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    <h2>Tarski Formalization Project Archives</h2>
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    <h3>Reflection in a point, midpoint of the base of an isosceles triangle, and the Krippenlemma</h3>
    <p></p>
    <p>
    The reflection of $a$ in $c$ is the unique point $x$ such that $T(a,c,x)$ and $ac = xc$.  It is
    written $x=s(c,a)$.  This chapter presents results from Gupta's thesis that are used in Chapter 8
    to help prove his famous results about the existence of perpendiculars.  Remember that in this chapter,
    we are not using the parallel axiom, any form of continuity (including line-circle and circle-circle continuity),
    and not even the upper dimension axiom.
     </p>
      
 

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      <tr>
        <td><a href="InputFiles/Satz7.2.in">Satz 7.2</a></td>
        <td><a href="Proofs/Satz7.2.prf">Satz7.2.prf</a><br>4 steps</td>
        <td> M(a,m,b) -> M(b,m,a)   </td>
      </tr>
      <tr>
        <td><a href="InputFiles/Satz7.3a.in">Satz 7.3a</a> </td>
        <td><a href="Proofs/Satz7.3a.prf">Satz7.3a.prf</a><br>2 steps</td>
        <td> M(a,m,b) -> m=a  </td>
      </tr>
      <tr>
        <td><a href="InputFiles/Satz7.3b.in">Satz 7.3b</a></td>
        <td><a href="Proofs/Satz7.3b.prf">Satz7.3b.prf</a><br>2 steps</td>
        <td> m=a -> M(a,m,b)  </td>
      </tr>
       <tr>
        <td><a href="InputFiles/Satz7.4a.in">Satz 7.4a</a></td>
        <td><a href="Proofs/Satz7.4a.prf">Satz7.4a.prf</a><br>2 steps</td>
        <td>   </td>
      </tr>
      <tr>
        <td><a href="InputFiles/Satz7.4b.in">Satz 7.4b</a></td>
        <td><a href="Proofs/Satz7.4b.prf">Satz7.4b.prf</a><br>3 steps</td>
        <td>     </td>
      </tr>
       <tr>
        <td><a href="InputFiles/Satz7.6.in">Satz 7.6</a></td>
        <td><a href="Proofs/Satz7.6.prf">Satz7.6.prf</a><br>1 step</td>
        <td>   </td>
      </tr>
       <tr>
        <td><a href="InputFiles/Satz7.7.in">Satz 7.7</a></td>
        <td><a href="Proofs/Satz7.7.prf">Satz7.7.prf</a><br>2 steps</td>
        <td> s(a,s(a,p)) = p (reflection is an involution)
            <br> In the book this is repeated as Satz 7.12 
        </td>
      </tr>
       <tr>
        <td><a href="InputFiles/Satz7.8.in">Satz 7.8</a></td>
        <td><a href="Proofs/Satz7.8.prf">Satz7.8.prf</a><br>5 steps</td>
        <td>   </td>
       <tr>
        <td><a href="InputFiles/Satz7.9.in">Satz 7.9</a></td>
        <td><a href="Proofs/Satz7.9.prf">Satz7.9.prf</a><br>3 steps</td>
        <td>   </td>
      </tr>
 
      <tr>
        <td><a href="InputFiles/Satz7.10a.in">Satz 7.10a</a></td>
        <td><a href="Proofs/Satz7.10a.prf">Satz7.10a.prf</a><br>3 steps</td>
        <td>   </td>
      </tr>
      <tr>
        <td><a href="InputFiles/Satz7.10b.in">Satz 7.10b</a></td>
        <td><a href="Proofs/Satz7.10b.prf">Satz7.10b.prf</a><br>2 steps</td>
        <td>   </td>
      </tr>
      <tr>
        <td><a href="InputFiles/Satz7.13.in">Satz 7.13</a></td>
        <td><a href="Proofs/Satz7.13.prf">Satz7.13.prf</a><br>77 steps</td>
        <td>   E(p,q,s(a,p),s(a,q)) </td>
      </tr>
       <tr>
        <td><a href="InputFiles/Satz7.15a.in">Satz 7.15a</a></td>
        <td><a href="Proofs/Satz7.15a.prf">Satz7.15a.prf</a><br>1 step</td>
        <td>   reflection preserves betweenness </td>
      </tr>
      </tr>
       <tr>
        <td><a href="InputFiles/Satz7.15b.in">Satz 7.15b</a></td>
        <td><a href="Proofs/Satz7.15b.prf">Satz7.15b.prf</a><br> 18 steps</td>
        <td>  converse of Satz 7.15a </td>
      </tr>
      <tr>
        <td><a href="InputFiles/Satz7.16a.in">Satz 7.16a</a></td>
        <td><a href="Proofs/Satz7.16a.prf">Satz7.16a.prf</a><br>2 steps</td>
        <td> reflection preserves congruence </td>
      </tr>
       <tr>
        <td><a href="InputFiles/Satz7.16b.in">Satz 7.16b</a></td>
        <td><a href="Proofs/Satz7.16b.prf">Satz7.16b.prf</a><br>22 steps</td>
        <td> converse of Satz 7.16a</td>
      </tr>
      <tr>
        <td><a href="InputFiles/Satz7.17.in">Satz 7.17</a></td>
        <td><a href="Proofs/Satz7.17.prf">Satz7.17.prf</a><br>22 steps</td>
        <td> uniqueness of midpoint </td>
      </tr>
       <tr>
        <td><a href="InputFiles/Satz7.18.in">Satz 7.18</a></td>
        <td><a href="Proofs/Satz7.18.prf">Satz7.18.prf</a><br>2 steps</td>
        <td>  s(a,p)=s(b,p)->a=b </td>
      </tr>
       <tr>
        <td><a href="InputFiles/Satz7.19.in">Satz 7.19</a></td>
        <td><a href="Proofs/Satz7.19.prf">Satz7.19.prf</a><br>9 steps</td>
        <td> s(a,s(b,p))=s(b,s(a,p))->a=b </td>
      </tr>
      <tr>
        <td><a href="InputFiles/Satz7.20.in">Satz 7.20</a></td>
        <td><a href="Proofs/Satz7.20.prf">Satz7.20.prf</a><br>33 steps</td>
        <td> Col(a,m,b) and E(m,a,m,b) ->a=b or M(a,m,b) </td>
      </tr>
       <tr>
        <td><a href="InputFiles/Satz7.21.in">Satz 7.21</a></td>
        <td><a href="Proofs/Satz7.21.prf">Satz7.21.prf</a><br>31 steps</td>
        <td> In a quadrilateral with opposite sides equal <br> the diagonals bisect each other. 
        (It is assumed that the diagonals meet, which is necessary since no dimension axiom is used.)
        <br> This is one of Quaife's challenge problems.</td>
      </tr>
      <tr>
        <td><a href="InputFiles/Satz7.22a.in">Satz 7.22a</a></td>
        <td><a href="Proofs/Satz7.22a.prf">Satz7.22a.prf</a>
            <br> 96 steps</td>
        <td> Gupta's  Krippenlemma <br> (an important lemma for Chapter 8)
            <br> Proved with additional assumption  ca1 &le; ca2.   The
              Krippenlemma is derived by two applications of this theorem,
              as in the book, but in the book this case is not stated separately.
        </td>
      </tr>

         <tr>
        <td><a href="InputFiles/Satz7.22.in">Satz 7.22</a></td>
        <td><a href="Proofs/Satz7.22.prf">Satz7.22.prf</a>
            <br> 12 steps</td>
        <td> Gupta's  Krippenlemma <br> (an important lemma for Chapter 8)</td>
      </tr>
          <tr>
        <td><a href="InputFiles/Satz7.25.in">Satz 7.25</a></td>
        <td><a href="Proofs/Satz7.25.prf">Satz7.25.prf</a><br>123 steps</td>
        <td> base of an isosceles triangle has a midpoint</td>
      </tr>
      
    </table>
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