\filename{null.tex}
\edited{last by stolfi on Fri May 29 21:45:15 1987}

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% \showboxbreadth=100
% \showboxdepth=1

\begintitles

    \title{Test}

    \shorttitle{Test}

    \titlerule

    \author{Fester Bestertester$\,{}^{\ast}$}

    \titlerule

    \support{\noindent
      ${}^{\ast}$ MAD Software Services, Inc..
    } 

\endtitles

       Without the vskip parsep in rfig:
      
      \setbox0\vbox{
        Without loss of generality, we can assume $x'<x$.  See figure
        \Fig{6}.  On the sweep line, $b''$ lies strictly between $b$ and
        $b'$.  On the sweep line, $b''$ lies strictly between $b$ and
        $b'$.

        \rfig 0.5cm by 8.0cm{}{\centered Figure \here\Fig{6}.}{\strut
          Without loss of generality, we can assume $x'<x$.  See figure
          \Fig{6}.  On the sweep line, $b''$ lies strictly between $b$ and
          $b'$. On the sweep line, $b''$ lies strictly between $b$ and
          $b'$.
        }

        \rfig 0.5cm by 8.0cm{}{\centered Figure \here\Fig{6}.}{\strut
          Without loss of generality, we can assume $x'<x$.  See figure
          \Fig{6}.  On the sweep line, $b''$ lies strictly between $b$ and
          $b'$.
        }
      }
      \tracebox{\box0}

      \setbox0\vbox{
        \rfig 0.5cm by 8.0cm{}{\centered Figure \here\Fig{6}.}{\strut
          Without loss of generality, we can assume $x'<x$.  See figure
          \Fig{6}.  On the sweep line, $b''$ lies strictly between $b$ and
          $b'$. On the sweep line, $b''$ lies strictly between $b$ and
          $b'$.
        }

        \rfig 0.5cm by 8.0cm{}{\centered Figure \here\Fig{6}.}{\strut
          Without loss of generality, we can assume $x'<x$.  See figure
          \Fig{6}.  On the sweep line, $b''$ lies strictly between $b$ and
          $b'$. On the sweep line, $b''$ lies strictly between $b$ and
          $b'$.
        }
      }
      \tracebox{\box0}

      \setbox0\vbox{
        \rfig 0.5cm by 8.0cm{}{\centered Figure \here\Fig{6}.}{\strut
          Without loss of generality, we can assume $x'<x$.  See figure
          \Fig{6}.  On the sweep line, $b''$ lies strictly between $b$ and
          $b'$.
        }
      Without loss of generality, we can assume $x'<x$.  See figure
      \Fig{6}.  On the sweep line, $b''$ lies strictly between $b$ and
      $b'$.
      }
      \tracebox{\box0}

      Without loss of generality, we can assume $x'<x$.  See figure
      \Fig{6}.  On the sweep line, $b''$ lies strictly between $b$ and
      $b'$.  What can it do between $x$ and that line?  It cannot cross
      $b$ or $b'$, since they are in the same list.  It cannot have its
      left endpoint there, otherwise it would sit either in a cone
      determined by $b$ and $a$ or by $a$ and $b'$, violating the
      invariant.  Therefore, $b''$ crosses the vertical line at $x$
      between $a$ and $b'$.  Since it cannot cross $b'$, and cannot have
      started between $a$ and $b'$, it must cross $a$ somewhere between
      $x'$ and $x$.
      
      \rfig 2cm by 6.5cm{}{\centered Figure \here\Fig{6}.}{\strut
        Without loss of generality, we can assume $x'<x$.  See figure
        \Fig{6}.  On the sweep line, $b''$ lies strictly between $b$ and
        $b'$.  What can it do between $x$ and that line?  It cannot cross
        $b$ or $b'$, since they are in the same list.  It cannot have its
        left endpoint there, otherwise it would sit either in a cone
        determined by $b$ and $a$ or by $a$ and $b'$, violating the
        invariant.  Therefore, $b''$ crosses the vertical line at $x$
        between $a$ and $b'$.  Since it cannot cross $b'$, and cannot have
        started between $a$ and $b'$, it must cross $a$ somewhere between
        $x'$ and $x$.
      }

      Without loss of generality, we can assume $x'<x$.  See figure
      \Fig{6}.  On the sweep line, $b''$ lies strictly between $b$ and
      $b'$.  What can it do between $x$ and that line?  It cannot cross
      $b$ or $b'$, since they are in the same list.  It cannot have its
      left endpoint there, otherwise it would sit either in a cone
      determined by $b$ and $a$ or by $a$ and $b'$, violating the
      invariant.  Therefore, $b''$ crosses the vertical line at $x$
      between $a$ and $b'$.  Since it cannot cross $b'$, and cannot have
      started between $a$ and $b'$, it must cross $a$ somewhere between
      $x'$ and $x$.
      
      Without loss of generality, we can assume $x'<x$.  See figure
      \Fig{6}.  On the sweep line, $b''$ lies strictly between $b$ and
      $b'$.  What can it do between $x$ and that line?  It cannot cross
      $b$ or $b'$, since they are in the same list.  It cannot have its
      left endpoint there, otherwise it would sit either in a cone
      determined by $b$ and $a$ or by $a$ and $b'$, violating the
      invariant.  Therefore, $b''$ crosses the vertical line at $x$
      between $a$ and $b'$.  Since it cannot cross $b'$, and cannot have
      started between $a$ and $b'$, it must cross $a$ somewhere between
      $x'$ and $x$.
      
      \rfig 2cm by 6.5cm{}{\centered Figure \here\Fig{6}.}{\strut
        Without loss of generality, we can assume $x'<x$.  See figure
        \Fig{6}.  On the sweep line, $b''$ lies strictly between $b$ and
        $b'$.  What can it do between $x$ and that line?  It cannot cross
        $b$ or $b'$, since they are in the same list.  It cannot have its
        left endpoint there, otherwise it would sit either in a cone
        determined by $b$ and $a$ or by $a$ and $b'$, violating the
        invariant.  Therefore, $b''$ crosses the vertical line at $x$
        between $a$ and $b'$.  Since it cannot cross $b'$, and cannot have
        started between $a$ and $b'$, it must cross $a$ somewhere between
        $x'$ and $x$.
      }

      \rfig 2cm by 6.5cm{}{\centered Figure \here\Fig{6}.}{\strut
        Without loss of generality, we can assume $x'<x$.  See figure
        \Fig{6}.  On the sweep line, $b''$ lies strictly between $b$ and
        $b'$.  What can it do between $x$ and that line?  It cannot cross
        $b$ or $b'$, since they are in the same list.  It cannot have its
        left endpoint there, otherwise it would sit either in a cone
        determined by $b$ and $a$ or by $a$ and $b'$, violating the
        invariant.  Therefore, $b''$ crosses the vertical line at $x$
        between $a$ and $b'$.  Since it cannot cross $b'$, and cannot have
        started between $a$ and $b'$, it must cross $a$ somewhere between
        $x'$ and $x$.
      }

      Without loss of generality, we can assume $x'<x$.  See figure
      \Fig{6}.  On the sweep line, $b''$ lies strictly between $b$ and
      $b'$.  What can it do between $x$ and that line?  It cannot cross
      $b$ or $b'$, since they are in the same list. 
      \rfig 2cm by 6.5cm{}{\centered Figure \here\Fig{6}.}{\strut
        Without loss of generality, we can assume $x'<x$.  See figure
        \Fig{6}.  On the sweep line, $b''$ lies strictly between $b$ and
        $b'$.  What can it do between $x$ and that line?  It cannot cross
        $b$ or $b'$, since they are in the same list.  It cannot have its
        left endpoint there, otherwise it would sit either in a cone
        determined by $b$ and $a$ or by $a$ and $b'$, violating the
        invariant.  Therefore, $b''$ crosses the vertical line at $x$
        between $a$ and $b'$.  Since it cannot cross $b'$, and cannot have
        started between $a$ and $b'$, it must cross $a$ somewhere between
        $x'$ and $x$.
      }
      It cannot have its left endpoint there, otherwise it would sit
      either in a cone determined by $b$ and $a$ or by $a$ and $b'$,
      violating the invariant.  Therefore, $b''$ crosses the vertical line
      at $x$ between $a$ and $b'$.  Since it cannot cross $b'$, and cannot
      have started between $a$ and $b'$, it must cross $a$ somewhere
      between $x'$ and $x$.  
      
  \subsection{Foo}
  
      \rfig 2cm by 6.5cm{}{\centered Figure \here\Fig{6}.}{\strut
        Without loss of generality, we can assume $x'<x$.  See figure
        \Fig{6}.  On the sweep line, $b''$ lies strictly between $b$ and
        $b'$.  What can it do between $x$ and that line?  It cannot cross
        $b$ or $b'$, since they are in the same list.  It cannot have its
        left endpoint there, otherwise it would sit either in a cone
        determined by $b$ and $a$ or by $a$ and $b'$, violating the
        invariant.  Therefore, $b''$ crosses the vertical line at $x$
        between $a$ and $b'$.  Since it cannot cross $b'$, and cannot have
        started between $a$ and $b'$, it must cross $a$ somewhere between
        $x'$ and $x$.
      }

      \rfig 2cm by 6.5cm{}{\centered Figure \here\Fig{6}.}{\strut
        Without loss of generality, we can assume $x'<x$.  See figure
        \Fig{6}.  On the sweep line, $b''$ lies strictly between $b$ and
        $b'$.  What can it do between $x$ and that line?  It cannot cross
        $b$ or $b'$, since they are in the same list.  It cannot have its
        left endpoint there, otherwise it would sit either in a cone
        determined by $b$ and $a$ or by $a$ and $b'$, violating the
        invariant.  Therefore, $b''$ crosses the vertical line at $x$
        between $a$ and $b'$.  Since it cannot cross $b'$, and cannot have
        started between $a$ and $b'$, it must cross $a$ somewhere between
        $x'$ and $x$.
      }
  

\vfill\bye

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