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We are trying to prove that in an arbitrary Hilbert plane (assuming Hilbert's axioms of Group I:Incidence, II:Order and III:Congruence https://en.wikipedia.org/wiki/Hilbert%27s_axioms), Tarski's axioms A1-A9 (https://en.wikipedia.org/wiki/Tarski%27s_axioms) for 2D neutral geometry hold.

It seems to be a folklore theorem, but we couldn't find a proper reference to a synthetic proof. We could prove most axioms but we have a problem with the five-segment axiom.

The five-segment axiom says that: $${(x \ne y \land Bxyz \land Bx'y'z' \land xy \equiv x'y' \land yz \equiv y'z' \land xu \equiv x'u' \land yu \equiv y'u')} \rightarrow zu \equiv z'u'$$

Using Hilbert's Foundations of Geometry theorem 14 (congruence of supplementary angles) and theorem 18 (Side-Side-Side) we can obtain the five segment axiom when u is not on line xy.

Does anyone know a reference for the proof of Tarski's five segment axiom in the degenerated case when u is on line xy ?

Julien Narboux
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We have found a proof and formalized it using the Coq proof assistant, the details are here: https://hal.inria.fr/hal-01332044

Julien Narboux
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  • Do you have the reference for the proof of Pythagoras's theorem using Tarski's axioms? (the proof is mentioned in the link, hence why I ask) – Chill2Macht Jul 02 '17 at 20:09
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    The proof of Pythagorean theorem from Tarski's axioms can be found in Chapter 15 of "Metamathematische Methoden in der Geometrie", by Schwabhäuser, Szmielew and Tarski. The formalization in Coq is here: http://geocoq.github.io/GeoCoq/html/GeoCoq.Tarski_dev.Ch15_lengths.html#pythagoras – Julien Narboux Jul 03 '17 at 07:35
  • This paper explains how you used Tarski's axioms to prove Pappus's theorem: https://hal.inria.fr/hal-01176508/file/Pappus.pdf -- which one shows how you used Tarski's axioms to prove Desargues's Theorem? I thought it might have been this one: https://hal.archives-ouvertes.fr/hal-01282550/file/arithmetization.pdf but it doesn't mention a proof of Desargues's Theorem. – Chill2Macht Jul 03 '17 at 08:43
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    There is no paper which describe our formalization of Desargues' theorem from Tarski's axioms. The proof that we formalized in the well known proof by Hessenberg which is also presented in "Metamathematische Methoden in der Geometrie". The formalization is here: http://geocoq.github.io/GeoCoq/html/GeoCoq.Tarski_dev.Ch13_6_Desargues_Hessenberg.html If you are curious about this proof, you can also have a look at : http://argo.matf.bg.ac.rs/downloads/HessenbergsTheorem/10.1.1.544.3402.pdf – Julien Narboux Jul 03 '17 at 09:16
  • Isn't that for projective geometry though, not Euclidean geometry? Or does the distinction not matter? I can look it up in "Metamathematische Methoden in der Geometrie" -- do you happen to know the page number? – Chill2Macht Jul 03 '17 at 09:47
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    The proof of Desargues' theorem can be found on pages 139-142. Remark 13.16 says that the versions of theorem proved are special cases of the projective version. We have to distinguish the case when the lines are parallel or not. – Julien Narboux Jul 03 '17 at 12:00