formalizing projective geometry in coq - unistra.frnarboux/slides/gdr-ltp-09.pdfformalizing...

Formalizing projective geometry in Coq

Nicolas Magaud Julien Narboux Pascal Schreck

Universite de Strasbourg, France

GDR LTP, 21 Octobre 2009, Universite Paris XII

Magaud Schreck Narboux (UdS) Formalizing projective geometry GDR LTP 1 / 23

Outline

1 Objectives

2 Formal proofs in geometry

3 Projective geometry

4 A case study : Desargues’ theorem

5 Formalization in Coq

6 Future work and conclusion


Geometric contraint solving and automated deduction in

geometry

1 Proofs of theorems in geometry are needed for solving geometriccontraint systems.

2 Most methods in automated deduction in geometry (Wu’s method,the area method) can only deal with geometric statements expressedas a ruler and compass construction.

We need a system to prove both mathematical theorems and programs!


Related work

Geometry

Hilbert’s Grundlagen [DDS00, MF03]

Tarski’s geometry [Nar07]

High-school geometry [Gui05]

Hessenberg’s theorem [BH08]

Convex-hulls algorithm [MF05] [PB01]

Image segmentation algorithm [Duf07]

Degenerated cases

Using Three-Valued Logic to Specify and Verify, Brandt andSchneider [BS05]

Using matroids

Incidence constraints, a combinatorial approach,Schreck-Michelucci[MS06]


Why projective geometry ?

Non degeneracy conditions

A simple framework


Axiom system (2D)

Line-Existence∀A B : Point, (∃l : Line,A ∈ l ∧ B ∈ l)

Point-Existence∀l m : Line, (∃A : Point,A ∈ l ∧ A ∈ m)

Axiom Uniqueness

∀A B : Point,∀l m : Line,A ∈ l ⇒ B ∈ l ⇒ A ∈ m ⇒ B ∈ m ⇒ A = B ∨ l = m

Axiom Four Points

∃A : Point,∃B : Point,∃C : Point,∃D : Point,distinct4A B C D ∧(∀l : Line, (A ∈ l ∧ B ∈ l ⇒ C /∈ l ∧ D /∈ l)∧(A ∈ l ∧ C ∈ l ⇒ B /∈ l ∧ D /∈ l)∧(A ∈ l ∧ D ∈ l ⇒ B /∈ l ∧ C /∈ l)∧(B ∈ l ∧ C ∈ l ⇒ A /∈ l ∧ D /∈ l)∧(B ∈ l ∧ D ∈ l ⇒ A /∈ l ∧ C /∈ l)∧(C ∈ l ∧ D ∈ l ⇒ A ∈ l ∧ B ∈ l))


Axiom system (≥ 3D)

Line-Existence ∀A B : Point,∃l : Line,A ∈ l ∧ B ∈ l

Pasch

∀A B C D : Point,∀lAB lCD lAC lBD : Line,A 6=B ∧ A 6=C ∧ A 6=D ∧ B 6=C ∧ B 6=D ∧ C 6=D∧A ∈ lAB ∧ B ∈ lAB ∧ C ∈ lCD ∧ D ∈ lCD∧A ∈ lAC ∧ C ∈ lAC ∧ B ∈ lBD ∧ D ∈ lBD∧(∃I : Point, I ∈ lAB ∧ I ∈ lCD) ⇒(∃J : Point, J ∈ lAC ∧ J ∈ lBD)

bA

bB

b Cb

D

bI

b J

Uniqueness ∀A B : Point,∀l m : Line,A ∈ l ∧ B ∈ l ∧ A ∈ m ∧ B ∈ m ⇒ A = B ∨ l = m

Three-Points ∀l : Line,∃A B C : Point,A 6= B ∧ B 6= C ∧ A 6= C ∧ A ∈ l ∧ B ∈ l ∧ C ∈ l

Lower-Dimension ∃l m : Line,∀p : Point, p 6∈ l ∨ p 6∈ m


Duality

Proving some theorems by duality:

3 points on a line ⇒ 3 lines through a point

A

BbC

b

b

⇒

Pb

1 point outside l1 and l2 ⇒ 1 line not through P1 and P2

b P

⇒

b P1

b P2

The solution: a functor from Projective Plane Geometry to itself


Implementation in Coq !

Module types (boxes) and Functors (arrows)

duality55

ProjectivePlane′ A1A2A′

3

back

44ProjectivePlane A1A2A3

forth

ss

Duality as a functorModule swap (M’: ProjectivePlane’) <: ProjectivePlane’.

Definition Point := M’.Line.

Definition Line := M’.Point.

Definition Incid := fun (x:Point) (y:Line) => M’.Incid y x.

[...]

Definition a1_exist := M’.a2_exist.

Definition a2_exist := M’.a1_exist.

[...]

ApplicationLemma outsider : forall l1 l2: Line, { P:Point | ~Incid P l1/\~Incid P l2}.

Proof. [...] Qed.

Lemma dual_example :

forall P1 P2 : Point,{l : Line | ~ Incid P1 l /\ ~ Incid P2 l}.

Proof. apply ProjectivePlaneFacts_m.outsider. Qed.


Models

Finite ModelsPG (dim, b) where given a point on a line, b is the number of otherlines through the point.

Fano’s plane ( = PG(2, 2) the smallest projective plane)

b

Ab

B

b

bC

b E

b

F

bDbG

PG(2,5) (31 points and as many lines, thus 923 521 cases)

Infinite Model: Homogeneous Coordinates


Desargues’ theorem

let E be a 3D or higher projective space, if the 3 lines joining the corres-ponding vertices of triangles ABC and A′B ′C ′ all meet in a point O, thenthe 3 intersections of pairs of corresponding sides α, β and γ lie on a line.

b

A

b

B

b

C

b O

b

C’

b

A’b

B’

b

b

b

β

γ

α


Why 3D ?

Desargue’s theorem is false in some 2D models.

Moulton’s plane

b b

b

b

b

b

b

bb

b A BC

A’B’

C’

O

α

γ

β

b b

b

b

b

b

b

b

b

b

β


Hessenberg’s theorem

If Pappus holds then Desargues holds as well.

Pappus

b

Ab

Bb

C

b

A’

b

B’

b

C’

b

Pb

Qb

R


The proof

Outline of the proof:

We prove the 3D versionof Desargues’ theorem.

We lift the 2D statementto a 3D version ofDesargues’ theoremwhich projects into the2D statement.

b

A

b

B

b

C

b O

b

C’

b

A’b

B’

b

b

b

β

γ

α


The proof




b

A

b

B

b

C

b O

b

C’

b

A’b

B’

bP

b

b

b

β

γ

α


The proof




b

A

b

B

b

C

b O

b

C’

b

A’b

B’

bP

bo

b c

ba

b b

b

b

b

β

γ

α


The proof




b

A

b

B

b

C

b O

b

C’

b

A’b

B’

bP

bo

b c

bab b

b

b

b

β

γ

α


The proof




b

A

b

B

b

C

b O

b

C’

b

A’b

B’

bP

bo

b c

ba b b

b

b

b

β

γ

α


The proof




b

A

b

B

b

C

b O

b

C’

b

A’b

B’

bP

bo

b cba b b

b

b

b

β

γ

α


Formalization in Coq

We use the concept of rank:

R1 ∀X ⊆E , 0≤ rk(X )≤|X | (nonnegative and subcardinal)

R2 ∀XY ⊆E ,X ⊆ Y ⇒ rk(X ) ≤ rk(Y ) (nondecreasing)

R3 ∀XY ⊆E , rk(X ∪ Y ) + rk(X ∩ Y ) ≤ rk(X ) + rk(Y )(submodular)


Example

rk{A,B} = 1 A = B

rk{A,B} = 2 A 6= B

rk{A,B ,C} = 2 A,B ,C are collinear

with at least two of them distinct

rk{A,B ,C} ≤ 2 A,B ,C are collinear

rk{A,B ,C} = 3 A,B ,C are not collinear

rk{A,B ,C ,D} = 3 A,B ,C ,D are co-planar,

not all collinear

rk{A,B ,C ,D} = 4 A,B ,C ,D are not co-planar

rk{A,B ,C ,D,E} ≤ 2 A,B ,C ,D,E are all collinear


Axiom system to capture projective at least 3-dimensional space

Rk-Singleton ∀P : Point, rk{P} ≥ 1

Rk-Couple ∀P Q : Point,P 6= Q ⇒ rk{P ,Q} ≥ 2

Rk-Pasch ∀A B C D, rk{A,B ,C ,D} ≤ 3 ⇒∃J, rk{A,B , J} = rk{C ,D, J} = 2

Rk-Three-Points∀A B ,∃C , rk{A,B ,C} = rk{B ,C} = rk{A,C} = 2

Rk-Lower-Dimension ∃A B C D, rk{A,B ,C ,D} ≥ 4


Outline of the proof

1 general lemmas (framework)

2 proof of the 3D statement

3 proof of the 2D statement by projection


Proof technique

Using axiom R3

∀XY , rk(X ∪ Y ) + rk(X ∩ Y ) ≤ rk(X ) + rk(Y )

⇓

rk{A,B ,C ,D, I} + rk{ I} ≤ rk{A,B , I} + rk{C ,D, I}


Proof technique

Using axiom R3


6⇓

rk{A,B ,C ,D, I} + rk{A, I} ≤ rk{A,B , I} + rk{C ,D, I}

This is false. If for instance A = C , {A,B , I} ∩ {C ,D, I} = {A, I}


Proof technique

Using axiom R3


6⇓

rk{A,B ,C ,D, I} + rk{A, I} ≤ rk{A,B , I} + rk{C ,D, I}

This is false. If for instance A = C , {A,B , I} ∩ {C ,D, I} = {A, I}

Lemma (R3-alt)

∀XYI , I ⊆ X ∩ Y ⇒ rk(X ∪ Y ) + rk(I ) ≤ rk(X ) + rk(Y )


Conclusion

A formalization of duality

Models

An axiom system to capture projective geometry based on ranks.

A way to express incidence relations thanks to ranks.

A case study: Desargues.

≥ 15k lines of Coq

Future work

Automatic proofs using hexamys.


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