Algebra Question

[David C. Ullrich]

Say P is a structure with an addition and
multiplication satisfying

(i) x+y = y+x
(ii) x+(y+z) = (x+y)+z
(iii) If x+z = y+z then x = y.
(iv) (P, multiplication) is an abelian group
(v) x(y+z) = xy+xz.

I gather that although the definitions are not
quite standard, this might be called a semi-ring
plus cancellation.

Now suppose that w + x' = w' + x and y + z' = y' + z.
Does it follow that

(*) wz + xy + x'y' + w'z' = w'z' + x'y' + xy + wz ?

If we had subtraction then this would be clear,
since (*) just says that (w-x)(y-z) = (w'-x')(y'-z')
and we're given that w-x = w'-x' and y-z = y'-z'
(btw if (*) is _not_ the same as (w-x)(y-z) = (w'-x')(y'-z')
then it's a typo in (*)). Seems like the cancellation
property (iii) should be a substitute for subtraction,
but I've gone around in circles with no luck.

(If anyone's curious, this has to do with showing
that the obvious definition of multiplication is
well-defined if we apply the Grothendiek construction
to try to get a ring from the semi-ring. The question
doesn't really matter, since in the context I have
in mind there are more axioms that help).

David C. Ullrich.

[Bastian Erdnuess]

David C Ullrich wrote:

> Say P is a structure with an addition and
> multiplication satisfying
>
>
> (i) x+y = y+x
> (ii) x+(y+z) = (x+y)+z
> (iii) If x+z = y+z then x = y.
> (iv) (P, multiplication) is an abelian group
> (v) x(y+z) = xy+xz.
>
> I gather that although the definitions are not
> quite standard, this might be called a semi-ring
> plus cancellation.
>
> Now suppose that w + x' = w' + x and y + z' = y' + z.
> Does it follow that
>
> (*) wz + xy + x'y' + w'z' = w'z' + x'y' + xy + wz ?
>
> If we had subtraction then this would be clear,
> since (*) just says that (w-x)(y-z) = (w'-x')(y'-z')
> and we're given that w-x = w'-x' and y-z = y'-z'
> (btw if (*) is _not_ the same as (w-x)(y-z) = (w'-x')(y'-z')
> then it's a typo in (*)).

So (*) should be

xz + yw + x'y' + z'w' = x'z' + y'w' + xy + zw ?

Add w'z and x'y on both sides. Factor for y and z on the left side and
for x' and w' on the right side. Replace all four terms in the
parentheses by the appropriate formula. Factor everything out and
compare.

Cheers,
Bastian

[Rob Johnson]

In article <r6cf16hkg4tv8a8dh...@4ax.com>,

David C. Ullrich <ull...@math.okstate.edu> wrote:
>Say P is a structure with an addition and
>multiplication satisfying
>
>
>(i) x+y = y+x
>(ii) x+(y+z) = (x+y)+z
>(iii) If x+z = y+z then x = y.
>(iv) (P, multiplication) is an abelian group
>(v) x(y+z) = xy+xz.
>
>I gather that although the definitions are not
>quite standard, this might be called a semi-ring
>plus cancellation.
>
>Now suppose that w + x' = w' + x and y + z' = y' + z.
>Does it follow that
>
>(*) wz + xy + x'y' + w'z' = w'z' + x'y' + xy + wz ?

Are you sure this is the equation you want to show? Perhaps I am
missing something, but it appears that the way this equation is
written, it can be shown simply using (i) and (ii).

However the grouping is assumed in (*), we can use (ii) to transform
the left-hand side of (*) to

wz + (xy + (x'y' + w'z'))

= (xy + (x'y' + w'z')) + wz using (i)

= ((x'y' + w'z') + xy) + wz using (i)

= ((w'z' + x'y') + xy) + wz using (i)

Now, we can use (ii) again to get the right-hand side of (*); for
example, to get the grouping the same as it was in the first line:

((w'z' + x'y') + xy) + wz

= (w'z' + x'y') + (xy + wz) using (ii)

= w'z' + (x'y' + (xy + wz)) using (ii)

>If we had subtraction then this would be clear,
>since (*) just says that (w-x)(y-z) = (w'-x')(y'-z')
>and we're given that w-x = w'-x' and y-z = y'-z'
>(btw if (*) is _not_ the same as (w-x)(y-z) = (w'-x')(y'-z')
>then it's a typo in (*)). Seems like the cancellation
>property (iii) should be a substitute for subtraction,
>but I've gone around in circles with no luck.
>
>(If anyone's curious, this has to do with showing
>that the obvious definition of multiplication is
>well-defined if we apply the Grothendiek construction
>to try to get a ring from the semi-ring. The question
>doesn't really matter, since in the context I have
>in mind there are more axioms that help).

Rob Johnson <r...@trash.whim.org>
take out the trash before replying
to view any ASCII art, display article in a monospaced font

[Bill Dubuque]

David C. Ullrich <ull...@math.okstate.edu> wrote:
>

> Say P is a structure with an addition and
> multiplication satisfying
>
> (i) x+y = y+x
> (ii) x+(y+z) = (x+y)+z
> (iii) If x+z = y+z then x = y.
> (iv) (P, multiplication) is an abelian group
> (v) x(y+z) = xy+xz.
>
> I gather that although the definitions are not
> quite standard, this might be called a semi-ring
> plus cancellation.
>

> Now suppose that w + X = W + x and y + Z = Y + z.
> Does it follow that
>
> (*) wy + xz + XY + WZ = WY + XZ + xy + wz [correcyed -wgd]

>
> If we had subtraction then this would be clear,

> since (*) just says that (w-x)(y-z) = (W-X)(Y-Z)
> and we're given that w-x = W-X and y-z = Y-Z
> (btw if (*) is _not_ the same as (w-x)(y-z) = (W-X)(Y-Z)

> then it's a typo in (*)). Seems like the cancellation
> property (iii) should be a substitute for subtraction,
> but I've gone around in circles with no luck.
>
> (If anyone's curious, this has to do with showing
> that the obvious definition of multiplication is
> well-defined if we apply the Grothendiek construction
> to try to get a ring from the semi-ring. The question
> doesn't really matter, since in the context I have
> in mind there are more axioms that help).

HINT wy+xz+xY+wZ = wY+xZ+xy+wz ie. (w-x)(y-z) = (w-x)(Y-Z)

+ wY+xZ+XY+WZ = WY+XZ+xY+wZ ie. (w-x)(Y-Z) = (W-X)(Y-Z)

=> wy+xz+XY+WZ + xY+wZ+wY+xZ = WY+XZ+xy+wz + xY+wZ+wY+xZ

=> wy+xz+XY+WZ = WY+XZ+xy+wz ie. (w-x)(y-z) = (W-X)(Y-Z)

That the summands hold true is easy, e.g.

wy+xz+xY+wZ = wY+xZ+xy+wz

<=> w(y+Z) + x(z+Y)
= w(Y+z) + x(Z+y)

--Bill Dubuque

[Rob Johnson]

In article <r6cf16hkg4tv8a8dh...@4ax.com>,

David C. Ullrich <ull...@math.okstate.edu> wrote:

Now that I have seen the reply from Bastian Erdnuess, I understand
that the question is really to show that

wy + xz + w'z' + x'y' = w'y' + x'z' + wz + xy (*)

After having worked this through, I realize that it is the samw as
Bill Dubuque's answer, but I will post it anyway since a second way
of looking at something is sometimes helpful.

(wy + xz + w'z' + x'y') + (x'y + w'z + w'y + x'z)

= (wy + x'y) + (xz + w'z) + (w'z' + w'y) + (x'y' + x'z)

= (w'y + xy) + (x'z + wz) + (w'z + w'y') + (x'y + x'z')

= (xy + wz + w'y' + x'z') + (w'y + x'z + w'z + x'y)

Now we just cancel the right summand from the far-left and far-right
sides to get (*):

wy + xz + w'z' + x'y' = xy + wz + w'y' + x'z'

[David C. Ullrich]

Thanks to evervone who helped. Sorry about the
typo (I was interrupted by Something Important
just as I was about to check things over) - glad
I included the explanation of where the problem
came from so you could figure out the actual question!

If you're curious how I could be so dense: The
basic idea, add something to both sides so you
can factor and then use the given identities, was
clear. But I was studiply overlooking the symmetry:
I added things to the left side, factored and
replaced, expecting to get something that looked
like the right side. Dumb - I should have realized
that similar manipulations on the right side would
also be needed.

Like I said, it was trivial from other axioms that
were going to be introduce anyway. The real
project is to characterize the set of positive
elements of an ordered field (no hints please,
I want to work it out myself). So: I could get
from semiring to ring using some ordered0fieldish
assumptions, but it seemed like it _should_ follow
just from the semiringish assumptions. It does.
Much nicer, thanks.

(Why? Because constructing the _positive_
reals using Dedekind cuts (sets of positive
rationals) works out much more elegantly
than constructing the reals using Dedekind
cuts; negative numbbers introduce unfortunate
complications in multiplication. No fair working
this out and publishing it, btw... you heard it
here first.)

[Rob Johnson]

In article <2010061...@whim.org>,

I started thinking about this as being a 4 component catalyst, where
x'y -> w'y -> w'z -> x'z -> x'y and realized that this could work
with one cancelling term and four steps instead of four cancelling
terms and one step:

(wy + xz + w'z' + x'y') + x'y

= (wy + x'y) + xz + w'z' + x'y'

= (w'y + xy) + xz + w'z' + x'y' [w + x' = w' + x]

= xy + xz + (w'z' + w'y) + x'y'

= xy + xz + (w'z + w'y') + x'y' [z' + y = w + y']

= xy + (xz + w'z) + w'y' + x'y'

= xy + (x'z + wz) + w'y' + x'y' [x + w' = x' + w]

= xy + wz + w'y' + (x'y' + x'z)

= xy + wz + w'y' + (x'y + x'z') [y' + z = y + z']

= (xy + wz + w'y' + x'z') + x'y

Cancel the x'y to get (*).

[Kaba]

> Rob Johnson <r...@trash.whim.org>
> take out the trash before replying
> to view any ASCII art, display article in a monospaced font

Hi Rob,

This is off-topic for the thread, but here goes. You wrote some time ago
the following text:

http://www.whim.org/nebula/math/conformalregress.html

It's a very nice one. What I was wondering though, is whether you have
some kind of software to produce the ascii art in your texts. Or do you
do it by hand?

--
http://kaba.hilvi.org

[Rob Johnson]

In article <MPG.2683350f9...@news.cc.tut.fi>,

I put it together by hand. It's not too difficult after a bit of
practice.

[Bastian Erdnuess]

Kaba wrote:

> This is off-topic for the thread, but here goes. You wrote some time ago
> the following text:
>
> http://www.whim.org/nebula/math/conformalregress.html
>
> It's a very nice one. What I was wondering though, is whether you have
> some kind of software to produce the ascii art in your texts. Or do you
> do it by hand?

The commando line versions of some math tools like Maple can do this.
Not that good for formulas but for a lot of other stuff is JavE
(www.jave.de).

Cheers,
Bastian

[Kaba]

Rob Johnson wrote:
> >http://www.whim.org/nebula/math/conformalregress.html
> >
> >It's a very nice one. What I was wondering though, is whether you have
> >some kind of software to produce the ascii art in your texts. Or do you
> >do it by hand?
>
> I put it together by hand. It's not too difficult after a bit of
> practice.

Ok:)

--
http://kaba.hilvi.org

[Kaba]

Bastian Erdnuess wrote:
> The commando line versions of some math tools like Maple can do this.

Yep..

> Not that good for formulas but for a lot of other stuff is JavE
> (www.jave.de).

Nice program that JavE.

--
http://kaba.hilvi.org

[Bill Dubuque]

By removing the motivational remarks from my proof, and by inlining
the lemmas, you've completely obfuscated the essence of the matter.
In case it was not clear, my proof is just a special case of the
well-known proof that congruences can be multiplied, e.g.

LEMMA a = A, b = B (mod n) => ab = AB (mod n)

PROOF n|a-A,b-B => n | (a-A)b + A(b-B) = ab - AB

As I mentioned in [1], this well-known identity also lies at
the foundation of the product rule for limits, derivatives,
differences, etc. Once one masters this basic idea it is a
purely mechanical task to translate it into other contexts,
just as I did in this thread. But stripping all this structure
from the proof - as you did above - leaves you pulling the proof
out of a hat. Why did you choose those particular expressions
to add together? Why does it work? Good proofs provide insight
not simply verification of a binary truth value.

Recall that I've stressed this point to you many times here
e.g. [2] when you inlined invocations of Bezout's Lemma into
an obfuscated proof of the Rational Root Test. As I said there:

Of course one can always "inline" lemmas, abstractions...to produce
completely "elementary" versions of theorems. A well-known case of
such is the completely self-contained proof of unique factorization
of integers by Klappauf, Lindemann, Zermelo [3]. This inlines the
Euclidean/Division algorithm descent into a direct inductive proof.

By eliminating all the internal structure supporting the argument
such proofs tremendously obfuscate the real essence of the matter.
In fact it was the discovery of such hidden structure (esp. ideals
and modules) in elementary number theory that led Dedekind to
invent most of the major classical algebraic structures. Without
the introduction of such abstraction there would be little hope
of conquering the complexity inherent in modern number theory

--Bill Dubuque

[1] sci.math, Mar 10 2004, Proof that Multiplication Modulo n is Associative?
http://groups.google.com/group/sci.math/msg/0d66427464e5b7a7
http://google.com/groups?selm=y8zbrn4xzfz.fsf%40nestle.ai.mit.edu

[2] sci.math, May 18, 2009, Irrationality of sqrt (n^2 - 1)
http://groups.google.com/group/sci.math/msg/5ce760473f5d4399
http://google.com/groups?selm=y8z3ab1hnm4.fsf%40nestle.csail.mit.edu

[3] http://www.math.uiuc.edu/~dan/ShortProofs/uf.html

[Rob Johnson]

In article <l2caaqu...@shaggy.csail.mit.edu>,

I removed nothing because I wrote my article based on the typo noted
by Bastian Erdnuess. I am sorry that you seem to find the need to
deprecate my work, but I was just trying to answer David's question.
I was not trying to detract from your post; in fact, I had not even
read your post until just before I posted, at which point, I added a
reference to your article. I am sorry that my post offended you so;
however, as soon as I saw that you had posted, I expected to get a
lashing for my post.

Obviously we have different approaches, and you find yours far
superior to mine.

Thanks for writing. I appreciate all opinions, even those which
find my articles so offensive.

[Gerry Myerson]

In article <g3nh16hp0mr5g3f0a...@4ax.com>,

David C. Ullrich <ull...@math.okstate.edu> wrote:

> (Why? Because constructing the _positive_
> reals using Dedekind cuts (sets of positive
> rationals) works out much more elegantly
> than constructing the reals using Dedekind

> cuts; negative numbers introduce unfortunate

> complications in multiplication. No fair working
> this out and publishing it, btw... you heard it
> here first.)

No, I read it (or something very much like it) first on pages 25-26 of
Conway's On Numbers And Games. He says the best path from the positive
integers to the reals is via the positive rationals and then the
positive reals, and the reason he gives is precisely that if you go
from the rationals to the reals by Dedekind cuts you have to make lots
of special cases depending on signs when you get to the multiplication.

--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)

[Bill Dubuque]

>>LEMMA a = A, b = B (mod n) => ab = AB (mod n)
>>
>>PROOF n|a-A,b-B => n | (a-A)b + A(b-B) = ab - AB
>

> I removed nothing because I wrote my article based on the typo noted

> by Bastian Erdnuess [...] in fact, I had not even read your post

> until just before I posted

Ok, then I'm quite interested in the motivation behind your proof.
Why did you choose those particular expressions? Why add them, etc?
As you said, a second way of looking at it may be helpful, so I'm
interested in understanding the key idea(s) of your "second way".

[David C. Ullrich]

Drat. Thanks for the information.

How'd he manage to steal my idea years before I had it?

[Bill Dubuque]

Gerry Myerson <ge...@maths.mq.edi.ai.i2u4email> wrote:
> David C. Ullrich <ull...@math.okstate.edu> wrote:
>>
>> (Why? Because constructing the _positive_
>> reals using Dedekind cuts (sets of positive
>> rationals) works out much more elegantly
>> than constructing the reals using Dedekind
>> cuts; negative numbers introduce unfortunate
>> complications in multiplication. No fair working
>> this out and publishing it, btw... you heard it
>> here first.)
>
> No, I read it (or something very much like it) first on pages 25-26 of
> Conway's On Numbers And Games. He says the best path from the positive
> integers to the reals is via the positive rationals and then the
> positive reals, and the reason he gives is precisely that if you go
> from the rationals to the reals by Dedekind cuts you have to make lots
> of special cases depending on signs when you get to the multiplication.

It's an interesting (though tedious) exercise to prove that the two
different ways yield isomorphic structures. I recall seeing the
details worked out in a textbook, but the title escapes me right now.

--Bill Dubuque

[master1729]

David Ullrich wrote :

since your a genius he must have a time machine.

[Rob Johnson]

In article <l2ck4px...@shaggy.csail.mit.edu>,

As I mentioned in a follow-up message, I had viewed this problem as
a catalytic process; that is, add something that effects a change but
itself remains unchanged. Then we can remove the catalyst using the
cancellation property (iii). In the equations above, I tried to
exhibit this process using parentheses. The four partial processes
(annotated in the follow-up) are given by

(wy + x'y) = (w + x')y = (w' + x)y = (w'y + xy) [1]

Equation [1] converts wy to xy while rotating catalyst x'y to w'y

(xz + w'z) = (x + w')z = (x' + w)z = (x'z + wz) [2]

Equation [2] converts xz to wz while rotating catalyst w'z to x'z

(w'z' + w'y) = w'(z' + y) = w'(z + y') = (w'z + w'y') [3]

Equation [3] converts w'z' to w'y' while rotating catalyst w'y to w'z

(x'y' + x'z) = x'(y' + z) = x'(y + z') = (x'y + x'z') [4]

Equation [4] converts x'y' to x'z' while rotating catalyst x'z to x'y

Combining these 4 equations, we convert wy + xz + w'z' + x'y' to
xy + wz + w'y' + x'z' while rotating the catalysts as follows
x'y -> w'y -> w'z -> x'z -> x'y. Therefore, the full catalyst,
x'y + w'z + w'y + x'z, remains unchanged.

Thus,

(wy + xz + w'z' + x'y') + (x'y + w'z + w'y + x'z) [5]

match elements from each group to get

= (wy + x'y) + (xz + w'z) + (w'z' + w'y) + (x'y' + x'z) [6]

use the catalytic processes described above to get

= (w'y + xy) + (x'z + wz) + (w'z + w'y') + (x'y + x'z') [7]

separate left and right elements from each group to get

= (xy + wz + w'y' + x'z') + (w'y + x'z + w'z + x'y) [8]

To get [6], use the commutativity and associativity of addition.
To get [7], use [1]-[4].
To get [8], use the commutativity and associativity of addition.

Those were the key ideas in what I had done.

[Bastian Erdnuess]

Rob Johnson wrote:

>>>>>> Now suppose that w + x' = w' + x and y + z' = y' + z.

>>>>> wy + xz + w'z' + x'y' = w'y' + x'z' + wz + xy (*)

> As I mentioned in a follow-up message, I had viewed this problem as

> a catalytic process; that is, add something that effects a change but
> itself remains unchanged. Then we can remove the catalyst using the
> cancellation property (iii).

That is, you do

(w y + x'y' + x z + w'z') + w'y
= w y + x'y' + x z + (w'z' + w'y )
= w y + x'y' + (x z + w'z ) + w'y'
= w y + (x'y' + x'z ) + w z + w'y'
= (w y + x'y ) + x'z' + w z + w'y'
= w'y + (x y + x'z' + w z + w'y')

while the "cat" cycles w'y -> w'z -> x'z -> x'y -> w'y and thus cancels.

Nice.

Cheers,
Bastian

[Rob Johnson]

In article <slrni1l58f....@wh36-e604.wh36.uni-karlsruhe.de>,

Indeed; that is the idea of the follow-up message I posted

<http://groups.google.com/group/sci.math/msg/6a226feb3c89597e>

>Nice.

Thanks.

[David C. Ullrich]

On 17 Jun 2010 15:04:47 -0400, Bill Dubuque <w...@nestle.csail.mit.edu>
wrote:

Of course they're both complete ordered fields (the wonderful thing
about Dedekind cuts is that completeness is almost obvious;
seems to me the most tedious thing is defining multiplication and
showing it works right with the original construction). Showing
that any two complete ordered fields are isomorphic doesn't
seem so hard or tedious, not that I've ever seen or written
down every detail. I've been intending to write down every
detail sometime soon, so just for practice:

[about ten minutes pass]

Ok, never mind. Its straightforward but a little tedious.

>--Bill Dubuque

[Transfer Principle]

On Jun 16, 5:32 pm, Gerry Myerson <ge...@maths.mq.edi.ai.i2u4email>
wrote:
> In article <g3nh16hp0mr5g3f0apdqnfq38b83750...@4ax.com>,

>  David C. Ullrich <ullr...@math.okstate.edu> wrote:
> > (Why? Because constructing the _positive_
> > reals using Dedekind cuts (sets of positive
> > rationals) works out much more elegantly
> > than constructing the reals using Dedekind
> > cuts; negative numbers introduce unfortunate
> > complications in multiplication. No fair working
> > this out and publishing it, btw... you heard it
> > here first.)
> No, I read it (or something very much like it) first on pages 25-26 of
> Conway's On Numbers And Games. He says the best path from the positive
> integers to the reals is via the positive rationals and then the
> positive reals, and the reason he gives is precisely that if you go
> from the rationals to the reals by Dedekind cuts you have to make lots
> of special cases depending on signs when you get to the multiplication.

Believe it or not, even _I_ have heard of constructing
the reals via D-cuts of _positive_ rationals rather
than all the rationals.

And where have I heard it from? Metamath (a computer
theorem prover created by Norm Megill). Megill writes:

"To construct the complex numbers, we start with the
finite ordinals (natural numbers) of set theory and
successively build temporary positive integers,
temporary positive rationals, temporary positive reals
(based on Dedekind cuts), temporary signed reals, and
finally the actual complex numbers."

Not only is multiplication easier to define on the
positive D-cuts rather than signed D-cuts, but it's
easier to define positive _rationals_ as equivalence
classes of ordered pairs of positive _naturals_, then
have to include special cases to ensure that the second
elements (denominator) of the rationals aren't zero. So
Megill avoids the problems with negatives and the
problem with zero until after the positive D-cuts.

(Of course, if we were to use Cauchy sequences instead
of D-cuts, we can at least avoid the problem of
defining multiplication for negative reals.)

[Bill Dubuque]

> [...]

But of course there are obvious symmetries that can be exploited
to streamline the proof. But that does not answer my questions.
Why did you choose those particular expression to start with?
Why did you add them? As-is your proof is pulled out of a hat
with no motivation at all for its genesis. That is why I had
assumed that you had started with one of the earlier posted
proofs when you mentioned that you saw them before posting.

When you said you had a "second way" I thought you meant
that you had a second conceptual way of viewing the proof,
not merely a syntactic rearrangement gotten by stripping
the motivational remarks and inlining the lemmas. Why
such obfuscations appeal to you continues to puzzle me,
just as it has in our many earlier exchanges on such.

> take out the trash before replying

I've tried everything, even teaching you "green" proof methods,
with energy-saving reusable lemmas, ROHS-compliant construction
materials, and insightful abstractions that promote optimal reuse.
However, you continue to generate more trash than I can take out.
Where did you learn those transfinite trash generation techniques?

[Rob Johnson]

In article <l2csk4j...@shaggy.csail.mit.edu>,

I didn't realize that David Ullrich needed a motivational proof; I
had the impression that he was simply looking for a justification
using the hypotheses he supplied. However, I did not pull my proof
from a hat. I started with the equations that convert terms from
wy + xz + w'z' + x'y' to terms in xy + wz + w'y' + x'z'.

Thus, I looked at the following equations (derived as in [1] - [4]
above):

(wy + x'y) = (w'y + xy) [converts wy to xy]

(xz + w'z) = (x'z + wz) [converts xz to wz]

(w'z' + w'y) = (w'z + w'y') [converts w'z' to w'y']

(x'y' + x'z) = (x'y + x'z') [converts x'y' to x'z']

These were the motivation for the following equations in my article

> = (wy + x'y) + (xz + w'z) + (w'z' + w'y) + (x'y' + x'z)
>
> = (w'y + xy) + (x'z + wz) + (w'z + w'y') + (x'y + x'z')

I figured that if the equation to be proved was true, the catalytic
terms would cancel.

>When you said you had a "second way" I thought you meant
>that you had a second conceptual way of viewing the proof,
>not merely a syntactic rearrangement gotten by stripping
>the motivational remarks and inlining the lemmas. Why
>such obfuscations appeal to you continues to puzzle me,
>just as it has in our many earlier exchanges on such.

If you won't take my word that I didn't just rearrange your proof and
remove your "motivational remarks", I don't know what I can say that
would change your mind. When I first saw your hint, it was not
immediately evident to me how the two hypotheses (w + x' = w' + x and
y + z' = y' + z) were used. That is why I decided to post what I had
done even though you had already posted your hint.

Of course, after a little thought, I saw that what you had done was
work from the motivational comment that David made regarding
subtraction (w-x = W-X and y-z = Y-Z). However, it still takes a bit
of work to get

> wy+xz+xY+wZ = wY+xZ+xy+wz ie. (w-x)(y-z) = (w-x)(Y-Z)

and

> wY+xZ+XY+WZ = WY+XZ+xY+wZ ie. (w-x)(Y-Z) = (W-X)(Y-Z)

(after changing primes to capitals) from the two hypotheses

w + X = W + x and y + Z = Y + z

I based my article (in particular, identities [1]-[4] above) on those
two hypotheses, not on your hint.

>> take out the trash before replying
>
>I've tried everything, even teaching you "green" proof methods,
>with energy-saving reusable lemmas, ROHS-compliant construction
>materials, and insightful abstractions that promote optimal reuse.
>However, you continue to generate more trash than I can take out.
>Where did you learn those transfinite trash generation techniques?

If undergraduate education at UCLA and graduate education at Princeton
have not forced me to prove things the way you like, then don't feel
bad that your educational efforts on usenet have failed. I guess I
am just a lost cause.

Rob Johnson <r...@trash.whim.org>

take out the trash before replying

[master1729]

timothy golden and myself have proposed that too.