[pdftex] destination with the same identifier

[Aleks Kleyn]

I creates short text which creates for me problem during compilation.

 

\documentclass{amsbook}

\usepackage[unicode]{hyperref}

\renewcommand\theHchapter{E.\arabic{chapter}}%

\numberwithin{section}{chapter}

\newtheorem{theorem}{Theorem}[section]

\makeatletter

\newcommand\StartLabelItem[1][theorem]%

{

%\counterwithin{enumi}{#1}%

\expandafter \def \csname%

theenumi\expandafter \endcsname \expandafter {\csname the#1\expandafter%

\endcsname.\expandafter \@arabic \csname c at enumi\endcsname }%

\def\labelenumi{\theenumi:}%

}

\usepackage{paracol}

\globalcounter{theorem}

\globalcounter{enumi}

\makeatother

 

\begin{document}

\chapter{C One}

\section{S One}

\begin{paracol}{2}

\begin{theorem}

\label{theorem one}

abc

\StartLabelItem

\begin{enumerate}

\item

the one

\label{item one}

\item

the two

\label{item two}

\end{enumerate}

\end{theorem}

\switchcolumn

\begin{theorem}

\label{theorem two}

abc

\StartLabelItem

\begin{enumerate}

\item

the one

\label{item one a}

\item

the two

\label{item two a}

\end{enumerate}

\end{theorem}

\end{paracol}

\end{document}

 

When I compile this text, I get error 

pdfTeX warning (ext4): destination with the same identifier (name{Item.1}) has 

been already used, duplicate ignored

My first impression was that report of error was mistake because all references work properly. But when I looked in aux file I saw following

\newlabel{theorem one}{{1.1.1}{1}{}{theorem.E.1.1.1}{}}

\newlabel{item one}{{1.1.1.1}{1}{}{Item.1}{}}

\newlabel{item two}{{1.1.1.2}{1}{}{Item.2}{}}

\newlabel{theorem two}{{1.1.2}{1}{}{theorem.E.1.1.2}{}}

\newlabel{item one a}{{1.1.2.1}{1}{}{Item.1}{}}

\newlabel{item two a}{{1.1.2.2}{1}{}{Item.2}{}}

 

In my original code instead of item.1 and item.2 was reference to number of theorem, and it postponed to later code which made harder to find source of problem. So to fix the problem, I need to make sure that last 2 lines has item.3 and item.4

 

The question is how I can do it

 

Thank you

 

Aleks Kleyn


 <http://alekskleyn.dyndns-home.com:4080/> http://alekskleyn.dyndns-home.com:4080/  

 <http://sites.google.com/site/alekskleyn/> http://sites.google.com/site/AleksKleyn/ 

 <http://arxiv.org/a/kleyn_a_1> http://arxiv.org/a/kleyn_a_1  

 <http://AleksKleyn.blogspot.com/> http://AleksKleyn.blogspot.com/

 <http://kleynaleks.blogspot.com/> http://KleynAleks.blogspot.com/

 

Derivative in Banach algebra


dx2

=1⊗ x+x⊗1


dx

  d x2 =(1⊗ x+x⊗1) ◦dx=dx x+x dx

 

 

 

 

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