A and Y in A; in that case, you use Separation to
obtain the set {X,Y} by using the property "p = X or p=Y").
So from X and Y, we get Z={X,Y}, and from Z we get the set
A = {x | exists y ( x in y and y in {X,Y})}.
From extension you have that "y in {X,Y}" is equivalent to "y=X or
y=Y". And then you have
A = {x | exists y ( x in y and (y=X or y=Y) )}
and from basic first order logic you get that this is the same as
A = {x | exists y ( (x is in y and y=X) or (x is in y and y=Y))}
and again, extension gives you that this is equivalent to
A = {x | x is in X or x is in Y}.
> Just think of models of ZFC in which addition (defined as it is
>in ZFC) is NOT associative.
If addition of cardinals is defined "as it is in ZFC", then it ->is<-
associated. If it is "not associative", then it is not "defined as it
is in ZFC". I would have thought that was obvious. Maybe it's just
because I tried to learn, instead of pontificating on what I think is
true, reality notwithstanding?
--
======================================================================
"I have yet to encounter a problem, no matter how complicated,
that when you look at it from the right angle does not become
still more complicated."
--- Poul Anderson
======================================================================
Arturo Magidin
magidin-at-member-ams-org
MoeBlee
Charlie-Boo wrote:
> You're using an Axiom of The Associativity of Disjunction, not ZFC.
> (Just like I described at the beginning of this thread.)
Association of disjunction is not usually an axiom of sentential logic
but is a theorem of sentential logic.
OF COURSE with formal Z set theory we make whatever use of first order
logic (which itself subsumes sentential logic) we want. That is on PAGE
ONE of many basic set theory textbooks. That you don't understand that
reveals what people have been saying all along - you know virually
nothing in the subject of set theory and mathematical logic. You are
saddled with basic misunderstandings that cause you to have a plainly
incompetent involvement in the subject.
MoeBlee
Charlie-Boo
Huh? You're simply using something not shown to be deducible from
ZFC.
> >> For all x, if x in U{A,B,C} <-> (x in U{A,B} or x in C)
> >> <-> (x in A or x in U{B,C}).
>
> > (The above is one possible axiom to add to ZFC to be able to prove
> > Associativity of Addition. You have not shown it to be a formal
> > theorem.)
>
> It's not an axiom of ZFC.
Therefore you are going outside of ZFC. (Your sentence is implied by
my next to the last sentence above.)
> It's a principle of logic.
Aren't we all?
C-B
Jesse F. Hughes
>> Associativity of union of disjoint sets is proved
>> from the theorem of logic that disjunction is associative.
>
> That's right. My question is whether you can do it using ZFC (axioms),
> as some claim implicitly.
Every proof in ZFC uses the axioms and rules of inference of logic.
And hence, when Herman refers to the above trivial theorem, he is not
going outside ZFC.
Your complaint is really utterly silly.
--
"People make mistakes. Better to live today and learn the truth, than
to be one of those poor saps who died deluded, thinking they knew
certain things that they just didn't. Thinking they had proofs that
they didn't." --James S. Harris, almost too sad for a .sig
MoeBlee
> Therefore you are going outside of ZFC.
Formal ZFC is a FIRST ORDER THEORY. The "ZFC axioms" are those that are
ADDED to "your favorite" axioms and/or inference rules for first order
logic. Anyone who has read the FIRST CHAPTER of many a basic textbook
in set theory understands that. Not you, though.
MoeBlee
Charlie-Boo
Jesse F. Hughes wrote:
> "CBT...@gmail.com" <CBT...@gmail.com> writes:
>
> >> Associativity of union of disjoint sets is proved
> >> from the theorem of logic that disjunction is associative.
> >
> > That's right. My question is whether you can do it using ZFC (axioms),
> > as some claim implicitly.
>
> Every proof in ZFC uses the axioms and rules of inference of logic.
Like I said, the ZFC axioms do not prove anything outside of which sets
exists. So, how many axioms would you like to have? Can you list
them? Then let's consider something in Mathematics outside their
domain and try to prove it.
You have axioms for defining sets and axioms for
Arithmetic/Union/Disjunction. But once you stop adding axioms, I can
tell you a simple theorem you cannot prove - because there are more
than aleph-0 theorems.
C-B
Charlie-Boo
Then it sounds like you (singular or plural) agree with me. You can't
derive essentially all of mathematics from ZFC alone?
C-B
> MoeBlee
lugi...@gmail.com
calculus. Propositional calculus is part of first-order logic. ZFC is
a first-order theory, i.e. first-order logic is part of ZFC. It is for
this reason that one can use the associativity of disjunction in ZFC.
lugi...@gmail.com
lugi...@gmail.com
> Jesse F. Hughes wrote:
> > "CBT...@gmail.com" <CBT...@gmail.com> writes:
> >
> > >> Associativity of union of disjoint sets is proved
> > >> from the theorem of logic that disjunction is associative.
> > >
> > > That's right. My question is whether you can do it using ZFC (axioms),
> > > as some claim implicitly.
> >
> > Every proof in ZFC uses the axioms and rules of inference of logic.
>
> Like I said, the ZFC axioms do not prove anything outside of which sets
> exists. So, how many axioms would you like to have? Can you list
> them? Then let's consider something in Mathematics outside their
> domain and try to prove it.
>
> You have axioms for defining sets and axioms for
> Arithmetic/Union/Disjunction. But once you stop adding axioms, I can
> tell you a simple theorem you cannot prove - because there are more
> than aleph-0 theorems.
>
of them. This is because there are countably many well-formed formulas
(wffs), and the theorems are a proper subset of the wffs.
MoeBlee
I don't want to debate what is "essentially all of mathematics".
But as to associativity of addition of natural numbers, it is a theorem
of Z.
Z set theory is the set of sentences in the language of Z that are
entailed by the axioms of Z. The sentence that is read as expressing
associativity of addition is a sentence entailed by the axioms of Z.
And, for first order, every sentence that is entailed by a set of
axioms is provable from the axioms. Of course, if you know any of the
common definitions of 'entailed' and 'provable' (specifically
'entailed' and 'provable' in the sense of first order), then you see
that first order logic is used for ANY first order theory.
Every sentence that is a theorem of first order logic with a language
of just one 2-place predicate symbol and the identity symbol is also a
theorem of Z set theory. That is true of ANY first order theory with
identity. The set of valid sentences is the least subset of ANY first
order theory.
When people say that a certain sentence is a theorem of a first order
theory, OBVIOUSLY they mean that from the non-logical axioms (such as
the ZFC axioms), using first order logic, the sentence is provable.
Nobody who understands anything about this would claim that the
non-logical axioms of ZFC alone, without axioms and/or inference rules
for first order logic, are sufficient to prove such theorems as
associativity of addition of natural numbers. And only someone who is
virtually clueless about the subject, such as you are, would think that
anyone who says "ZFC proves associativity of addition of natural
numbers" is denying or overlooking or hiding the obvious element that
ZFC subsumes first order logic for the language of set theory.
MoeBlee
MoeBlee
> You have axioms for defining sets and axioms for
> Arithmetic/Union/Disjunction. But once you stop adding axioms, I can
> tell you a simple theorem you cannot prove - because there are more
> than aleph-0 theorems.
That is just a stupid thing to say. A theorem, BY DEFINITION, is a
formula that can be proven. More precisely, a theorem of a set of
sentences is a formula that can be proven from that set of sentences.
And in context of formal first order ZFC that will be first order
provability.
You know virtually nothing about mathematical logic. What a joke you
are.
MoeBlee
Darren
Charlie-Boo wrote:
> Many people say that all of Mathematics can be done using ZFC. Ok, how
> about proving that addition is associative, you know, A+(B+C) = (A+B)+C
> for all numbers A,B,C? It is a simple statement so it shouldn't take
> more than a few steps. Could someone show it (the formal derivation)
> here?
>
The four color problem and FLT are also simple statements (in fact,
many of the important open problems in mathematics are simple
statements). How does the simplicity of a statement imply that it
should have a short proof?
Darren
Jesse F. Hughes
> Jesse F. Hughes wrote:
>>
>> Every proof in ZFC uses the axioms and rules of inference of logic.
>
> Like I said, the ZFC axioms do not prove anything outside of which sets
> exists.
A fairly meaningless statement. ZFC proves that A u B = B u A. Is
that a proof of "which sets exist"? I wouldn't say so.
> So, how many axioms would you like to have? Can you list
> them? Then let's consider something in Mathematics outside their
> domain and try to prove it.
Take any standard axiomatization of FOL and that will do nicely,
thanks. It is silly to pretend anything substantial can be proved
without FOL, since ZFC has *no rules of inference*.
> You have axioms for defining sets and axioms for
> Arithmetic/Union/Disjunction. But once you stop adding axioms, I can
> tell you a simple theorem you cannot prove - because there are more
> than aleph-0 theorems.
More than aleph-0 theorems! *Wow*! That is utterly amazing, since
the language of ZFC has only countably many formulas.
I feel like I have glimpsed God. Tell me more, Charlie! Tell me
more!
--
"And that's what's wrong with Usenet. You people [...] can reply to a
post [...] and then convince yourselves that you're great. Because
you can open your mouths you think what you have is worth saying that
you have proven your value." -- James S. Harris masters self-diagnosis
Charlie-Boo
But axioms and rules of inference would have to be presented for
Propositional calculus, First-order logic, and the Associativity of
Disjunction. Where does ZFC do that?
Answer: It doesn't. You are again adding axioms and rules to ZFC in
order to prove anything outside of the existence of certain sets (the
only question answered by ZFC.)
(In CBL, ZFC is expression x/SE which means to list all wffs which
define sets, which is all ZFC is doing.)
C-B
MoeBlee
> But axioms and rules of inference would have to be presented for
> Propositional calculus, First-order logic, and the Associativity of
> Disjunction. Where does ZFC do that?
We do that when we declare ZFC to be a first order theory. That is in
chapter one of many a basic textbook of set theory. How many times do
you need to be told this?
> Answer: It doesn't. You are again adding axioms and rules to ZFC in
> order to prove anything outside of the existence of certain sets (the
> only question answered by ZFC.)
No, you ignoramus, we add the axioms of ZFC to the axioms and inference
rules of first order logic, not vice versa.
> (In CBL, ZFC is expression x/SE which means to list all wffs which
> define sets, which is all ZFC is doing.)
CBL is the work of an ignoramus as to mathematical logic and set
theory. Granted, it's possible that an ignoramus might produce
something of value, but not bloody likely.
MoeBlee
Charlie-Boo
Ok, then how about proving the Pythagorean Theorem in ZFC? Here's a
link to 72 different proofs of the Pythagorean Theorem:
http://www.cut-the-knot.org/pythagoras/index.shtml
Can ZFC do ANY of them?
What do you really mean by "interpretable in ZFC"? All the above is
doing is to make statements concerning natural numbers but use sets
expressions to represent them rather than 0, 1, 2, 3, . . . or whatever
(Set Theory expressions are the whatever.)
You can do that with directed graphs instead of sets. Or any infinite
r.e. set. What about ZFC makes PA "interpretable"? The ZFC axioms -
are they instrumental in PA being "interpretable"?
Answer: No, the only aspect of Set Theory being used is the fact that
there is a set for each natural number ({n} for natural number n) so
you can use set theory expressions for numbers. That doesn't make the
PA axioms and definitions any part of ZFC. The ZFC axioms only tell us
what sets exist - nothing about Arithmentic or PA in particular.
Got it? :)
G. Frege
>
> Charlie-Boo wrote:
>>
>> But axioms and rules of inference would have to be presented for
>> Propositional calculus, First-order logic, and the Associativity of
>> Disjunction. Where does ZFC do that?
>>
> We do that when we declare ZFC to be a first order theory. That is in
> chapter one of many a basic textbook of set theory. How many times do
> you need to be told this?
>
1, 2, 3, ...
>
> CBL is the work of an ignoramus as to mathematical logic and set
> theory. Granted, it's possible that an ignoramus might produce
> something of value, but not bloody likely.
>
Agree. I guess "CBL" actually means "Charlie-Boo['s] Logic". :-)
F.
--
E-mail: info<at>simple-line<dot>de
Chris Menzel
>> ...
>> Let I be the set {A,B,C}. Then (AUB)UC = AU(BUC) = UI, where the
>> existence of UI is guaranteed by the union axiom of ZFC.
>
> Yes. Again, ZFC says which sets exist, but where does it say that
> (AUB)UC = AU(BUC) - how do we establish that?
>
> (Answer: With additional axioms relating to union, intersection,
> subsets et. al., which parallel Arithmetic.)
Additional axioms? You mean axioms that aren't already in ZFC? There
is indeed an axiom that says, in effect, that the union of any two sets
exists -- though it is expressed solely in terms of membership. The
associativity of union follows almost immediately from the axiom. There
are no axioms "relating" directly to intersection and subset. Facts
about these relations follow from their definitions and the other axioms
of ZFC -- all of which involve only the membership relation.
Charlie-Boo
lugi...@gmail.com wrote:
> Charlie-Boo wrote:
> > Jesse F. Hughes wrote:
> > > "CBT...@gmail.com" <CBT...@gmail.com> writes:
> > >
> > > >> Associativity of union of disjoint sets is proved
> > > >> from the theorem of logic that disjunction is associative.
> > > >
> > > > That's right. My question is whether you can do it using ZFC (axioms),
> > > > as some claim implicitly.
> > >
> > > Every proof in ZFC uses the axioms and rules of inference of logic.
> >
> > Like I said, the ZFC axioms do not prove anything outside of which sets
> > exists. So, how many axioms would you like to have? Can you list
> > them? Then let's consider something in Mathematics outside their
> > domain and try to prove it.
> >
> > You have axioms for defining sets and axioms for
> > Arithmetic/Union/Disjunction. But once you stop adding axioms, I can
> > tell you a simple theorem you cannot prove - because there are more
> > than aleph-0 theorems.
> >
> There are only countable many theorems, and thus there are only aleph-0
> of them. This is because there are countably many well-formed formulas
> (wffs), and the theorems are a proper subset of the wffs.
I am referring to all theorems, not just those generated by the Logic.
Godel proved there are more of them than what the Logic generates.
(See his 1931 article.)
Chris Menzel
This is nothing but a silly straw man. NOBODY has ever claimed you
could derive all of mathematics from ZFC alone that sense -- indeed, the
claim is incoherent, as "derive" implies an underlying logical system.
The claim in question about ZFC is universally understood (except by
you) to mean that all of classical mathematics can be derived from ZFC
in first-order logic.
Charlie-Boo
MoeBlee wrote:
> Charlie-Boo wrote:
> > You have axioms for defining sets and axioms for
> > Arithmetic/Union/Disjunction. But once you stop adding axioms, I can
> > tell you a simple theorem you cannot prove - because there are more
> > than aleph-0 theorems.
>
> That is just a stupid thing to say. A theorem, BY DEFINITION, is a
> formula that can be proven.
1 = The Theorems of Mathematics that people say are all included in
ZFC.
2 = The Theorems of ZFC.
You are confusing 1 and 2.
C-B
Charlie-Boo
Since we refer to formal proofs, it shouldn't take too many
applications of the wff transformations to turn the axioms into the
simple (short) wff that expresses the theorem.
It expresses a rejection of the notion that a proof might exist but be
too long to show.
C-B
> Darren
Charlie-Boo
Jesse F. Hughes wrote:
> "Charlie-Boo" <shyma...@gmail.com> writes:
>
> > Jesse F. Hughes wrote:
> >>
> >> Every proof in ZFC uses the axioms and rules of inference of logic.
> >
> > Like I said, the ZFC axioms do not prove anything outside of which sets
> > exists.
>
> A fairly meaningless statement. ZFC proves that A u B = B u A.
No its doesn't, no more than it proves A u (B u C) = (A u B) u C.
> Is that a proof of "which sets exist"? I wouldn't say so.
No, it is the assertion that there exists a proof of commuttativity of
union in ZFC (which is BS.)
> > So, how many axioms would you like to have? Can you list
> > them? Then let's consider something in Mathematics outside their
> > domain and try to prove it.
>
> Take any standard axiomatization of FOL and that will do nicely,
> thanks. It is silly to pretend anything substantial can be proved
> without FOL, since ZFC has *no rules of inference*.
What do these people mean when they say what I describe in the OP?
C-B
Charlie-Boo
Jesse F. Hughes wrote:
> "Charlie-Boo" <shyma...@gmail.com> writes:
>
> > Jesse F. Hughes wrote:
> >>
> >> Every proof in ZFC uses the axioms and rules of inference of logic.
> >
> > Like I said, the ZFC axioms do not prove anything outside of which sets
> > exists.
>
> A fairly meaningless statement. ZFC proves that A u B = B u A. Is
> that a proof of "which sets exist"? I wouldn't say so.
>
> > So, how many axioms would you like to have? Can you list
> > them? Then let's consider something in Mathematics outside their
> > domain and try to prove it.
>
> Take any standard axiomatization of FOL and that will do nicely,
> thanks. It is silly to pretend anything substantial can be proved
> without FOL, since ZFC has *no rules of inference*.
>
> > You have axioms for defining sets and axioms for
> > Arithmetic/Union/Disjunction. But once you stop adding axioms, I can
> > tell you a simple theorem you cannot prove - because there are more
> > than aleph-0 theorems.
>
> More than aleph-0 theorems! *Wow*! That is utterly amazing, since
> the language of ZFC has only countably many formulas.
>
> I feel like I have glimpsed God. Tell me more, Charlie! Tell me
> more!
The number of theorems of Mathematics is not aleph-0.
C-B
Charlie-Boo
MoeBlee wrote:
> Charlie-Boo wrote:
> > MoeBlee wrote:
> > > Charlie-Boo wrote:
> > > > Therefore you are going outside of ZFC.
> > >
> > > Formal ZFC is a FIRST ORDER THEORY. The "ZFC axioms" are those that are
> > > ADDED to "your favorite" axioms and/or inference rules for first order
> > > logic. Anyone who has read the FIRST CHAPTER of many a basic textbook
> > > in set theory understands that. Not you, though.
> >
> > Then it sounds like you (singular or plural) agree with me. You can't
> > derive essentially all of mathematics from ZFC alone?
>
> I don't want to debate what is "essentially all of mathematics".
You don't have to. The ZFC axioms don't prove anything if you have
to use "your favorite axioms". Then what good is ZFC? These other
sets of axioms and rules worked fine without ZFC. Why add ZFC?
ZFC can prove things about what sets exist [x/SE in CBL]. But when you
go outside of that question, you have to (above):
1. Do it in its own system - e.g. Number Theory in PA.
2. Express the natural numbers as set expressions.
3. Rewrite the operations as being set operations rather than
arithmetic operations.
4. Declare ZFC to axiomatize this area and sometimes even say things
like the following about 7 to 10 axioms given as being ZFC:
"There you have it, the axioms for (essentially) all of mathematics!
If you keep a copy in your wallet, you will carry with you the encoding
for all theorems ever proved and that ever will be proved, from the
most mundane to the most profound."
Do you agree with the above?
> But as to associativity of addition of natural numbers, it is a theorem
> of Z.
>
> Z set theory is the set of sentences in the language of Z that are
> entailed by the axioms of Z. The sentence that is read as expressing
> associativity of addition is a sentence entailed by the axioms of Z.
>
> And, for first order, every sentence that is entailed by a set of
> axioms is provable from the axioms. Of course, if you know any of the
> common definitions of 'entailed' and 'provable' (specifically
> 'entailed' and 'provable' in the sense of first order), then you see
> that first order logic is used for ANY first order theory.
>
> Every sentence that is a theorem of first order logic with a language
> of just one 2-place predicate symbol and the identity symbol is also a
> theorem of Z set theory. That is true of ANY first order theory with
> identity. The set of valid sentences is the least subset of ANY first
> order theory.
>
> When people say that a certain sentence is a theorem of a first order
> theory, OBVIOUSLY they mean that from the non-logical axioms (such as
> the ZFC axioms), using first order logic, the sentence is provable.
>
> Nobody who understands anything about this would claim that the
> non-logical axioms of ZFC alone, without axioms and/or inference rules
> for first order logic, are sufficient to prove such theorems as
> associativity of addition of natural numbers.
Where does ZFC give an axiomatization of first order logic? Someone
has to propose, construct and explain that particular axiomatization.
ZFC gives about 10 axioms concerning what sets exist. That's all.
You are adding axioms and rules not given in ZFC and saying that it is
already in ZFC.
How about proving the Pythagorean Theorem?
It is just another example of nothing having been formalized in
Computer Science outside of almost everything (see 37 branches
elsewhere) in CBL.
C-B
Virgil
"Charlie-Boo" <shyma...@gmail.com> wrote:
> > Yes PA is part of ZFC, or more formally, it is interpretable in ZFC.
>
> Ok, then how about proving the Pythagorean Theorem in ZFC? Here's a
> link to 72 different proofs of the Pythagorean Theorem:
>
> http://www.cut-the-knot.org/pythagoras/index.shtml
>
> Can ZFC do ANY of them?
Probably all of them, but if you do not accept Arturo Magidin's model of
natural number arithmetic, no one is going to bother constructing the
much more complicated model of geometry for you in which to do it.
>
> What do you really mean by "interpretable in ZFC"? All the above is
> doing is to make statements concerning natural numbers but use sets
> expressions to represent them rather than 0, 1, 2, 3, . . . or whatever
> (Set Theory expressions are the whatever.)
If one can identify everything that holds in natural number arithmetic
with something that holds in the same way in ZFC, then one has
represented that arithmetic.
>
> You can do that with directed graphs instead of sets. Or any infinite
> r.e. set. What about ZFC makes PA "interpretable"?
Just lucky, I guess!
> The ZFC axioms -
> are they instrumental in PA being "interpretable"?
Well, there re other axiom systems, Like NBG, for example, which also
work.
>
> Answer: No, the only aspect of Set Theory being used is the fact that
> there is a set for each natural number ({n} for natural number n) so
> you can use set theory expressions for numbers. That doesn't make the
> PA axioms and definitions any part of ZFC. The ZFC axioms only tell us
> what sets exist - nothing about Arithmentic or PA in particular.
So that ZFC can be used for other things as well.
>
> Got it? :)
>
> C-B
The question must be whether the version of arithmetic that is
constructable in ZFC can be shown to be associative within ZFC.
And the short answer is "Yes!"
Arturo Magidin provided a much longer answer to the same effect.
Virgil
Chris Menzel <cme...@remove-this.tamu.edu> wrote:
> On 15 Dec 2006 08:52:06 -0800, CBT...@gmail.com <CBT...@gmail.com> said:
> >> ...
> >> Let I be the set {A,B,C}. Then (AUB)UC = AU(BUC) = UI, where the
> >> existence of UI is guaranteed by the union axiom of ZFC.
> >
> > Yes. Again, ZFC says which sets exist, but where does it say that
> > (AUB)UC = AU(BUC) - how do we establish that?
The common way of establishing the equality of two sets is to show that
every member of each is also a member of the other.
Then by the axiom of extentionality, ...
Charlie-Boo
MoeBlee wrote:
> Charlie-Boo wrote:
> > But axioms and rules of inference would have to be presented for
> > Propositional calculus, First-order logic, and the Associativity of
> > Disjunction. Where does ZFC do that?
>
> We do that when we declare ZFC to be a first order theory.
You can't add to ZFC by declaration. That just proves my assertion
that you are adding to ZFC - you call it "declare".
> That is in
> chapter one of many a basic textbook of set theory. How many times do
> you need to be told this?
>
> > Answer: It doesn't. You are again adding axioms and rules to ZFC in
> > order to prove anything outside of the existence of certain sets (the
> > only question answered by ZFC.)
>
> No, you ignoramus, we add the axioms of ZFC to the axioms and
inference
> rules of first order logic, not vice versa.
But addition is commutative!! Such sweet irony.
> > (In CBL, ZFC is expression x/SE which means to list all wffs which
> > define sets, which is all ZFC is doing.)
>
> CBL is the work of an ignoramus as to mathematical logic and set
> theory.
You are attacking the messenger.
C-B
Chip Eastham
Extreme non sequitur. Your question was about arithmethic,
not geometry.
> What do you really mean by "interpretable in ZFC"? All the above is
> doing is to make statements concerning natural numbers but use sets
> expressions to represent them rather than 0, 1, 2, 3, . . . or whatever
> (Set Theory expressions are the whatever.)
Yes, that's more or less what "PA is interpretable in ZFC" means,
that the formal first-order theory of Peano arithmetic can be given
an interpretation, or model, constructed (from sets) in ZFC.
> You can do that with directed graphs instead of sets. Or any infinite
> r.e. set. What about ZFC makes PA "interpretable"? The ZFC axioms -
> are they instrumental in PA being "interpretable"?
The predicate here is a relationship between the theory of sets
called ZFC and the theory of arithmetic called PA. Both are
formal first order theories in and of themselves. Nobody is
painting a picture that ZFC "makes" PA something that it
otherwise isn't. The relationship is that PA is interpretable
in ZFC. The axioms of ZFC are "instrumental" in proofs of
the satisfiability of the axioms of PA under an interpretation
consisting of a model "internal" to ZFC, that is a construction
of sets. This modeling (or interpretation) can be handled in
a variety of ways. As someone else pointed out to you, one
of the axioms in ZFC is an Axiom of Infinity, which taken with
the machinery of sets gives one a really quick start towards
constructing (a model of) "the" natural numbers.
> Answer: No, the only aspect of Set Theory being used is the fact that
> there is a set for each natural number ({n} for natural number n) so
> you can use set theory expressions for numbers. That doesn't make the
> PA axioms and definitions any part of ZFC. The ZFC axioms only tell us
> what sets exist - nothing about Arithmentic or PA in particular.
You seem here to have no comprehension of the foundations
of mathematics.
> Got it? :)
>
> C-B
If by that you mean, buzz off, yes, I got it.
--chip
Charlie-Boo
No it doesn't. CBL is a Computation Based Logic. You show that you
know little about the subject about which you attempt to speak.
C-B
Jesse F. Hughes
>> I feel like I have glimpsed God. Tell me more, Charlie! Tell me
>> more!
>
> The number of theorems of Mathematics is not aleph-0.
That's not more. That's what you said before.
You're holding out on me, man. You know stuff. I want to know stuff
too. Come on, what is it?
--
"I believe that my job is to go out and explain to the people what's on
my mind. That's why I'm having this press conference, see? I'm telling
you what's on my mind. And what's on my mind is winning the war on
terror." --- George W. Bush
Jesse F. Hughes
> Jesse F. Hughes wrote:
>>
>> Take any standard axiomatization of FOL and that will do nicely,
>> thanks. It is silly to pretend anything substantial can be proved
>> without FOL, since ZFC has *no rules of inference*.
>
> What do these people mean when they say what I describe in the OP?
>
I know you're having problems with this, so I will type it slowly.
ZFC has no rules of inference. Therefore, if we *don't* consider the
axioms and rules of first order logic part of ZFC, then ZFC proves
damn near nothing.
But that's okay. EVERY STANDARD MATHEMATICAL THEORY INCLUDES THE
AXIOMS AND RULES OF FOL. (Which set of axioms and rules? Pick any of
the many, many provably equivalent presentations of FOL. There is no
reason to care which one.)
--
"Run mathematicians, RUN!!! I'm coming for you. It may take a few
months, but I'll get [computer verification of my proof] and then your
lives will be ended as you previously knew it." -- JSH meets PVS
Charlie-Boo
Chris Menzel wrote:
> On 15 Dec 2006 08:52:06 -0800, CBT...@gmail.com <CBT...@gmail.com> said:
> >> ...
> >> Let I be the set {A,B,C}. Then (AUB)UC = AU(BUC) = UI, where the
> >> existence of UI is guaranteed by the union axiom of ZFC.
> >
> > Yes. Again, ZFC says which sets exist, but where does it say that
> > (AUB)UC = AU(BUC) - how do we establish that?
> >
> > (Answer: With additional axioms relating to union, intersection,
> > subsets et. al., which parallel Arithmetic.)
>
> Additional axioms? You mean axioms that aren't already in ZFC? There
> is indeed an axiom that says, in effect, that the union of any two sets
> exists -- though it is expressed solely in terms of membership. The
> associativity of union follows almost immediately from the axiom.
Do it, please.
> There
> are no axioms "relating" directly to intersection and subset. Facts
> about these relations follow from their definitions
It's supposed to come from its axioms, not ad hoc observations. That
is not an axiomatization.
C-B
Virgil
"Charlie-Boo" <shyma...@gmail.com> wrote:
> 1 = The Theorems of Mathematics that people say are all included in
> ZFC.
> 2 = The Theorems of ZFC.
>
> You are confusing 1 and 2.
Your 2 includes your 1, provided you are only listening to people who
know what they are talking about.
If, as appears likely, you are listening to all sorts of people who do
not know what they are talking about, then your notion of 1 may be quite
remote from reality.
Virgil
"Charlie-Boo" <shyma...@gmail.com> wrote:
While I suppose it is theoretically possible to show the entire proof of
the 4 color theorem, I will let CB test its practicality.
Virgil
"Charlie-Boo" <shyma...@gmail.com> wrote:
> Jesse F. Hughes wrote:
> > "Charlie-Boo" <shyma...@gmail.com> writes:
> >
> > > Jesse F. Hughes wrote:
> > >>
> > >> Every proof in ZFC uses the axioms and rules of inference of logic.
> > >
> > > Like I said, the ZFC axioms do not prove anything outside of which sets
> > > exists.
> >
> > A fairly meaningless statement. ZFC proves that A u B = B u A.
>
> No its doesn't, no more than it proves A u (B u C) = (A u B) u C.
It 'proves' those equalities in the sense that one can construct a
formal proof within ZFC of either of them
>
> > Is that a proof of "which sets exist"? I wouldn't say so.
>
> No, it is the assertion that there exists a proof of commuttativity of
> union in ZFC (which is BS.)
Does CB claim that such a proof within ZFC is NOT possible?
Jesse F. Hughes
>> There are only countable many theorems, and thus there are only aleph-0
>> of them. This is because there are countably many well-formed formulas
>> (wffs), and the theorems are a proper subset of the wffs.
>
> I am referring to all theorems, not just those generated by the Logic.
> Godel proved there are more of them than what the Logic generates.
> (See his 1931 article.)
Great. But I guess some of these theorems cannot be expressed even in
any idealized human language, huh? After all, in all the languages we
have, there is a finite alphabet and words and sentences are formed by
finite strings of those alphabet.
And the set of sentences (or paragraphs) expressible in human
languages are countable.
Now, some folks *do* deal with bigger languages in mathematical
logic. It's okay to imagine that we could, say, write down a sentence
that involves the decimal representation of pi in its entirety. But
you're being purposefully misleading when you play these dreamy little
games but forget to tell people.
--
Jesse F. Hughes
"Well, you know as soon as you have a new number I will be happy to
add it to the list. Don't try those childish tit-for-tat games with
me." -- Ross Finlayson on Cantor's theorem.
Virgil
"Charlie-Boo" <shyma...@gmail.com> wrote:
> > Agree. I guess "CBL" actually means "Charlie-Boo['s] Logic". :-)
>
> No it doesn't. CBL is a Computation Based Logic. You show that you
> know little about the subject about which you attempt to speak.
Apparently nobody knows much about CBL,as "Computation Based Logic" has
only 4 Google hits, and they all are dubious about the worth of
Computation Based Logic.
Chris Menzel
>
> Chris Menzel wrote:
>> On 15 Dec 2006 08:52:06 -0800, CBT...@gmail.com <CBT...@gmail.com> said:
>> >> ...
>> >> Let I be the set {A,B,C}. Then (AUB)UC = AU(BUC) = UI, where the
>> >> existence of UI is guaranteed by the union axiom of ZFC.
>> >
>> > Yes. Again, ZFC says which sets exist, but where does it say that
>> > (AUB)UC = AU(BUC) - how do we establish that?
>> >
>> > (Answer: With additional axioms relating to union, intersection,
>> > subsets et. al., which parallel Arithmetic.)
>>
>> Additional axioms? You mean axioms that aren't already in ZFC?
>> There is indeed an axiom that says, in effect, that the union of any
>> two sets exists -- though it is expressed solely in terms of
>> membership. The associativity of union follows almost immediately
>> from the axiom.
>
> Do it, please.
But this has already been done for you in detail. Why are you asking
again? And really, you should give it a go yourself; you just might
learn something about ZFC.
>> There are no axioms "relating" directly to intersection and subset.
>> Facts about these relations follow from their definitions
>
> It's supposed to come from its axioms, not ad hoc observations. That
> is not an axiomatization.
What are you talking about? What ad hoc observations?
aatu.kos...@xortec.fi
> You show that you know little about the subject about which you attempt to speak.
It's good you recognize your shortcomings; it is indeed apparent you
know little about the subject about which you attempt to speak.
Happily, this condition is curable: just go through a few introductory
textbooks.
--
Aatu Koskensilta (aatu.kos...@xortec.fi)
"Wovon man nicht sprechen kann, daruber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
aatu.kos...@xortec.fi
> But once you stop adding axioms, I can tell you a simple theorem you cannot
> prove - because there are more than aleph-0 theorems.
A remarkable discovery! Alas, you're not the first to draw this
startling implication from Gödel's theorems. But perhaps you can show
us a proof of your wonderful result in CBL?
Charlie-Boo
Virgil wrote:
> In article <1166239723.1...@n67g2000cwd.googlegroups.com>,
> "Charlie-Boo" <shyma...@gmail.com> wrote:
> > 1 = The Theorems of Mathematics that people say are all included in
> > ZFC.
> > 2 = The Theorems of ZFC.
> >
> > You are confusing 1 and 2.
>
> Your 2 includes your 1, provided you are only listening to people who
> know what they are talking about.
Are 1 and 2 not well-defined (for our purposes at least)? The phrase
"that people say are all included in ZFC" is meant to be an aside, not
qualifying "The Theorems of Mathematics".
1 = The Theorems of Mathematics
2 = The Theorems of ZFC
2 is said to include 1. I referred to 1 as theorems. Do you think 1
is aleph-0? Is 2?
> If, as appears likely, you are listening to all sorts of people who do
> not know what they are talking about, then your notion of 1 may be quite
> remote from reality.
No. 1 and 2 are self-contained. Their properties are not a function
of this author.
C-B
Charlie-Boo
That subterfuge (obfuscation) is why I chose something as simple as
Associativity of Addition, and mentioned early on that the formal proof
shouldn't be too long - meaning not too long to present. (No excuses.)
C-B
Charlie-Boo
That is my belief, yes. Can you actually show otherwise?
C-B
Charlie-Boo
Jesse F. Hughes wrote:
> "Charlie-Boo" <shyma...@gmail.com> writes:
>
> >> There are only countable many theorems, and thus there are only aleph-0
> >> of them. This is because there are countably many well-formed formulas
> >> (wffs), and the theorems are a proper subset of the wffs.
> >
> > I am referring to all theorems, not just those generated by the Logic.
> > Godel proved there are more of them than what the Logic generates.
> > (See his 1931 article.)
>
> Great. But I guess some of these theorems cannot be expressed even in
> any idealized human language, huh? After all, in all the languages we
> have, there is a finite alphabet and words and sentences are formed by
> finite strings of those alphabet.
There you go. Bingo!
> And the set of sentences (or paragraphs) expressible in human
> languages are countable.
>
> Now, some folks *do* deal with bigger languages in mathematical
> logic. It's okay to imagine that we could, say, write down a sentence
> that involves the decimal representation of pi in its entirety. But
> you're being purposefully misleading when you play these dreamy little
> games but forget to tell people.
How can forgetfulness be on purpose?
C-B
> --
> Jesse F. Hughes
> "Well, you know as soon as you have a new number I will be happy to
> add it to the list. Don't try those childish tit-for-tat games with
> me." -- Ross Finlayson on Cantor's theorem.
Reminds me of the mother of 3 children, tit, tat and toe.
Charlie-Boo
It sounds like you're searching Google Images, or perhaps Google Video.
(Actually, you might find a lecture on CBL in the latter.)
Did you try searching Google Groups? Or Google Scholar?
http://groups.google.com/group/sci.logic/msg/42506a32dbc6afbc?hl=en&
http://groups.google.com/group/sci.logic/msg/0ac686a05f29e973?hl=en&
C-B
Charlie-Boo
Chris Menzel wrote:
> On 15 Dec 2006 20:01:52 -0800, Charlie-Boo <shyma...@gmail.com> said:
> >
> > Chris Menzel wrote:
> >> On 15 Dec 2006 08:52:06 -0800, CBT...@gmail.com <CBT...@gmail.com> said:
> >> >> ...
> >> >> Let I be the set {A,B,C}. Then (AUB)UC = AU(BUC) = UI, where the
> >> >> existence of UI is guaranteed by the union axiom of ZFC.
> >> >
> >> > Yes. Again, ZFC says which sets exist, but where does it say that
> >> > (AUB)UC = AU(BUC) - how do we establish that?
> >> >
> >> > (Answer: With additional axioms relating to union, intersection,
> >> > subsets et. al., which parallel Arithmetic.)
> >>
> >> Additional axioms? You mean axioms that aren't already in ZFC?
> >> There is indeed an axiom that says, in effect, that the union of any
> >> two sets exists -- though it is expressed solely in terms of
> >> membership. The associativity of union follows almost immediately
> >> from the axiom.
> >
> > Do it, please.
>
> But this has already been done for you in detail. Why are you asking
> again? And really, you should give it a go yourself; you just might
> learn something about ZFC.
First you raise your hand and say, "I can do it. I can do it." Then
when I say, "Ok. Let's see it." you lament, "Oh, someone else already
did." Shy in front of the class?
> >> There are no axioms "relating" directly to intersection and subset.
> >> Facts about these relations follow from their definitions
> >
> > It's supposed to come from its axioms, not ad hoc observations. That
> > is not an axiomatization.
>
> What are you talking about? What ad hoc observations?
Facts that follow from their definitions. The axioms are supposed to
capture that formally and once and for all. That's the whole point of
an axiomatization.
C-B
Charlie-Boo
aatu.kos...@xortec.fi wrote:
> Charlie-Boo wrote:
> > You show that you know little about the subject about which you attempt to speak.
>
> It's good you recognize your shortcomings; it is indeed apparent you
> know little about the subject about which you attempt to speak.
> Happily, this condition is curable: just go through a few introductory
> textbooks.
You apparently forget that I am the one who wrote the book on CBL. Do
you doubt that I read the books that I write? (Or maybe I leave that
up to the proofreaders?)
If you can show that ZFC can prove the Associativity of Addition, that
would be useful. (Big IF.)
C-B
Charlie-Boo
> Charlie-Boo wrote:
> > But once you stop adding axioms, I can tell you a simple theorem you cannot
> > prove - because there are more than aleph-0 theorems.
>
> A remarkable discovery! Alas, you're not the first to draw this
> startling implication from Gödel's theorems.
I should introduce you to the people who disagree - though they do
agree with the "startling" part.
But that does show that my observations are publishable grade material.
Now if I could just get up a little earlier in the morning . . .
> But perhaps you can show
> us a proof of your wonderful result in CBL?
Sure - which result? I have posted at least a dozen proofs generated
by CBL (all demonstrably superior to conventional wisdom i.e. what has
been published - as far as my eye can see, and appraently the eyes of
others.)
Are you interested in Incompleteness in Logic? Theory of Computation?
Recursion Theory? Set Theory? Program Synthesis? Did you peruse the
list of the first 37 branches of Computer Science generated by CBL?
http://groups.google.com/group/sci.logic/msg/0ac686a05f29e973?hl=en&
C-B
aatu.kos...@xortec.fi
> You apparently forget that I am the one who wrote the book on CBL.
You were asking about ZFC. I have no doubt you're the leading expert on
CBL.
> If you can show that ZFC can prove the Associativity of Addition, that
> would be useful. (Big IF.)
Just go through any introductory textbook and you'll find the proof.
aatu.kos...@xortec.fi
> aatu.kos...@xortec.fi wrote:
> > But perhaps you can show
> > us a proof of your wonderful result in CBL?
>
> Sure - which result?
Why, the uncountability of the set of theorems, of course!
> I have posted at least a dozen proofs generated
> by CBL (all demonstrably superior to conventional wisdom i.e. what has
> been published - as far as my eye can see, and appraently the eyes of
> others.)
I'm sure you're eminently pleased with CBL. But who are these others
you mention?
Charlie-Boo
aatu.kos...@xortec.fi wrote:
> Charlie-Boo wrote:
> > You apparently forget that I am the one who wrote the book on CBL.
>
> You were asking about ZFC. I have no doubt you're the leading expert on
> CBL.
>
> > If you can show that ZFC can prove the Associativity of Addition, that
> > would be useful. (Big IF.)
>
> Just go through any introductory textbook and you'll find the proof.
Unsubstantiated fantasy.
C-B
aatu.kos...@xortec.fi
> Unsubstantiated fantasy.
It is commendable you so clearly understand the nature of your bizarre
claims. Still, why not have a look at some decent introductory textbook
on the subject?
Jesse F. Hughes
> aatu.kos...@xortec.fi wrote:
>> Charlie-Boo wrote:
>> > If you can show that ZFC can prove the Associativity of Addition, that
>> > would be useful. (Big IF.)
>>
>> Just go through any introductory textbook and you'll find the proof.
>
> Unsubstantiated fantasy.
The proof for ordinal addition is given on page 52 of Takeuti and
Zaring's "Introduction to Axiomatic Set Theory". It's a simple
transfinite induction.
Note: on page 5 they explicitly give the axiomatization of FOL that
they're using. Because everyone aside from Charlie understands that
ZFC is an FOL theory.
--
Jesse F. Hughes
"If the world weren't rather strange, by now I should at least be with
some research group talking about my number theory research."
-- James S. Harris learns the world is a funny place
abo
aatu.kos...@xortec.fi wrote:
> Charlie-Boo wrote:
> > Unsubstantiated fantasy.
>
> It is commendable you so clearly understand the nature of your bizarre
> claims.
>
May I ask if your body has been taken over by Torkel's ghost? I don't
recall you ever using this gambit - purposefully misunderstanding the
poster and repeating his declaration as if it applied to him - until a
few months ago, and now you recently seem to use it on several
occasions. This was one of Torkel's favorite moves - indeed one which
he proudly declared on his web site about how to be a pest on the
internet, to which I took a rather great dislike. I wouldn't mention
this except that you have also acquired other intonations which I
usually associated with him. Is he channelling through you?
Mitch
MoeBlee wrote:
> Mitch wrote:
> > MoeBlee wrote:
> > > It's simply a plain stone cold fact that we do prove the associativity
> > > of addition of natural numbers from the axioms of Z set theory alone.
> > >
> >
> > So associativity of addition (of suitably defined numbers in ZFC)
> > depends on the associativity of conjunction in FOL (which is
> > understood as part of the specification of ZFC).
>
> Sentential DISjunction (not conjunction, though it is also a sentential
> theorem), I think you meant.
Gahhh! Yes. Worse than left and right.
> And, as you and I understand, of course Z set theory subsumes
> sentential logic. But even then, I don't recall (I'd have to look back
> at all the steps) that the other proof that comes from the recursive
> definition of addition (as opposed to the definition based on unions of
> disjunct sets) uses associativity of disjunction (though it might, as I
> mentioned, if we looked back at all steps).
Hmmm....yes. arturo's proof in PA doesn't seem to use associativity of
disjunction.
But the other more explicitly ZFC proof (Dave Seaman) explicitly
involves associativity of disjunction. Necessarily? Necessarily as a
proof step? Well, without thinking too hard I'd guess if 'or' were
-not- associative then a whole bunch of things would not happen
(including a whole bunch of set theory.
But that point leads to an answer for CB: if you didn't have first
order logical rules as a basis upon which the ZFC axioms are added,
then really you just can't do anything -at all- in ZFC. No gas in the
engine. or really no carburetor.
As an aside, I distinctly remember that commutativity of addition is
necessarily inductive. Is associativity also?
Mitch
David C. Ullrich
wrote:
>
>Charlie-Boo wrote:
>>[...]
>
>You seem here to have no comprehension of the foundations
>of mathematics.
That wasn't clear from the title of the thread?
>> Got it? :)
>>
>> C-B
>
>If by that you mean, buzz off, yes, I got it.
>
>--chip
************************
David C. Ullrich