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Re: [AUCTeX] Adding \colon to LaTeX-math-default
From: |
Plamen Tanovski |
Subject: |
Re: [AUCTeX] Adding \colon to LaTeX-math-default |
Date: |
Tue, 5 Apr 2005 23:05:52 +0200 |
User-agent: |
Mutt/1.4i |
Ralf Angeli wrote:
> * Plamen Tanovski (2005-04-03) writes:
>
> > is it reasonable to add \colon to the LaTeX-math-default list and
> > bound it to (surprise!) `?:'?
>
> As nobody else seems to have an opinion here and it looks useful to me
> (even though I barely use math myself), I installed it. Thanks for
> the suggestion. Addtionally I cleaned up a few things in the math
> menu.
Thanks. AucTeX math mode is really a good thing, but I think for a
(real big) math typesetting, one need either a second keyboard or
using more short-cuts and keys like the Windos keys and the numeric
keypad on a typical PC-keyboard. Typping the following (on a german
keyboard) is quite slow, even with AucTeX.
Best regards,
P.
and therefore $x_0=\Sigma y_i'\in C_0^\bot$. Since
$x_0=t_1t_2^{-1}(0)$, we have proved that $t_1t_2^{-1}\in T^{+}$, as
required.
Let $t\in T^{++}$ and let $g\in KtK\cap KtU^-$. Then
by~\ref{lemma:2.2.1} as before we have $g^{-1}\in
P_{(C'_0)}t^{-1}K\cap Kt^{-1}K$ for some translate~$C_0'$ of the
cone~$-C_0$, and hence $g_1=g^{-1}t\in P_{(C_0)}P_{(x)}\cap
P_{(0)}P_(x)$, where $x=t^{-1}(0)$. Now $x\in -\bar{C}_0$, and
$-\bar{C}_0$ intersects~$C_0'$. Hence there exists $y\in C_0'$ such
that $x$ lies in the segment~$[0y]$.
Let $g_1(x)=x'\in\mathcal{P}$. Since $g_1\in pP_{(x)}$ for some $p\in
P_{(0)=K}$, we have $x'=p(x)$ and $p(0)=0$.