{{interfaces | imports = Interface:Line segment congruence | exports = Interface:Betweenness of points }}

This is part of a series of modules which prove a variety of geometrical theorems starting with Tarski's axioms for geometry. We follow the formalization of Julien Narboux<ref>The formal proofs are at http://www.lix.polytechnique.fr/Labo/Julien.Narboux/tarski.html Formalization of Tarski's geometry in the Coq proof assistant and are described in Julien Narboux (2007), "http://citeseerx.ist.psu.edu/viewdoc/summary?doi= Mechanical Theorem Proving in Tarski’s Geometry", F. Botana and T. Recio (Eds.): ADG 2006, LNAI 4869, pp. 139–156</ref> which itself closely follows a treatise by Schwabhäuser, Szmielew, and Tarski.<ref>W. Schwabhäuser, W Szmielew, and A. Tarski (1983), ''Metamathematische Methoden in der Geometrie'', ISBN 0387129588</ref>

We import the theorems of propositional logic and predicate logic, and the geometry results so far and define some variables:


We start with a degenerate case of betweenness. It illustrates that we follow Tarski's definition of betweenness, in which the points need not be distinct, rather than Hilbert's, in which they do need to be.<ref>Discussion at Bet in Narboux</ref>

The proof extends the line segment A B beyond B to a point a such that B a ≡ B B. But B a ≡ B B implies B = a, so we have in fact shown between A B B.<ref>beetween_trivial in Narboux</ref>

Betweenness symmetry

Exchanging the endpoints of a betweenness formula has no effect. In other words, B is between A and C implies B is between C and A.<ref>between_symmetry in Narboux</ref><ref>Symmetry Axiom for Betweenness in Givant and Tarski (1999)</ref>

We move much of the proof into a lemma because it will be useful in other similar proofs.

Symmetry lets us easily prove the other degenerate case.<ref>beetween_trivial2 in Narboux</ref>


If B is between A and C, and A is between B and C, then A and B must be equal.<ref>between_egality in Narboux</ref>

This property is analogous to antisymmetry for an order relation.


In this section we prove six similar theorems which involve four points on a line. Given two betweenness relationships, we conclude a third. Three of the six are just versions of the other three which are flipped left-to-right.

Inner transitivity, regular and flipped

The first one is known as inner transitivity.<ref>Inner Transitivity Axiom for Betweenness in Givant and Tarski, 1999</ref><ref>between_inner_transitivity in Narboux</ref>

Reflecting this left-to-right gives our next theorem.<ref>between_exchange3 in Narboux</ref>

Outer transitivity, flipped

For outer transitivity, we prove the "flipped" version first (where we consider the "non-flipped" one to be the one in Givant and Tarski (1999)).

To be specific, the flipped outer transitivity is between A B C ∧ between B C D ∧ B ≠ C → between A C D. First we extend A C beyond C to a point a such that C a ≡ C D. Then between A B C and between A C a give us between B C a by BetweennessInnerTransitivityFlipped. That's part of what we need to apply uniqueness of segment construction, which gives us a = D. Then we can substitute between A C a to get between A C D, which is what we want.<ref>outer_transitivity_between2 in Narboux</ref>

The first two lemmas get as far as between B C a in the proof sketch above.

Middle transitivity

The next result is between A B D ∧ between B C D → between A C D. The proof has two cases. If B = C, then we can substitute between A B D into between A C D and we are done. If B ≠ C, then we first apply inner transitivity to get between A B C, and then apply outer transitivity (flipped) to get between A C D.<ref>between_exchange2 in Narboux</ref>

Outer transitivity, non-flipped

The non-flipped outer transitivity is between A B C ∧ between B C D ∧ B ≠ C → between A B D.<ref>outer_transitivity_between in Narboux</ref><ref>Axiom 16 in Givant and Tarski, 1999</ref> Our proof just takes BetweennessOuterTransitivityFlipped and flips it, whereas Narboux's proof also invokes BetweennessMiddleTransitivity.

Middle transitivity, flipped

Here we prove between A B C ∧ between A C D → between A B D.<ref>between_exchange4 in Narboux</ref>

Betweenness of more than three points

Narboux defines the concept of a betweenness relationship between four points, but we omit it here as Narboux does not use it anywhere.<ref>Bet_4 and l_3_9_4 in Narboux</ref>


There exist two distinct points. In symbols, ∃ a ∃ b a ≠ b.<ref>two_distinct_points in Narboux</ref> The proof is based on the LowerDimension axiom. There are two cases. If a ≠ b, we are done. If a = b, then we can take ¬ between b c a from the axiom and turn it into ¬ between b c b which in turn implies b ≠ c.

Formalizing this proof has two parts. The first one follows the informal proof above and gets us as far as showing a ≠ b or b ≠ c. The second part does the predicate logic needed to switch variables and combine the two inequalities.

The first thing is a lemma biconditionalizing Indivisibility.

Extending a line segment to a distinct point

Any line segment A B can be extended beyond B to a point c which does not equal B.<ref>point_construction_different in Narboux</ref>

First we prove a lemma which just rearranges CongruenceIdentityFromEquality.

A point distinct from a given point exists

Given a point A, there exists a point b distinct from it.<ref>another_point in Narboux</ref> Here's the sketch of Narboux's proof: apply Twoness to get x and y such that x ≠ y. If A ≠ x, we are done (x is the desired point). If A = x, then subsituting into x ≠ y gives us A ≠ y, and y is the desired point. It would be possible (although somewhat cumbersome) to formalize this argument here, but it will be easier just to prove this theorem as a special case of PointConstructionDifferent.

Lemma l3_17 (Pasch variant)

thumb|right|Top: statement of the lemma; Middle: construction of {{{q′}}}; Bottom: construction of {{{q}}} This lemma is a variant of Pasch's axiom. The lemma is between A B C ∧ between A′ B′ C ∧ between A P A′ → ∃ q (between P q C ∧ between B q B′). The proof consists of two applications of Pasch's axiom. The first one uses the triangle A A′ C (with points P and B′) to construct a point q′. The second uses the triangle A C B′ (with points B and q′) to construct the point q.

BetweenBuilder is deprecated

{{deprecated|renaming BetweenBuilder to BetweennessBuilder everywhere. During the transition it is probably best to ensure that both are defined (at least where they are still both used)}}

We are in the process of renaming BetweenBuilder to BetweennessBuilder. Until it is renamed everywhere, just prove one in terms of the other (they are identical except the name).


We now export to Interface:Betweenness of points.



{{DEFAULTSORT:{{PAGENAME}}}} Category:Euclidean geometries (general) and generalizations Category:Foundations of classical theories (including reverse mathematics)

Login to edit