Can ZFC prove Addition is Associative?

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Charlie-Boo

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Dec 15, 2006, 9:27:08 AM12/15/06
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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?

C-B

DmitryKovtun

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Dec 15, 2006, 9:36:16 AM12/15/06
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"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

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Dec 15, 2006, 9:48:21 AM12/15/06
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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

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Dec 15, 2006, 9:50:41 AM12/15/06
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Charlie-Boo wrote:

Showing that addition is associative using just PA takes more than a few
steps.

Bob Kolker

Charlie-Boo

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Dec 15, 2006, 9:55:08 AM12/15/06
to
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.)

> 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

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Dec 15, 2006, 10:01:45 AM12/15/06
to

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

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Dec 15, 2006, 10:03:10 AM12/15/06
to

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

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Dec 15, 2006, 10:23:30 AM12/15/06
to

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

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Dec 15, 2006, 11:05:14 AM12/15/06
to

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

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Dec 15, 2006, 11:20:15 AM12/15/06
to

>> = | (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

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Dec 15, 2006, 11:26:11 AM12/15/06
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LOL : )

Herman Rubin

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Dec 15, 2006, 11:33:58 AM12/15/06
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In article <4ufr24F...@mid.individual.net>,

>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

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Dec 15, 2006, 11:43:44 AM12/15/06
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Charlie-Boo wrote:

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

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Dec 15, 2006, 11:45:13 AM12/15/06
to
In article <1166198714.7...@73g2000cwn.googlegroups.com>,
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

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Dec 15, 2006, 11:46:29 AM12/15/06
to

LOL : )

CBT...@gmail.com

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Dec 15, 2006, 11:52:06 AM12/15/06
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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

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Dec 15, 2006, 12:02:13 PM12/15/06
to

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

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Dec 15, 2006, 12:04:10 PM12/15/06
to
Charlie-Boo wrote:

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

unread,
Dec 15, 2006, 12:05:37 PM12/15/06
to
Charlie-Boo wrote:

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

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Dec 15, 2006, 12:06:57 PM12/15/06
to
Charlie-Boo wrote:

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

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Dec 15, 2006, 12:12:37 PM12/15/06
to
In article <1166194508.3...@t46g2000cwa.googlegroups.com>,

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

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Dec 15, 2006, 12:12:46 PM12/15/06
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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

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Dec 15, 2006, 12:14:06 PM12/15/06
to

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

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Dec 15, 2006, 12:19:10 PM12/15/06
to
In article <1166201526....@73g2000cwn.googlegroups.com>,

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

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Dec 15, 2006, 12:21:31 PM12/15/06
to

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

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Dec 15, 2006, 12:32:11 PM12/15/06
to

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

unread,
Dec 15, 2006, 12:33:39 PM12/15/06
to

That's right. My question is whether you can do it using ZFC (axioms),
as some claim implicitly.

C-B

CBT...@gmail.com

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Dec 15, 2006, 12:36:22 PM12/15/06
to

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

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Dec 15, 2006, 12:51:06 PM12/15/06
to

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

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Dec 15, 2006, 1:09:20 PM12/15/06
to
Charlie-Boo wrote:
> 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

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Dec 15, 2006, 1:11:36 PM12/15/06
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MoeBlee wrote:
> 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

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Dec 15, 2006, 1:15:57 PM12/15/06
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Charlie-Boo wrote:
> 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

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Dec 15, 2006, 1:19:07 PM12/15/06
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CBT...@gmail.com wrote:
> 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

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Dec 15, 2006, 1:22:11 PM12/15/06
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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 th