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<h2>Tarski Formalization Project Archives</h2>
<p> </p>
<h3>Chapter 8: right angles and perpendiculars</h3>
<p>The posted input files are now all mechanically generated from a master list of theorems.
For more information about our methodology see the <a href="http://www.michaelbeeson.com/research/FormalTarski/index.php"> top page of this project</a>.
</p>
<p> The three main results in this chapter are all due to Gupta (1965): the existence of dropped
and erected perpendiculars, in Satz 8.18 and Satz 8.22, and the application of those results
to the construction of a midpoint of any segment in Satz 8.22.
</p>
<p> These proofs by Gupta make no use of any form of the continuity axiom (in particular no circles are involved),
and they also make no use of the parallel axiom, i.e., they use only neutral geometry; and they do not
use the upper dimension axiom either. The proofs rely on Gupta's work in Chapter 7, specifically the fact
that the base of an isosceles triangle has a midpoint and the Krippenlemma.
</p>
<p>The last theorem in the chapter,
Satz 8.24, is a more detailed statement about the midpoint, which actually gives a term to construct
the midpoint, in terms of the perpendiculars. It is worth noting, however, that since the perpendiculars
might be dropped or erected, SST does not give a single term that always constructs the midpoint. In
addition there is another case distinction in the proof of that theorem, about which of two segments is
the shorter. By "symmetry" it suffices "without loss of generality" to prove just one case, according
to the book, and that is what we do in Satz 8.24a and Satz 8.24b. (Since 8.24 has a conclusion that is
a conjunction, it has to become two files in resolution form.)
</p>
<table width="635" border="1">
<tr>
<th width="104" scope="col">Input File</th>
<th width="123" scope="col">Proof</th>
<th width="300" scope="col">Commentary</th>
</tr>
<tr>
<td><a href="InputFiles/Satz8.2.in">Satz 8.2</a></td>
<td><a href="Proofs/Satz8.2.prf">Satz8.2.prf</a><br>6 steps</td>
<td> R(a,b,c) -> R(c,b,a) </td>
</tr>
<tr>
<td><a href="InputFiles/Satz8.3.in">Satz 8.3</a></td>
<td><a href="Proofs/Satz8.3.prf">Satz8.3.prf</a><br>8 steps</td>
<td> R(a,b,c) and a != b and Col(b,a,d) -> R(d,b,c) </td>
</tr>
<tr>
<td><a href="InputFiles/Satz8.4.in">Satz 8.4</a></td>
<td><a href="Proofs/Satz8.4.prf">Satz8.4.prf</a><br>4 steps</td>
<td> R(a,b,c) -> R(a,b,s(b,c)) </td>
</tr>
<tr>
<td><a href="InputFiles/Satz8.5.in">Satz 8.5</a></td>
<td><a href="Proofs/Satz8.5.prf">Satz8.5.prf</a><br>3 steps</td>
<td> R(a,b,b) </td>
</tr>
<tr>
<td><a href="InputFiles/Satz8.6.in">Satz 8.6</a></td>
<td><a href="Proofs/Satz8.6.prf">Satz8.6.prf</a><br>5 steps</td>
<td> R(a,b,c) and T(p,b,c) and T(a,c,p) -> b=c </td>
</tr>
<tr>
<td><a href="InputFiles/Satz8.7.in">Satz 8.7</a></td>
<td><a href="Proofs/Satz8.7.prf">Satz8.7.prf</a><br>43 steps</td>
<td> R(a,b,c) and R(a,c,b)-> b=c </td>
</tr>
<tr>
<td><a href="InputFiles/Satz8.8.in">Satz 8.8</a></td>
<td><a href="Proofs/Satz8.8.prf">Satz8.8.prf</a><br>3 steps</td>
<td> R(a,b,a)-> a=b </td>
</tr>
<tr>
<td><a href="InputFiles/Satz8.9.in">Satz 8.9</a></td>
<td><a href="Proofs/Satz8.9.prf">Satz8.9.prf</a><br>2 steps</td>
<td> R(a,b,c) and Col(a,b,c) -> a=b or c=b </td>
</tr>
<tr>
<td><a href="InputFiles/Satz8.10.in">Satz 8.10</a></td>
<td><a href="Proofs/Satz8.10.prf">Satz8.10.prf</a><br>28 steps</td>
<td> R(a,b,c) and (abc) congruent (pqr) -> R(pqr) </td>
</tr>
<tr>
<td><a href="InputFiles/Satz8.12a.in">Satz 8.12a</a></td>
<td><a href="Proofs/Satz8.12a.prf">Satz8.12a.prf</a><br>10 steps</td>
<td> perpAt(a,b,x,p,q) ->perpAt(p,q,x,a,b) </td>
</tr>
<tr>
<td><a href="InputFiles/Satz8.12b.in">Satz 8.12b</a></td>
<td><a href="Proofs/Satz8.12b.prf">Satz8.12b.prf</a><br>3 steps</td>
<td> perp(a,b,p,q) ->perp(p,q,a,b) </td>
</tr>
<tr>
<td><a href="InputFiles/Satz8.13a.in">Satz 8.13a</a></td>
<td><a href="Proofs/Satz8.13a.prf">Satz8.13a.prf</a><br>33 steps</td>
<td> Left to right of 8.13 </td>
</tr>
<tr>
<td><a href="InputFiles/Satz8.13b.in">Satz 8.13b</a></td>
<td><a href="Proofs/Satz8.13b.prf">Satz8.13b.prf</a><br>16 steps</td>
<td> Right to left of 8.13. In a pure first-order treatment this is
much more difficult than it appears in Szmielew. </td>
</tr>
<tr>
<td><a href="InputFiles/Satz8.14a.in">Satz 8.14a</a></td>
<td><a href="Proofs/Satz8.14a.prf">Satz8.14a.prf</a><br>18 steps</td>
<td> perpendicular lines are distinct </td>
</tr>
<tr>
<td><a href="InputFiles/Satz8.14b.in">Satz 8.14b</a></td>
<td><a href="Proofs/Satz8.14b.prf">Satz8.14b.prf</a><br>7 steps</td>
<td> if c lies on both Line(a,b) and Line(p,q) and perp(a,b,p,q) then perpAt(a,b,c,p,q) </td>
</tr>
<tr>
<td><a href="InputFiles/Satz8.14c.in">Satz 8.14c</a></td>
<td><a href="Proofs/Satz8.14c.prf">Satz8.14c.prf</a><br>6 steps</td>
<td> perpendicular lines are perpendicular at only one point </td>
</tr>
<tr>
<td><a href="InputFiles/Satz8.15.in">Satz 8.15</a></td>
<td><a href="Proofs/Satz8.15.prf">Satz8.15.prf</a><br>10 steps</td>
<td> a!=b and Col(a,b,x) and perp(a,b,c,x) -> perp(a,b,x,c,x) </td>
</tr>
<tr>
<td><a href="InputFiles/Satz8.16a.in">Satz 8.16a</a></td>
<td><a href="Proofs/Satz8.16a.prf">Satz8.16a.prf</a><br>7 steps</td>
<td> a!=b and Col(a,b,x) and Col(a,b,u) and u != x and perp(a,b,c,x)<br>
-> not Col(a,b,c) and R(c,x,u) </td>
</tr>
<tr>
<td><a href="InputFiles/Satz8.16b.in">Satz 8.16b</a></td>
<td><a href="Proofs/Satz8.16b.prf">Satz8.16b.prf</a><br>17 steps</td>
<td> a!=b and Col(a,b,x) and Col(a,b,u) and u != x and <br>
not Col(a,b,c) and R(c,x,u) -> perp(a,b,c,x) </td>
</tr>
<tr>
<td><a href="InputFiles/Satz8.18a.in">Satz 8.18a</a></td>
<td><a href="Proofs/Satz8.18a.prf">Satz8.18a.prf</a><br>8 steps</td>
<td> uniqueness of the perpendicular to a line from a point not on the line.</td>
</tr>
<tr>
<td><a href="InputFiles/Satz8.18.in">Satz 8.18</a></td>
<td><a href="Proofs/Satz8.18.prf">Satz8.18.prf</a>
<br> 265 steps</td>
<td> Lotsatz (dropped perpendicular) <br> There is a perpendicular to a line from a point not on the line.</td>
</tr>
<tr>
<td><a href="InputFiles/Satz8.20a.in">Satz 8.20a</a></td>
<td><a href="Proofs/Satz8.20a.prf">Satz8.20a.prf</a><br>36 steps</td>
<td>R(a,b,c) and M(s(a,c),p,s(b,c)) -> R(b,a,p) </td>
</tr>
<tr>
<td><a href="InputFiles/Satz8.20b.in">Satz 8.20b</a></td>
<td><a href="Proofs/Satz8.20b.prf">Satz8.20b.prf</a><br>15 steps</td>
<td>R(a,b,c) and M(s(a,c),p,s(b,c)) and b != c -> a != p)</td>
</tr>
<tr>
<td><a href="InputFiles/perp1.in"> perp1.in</a></td>
<td><a href="Proofs/perp1.prf">perp1.prf</a><br>16 steps</td>
<td> perp(a,b,c,d) -> perp(b,a,c,d)
<br>Szmielew's treatment of lines is partly second order.
</td>
</tr>
<tr>
<td><a href="InputFiles/Satz8.21a.in">Satz 8.21a</a><br>Case 1 of Satz 8.21</td>
<td><a href="Proofs/Satz8.21a.prf">Satz8.21a.prf</a>
<br> 127 steps</td>
<td> (Erected perpendicular) <br> There is a perpendicular to a line through a point on the line <br>
on the opposite side from a given point not on the line.
</td>
</tr>
<tr>
<td><a href="InputFiles/Satz8.21.in">Satz 8.21</a></td>
<td><a href="Proofs/Satz8.21.prf">Satz8.21.prf</a><br>7 steps</td>
<td> Given point c and line ab, there is a perpendicular ap to ab <br>
and a point t on line ab between p and c.
<br> This differs from 8.21a only in that c might lie on Line(a,b).</td>
</td>
</tr>
<tr>
<td><a href="InputFiles/Satz8.22b.in">Satz 8.22b</a></td>
<td><a href="Proofs/Satz8.22b.prf">Satz8.22b.prf</a>
<br> 224 steps
</td>
<td> Given segment ab and perpendiculars ap and qb, and point t on line ab between p and q,
with ap <= qb, then segment ab has a midpoint.
<br> This lemma is applied twice to prove 8.22, with different instantiations.
In the book, the proof of 8.22 says "by the symmetry of the
hypotheses" we can assume ap <=qb. We handle that by applying this lemma twice.
</td>
</tr>
<tr>
<td><a href="InputFiles/Satz8.22.in">Satz 8.22</a></td>
<td><a href="Proofs/Satz8.22.prf">Satz8.22.prf</a><br>23 steps</td>
<td> Every segment has a midpoint (Gupta 1965)</td>
</tr>
<tr>
<td><a href="InputFiles/Satz8.24a.in">Satz 8.24a</a>
</td>
<td><a href="Proofs/Satz8.24a.prf">Satz8.24a.prf </a><br>171 steps
<br>
</td>
<td> Given segment ab and perpendiculars ap and qb, and point t on line ab between p and q,
with ap <= qb as witnessed by T(b,r,q) and ap=br, then segment ab has a midpoint x,
and moreover x is also the midpoint of pr. The point x is defined
using the Skolem function for inner Pasch.
<br>
This is very similar to 8.22b, except it assumes the existence of the perpendiculars that are
constructed in the proof of 8.22b, and in terms of those one can give a term for the midpoint.
<br> SST gives it a one-line "proof" referring to the proof of 8.22, which is really the proof of 8.22b.
Technically you can refer only to a theorem; the proof must be repeated.
<br>
</td>
</tr>
<tr>
<td><a href="InputFiles/Satz8.24b.in">Satz 8.24b</a>
</td>
<td><a href="Proofs/Satz8.24b.prf">Satz8.24b.prf </a><br>163 steps
<br>
</td>
<td> This is the second conclusion of Satz 8.24, under an additional inequality.
</td>
</tr>
</table>
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