Connectivity for betweenness

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 which itself closely follows a treatise by Schwabhäuser, Szmielew, and Tarski.

This page is devoted to the proof of one theorem,. This is called outer connectivity for betweenness by Narboux and Givant. It was once considered an axiom of Tarski's system, but was proved from the other axioms by Gupta in 1965

The proof is not simple; Givant calls it "rather involved" and Narboux devotes two and a half pages to an informal proof (and an explanation of part of the formal proof).

We import the theorems of propositional logic and predicate logic, and the geometry results so far and define some variables:  import (CLASSICAL Interface:Classical_propositional_calculus ) import (FIRSTORDER Interface:First-order_logic_with_quantifiability (CLASSICAL) ) import (START Interface:Collinearity (CLASSICAL FIRSTORDER) )

var (point A B C D E P Q R A′ B′ C′ D′ B″ C″) var (variable a b b′ c c′ d′ e p q r b″ c″) 

Summary


Here's an outline of the proof, illustrating which points we construct but leaving most of the detailed line segment congruences for the detailed formal proof.

We first construct  extending the line segments   and   by a distance of   in both cases. Eventually, the proof will show that  or.

Then extend the line segments  and   to points   and   by distances of   and. The result of this construction is that the distance from  is the same to either   or   and thus we can conclude that.

The point  lies at the intersection of   and.

At this point we assume  and will deduce from that.

Extend  by the distance   to a point. Extend  by the distance   to a point. Extend  by the distance   to a point.

By some congruences we can show first, which we'll need at the end of the proof, and  , which leads to the next step.

At this point we will apply  five times. In every case we will show that  (based on two such existing congruences). For the starting point, we already have  and. In turn, we will prove a similar congruence for,  ,  , and. Those cases easily fit the pictures we have been drawing so far. But we don't stop there: because,  , and   are collinear, we can also conclude   (this last one doesn't quite as readily lend itself to being illustrated in diagram form, but the first four are illustrated in the adjacent figures).

Of course,  implies , which in turn implies   (via  ).

We have shown that  implies , or in other words. The former implies  and the latter implies , so we reach the conclusion that.

Construction of C′, D′, B′, and B″
We construct  such that   and   such that.

Then we construct  such that   and   such that.

The following lemma embodies these constructions, and a bit of rearrangement of the quantifiers.  thm (BetweennessOuterConnectivity-cdbb ((A c′ d′ b′ b″) (B c′ d′ b′ b″) (C c′ d′ b′ b″) (D c′ d′ b′ b″))   ((((A ≠ B) ∧ (between A B C)) ∧ (between A B D)) → (∃ c′ (∃ d′ (∃ b′ (∃ b″     ((((((between A D (value c′)) ∧ (D (value c′) ≡ C D)) ∧ ((between A C (value d′)) ∧ (C (value d′) ≡ C D))) ∧     ((between A (value c′) (value b′)) ∧ ((value c′) (value b′) ≡ C B))) ∧ ((between A (value d′) (value b″)) ∧ ((value d′) (value b″) ≡ D B))) ∧     (((A ≠ B) ∧ (between A B C)) ∧ (between A B D)))))))) ( c′ A D C D SegmentConstruction d′ A C C D SegmentConstruction introduceConjunction

c′ d′ ((between A D (value c′)) ∧ (D (value c′) ≡ C D)) ((between A C (value d′)) ∧ (C (value d′) ≡ C D)) ThereExistsScattering eliminateBiconditionalForward

applyModusPonens  So far we have the construction of  and , specifically. We now turn to.  b′ A (value c′) C B SegmentConstruction d′ generalize c′ generalize

combineThereExistsForAll

d′         (((between A D (value c′)) ∧ (D (value c′) ≡ C D)) ∧            ((between A C (value d′)) ∧ (C (value d′) ≡ C D))) (∃ b′ ((between A (value c′) (value b′)) ∧ ((value c′) (value b′) ≡ C B))) ThereExistsConjunctionRightCombining c′ addThereExists applyModusPonens  That gives us. The next step is to move  to the front.  b′         (((between A D (value c′)) ∧ (D (value c′) ≡ C D)) ∧            ((between A C (value d′)) ∧ (C (value d′) ≡ C D))) ((between A (value c′) (value b′))           ∧ ((value c′) (value b′) ≡ C B)) ThereExistsConjunctionMovement eliminateBiconditionalForward d′ addThereExists c′ addThereExists applyModusPonens  Now we construct, which starts with   and then moves the quantifiers to the front.  b″ A (value d′) D B SegmentConstruction d′ generalize

introduceConjunction

c′         (∃ d′ (∃ b′            ((((between A D (value c′)) ∧ (D (value c′) ≡ C D)) ∧ ((between A C (value d′)) ∧ (C (value d′) ≡ C D))) ∧             ((between A (value c′) (value b′)) ∧ ((value c′) (value b′) ≡ C B))) ))         (∀ d′ (∃ b″ ((between A (value d′) (value b″)) ∧ ((value d′) (value b″) ≡ D B)))) ThereExistsConjunctionRightMovement eliminateBiconditionalForward applyModusPonens

d′         (∃ b′            ((((between A D (value c′)) ∧ (D (value c′) ≡ C D)) ∧              ((between A C (value d′)) ∧ (C (value d′) ≡ C D))) ∧ ((between A (value c′) (value b′))               ∧ ((value c′) (value b′) ≡ C B)))          ) (∃ b″ ((between A (value d′) (value b″)) ∧ ((value d′) (value b″) ≡ D B))) ThereExistsConjunctionRightCombining c′ addThereExists applyModusPonens

b′ b″         ((((between A D (value c′)) ∧ (D (value c′) ≡ C D)) ∧ ((between A C (value d′)) ∧ (C (value d′) ≡ C D))) ∧           ((between A (value c′) (value b′)) ∧ ((value c′) (value b′) ≡ C B))) ((between A (value d′) (value b″)) ∧ ((value d′) (value b″) ≡ D B)) ThereExistsScattering eliminateBiconditionalForward d′ addThereExists c′ addThereExists applyModusPonens  Now we need to add in, and move it inside all four quantifiers.  (((A ≠ B) ∧ (between A B C)) ∧ (between A B D)) introduceAntecedent

(((A ≠ B) ∧ (between A B C)) ∧ (between A B D)) ImplicationReflexivity composeConjunction

c′         (∃ d′ (∃ b′ (∃ b″            (((((between A D (value c′)) ∧ (D (value c′) ≡ C D)) ∧ ((between A C (value d′)) ∧ (C (value d′) ≡ C D))) ∧           ((between A (value c′) (value b′)) ∧ ((value c′) (value b′) ≡ C B))) ∧ ((between A (value d′) (value b″))             ∧ ((value d′) (value b″) ≡ D B)))          ))) (((A ≠ B) ∧ (between A B C)) ∧ (between A B D)) ThereExistsConjunctionRightMovement eliminateBiconditionalForward applySyllogism

d′         (∃ b′ (∃ b″            (((((between A D (value c′)) ∧ (D (value c′) ≡ C D)) ∧            ((between A C (value d′)) ∧ (C (value d′) ≡ C D))) ∧ ((between A (value c′) (value b′))             ∧ ((value c′) (value b′) ≡ C B))) ∧            ((between A (value d′) (value b″)) ∧ ((value d′) (value b″) ≡ D B))) ))         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D)) ThereExistsConjunctionRightMovement eliminateBiconditionalForward c′ addThereExists applySyllogism

b′         (∃ b″            (((((between A D (value c′)) ∧ (D (value c′) ≡ C D)) ∧ ((between A C (value d′)) ∧ (C (value d′) ≡ C D))) ∧           ((between A (value c′) (value b′)) ∧ ((value c′) (value b′) ≡ C B))) ∧ ((between A (value d′) (value b″))             ∧ ((value d′) (value b″) ≡ D B)))          ) (((A ≠ B) ∧ (between A B C)) ∧ (between A B D)) ThereExistsConjunctionRightMovement eliminateBiconditionalForward d′ addThereExists c′ addThereExists applySyllogism

b″         (((((between A D (value c′)) ∧ (D (value c′) ≡ C D)) ∧            ((between A C (value d′)) ∧ (C (value d′) ≡ C D))) ∧ ((between A (value c′) (value b′))             ∧ ((value c′) (value b′) ≡ C B))) ∧            ((between A (value d′) (value b″)) ∧ ((value d′) (value b″) ≡ D B))) (((A ≠ B) ∧ (between A B C)) ∧ (between A B D)) ThereExistsConjunctionRightMovement eliminateBiconditionalForward b′ addThereExists d′ addThereExists c′ addThereExists applySyllogism )) 

B′ is equal to B″
In this section we'll show that. The idea of the proof is simple: to get from  to   we go through   and , and to get from   to   we go through   and. As can be seen from the way we constructed the points, the three segments in the one case are congruent to the three segments in the other (although in reverse order), so the distance from  to either   or   is the same. Because, this implies that   by the uniqueness of segment construction (applied to   and  ).

B C′ ≡ B″ C
Here we prove, starting with a few lemmas which we'll need.  thm (b-d-cprime    (((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧   ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧   (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) → (between B D C′)) (  First is.  (((((between A D C′) ∧ (D C′ ≡ C D)) ∧           ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧           ((between A D′ B″) ∧ (D′ B″ ≡ D B))) (((A ≠ B) ∧ (between A B C)) ∧ (between A B D)) ConjunctionLeftElimination eliminateLeftConjunctInConsequent  Second is  (((((between A D C′) ∧ (D C′ ≡ C D)) ∧           ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧           ((between A D′ B″) ∧ (D′ B″ ≡ D B))) (((A ≠ B) ∧ (between A B C)) ∧ (between A B D)) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent

composeConjunction </jh> The result follows from transitivity.  A B D C′ BetweennessInnerTransitivityFlipped applySyllogism ))

thm (bdoubleprime-dprime-c    (((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧   ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧   (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) → (between B″ D′ C)) ( </jh> This is another application of transitivity, starting with.  (((((between A D C′) ∧ (D C′ ≡ C D)) ∧           ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧           ((between A D′ B″) ∧ (D′ B″ ≡ D B))) (((A ≠ B) ∧ (between A B C)) ∧ (between A B D)) ConjunctionRightElimination eliminateLeftConjunctInConsequent eliminateRightConjunctInConsequent

A D′ B″ BetweennessSymmetry eliminateBiconditionalReverse applySyllogism </jh> The other thing we need is.  (((((between A D C′) ∧ (D C′ ≡ C D)) ∧           ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧           ((between A D′ B″) ∧ (D′ B″ ≡ D B))) (((A ≠ B) ∧ (between A B C)) ∧ (between A B D)) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent eliminateRightConjunctInConsequent

A C D′ BetweennessSymmetry eliminateBiconditionalReverse applySyllogism

composeConjunction

B″ D′ A C BetweennessInnerTransitivity applySyllogism ))

thm (d-cprime-dprime-c    (((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧   ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧   (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) → (D C′ ≡ D′ C)) ( </jh> Each of the segments  and   is congruent to , so we just need to apply symmetry and transitivity.  (((((between A D C′) ∧ (D C′ ≡ C D)) ∧           ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧           ((between A D′ B″) ∧ (D′ B″ ≡ D B))) (((A ≠ B) ∧ (between A B C)) ∧ (between A B D)) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent

(((((between A D C′) ∧ (D C′ ≡ C D)) ∧           ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧           ((between A D′ B″) ∧ (D′ B″ ≡ D B))) (((A ≠ B) ∧ (between A B C)) ∧ (between A B D)) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent eliminateLeftConjunctInConsequent

C D′ C D CongruenceSymmetry eliminateBiconditionalReverse applySyllogism

composeConjunction

D C′ C D C D′ CongruenceTransitivity applySyllogism

D C′ C D′ CongruenceRightCommutativity eliminateBiconditionalReverse applySyllogism ))

thm (b-cprime-bdoubleprime-c    (((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧   ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧   (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) → (B C′ ≡ B″ C)) ( </jh> We prove this theorem by applying  to   and. The first step is.  A D C′ C D′ B′ B B″ b-d-cprime </jh> The next is.  A D C′ C D′ B′ B B″ bdoubleprime-dprime-c composeConjunction </jh> The next is.  (((((between A D C′) ∧ (D C′ ≡ C D)) ∧           ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧           ((between A D′ B″) ∧ (D′ B″ ≡ D B))) (((A ≠ B) ∧ (between A B C)) ∧ (between A B D)) ConjunctionRightElimination eliminateLeftConjunctInConsequent eliminateLeftConjunctInConsequent

D′ B″ D B CongruenceSymmetry eliminateBiconditionalReverse applySyllogism

D B D′ B″ CongruenceCommutativity eliminateBiconditionalReverse applySyllogism

composeConjunction </jh> The last is  A D C′ C D′ B′ B B″ d-cprime-dprime-c composeConjunction </jh> Three segment gives us, the desired result.  B D C′ B″ D′ C OuterThreeSegment applySyllogism )) </jh>

B B′ ≡ B″ B
Here we extend the congruence from the previous section one more point. This is also an application of, and we again break out some of the pieces into lemmas.  thm (a-b-cprime    (((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧   ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧   (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) → (between A B C′)) ( </jh> This is an application of transitivity to  (((((between A D C′) ∧ (D C′ ≡ C D)) ∧           ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧           ((between A D′ B″) ∧ (D′ B″ ≡ D B))) (((A ≠ B) ∧ (between A B C)) ∧ (between A B D)) ConjunctionLeftElimination eliminateLeftConjunctInConsequent </jh> and.  (((((between A D C′) ∧ (D C′ ≡ C D)) ∧           ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧           ((between A D′ B″) ∧ (D′ B″ ≡ D B))) (((A ≠ B) ∧ (between A B C)) ∧ (between A B D)) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent

composeConjunction

A B D C′ BetweennessMiddleTransitivityFlipped applySyllogism ))

thm (b-cprime-bprime    (((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧   ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧   (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) → (between B C′ B′)) ( </jh> Here we apply transitivity to.  A D C′ C D′ B′ B B″ a-b-cprime </jh> and.  (((((between A D C′) ∧ (D C′ ≡ C D)) ∧           ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧           ((between A D′ B″) ∧ (D′ B″ ≡ D B))) (((A ≠ B) ∧ (between A B C)) ∧ (between A B D)) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent eliminateRightConjunctInConsequent

composeConjunction

A B C′ B′ BetweennessInnerTransitivityFlipped applySyllogism ))

thm (a-c-bdoubleprime    (((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧   ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧   (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) → (between A C B″)) ( </jh> Here we apply transitivity to  (((((between A D C′) ∧ (D C′ ≡ C D)) ∧           ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧           ((between A D′ B″) ∧ (D′ B″ ≡ D B))) (((A ≠ B) ∧ (between A B C)) ∧ (between A B D)) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent eliminateRightConjunctInConsequent </jh> and <jh> (((((between A D C′) ∧ (D C′ ≡ C D)) ∧           ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧           ((between A D′ B″) ∧ (D′ B″ ≡ D B))) (((A ≠ B) ∧ (between A B C)) ∧ (between A B D)) ConjunctionRightElimination eliminateLeftConjunctInConsequent eliminateRightConjunctInConsequent

composeConjunction

A C D′ B″ BetweennessMiddleTransitivityFlipped applySyllogism ))

thm (bdoubleprime-c-b    (((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧   ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧   (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) → (between B″ C B)) ( </jh> Here we apply transitivity to <jh> (((((between A D C′) ∧ (D C′ ≡ C D)) ∧           ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧           ((between A D′ B″) ∧ (D′ B″ ≡ D B))) (((A ≠ B) ∧ (between A B C)) ∧ (between A B D)) ConjunctionLeftElimination eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent </jh> and the previous lemma <jh> A D C′ C D′ B′ B B″ a-c-bdoubleprime composeConjunction </jh> to produce. Then we just need to switch the order of the endpoints. <jh> A B C B″ BetweennessInnerTransitivityFlipped applySyllogism

B C B″ BetweennessSymmetry eliminateBiconditionalReverse applySyllogism ))

thm (b-bprime-bdoubleprime-b    (((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧   ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧   (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) → (B B′ ≡ B″ B)) ( </jh> First we need. <jh> A D C′ C D′ B′ B B″ b-cprime-bprime </jh> Next we need. <jh> A D C′ C D′ B′ B B″ bdoubleprime-c-b composeConjunction </jh> : <jh> A D C′ C D′ B′ B B″ b-cprime-bdoubleprime-c composeConjunction </jh> : <jh> (((((between A D C′) ∧ (D C′ ≡ C D)) ∧           ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧           ((between A D′ B″) ∧ (D′ B″ ≡ D B))) (((A ≠ B) ∧ (between A B C)) ∧ (between A B D)) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent eliminateLeftConjunctInConsequent

composeConjunction

B C′ B′ B″ C B OuterThreeSegment applySyllogism )) </jh>

B″ = B′
In this section we apply  to prove that   and   are in fact the same point. We start with some more betweenness transitivity lemmas. <jh> thm (a-b-dprime   (((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧   ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧   (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) → (between A B D′)) ( </jh> : <jh> (((((between A D C′) ∧ (D C′ ≡ C D)) ∧           ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧           ((between A D′ B″) ∧ (D′ B″ ≡ D B))) (((A ≠ B) ∧ (between A B C)) ∧ (between A B D)) ConjunctionLeftElimination eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent

</jh> : <jh> (((((between A D C′) ∧ (D C′ ≡ C D)) ∧           ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧           ((between A D′ B″) ∧ (D′ B″ ≡ D B))) (((A ≠ B) ∧ (between A B C)) ∧ (between A B D)) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent eliminateRightConjunctInConsequent

composeConjunction

A B C D′ BetweennessMiddleTransitivityFlipped applySyllogism ))

thm (bdoubleprime-bprime   (((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧   ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧   (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) → (B″ = B′)) ( </jh> We'll prove  (and flip the order later). We first need. <jh> (((((between A D C′) ∧ (D C′ ≡ C D)) ∧           ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧           ((between A D′ B″) ∧ (D′ B″ ≡ D B))) (((A ≠ B) ∧ (between A B C)) ∧ (between A B D)) ConjunctionLeftElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent </jh> Next is, which follows from   and   by betweenness transitivity. <jh> A D C′ C D′ B′ B B″ a-b-cprime

(((((between A D C′) ∧ (D C′ ≡ C D)) ∧           ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧           ((between A D′ B″) ∧ (D′ B″ ≡ D B))) (((A ≠ B) ∧ (between A B C)) ∧ (between A B D)) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent eliminateRightConjunctInConsequent composeConjunction

A B C′ B′ BetweennessMiddleTransitivityFlipped applySyllogism

composeConjunction </jh> Next is. <jh> A D C′ C D′ B′ B B″ b-bprime-bdoubleprime-b composeConjunction </jh> Next is, which follows from   and   by betweenness transitivity. <jh> A D C′ C D′ B′ B B″ a-b-dprime

(((((between A D C′) ∧ (D C′ ≡ C D)) ∧           ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧           ((between A D′ B″) ∧ (D′ B″ ≡ D B))) (((A ≠ B) ∧ (between A B C)) ∧ (between A B D)) ConjunctionRightElimination eliminateLeftConjunctInConsequent eliminateRightConjunctInConsequent composeConjunction

A B D′ B″ BetweennessMiddleTransitivityFlipped applySyllogism

composeConjunction </jh> Finally we need. <jh> B B″ CongruenceABBA

((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) introduceAntecedent composeConjunction </jh> Applying segment construction uniqueness, and then flipping the order of the equality, we are done. <jh> A B B′ B″ B B″ SegmentConstructionUniqueness applySyllogism

B′ B″ EqualitySymmetry eliminateBiconditionalReverse applySyllogism )) </jh>

Construction of E
In this section we construct the point  and also prove a number of congruences. First (even before we construct E), we can apply outer five segment to the line segments  and   with points   and   to give us. Then we construct  as the intersection of   and   (the axiom of Pasch ensures its existence).

The congruences  and   may remind us of the familiar theorem that the diagonals of a rhombus bisect each other, but of course to say it that way would get ahead of ourselves. We can readily prove each of those two congruences, however, from a simple invocation of inner five segment.

C′ D′ ≡ C D
We just said that this follows from inner five segment, but that's a slight oversimplification. That's true for the  case (which we'll prove first), but there's a separate   case. We'll start with the  case, and before the case itself we'll prove a few lemmas which represent the various antecedents of. <jh> thm (b-c-dprime   (((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧   ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧   (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) → (between B C D′)) ( </jh> We prove this via betweenness transitivity from <jh> A D C′ C D′ B′ B B″ bdoubleprime-c-b

B″ C B BetweennessSymmetry eliminateBiconditionalReverse applySyllogism </jh> and. <jh> A D C′ C D′ B′ B B″ bdoubleprime-dprime-c

B″ D′ C BetweennessSymmetry eliminateBiconditionalReverse applySyllogism

composeConjunction

B C B″ D′ BetweennessInnerTransitivity applySyllogism ))

thm (bprime-cprime-d   (((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧   ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧   (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) → (between B′ C′ D)) ( </jh> This follows by betweenness transitivity from <jh> A D C′ C D′ B′ B B″ b-cprime-bprime

B C′ B′ BetweennessSymmetry eliminateBiconditionalReverse applySyllogism </jh> and <jh> A D C′ C D′ B′ B B″ b-d-cprime

B D C′ BetweennessSymmetry eliminateBiconditionalReverse applySyllogism

composeConjunction

B′ C′ B D BetweennessInnerTransitivity applySyllogism ))

</jh> The next two lemmas just apply a substitution to turn, which we have already proved, into. <jh> thm (b-cprime-bprime-c-lemma   (((B″ = B′) ∧ (B C′ ≡ B″ C)) → (B C′ ≡ B′ C)) ( B EqualityReflexivity C′ EqualityReflexivity C EqualityReflexivity B B C′ C′ B″ B′ C C CongruenceBuilder detach2of2 detach2of3 detach1of2

eliminateBiconditionalReverseInConsequent

import ))

thm (b-cprime-bprime-c   (((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧   ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧   (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) → (B C′ ≡ B′ C)) ( A D C′ C D′ B′ B B″ bdoubleprime-bprime A D C′ C D′ B′ B B″ b-cprime-bdoubleprime-c composeConjunction

B″ B′ B C′ C b-cprime-bprime-c-lemma applySyllogism ))

thm (cprime-dprime-c-d-case-bnotc   (((B ≠ C) ∧    ((((((between A D C′) ∧ (D C′ ≡ C D)) ∧    ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧   ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D)))) → (C′ D′ ≡ C D)) ( </jh> For outer five segment, we first need. <jh> (B ≠ C)         ((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧           ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧           (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ConjunctionRightElimination </jh> Next is. <jh> (B ≠ C)         ((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧           ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧           (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ConjunctionLeftElimination

A D C′ C D′ B′ B B″ b-c-dprime applySyllogism

composeConjunction </jh> Next is. <jh> (B ≠ C)         ((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧           ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧           (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ConjunctionLeftElimination

A D C′ C D′ B′ B B″ bprime-cprime-d applySyllogism

composeConjunction </jh> Next is. <jh> (B ≠ C)         ((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧           ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧           (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ConjunctionLeftElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent eliminateLeftConjunctInConsequent

C′ B′ C B CongruenceSymmetry eliminateBiconditionalReverse applySyllogism

C B C′ B′ CongruenceCommutativity eliminateBiconditionalReverse applySyllogism

composeConjunction </jh> Next is. <jh> (B ≠ C)         ((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧           ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧           (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ConjunctionLeftElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent eliminateLeftConjunctInConsequent

(B ≠ C)         ((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧           ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧           (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ConjunctionLeftElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent

D C′ C D CongruenceSymmetry eliminateBiconditionalReverse applySyllogism

C D D C′ CongruenceRightCommutativity eliminateBiconditionalReverse applySyllogism

composeConjunction

C D′ C D C′ D CongruenceTransitivity applySyllogism

composeConjunction </jh> Next is. <jh> (B ≠ C)         ((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧           ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧           (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ConjunctionLeftElimination

A D C′ C D′ B′ B B″ b-cprime-bprime-c applySyllogism

composeConjunction </jh> Last is. <jh> C C′ CongruenceABBA

((B ≠ C) ∧         ((((((between A D C′) ∧ (D C′ ≡ C D)) ∧          ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧         ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D)))) introduceAntecedent

composeConjunction </jh> Applying outer five segment gives us. We flip the order of the points to turn that into. <jh> B C D′ B′ C′ D C′ C OuterFiveSegment applySyllogism

D′ C′ D C CongruenceCommutativity eliminateBiconditionalReverse applySyllogism )) </jh>

We now turn to the  case. We first note that in this case  implies <jh> thm (bprime-cprime   (((B = C) ∧    ((((((between A D C′) ∧ (D C′ ≡ C D)) ∧    ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧   ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D)))) → (B′ = C′)) ( (B = C)         ((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧           ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧           (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ConjunctionRightElimination

(B = C)         ((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧           ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧           (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ConjunctionLeftElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent eliminateLeftConjunctInConsequent

C′ B′ C B CongruenceSymmetry eliminateBiconditionalReverse applySyllogism

C B C′ B′ CongruenceCommutativity eliminateBiconditionalReverse applySyllogism

composeConjunction

B C B′ C′ CongruenceIdentityFromEquality import applySyllogism ))

thm (cprime-dprime-c-d-case-bc-1   ((((B″ = B′) ∧ (B′ = C′)) ∧ (D′ B″ ≡ D B)) → (C′ D′ ≡ B D)) ( B″ B′ C′ EqualityTransitivity

D′ EqualityReflexivity D EqualityReflexivity B EqualityReflexivity D′ D′ B″ C′ D D B B CongruenceBuilder detach2of2 detach2of2 detach1of2

eliminateBiconditionalReverseInConsequent </jh> The top thing on the proof stack is now. We just need to combine it with the other thing on the proof stack, import, and adjust the consequent. <jh> applySyllogism

import

D′ C′ D B CongruenceCommutativity eliminateBiconditionalReverse applySyllogism ))

thm (cprime-dprime-c-d-case-bc-2   (((((B″ = B′) ∧ (B′ = C′)) ∧ (D′ B″ ≡ D B)) ∧ (B = C)) → (C′ D′ ≡ C D)) ( (((B″ = B′) ∧ (B′ = C′)) ∧ (D′ B″ ≡ D B)) (B = C)         ConjunctionRightElimination B″ B′ C′ D′ D B cprime-dprime-c-d-case-bc-1 applySyllogism

(((B″ = B′) ∧ (B′ = C′)) ∧ (D′ B″ ≡ D B)) (B = C)         ConjunctionLeftElimination C′ EqualityReflexivity D′ EqualityReflexivity D EqualityReflexivity C′ C′ D′ D′ B C D D CongruenceBuilder detach2of2 detach2of3 detach1of2 applySyllogism

eliminateBiconditionalReverseInConsequent

composeConjunction

(C′ D′ ≡ B D) (C′ D′ ≡ C D) ModusPonens applySyllogism ))

thm (cprime-dprime-c-d-case-bc   (((B = C) ∧    ((((((between A D C′) ∧ (D C′ ≡ C D)) ∧    ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧   ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D)))) → (C′ D′ ≡ C D)) ( </jh> We've proved all the interesting parts. We just need to pick out the various antecedents for. : <jh> (B = C)         ((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧           ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧           (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ConjunctionLeftElimination A D C′ C D′ B′ B B″ bdoubleprime-bprime applySyllogism </jh> <jh> B C A D C′ D′ B′ B″ bprime-cprime composeConjunction </jh> <jh> (B = C)         ((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧           ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧           (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ConjunctionLeftElimination eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent eliminateLeftConjunctInConsequent composeConjunction </jh> : <jh> (B = C)         ((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧           ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧           (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ConjunctionRightElimination composeConjunction

B″ B′ C′ D′ D B C cprime-dprime-c-d-case-bc-2 applySyllogism )) </jh>

Finishing the proof of  just involves assembling the two cases.

<jh> thm (cprime-dprime-c-d   (((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧   ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧   (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) → (C′ D′ ≡ C D)) ( B C A D C′ D′ B′ B″ cprime-dprime-c-d-case-bc export

B C A D C′ D′ B′ B″ cprime-dprime-c-d-case-bnotc export

eliminateCases )) </jh>

Construction of E
We're now ready to construct the point. We already have the betweenness relationships we need for the axiom of Pasch,  and. Applying, we get   and. <jh> thm (e-exists-1 ((e A) (e D) (e C′) (e C) (e D′) (e B′) (e B) (e B″))   (((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧   ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧   (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) → (∃ e ((between C (value e) C′) ∧ (between D (value e) D′)))) ( (((((between A D C′) ∧ (D C′ ≡ C D)) ∧           ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧           ((between A D′ B″) ∧ (D′ B″ ≡ D B))) (((A ≠ B) ∧ (between A B C)) ∧ (between A B D)) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent eliminateRightConjunctInConsequent

A C D′ BetweennessSymmetry eliminateBiconditionalReverse applySyllogism

(((((between A D C′) ∧ (D C′ ≡ C D)) ∧           ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧           ((between A D′ B″) ∧ (D′ B″ ≡ D B))) (((A ≠ B) ∧ (between A B C)) ∧ (between A B D)) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent

A D C′ BetweennessSymmetry eliminateBiconditionalReverse applySyllogism

composeConjunction

D′ C A C′ D e Pasch applySyllogism )) </jh>

Here's a variant of the previous theorem with the antecedent repeated inside the quantifier. <jh> thm (e-exists ((e A) (e D) (e C′) (e C) (e D′) (e B′) (e B) (e B″))   (((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧   ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧   (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) → (∃ e   (((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧     ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧     (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C (value e) C′) ∧ (between D (value e) D′))))) ( ((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ImplicationReflexivity

A D C′ C D′ B′ B B″ e e-exists-1

composeConjunction

e         ((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧           ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧           (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ((between C (value e) C′) ∧ (between D (value e) D′)) ThereExistsConjunctionMovement eliminateBiconditionalForward

applySyllogism )) </jh>

E C ≡ E C′
Here we apply inner five segment to prove. We are applying inner five segment to the line segments  and   and points   and. <jh> thm (e-c-e-cprime   ((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧    ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧   ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧   ((between C E C′) ∧ (between D E D′))) → (E C ≡ E C′)) ( </jh> First we need, which is one of our antecedents. <jh> ((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ((between C E C′) ∧ (between D E D′)) ConjunctionLeftElimination eliminateLeftConjunctInConsequent </jh> Secondly we need a second copy of. <jh> (between D E D′) ConjunctionIdempotence eliminateBiconditionalReverse applySyllogism </jh> Thirdly we need, which is an identity. <jh> D D′ CongruenceReflexivity (((((((between A D C′) ∧ (D C′ ≡ C D)) ∧         ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧         ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧         ((between C E C′) ∧ (between D E D′))) introduceAntecedent composeConjunction </jh> Fourth we need, which is an identity. <jh> E D′ CongruenceReflexivity (((((((between A D C′) ∧ (D C′ ≡ C D)) ∧         ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧         ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧         ((between C E C′) ∧ (between D E D′))) introduceAntecedent composeConjunction </jh> Next we need, which is one of our antecedents (slightly rearranged). <jh> ((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ((between C E C′) ∧ (between D E D′)) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent

D C′ C D CongruenceRightCommutativity eliminateBiconditionalReverse applySyllogism

D C′ D C CongruenceSymmetry eliminateBiconditionalReverse applySyllogism

composeConjunction </jh> Finally we need, which follows from the antecedent   and  , which we proved previously. <jh> ((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ((between C E C′) ∧ (between D E D′)) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent eliminateLeftConjunctInConsequent

((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ((between C E C′) ∧ (between D E D′)) ConjunctionRightElimination A D C′ C D′ B′ B B″ cprime-dprime-c-d applySyllogism

C′ D′ C D CongruenceSymmetry eliminateBiconditionalReverse applySyllogism

composeConjunction

C D′ C D C′ D′ CongruenceTransitivity applySyllogism

C D′ C′ D′ CongruenceCommutativity eliminateBiconditionalReverse applySyllogism

composeConjunction </jh> Applying inner five segment we get <jh> D E D′ D E D′ C C′ InnerFiveSegment applySyllogism )) </jh>

E D ≡ E D′
The proof that  is exactly analogous, but applies inner five segment to the line segments   and   and points   and. <jh> thm (e-d-e-dprime   ((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧    ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧   ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧   ((between C E C′) ∧ (between D E D′))) → (E D ≡ E D′)) ( </jh> First we need, which is one of our antecedents. <jh> ((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ((between C E C′) ∧ (between D E D′)) ConjunctionLeftElimination eliminateRightConjunctInConsequent </jh> Secondly we need a second copy of. <jh> (between C E C′) ConjunctionIdempotence eliminateBiconditionalReverse applySyllogism </jh> Thirdly we need, which is an identity. <jh> C C′ CongruenceReflexivity (((((((between A D C′) ∧ (D C′ ≡ C D)) ∧         ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧         ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧         ((between C E C′) ∧ (between D E D′))) introduceAntecedent composeConjunction </jh> Fourth we need, which is an identity. <jh> E C′ CongruenceReflexivity (((((((between A D C′) ∧ (D C′ ≡ C D)) ∧         ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧         ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧         ((between C E C′) ∧ (between D E D′))) introduceAntecedent composeConjunction </jh> Next we need, which is one of our antecedents (slightly rearranged). <jh> ((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ((between C E C′) ∧ (between D E D′)) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent eliminateLeftConjunctInConsequent

C D′ C D CongruenceSymmetry eliminateBiconditionalReverse applySyllogism

composeConjunction </jh> Finally we need, which follows from the antecedent   and  , which we proved previously. <jh> ((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ((between C E C′) ∧ (between D E D′)) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent

((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ((between C E C′) ∧ (between D E D′)) ConjunctionRightElimination A D C′ C D′ B′ B B″ cprime-dprime-c-d applySyllogism

C′ D′ C D CongruenceSymmetry eliminateBiconditionalReverse applySyllogism

composeConjunction

D C′ C D C′ D′ CongruenceTransitivity applySyllogism

D C′ C′ D′ CongruenceLeftCommutativity eliminateBiconditionalReverse applySyllogism

composeConjunction </jh> Applying inner five segment we get <jh> C E C′ C E C′ D D′ InnerFiveSegment applySyllogism )) </jh>

Construction of P, Q, and R
Extend  by the distance   to a point. Extend  by the distance   to a point. Extend  by the distance   to a point.

In other words, we are adding <jh> thm (p-q-r-exist ((p q r A) (p q r D) (p q r C′) (p q r C) (p q r D′) (p q r B′) (p q r B) (p q r B″) (p q r E))   ((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧    ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧   ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧   ((between C E C′) ∧ (between D E D′))) → (∃ r (∃ p (∃ q ( (((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧   ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧   ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧   ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C (value p)) ∧ (C (value p) ≡ C D′))) ∧   ((between D′ C (value r)) ∧ (C (value r) ≡ C E))) ∧ ((between (value p) (value r) (value q)) ∧ ((value r) (value q) ≡ (value r) (value p))) ))))) (       (((((((between A D C′) ∧ (D C′ ≡ C D)) ∧          ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧         ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧         ((between C E C′) ∧ (between D E D′))) ImplicationReflexivity </jh> We're ready to construct. <jh> p C′ C C D′ SegmentConstruction introduceRightConjunctToConsequent

p         (((((((between A D C′) ∧ (D C′ ≡ C D)) ∧            ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧           ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧           ((between C E C′) ∧ (between D E D′))) ((between C′ C (value p)) ∧ (C (value p) ≡ C D′)) ThereExistsConjunctionMovement eliminateBiconditionalForward applySyllogism

</jh> The construction of  is similar. <jh> r D′ C C E SegmentConstruction introduceRightConjunctToConsequent

# TODO: this can be simplified via ThereExistsScattering, right? r         (∃ p             ((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧            ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧           ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧           ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C (value p)) ∧ (C (value p) ≡ C D′)))) ((between D′ C (value r)) ∧ (C (value r) ≡ C E)) ThereExistsConjunctionMovement eliminateBiconditionalForward applySyllogism

p         ((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧           ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧           (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧           ((between C′ C (value p)) ∧ (C (value p) ≡ C D′))) ((between D′ C (value r)) ∧ (C (value r) ≡ C E)) ThereExistsConjunctionRightMovement eliminateBiconditionalForward r addThereExists applySyllogism </jh> The construction of  is only slightly different (with the most obvious difference in the logic being that it depends on   and  ). <jh> q (value p) (value r) (value r) (value p) SegmentConstruction p generalize r generalize introduceRightConjunctToConsequent

r         (∃ p (((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧           ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧           (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧           ((between C′ C (value p)) ∧ (C (value p) ≡ C D′))) ∧ ((between D′ C (value r)) ∧ (C (value r) ≡ C E)))) (∀ p (∃ q ((between (value p) (value r) (value q)) ∧ ((value r) (value q) ≡ (value r) (value p))))) ThereExistsConjunctionRightCombining applySyllogism

p         (((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧            ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧           ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧           ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C (value p)) ∧ (C (value p) ≡ C D′))) ∧           ((between D′ C (value r)) ∧ (C (value r) ≡ C E))) (∃ q ((between (value p) (value r) (value q)) ∧ ((value r) (value q) ≡ (value r) (value p)))) ThereExistsConjunctionRightCombining r addThereExists applySyllogism

q         (((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧            ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧           ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧           ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C (value p)) ∧ (C (value p) ≡ C D′))) ∧           ((between D′ C (value r)) ∧ (C (value r) ≡ C E))) ((between (value p) (value r) (value q)) ∧ ((value r) (value q) ≡ (value r) (value p))) ThereExistsConjunctionMovement eliminateBiconditionalForward p addThereExists r addThereExists applySyllogism )) </jh>

C ≠ D′
Our goal is to prove. We do this by adding  to our antecedent and then showing it implies. The first step is to show that. This is a straightforward consequence of  and the congruence of   and , but it will take a few lemmas to do all the needed arranging. <jh> thm (dprime-cprime   ( (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧   ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧   ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧   ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧   ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧   (C = D′)) → (D′ = C′)) ( </jh> First we need. <jh> ((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧         ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧         ((between P R Q) ∧ (R Q ≡ R P))) (C = D′) ConjunctionLeftElimination </jh> Secondly we need  which we derive from   (an antecedent),   (previously proved), and a bit of rearrangement. <jh> ((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧         ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧         ((between P R Q) ∧ (R Q ≡ R P))) (C = D′) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent eliminateLeftConjunctInConsequent

((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧         ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧         ((between P R Q) ∧ (R Q ≡ R P))) (C = D′) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent

A D C′ C D′ B′ B B″ cprime-dprime-c-d applySyllogism

C′ D′ C D CongruenceSymmetry eliminateBiconditionalReverse applySyllogism

composeConjunction

C D′ C D C′ D′ CongruenceTransitivity applySyllogism

C D′ C′ D′ CongruenceRightCommutativity eliminateBiconditionalReverse applySyllogism

composeConjunction </jh> Now we can show. <jh> C D′ D′ C′ CongruenceIdentityFromEquality import applySyllogism ))

thm (c-cprime   ( (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧   ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧   ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧   ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧   ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧   (C = D′)) → (C = C′)) ( ((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧         ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧         ((between P R Q) ∧ (R Q ≡ R P))) (C = D′) ConjunctionLeftElimination

A D C′ C D′ B′ B B″ E P R Q dprime-cprime

composeConjunction

C D′ C′ EqualityTransitivity applySyllogism ))

thm (c-dprime   ( (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧   ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧   ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧   ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧   ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧   (C ≠ C′)) → (C ≠ D′)) ( A D C′ C D′ B′ B B″ E P R Q c-cprime export

(C = D′) (C = C′) Transposition eliminateBiconditionalReverse applySyllogism

import )) </jh>

R P ≡ E D′
The proof is by outer five segment on  and   and points   and. We pretty much have everything we need to apply outer five segment, although  is complicated enough to break out into its own lemma. This lemma follows from  and.

<jh> thm (p-c-e   ( (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧   ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧   ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧   ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧   ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧   (C ≠ C′)) → (between P C E)) ( ((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧         ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧         ((between P R Q) ∧ (R Q ≡ R P))) (C ≠ C′) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent eliminateRightConjunctInConsequent

C′ C P BetweennessSymmetry eliminateBiconditionalReverse applySyllogism

((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧         ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧         ((between P R Q) ∧ (R Q ≡ R P))) (C ≠ C′) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent eliminateRightConjunctInConsequent

composeConjunction

P C C′ E BetweennessInnerTransitivity applySyllogism ))

thm (r-p-e-dprime   ( (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧   ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧   ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧   ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧   ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧   (C ≠ C′)) → (R P ≡ E D′)) ( </jh> First we need. We proved it in the previous section. <jh> A D C′ C D′ B′ B B″ E P R Q c-dprime

C D′ EqualitySymmetry addNegation eliminateBiconditionalReverse applySyllogism </jh> Next is, from the construction of. <jh> ((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧         ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧         ((between P R Q) ∧ (R Q ≡ R P))) (C ≠ C′) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent eliminateRightConjunctInConsequent

composeConjunction </jh> Next is. <jh> A D C′ C D′ B′ B B″ E P R Q p-c-e composeConjunction </jh> Next is, from the construction of. <jh> ((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧         ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧         ((between P R Q) ∧ (R Q ≡ R P))) (C ≠ C′) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent eliminateLeftConjunctInConsequent

C P C D′ CongruenceCommutativity eliminateBiconditionalReverse applySyllogism

P C D′ C CongruenceSymmetry eliminateBiconditionalReverse applySyllogism

composeConjunction </jh> Next is, from the construction of. <jh> ((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧         ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧         ((between P R Q) ∧ (R Q ≡ R P))) (C ≠ C′) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent eliminateLeftConjunctInConsequent

composeConjunction </jh> Next is, which holds trivially. <jh> D′ P CongruenceABBA introduceRightConjunctToConsequent </jh> Next is, from the construction of. <jh> ((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧         ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧         ((between P R Q) ∧ (R Q ≡ R P))) (C ≠ C′) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent eliminateLeftConjunctInConsequent

composeConjunction </jh> We can then conclude. <jh> D′ C R P C E P D′ OuterFiveSegment applySyllogism )) </jh>

R Q ≡ E D
This is an easy consequence of. <jh> thm (r-q-e-d   ( (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧   ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧   ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧   ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧   ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧   (C ≠ C′)) → (R Q ≡ E D)) ( </jh> We start with  (from the construction of  ). <jh> ((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧         ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧         ((between P R Q) ∧ (R Q ≡ R P))) (C ≠ C′) ConjunctionRightElimination eliminateLeftConjunctInConsequent eliminateLeftConjunctInConsequent </jh> Next is  (from the previous section). <jh> A D C′ C D′ B′ B B″ E P R Q r-p-e-dprime composeConjunction

R Q R P E D′ CongruenceTransitivity applySyllogism </jh> Finally is  (which we proved a few sections back). <jh> ((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧         ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧         ((between P R Q) ∧ (R Q ≡ R P))) (C ≠ C′) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent

A D C′ C D′ B′ B B″ E e-d-e-dprime applySyllogism

E D E D′ CongruenceSymmetry eliminateBiconditionalReverse applySyllogism

composeConjunction

R Q E D′ E D CongruenceTransitivity applySyllogism )) </jh>

D′ D ≡ P Q
The congruences from the previous section,  and , lead to  , which is a congruence we'll need later on. The proof is by outer three segment on  and. <jh> thm (dprime-d-p-q   ( (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧   ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧   ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧   ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧   ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧   (C ≠ C′)) → (D′ D ≡ P Q)) ( </jh> First is <jh> ((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧         ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧         ((between P R Q) ∧ (R Q ≡ R P))) (C ≠ C′) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent eliminateLeftConjunctInConsequent

D E D′ BetweennessSymmetry eliminateBiconditionalReverse applySyllogism </jh> Next is <jh> ((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧         ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧         ((between P R Q) ∧ (R Q ≡ R P))) (C ≠ C′) ConjunctionRightElimination eliminateLeftConjunctInConsequent eliminateRightConjunctInConsequent

composeConjunction </jh> Next is. <jh> A D C′ C D′ B′ B B″ E P R Q r-p-e-dprime

R P E D′ CongruenceSymmetry eliminateBiconditionalReverse applySyllogism

E D′ R P CongruenceCommutativity eliminateBiconditionalReverse applySyllogism

composeConjunction </jh> Last is. <jh> A D C′ C D′ B′ B B″ E P R Q r-q-e-d

R Q E D CongruenceSymmetry eliminateBiconditionalReverse applySyllogism

composeConjunction

D′ E D P R Q OuterThreeSegment applySyllogism )) </jh>

C Q ≡ C D
To prove, there are two cases, based on whether   or not.

case
We start with the case in which.

First is, which follows via   from. <jh> thm (c-q-c-d-case-dprime-e-e-d   ( ((D′ = E) ∧   (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧   ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧   (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧   ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧   ((between P R Q) ∧ (R Q ≡ R P))) ∧ (C ≠ C′))) → (E = D)) ( (D′ = E)         (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧          ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧         ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧         ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧         ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧         (C ≠ C′)) ConjunctionRightElimination

(D′ = E)         (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧          ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧         ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧         ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧         ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧         (C ≠ C′)) ConjunctionLeftElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent

A D C′ C D′ B′ B B″ E e-d-e-dprime applySyllogism

E D E D′ CongruenceSymmetry eliminateBiconditionalReverse applySyllogism

E D′ E D CongruenceLeftCommutativity eliminateBiconditionalReverse applySyllogism

composeConjunction

D′ E E D CongruenceIdentityFromEquality import

applySyllogism ))

thm (c-q-c-d-case-dprime-e-dprime-d   ( ((D′ = E) ∧   (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧   ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧   (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧   ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧   ((between P R Q) ∧ (R Q ≡ R P))) ∧ (C ≠ C′))) → (D′ = D)) ( (D′ = E)         (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧          ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧         ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧         ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧         ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧         (C ≠ C′)) ConjunctionRightElimination

D′ E A D C′ C B′ B B″ P R Q c-q-c-d-case-dprime-e-e-d composeConjunction

D′ E D EqualityTransitivity applySyllogism )) </jh>

It might seem that having proved  we are done with this case (since after all,   is the long-term goal of this proof). Although we could, of course, proceed this way, it would entail carrying along the two cases to almost the end of the proof. So we instead will take a few more steps to prove. The next one is.

<jh> thm (c-q-c-d-case-dprime-e-p-q   ( ((D′ = E) ∧   (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧   ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧   (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧   ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧   ((between P R Q) ∧ (R Q ≡ R P))) ∧ (C ≠ C′))) → (P = Q)) ( D′ E A D C′ C B′ B B″ P R Q c-q-c-d-case-dprime-e-dprime-d

(D′ = E)         (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧          ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧         ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧         ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧         ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧         (C ≠ C′)) ConjunctionLeftElimination A D C′ C D′ B′ B B″ E P R Q dprime-d-p-q applySyllogism

composeConjunction

D′ D P Q CongruenceIdentityFromEquality import

applySyllogism )) </jh>

Now that we have, we just need to apply transitivity and a substitution to   and   to get. <jh> thm (c-q-c-d-case-dprime-e   ( ((D′ = E) ∧   (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧   ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧   (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧   ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧   ((between P R Q) ∧ (R Q ≡ R P))) ∧ (C ≠ C′))) → (C Q ≡ C D)) ( D′ E A D C′ C B′ B B″ P R Q c-q-c-d-case-dprime-e-p-q </jh> Now we need. <jh> (D′ = E)         (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧          ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧         ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧         ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧         ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧         (C ≠ C′)) ConjunctionLeftElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent eliminateLeftConjunctInConsequent

(D′ = E)         (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧          ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧         ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧         ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧         ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧         (C ≠ C′)) ConjunctionLeftElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent eliminateLeftConjunctInConsequent

composeConjunction

C P C D′ C D CongruenceTransitivity applySyllogism

composeConjunction </jh> <jh> C EqualityReflexivity C EqualityReflexivity D EqualityReflexivity C C P Q C C D D CongruenceBuilder detach2of2 detach2of2 detach1of2 </jh> A bit of rearranging and we are done. <jh> eliminateBiconditionalReverseInConsequent import applySyllogism )) </jh>

case
The more interesting case is. It follows from outer five segment on line segments  and   and points   and. <jh> thm (c-q-c-d-case-dprime-not-e   ( ((D′ ≠ E) ∧   (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧   ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧   (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧   ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧   ((between P R Q) ∧ (R Q ≡ R P))) ∧ (C ≠ C′))) → (C Q ≡ C D)) ( </jh> First we need. <jh> (D′ ≠ E)         (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧          ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧         ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧         ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧         ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧         (C ≠ C′)) ConjunctionRightElimination </jh> Next we need. <jh> (D′ ≠ E)         (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧          ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧         ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧         ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧         ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧         (C ≠ C′)) ConjunctionLeftElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent eliminateLeftConjunctInConsequent

D E D′ BetweennessSymmetry eliminateBiconditionalReverse applySyllogism

composeConjunction </jh> Next is. <jh> (D′ ≠ E)         (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧          ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧         ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧         ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧         ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧         (C ≠ C′)) ConjunctionLeftElimination eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent eliminateRightConjunctInConsequent

composeConjunction </jh> Next is, which we proved a few sections back. <jh> (D′ ≠ E)         (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧          ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧         ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧         ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧         ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧         (C ≠ C′)) ConjunctionLeftElimination

A D C′ C D′ B′ B B″ E P R Q r-p-e-dprime applySyllogism

R P E D′ CongruenceSymmetry eliminateBiconditionalReverse applySyllogism

E D′ R P CongruenceCommutativity eliminateBiconditionalReverse applySyllogism

composeConjunction </jh> Next is, which we proved a few sections back. <jh> (D′ ≠ E)         (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧          ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧         ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧         ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧         ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧         (C ≠ C′)) ConjunctionLeftElimination

A D C′ C D′ B′ B B″ E P R Q r-q-e-d applySyllogism

R Q E D CongruenceSymmetry eliminateBiconditionalReverse applySyllogism

composeConjunction </jh> Here is, from the construction of. <jh> (D′ ≠ E)         (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧          ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧         ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧         ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧         ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧         (C ≠ C′)) ConjunctionLeftElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent eliminateLeftConjunctInConsequent

C P C D′ CongruenceSymmetry eliminateBiconditionalReverse applySyllogism

C D′ C P CongruenceCommutativity eliminateBiconditionalReverse applySyllogism

composeConjunction </jh> Finally we need, which comes from the construction of. <jh> (D′ ≠ E)         (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧          ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧         ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧         ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧         ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧         (C ≠ C′)) ConjunctionLeftElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent eliminateLeftConjunctInConsequent

C R C E CongruenceSymmetry eliminateBiconditionalReverse applySyllogism

C E C R CongruenceCommutativity eliminateBiconditionalReverse applySyllogism

composeConjunction </jh> From outer five segment we can conclude. <jh> D′ E D P R Q C C OuterFiveSegment applySyllogism

D C Q C CongruenceSymmetry eliminateBiconditionalReverse applySyllogism

Q C D C CongruenceCommutativity eliminateBiconditionalReverse applySyllogism )) </jh>

Combining the two cases
<jh> thm (c-q-c-d   ( (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧   ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧   ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧   ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧   ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧   (C ≠ C′)) → (C Q ≡ C D)) ( D′ E A D C′ C B′ B B″ P R Q c-q-c-d-case-dprime-e export

D′ E A D C′ C B′ B B″ P R Q c-q-c-d-case-dprime-not-e export

eliminateCases )) </jh>

A Corollary: C P ≡ C Q
That the previous result implies  is apparent from looking at the single ticked line segments in the diagram. The proof is just some congruence transitivity. <jh> thm (c-p-c-q   ( (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧   ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧   ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧   ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧   ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧   (C ≠ C′)) → (C P ≡ C Q)) ( A D C′ C D′ B′ B B″ E P R Q c-q-c-d

((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧         ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧         ((between P R Q) ∧ (R Q ≡ R P))) (C ≠ C′) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent eliminateLeftConjunctInConsequent

C D′ C D CongruenceSymmetry eliminateBiconditionalReverse applySyllogism

composeConjunction

C Q C D C D′ CongruenceTransitivity applySyllogism

((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧         ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧         ((between P R Q) ∧ (R Q ≡ R P))) (C ≠ C′) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent eliminateLeftConjunctInConsequent

C P C D′ CongruenceSymmetry eliminateBiconditionalReverse applySyllogism

composeConjunction

C Q C D′ C P CongruenceTransitivity applySyllogism

C Q C P CongruenceSymmetry eliminateBiconditionalReverse applySyllogism )) </jh>

The line containing C, R, D′, B, B′, C′, and P
We've set up the dominos, and in this section we push them over. For a domino (point)  to "fall" means that we prove   (via  ). So we need three collinear points, the first two of which are not equal, and the first two of which already have their congruences with  and   proven. That gives us the third congruence, and then this domino is ready to push on one of the following ones.

D′
We start with,  , and.

The first part is. We will show that  implies , so a transposition gives us what we want. <jh> thm (c-e-lemma   (((C R ≡ C E) ∧ (R = C)) → (C = E)) ( C R C E CongruenceLeftCommutativity eliminateBiconditionalReverse

R C C E CongruenceIdentityFromEquality applyComm applySyllogism

import ))

thm (c-e   ( (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧   ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧   ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧   ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧   ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧   (R = C)) → (C = E)) ( ((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧         ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧         ((between P R Q) ∧ (R Q ≡ R P))) (R = C)         ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent eliminateLeftConjunctInConsequent

((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧         ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧         ((between P R Q) ∧ (R Q ≡ R P))) (R = C)         ConjunctionLeftElimination

composeConjunction

C R E c-e-lemma

applySyllogism ))

thm (e-cprime   ( (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧   ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧   ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧   ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧   ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧   (R = C)) → (E = C′)) ( A D C′ C D′ B′ B B″ E P R Q c-e

C E EqualitySymmetry eliminateBiconditionalReverse applySyllogism

((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧         ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧         ((between P R Q) ∧ (R Q ≡ R P))) (R = C)         ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent

A D C′ C D′ B′ B B″ E e-c-e-cprime applySyllogism

composeConjunction

E C E C′ CongruenceIdentityFromEquality import

applySyllogism ))

thm (r-c-implies-c-cprime   ( (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧   ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧   ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧   ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧   ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧   (R = C)) → (C = C′)) ( A D C′ C D′ B′ B B″ E P R Q c-e A D C′ C D′ B′ B B″ E P R Q e-cprime composeConjunction

C E C′ EqualityTransitivity applySyllogism ))

thm (r-not-c   ( (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧   ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧   ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧   ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧   ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧   (C ≠ C′)) → (R ≠ C)) ( A D C′ C D′ B′ B B″ E P R Q r-c-implies-c-cprime export

(R = C) (C = C′) Transposition eliminateBiconditionalReverse applySyllogism

import )) </jh>

Now we are ready to prove. <jh> thm (dprime-p-dprime-q   ( (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧   ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧   ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧   ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧   ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧   (C ≠ C′)) → (D′ P ≡ D′ Q)) ( </jh> First we need. <jh> A D C′ C D′ B′ B B″ E P R Q r-not-c </jh> Next is. <jh> ((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧         ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧         ((between P R Q) ∧ (R Q ≡ R P))) (C ≠ C′) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent eliminateRightConjunctInConsequent

D′ C R BetweennessSymmetry eliminateBiconditionalReverse applySyllogism

R C D′ BetweennessCollinearity applySyllogism

composeConjunction </jh> Next is. <jh> ((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧         ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧         ((between P R Q) ∧ (R Q ≡ R P))) (C ≠ C′) ConjunctionRightElimination eliminateLeftConjunctInConsequent eliminateLeftConjunctInConsequent

R Q R P CongruenceSymmetry eliminateBiconditionalReverse applySyllogism

composeConjunction </jh> Finally is, which lets us apply. <jh> A D C′ C D′ B′ B B″ E P R Q c-p-c-q

composeConjunction

R C D′ P Q EquidistantLine applySyllogism )) </jh>

B′
The next point is, which follows from the results for   and. We break out one substitution into a lemma, but other than that we have everything we need.

<jh> thm (c-dprime-bprime   ( (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧   ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧   ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧   ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧   ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧   (C ≠ C′)) → (between C D′ B′)) ( </jh> First we put on the proof stack two things:  and. <jh> ((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧         ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧         ((between P R Q) ∧ (R Q ≡ R P))) (C ≠ C′) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent

A D C′ C D′ B′ B B″ bdoubleprime-dprime-c applySyllogism

((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧         ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧         ((between P R Q) ∧ (R Q ≡ R P))) (C ≠ C′) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent

A D C′ C D′ B′ B B″ bdoubleprime-bprime applySyllogism </jh> Next comes our substitution:. <jh> D′ EqualityReflexivity C EqualityReflexivity B″ B′ D′ D′ C C BetweennessBuilder detach2of2 detach2of2 </jh> We detach that twice to get. <jh> eliminateBiconditionalReverseInConsequent applySyllogism

applyModusPonensInConsequent </jh> We swap the endpoints, as the swapped version will be more convenient later. <jh> B′ D′ C BetweennessSymmetry eliminateBiconditionalReverse applySyllogism ))

thm (bprime-p-bprime-q   ( (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧   ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧   ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧   ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧   ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧   (C ≠ C′)) → (B′ P ≡ B′ Q)) ( </jh> The first part is. <jh> A D C′ C D′ B′ B B″ E P R Q c-dprime </jh> Next is <jh> A D C′ C D′ B′ B B″ E P R Q c-dprime-bprime C D′ B′ BetweennessCollinearity applySyllogism

composeConjunction </jh> Next is <jh> A D C′ C D′ B′ B B″ E P R Q c-p-c-q

composeConjunction </jh> Last is <jh> A D C′ C D′ B′ B B″ E P R Q dprime-p-dprime-q

composeConjunction

C D′ B′ P Q EquidistantLine applySyllogism )) </jh>

B
The next point is, which also follows from the results for   and. We have everything we need here with even less work than in the previous section.

<jh> thm (b-p-b-q   ( (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧   ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧   ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧   ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧   ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧   (C ≠ C′)) → (B P ≡ B Q)) ( </jh> The first part is. <jh> A D C′ C D′ B′ B B″ E P R Q c-dprime </jh> Next is <jh> ((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧         ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧         ((between P R Q) ∧ (R Q ≡ R P))) (C ≠ C′) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent

A D C′ C D′ B′ B B″ b-c-dprime applySyllogism

B C D′ BetweennessCollinearity applySyllogism

B C D′ CollinearityRotation eliminateBiconditionalReverse applySyllogism

composeConjunction </jh> Next is <jh> A D C′ C D′ B′ B B″ E P R Q c-p-c-q

composeConjunction </jh> Last is <jh> A D C′ C D′ B′ B B″ E P R Q dprime-p-dprime-q

composeConjunction

C D′ B P Q EquidistantLine applySyllogism )) </jh>

C′
The next point is, which follows from the results for   and.

Before we get to the main proof of this, we prove a few lemmas. The first one we need is. As we already have , this is just a substitution.

<jh> thm (bprime-c-b    (((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧   ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧   (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) → (between B′ C B)) ( A D C′ C D′ B′ B B″ bdoubleprime-c-b

A D C′ C D′ B′ B B″ bdoubleprime-bprime

C EqualityReflexivity B EqualityReflexivity B″ B′ C C B B BetweennessBuilder detach2of2 detach2of2

eliminateBiconditionalReverseInConsequent

applySyllogism

applyModusPonensInConsequent )) </jh>

The next lemma is. As usual, this will be via the contrapositive.

<jh> thm (b-bprime-c-cprime   ( (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧   ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧   ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧   ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧   ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧   (B = B′)) → (C = C′)) ( </jh> First we prove, by means of. <jh> ((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧         ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧         ((between P R Q) ∧ (R Q ≡ R P))) (B = B′) ConjunctionLeftElimination

((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧         ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧         ((between P R Q) ∧ (R Q ≡ R P))) (B = B′) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent

A D C′ C D′ B′ B B″ bprime-c-b applySyllogism

B′ C B BetweennessSymmetry eliminateBiconditionalReverse applySyllogism

composeConjunction

B B′ C IndivisibilityFromEquality applySyllogism

B C EqualitySymmetry eliminateBiconditionalReverse applySyllogism </jh> Then we prove, by means of. <jh> ((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧         ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧         ((between P R Q) ∧ (R Q ≡ R P))) (B = B′) ConjunctionLeftElimination

((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧         ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧         ((between P R Q) ∧ (R Q ≡ R P))) (B = B′) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent A D C′ C D′ B′ B B″ b-cprime-bprime applySyllogism

composeConjunction

B B′ C′ IndivisibilityFromEquality applySyllogism </jh> Now we combine  and   to produce. <jh> composeConjunction

C B C′ EqualityTransitivity applySyllogism ))

thm (b-bprime   ( (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧   ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧   ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧   ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧   ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧   (C ≠ C′)) → (B ≠ B′)) ( A D C′ C D′ B′ B B″ E P R Q b-bprime-c-cprime export

(B = B′) (C = C′) Transposition eliminateBiconditionalReverse applySyllogism

import ))

thm (cprime-p-cprime-q   ( (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧   ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧   ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧   ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧   ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧   (C ≠ C′)) → (C′ P ≡ C′ Q)) ( </jh> First is. <jh> A D C′ C D′ B′ B B″ E P R Q b-bprime </jh> Next is. <jh> ((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧         ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧         ((between P R Q) ∧ (R Q ≡ R P))) (C ≠ C′) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent

A D C′ C D′ B′ B B″ b-cprime-bprime applySyllogism

B C′ B′ BetweennessCollinearity applySyllogism

B C′ B′ Collinearity23 eliminateBiconditionalReverse applySyllogism

composeConjunction </jh> Next is <jh> A D C′ C D′ B′ B B″ E P R Q b-p-b-q

composeConjunction </jh> Last is <jh> A D C′ C D′ B′ B B″ E P R Q bprime-p-bprime-q

composeConjunction

B B′ C′ P Q EquidistantLine applySyllogism )) </jh>

P
The next point is, which follows from the results for   and. It would be harder to draw a diagram for this one (if the diagram were to show the congruences as well as just,  , and  ), but the proof is much the same as the ones we've done so far.

<jh> thm (p-p-p-q   ( (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧   ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧   ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧   ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧   ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧   (C ≠ C′)) → (P P ≡ P Q)) ( </jh> First is. <jh> ((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧         ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧         ((between P R Q) ∧ (R Q ≡ R P))) (C ≠ C′) ConjunctionLeftElimination

C C′ EqualitySymmetry addNegation eliminateBiconditionalReverse applySyllogism </jh> Next is. <jh> ((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧         ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧         (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧         ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧         ((between P R Q) ∧ (R Q ≡ R P))) (C ≠ C′) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent eliminateRightConjunctInConsequent

C′ C P BetweennessCollinearity applySyllogism

composeConjunction </jh> Next is <jh> A D C′ C D′ B′ B B″ E P R Q cprime-p-cprime-q

composeConjunction </jh> Last is <jh> A D C′ C D′ B′ B B″ E P R Q c-p-c-q

composeConjunction

C′ C P P Q EquidistantLine applySyllogism )) </jh>

Assembling the result
Having proved, we are most of the way done.

P = Q
<jh> thm (p-q   ( (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧   ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧   ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧   ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧   ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧   (C ≠ C′)) → (P = Q)) ( A D C′ C D′ B′ B B″ E P R Q p-p-p-q P P Q CongruenceIdentityLeft applySyllogism )) </jh>

D = D′
Here's a variant of. Because we (almost?) always want the imported one, the plan is to phase out the exported version in favor of the imported one. <jh> var (point x y z w) thm (CongruenceIdentityFromEqualityImported  (((x = y) ∧ (x y ≡ z w)) → (z = w)) ( x y z w CongruenceIdentityFromEquality import )) </jh>

And  follows from   and. <jh> thm (d-dprime   ( (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧   ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧   ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧   ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧   ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧   (C ≠ C′)) → (D = D′)) ( A D C′ C D′ B′ B B″ E P R Q p-q

A D C′ C D′ B′ B B″ E P R Q dprime-d-p-q

D′ D P Q CongruenceSymmetry eliminateBiconditionalReverse applySyllogism

composeConjunction

P Q D′ D CongruenceIdentityFromEqualityImported applySyllogism

D′ D EqualitySymmetry eliminateBiconditionalReverse applySyllogism )) </jh>

between A C D ∨ between A D C
In this section we prove that the long list of antecedents we have been working with implies. There are a few steps. <jh> thm (c-cprime-or-d-dprime   ( ((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧   ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧   (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧   ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧   ((between P R Q) ∧ (R Q ≡ R P))) → ((C = C′) ∨ (D = D′))) ( A D C′ C D′ B′ B B″ E P R Q d-dprime export

(C = C′) (D = D′) DisjunctionImplication eliminateBiconditionalForward applySyllogism ))

thm (a-d-c   ( (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧   ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧   ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧   ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧   ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧   (C = C′)) → (between A D C)) ( (((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧         ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧         ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧         ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧         ((between D′ C R) ∧ (C R ≡ C E))) ((between P R Q) ∧ (R Q ≡ R P)) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent </jh> The substitution is, which we then apply to prove the desired result. <jh> A EqualityReflexivity D EqualityReflexivity A A D D C C′ BetweennessBuilder detach2of3 detach1of2

eliminateBiconditionalForwardInConsequent applyComm

applySyllogism import ))

thm (a-c-d   ( (((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧   ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧   ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧   ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧   ((between D′ C R) ∧ (C R ≡ C E))) ∧ ((between P R Q) ∧ (R Q ≡ R P))) ∧   (D = D′)) → (between A C D)) ( (((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧         ((between A C D′) ∧ (C D′ ≡ C D))) ∧ ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧         ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧ (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧         ((between C E C′) ∧ (between D E D′))) ∧ ((between C′ C P) ∧ (C P ≡ C D′))) ∧         ((between D′ C R) ∧ (C R ≡ C E))) ((between P R Q) ∧ (R Q ≡ R P)) ConjunctionRightElimination eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateRightConjunctInConsequent eliminateLeftConjunctInConsequent eliminateRightConjunctInConsequent </jh> The substitution is, which we then apply to prove the desired result. <jh> A EqualityReflexivity C EqualityReflexivity A A C C D D′ BetweennessBuilder detach2of3 detach1of2

eliminateBiconditionalForwardInConsequent applyComm

applySyllogism import ))

thm (a-c-d-or-a-d-c   ( ((((((((((between A D C′) ∧ (D C′ ≡ C D)) ∧ ((between A C D′) ∧ (C D′ ≡ C D))) ∧   ((between A C′ B′) ∧ (C′ B′ ≡ C B))) ∧ ((between A D′ B″) ∧ (D′ B″ ≡ D B))) ∧   (((A ≠ B) ∧ (between A B C)) ∧ (between A B D))) ∧ ((between C E C′) ∧ (between D E D′))) ∧   ((between C′ C P) ∧ (C P ≡ C D′))) ∧ ((between D′ C R) ∧ (C R ≡ C E))) ∧   ((between P R Q) ∧ (R Q ≡ R P))) → ((between A C D) ∨ (between A D C))) ( A D C′ C D′ B′ B B″ E P R Q c-cprime-or-d-dprime

A D C′ C D′ B′ B B″ E P R Q a-d-c (between A D C) (between A C D) DisjunctionLeftIntroduction applySyllogism export

A D C′ C D′ B′ B B″ E P R Q a-c-d (between A C D) (between A D C) DisjunctionRightIntroduction applySyllogism export

composeConjunction

(C = C′) ((between A C D) ∨ (between A D C)) (D = D′) DisjunctionComposition eliminateBiconditionalReverse applySyllogism

applyModusPonensInConsequent )) </jh>

Betweenness outer connectivity
We now are prepared to take the various lemmas which represent the construction of the various points, tie them together (basically just syllogisms, with suitable handling of ∃), and then remove the quantifiers once they are no longer needed. <jh> thm (BetweennessOuterConnectivity ( (A c′ d′ b′ b″ e r p q)   (B c′ d′ b′ b″ e r p q)   (C c′ d′ b′ b″ e r p q)   (D c′ d′ b′ b″ e r p q))    ((((A ≠ B) ∧ (between A B C)) ∧ (between A B D)) → ((between A C D) ∨ (between A D C)))  ( A B C D c′ d′ b′ b″ BetweennessOuterConnectivity-cdbb

A D (value c′) C (value d′) (value b′) B (value b″) e e-exists b″ addThereExists b′ addThereExists d′ addThereExists c′ addThereExists applySyllogism

A D (value c′) C (value d′) (value b′) B (value b″) (value e) r p q p-q-r-exist e addThereExists b″ addThereExists b′ addThereExists d′ addThereExists c′ addThereExists applySyllogism

A D (value c′) C (value d′) (value b′) B (value b″) (value e) (value p) (value r) (value q) a-c-d-or-a-d-c q addThereExists p addThereExists r addThereExists e addThereExists b″ addThereExists b′ addThereExists d′ addThereExists c′ addThereExists applySyllogism

removeThereExistsInConsequent removeThereExistsInConsequent removeThereExistsInConsequent removeThereExistsInConsequent removeThereExistsInConsequent removeThereExistsInConsequent removeThereExistsInConsequent removeThereExistsInConsequent )) </jh>

Export
We now export to Interface:Connectivity for betweenness. <jh> export (RESULTS Interface:Connectivity_for_betweenness (CLASSICAL FIRSTORDER) ) </jh>