\relax 
\@writefile{lof}{\contentsline {figure}{\numberline {1}{\ignorespaces The case when $i_1 \not = j_1$. We show only the job $i_1$ in $A$. Black dots indicate intermediate jobs. (a) Can the finish time of $i_1$ be larger than the finish time of $j_1$ (potentially causing $i_1$ to conflict with $j_2$)? (b) No! The reason is that $i_1$ is the first job selected by the EFT algorithm. Hence, its finish time must be less than or equal to the finish time of $j_1$. (c) Therefore, if we replace $j_1$ with $i_1$ in $O$, all the jobs in $O$ continue to be mutually compatible.}}{1}\protected@file@percent }
\newlabel{fig:i1-ji}{{1}{1}}
\@writefile{toc}{\contentsline {paragraph}{Case 1: $i_1 \not = j_1$.}{1}\protected@file@percent }
\@writefile{lof}{\contentsline {figure}{\numberline {2}{\ignorespaces The case when $i_p \not = j_p$, for some $p > 1$. We show only the jobs $i_{p-1}$ and $i_p$ in $A$. Black dots indicate earlier, intermediate, or later jobs. (a) Can the finish time of $i_p$ be larger than the finish time of $j_p$ (potentially causing $i_p$ and $j_{p+1}$ to conflict)? (b) No! The reason is that both $i_p$ and $j_p$ start after $i_{p-1}$ finishes. Therefore, after the EFT algorithm has selected $i_{p-1}$ and included it in $A$, both $i_p$ and $j_p$ (which are compatible with $i_{p-1}$) were available for being chosen as the next job in $A$, However, the EFT algorithm selected the job $i_p$. Hence, its finish time must be less than or equal to the finish time of $j_p$. (c) Therefore, if we replace $j_p$ with $i_p$ in $O$, all the jobs in $O$ continue to be mutually compatible.}}{2}\protected@file@percent }
\newlabel{fig:ip-jp}{{2}{2}}
\@writefile{toc}{\contentsline {paragraph}{Case 2: $i_p \not = j_p$, for some $1 < p \leq k$.}{2}\protected@file@percent }
\@writefile{toc}{\contentsline {paragraph}{Case 3: $i_p = j_p$ for all $1 \leq p \leq k$ but $m > k$.}{3}\protected@file@percent }