Ordering Binary Relations (publishing) - sci.math

Message from discussion Ordering Binary Relations (publishing)

William Elliot

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More options Sep 19 2008, 6:43 am

Newsgroups: sci.math

From: William Elliot <ma...@hevanet.remove.com>

Date: Fri, 19 Sep 2008 03:43:12 -0700

Local: Fri, Sep 19 2008 6:43 am

Subject: Re: Ordering Binary Relations (publishing)

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On Fri, 19 Sep 2008, Victor Porton wrote:
> On Sep 18, 11:58 am, William Elliot <ma...@hevanet.remove.com> wrote:
> > I have found an astonishingly simple and direct way of proving the major
> > lemma of that paper.

> > First I show for all f,g
> > x in dom f /\ dom g ==> f*(x) x g*(x) subset gf^-1

> > Hence immediately
> > x in dom f /\ dom g ==> f*(x) x g*(x) /\ gf^-1 = f*(x) x g*(x)

> > Consequently
> > ff^-1 = gf^-1 ==> for all x in dom f /\ dom g, f*(x) = g*(x)

> > which shows
> > ff^-1 = gf^-1, dom f subset dom g ==> for all x in dom f, f*(x) = g*(x)
> > and finally
> > ff^-1 = gf^-1, dom f subset dom g ==> f <= g

> "f*(x) x g*(x) subset gf^-1"
> What you mean by "x"?

The cross product
. . AxB = { (x,y) | x in A, y in B }

However, if you read my other recent post, you will see that I retracked
that proof. Is this a counter example to the lemma? N is positive
integers.

f = { (n,m) in NxN | n <= m }
g = NxN; f subset g; dom f = dom g

ff^-1 = NxN = gf^-1

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