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Topic: Conclusion
Replies: 2   Last Post: Nov 7, 2012 9:51 AM

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Dan Christensen

Posts: 2,796
Registered: 7/9/08
Re: Conclusion
Posted: Nov 7, 2012 9:51 AM
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On Nov 7, 9:28 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
> > On Nov 7, 6:53 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> >> Dan Christensen <Dan_Christen...@sympatico.ca> writes:
> >> > On Nov 6, 6:18 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> >> >> You didn't say that.  You said earlier today that composition was not
> >> >> functional.

>
> >> > [snip]
>
> >> > Pay attention, Jesse. In every version of my definition of category,
> >> > composition was presented as a function. For every ordered pair of
> >> > compatible morphisms, there would exist a unique morphism that was
> >> > their composition. After posting my latest version, I began to openly
> >> > question it here, wondering if indeed such a unique morphism always
> >> > existed. It doesn't always, but, as Aatu pointed out in effect, you
> >> > arbitrarily pick one of the alternatives for your definition of
> >> > compositions to maintain functionality -- "We flip a coin." So, my
> >> > latest definition of stood.

>
> >> Wow.  You are either a liar or have a mental condition that represses
> >> all memory of your errors.

>
> >> Look at message https://groups.google.com/group/sci.math/msg/87d7499de8ad28de?dmode=s...
> >> Message-ID: <60c69813-35f2-4bc7-8fb2-f81b8e0ea...@h9g2000yqd.googlegroups.com>

>
> >> Here, we find axiom 2:
>
> >> 2  ALL(f):ALL(g):ALL(h):[f @ mor & g @ mor & h @ mor => [(f,g,h) @
> >> comp
> >>    <=> cod(f)=dom(g) & dom(f)=dom(h) & cod(h)=cod(g)]]

>
> > A little creative editing, Jesse? In the text just prior to that (see
> > link), I wrote:

>
> > "I also think there may be a larger problem with the functionality of
> > composition. Suppose, for example, that f is morphism from object A to
> > object B, that g is morphism from B to C, and that h1 and h2 are
> > distinct morphisms from A to C. Then comp(g,f) as defined here could
> > be either h1 or h2, could it not? Should we define some equivalence
> > relation on mor. Should comp be seen as a non-functional mapping from
> > mor x mor to mor? How about something like..."

>
> What the fuck are you on about?  Axiom 2 stated above is clearly about a
> non-functional relation called comp.  This excerpt supports that claim.
> You gave an explicit axiom in which composition was non-functional.
>


Poor lad...


> > Aatu, it seems, kicked your ill-informed butt with his "We flip a
> > coin" remark. You no longer claim I was "lying" about that. That's
> > real progress. But here, it seems you are reduced to deliberately
> > taking my words out of context and, again, calling me a liar!

>
> Aatu kicked my ill-informed butt?



Apparently.

> I think not.  Why not ask him if
> that's what he thinks?
>
> I think that his comment is utterly misleading (though, I have not
> doublechecked the context, so I am prepared to eat those words).



A very healthy attitude, Jesse! Try https://groups.google.com/group/sci.logic/msg/9c0fbf4e60780668

> We do
> not "flip a coin" to define composition.



Again, that was his reply to my question, "What to do when there are
multiple, distinct morphisms from which to choose the
composition?" (see link above)

> There are two distinct
> categories involving the objects and morphisms you describe.  They are
> distinct because they have different rules of composition (in fact, I
> recall Aatu said something like this, so coin-flipping seems mighty odd
> to me).  But I'll let Aatu explain his use of that term if he cares to.
>
> Why not ask Aatu if he thinks that he and I have any disagreement on the
> role of composition in defining a category?
>
> As far as taking your words out of context, you said:
>
>   In every version of my definition of category, composition was
>   presented as a function.
>
> Axiom 2 is a version of your definition of category in which composition
> was explicitly non-functional.
>



Pay attention, Jesse. Once again, this was just part of my question
that you so unwisely snipped in your vain attempt to discredit me.


> > In your desperate, never-ending attempts to make me look stupid, it
> > seems to have backfired on you, Jesse.

>
> Oh, yeah.  It's mighty embarrassing for me, I can tell you.
>


Very good! Now, that didn't hurt so much, did it?

Dan
Download my DC Proof 2.0 software at http://www.dcproof.com


Date Subject Author
11/7/12
Read Re: Conclusion
Jesse F. Hughes
11/7/12
Read Re: Conclusion
Dan Christensen

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