Formalizing Desargues' theorem in Coq using ranks in Coq

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Formalizing Desargues’ theorem in Coq using ranks in

Coq

Nicolas Magaud, Julien Narboux, Pascal Schreck

To cite this version:

Nicolas Magaud, Julien Narboux, Pascal Schreck. Formalizing Desargues’ theorem in Coq using ranks

in Coq. 24th Annual ACM Symposium on Applied Computing, Xiao-Shan Gao, Robert Joan-Arinyo,

Dominique Michelucci, Mar 2009, Honolulu, United States. pp.1110-1115, �10.1145/1529282.1529527�.

�inria-00335719�

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Formalizing Desargues’ theorem in Coq using ranks

∗

Nicolas Magaud

LSIIT UMR 7005

CNRS-ULP

Université de Strasbourg

Nicolas.Magaud@dpt-info.u-strasbg.fr

Julien Narboux

LSIIT UMR 7005

CNRS-ULP

Université de Strasbourg

Julien.Narboux@dpt-info.u-strasbg.fr

Pascal Schreck

LSIIT UMR 7005

CNRS-ULP

Université de Strasbourg

Pascal.Schreck@dpt-info.u-strasbg.fr

ABSTRACT

FormalizinggeometrytheoremsinaproofassistantlikeCoq is hallenging. As emphasized inthe literature, the non-degenera y onditions leads to te hni al proofs. In addi-tion,when onsideringhigher-dimensions,theamountof in- iden erelations(e.g. point-line,point-plane,line-plane) in-du enumerouste hni allemmas. Inthispaper,wepresent anoriginalapproa hbasedonthenotionofrankwhi h al-lowsto des ribein iden eandnon-in iden erelations su h asequality, ollinearity and oplanarity homogeneously. It allowsto arryoutproofsinamoresystemati way. To val-idatethisapproa h,weformalizeinCoq(usingonlyranks) one of the fundamental theorems of the proje tive spa e, namelyDesargues'theorem.

Categories and Subject Descriptors

I.2.4 [COMPUTING METHODOLOGIES℄: Knowl-edge Representation Formalisms and Methods; G.4 [MA-THEMATICAL SOFTWARE℄: Certi ation and test-ing

Keywords

formalization,Desargues,rank,proje tivegeometry,Coq

1. INTRODUCTION

ThispaperdealswithformallyprovinginCoqDesargues' theorem inat least3-dimensional proje tive geometry us-ingthe on eptofrank. Inthelongerterm,theunderlying obje tiveofthepresentedwork onsistsindesigninga geo-metri formalproverabletohandlethenon-degenera y on-ditionsingeometry, andespe iallyingeometri onstraint solving[12,11℄.

Indeed,whendesigninganobje tbythemeanof geomet-ri onstraints,onehasto orre tlyand ompletelyspe ify theshapeofthe obje t. The orresponding onstraint sys-tem anlead to ontradi tions and degenerate aseswhi h

∗

partiallysupportedbytheresear hproje tGalapagos.

This is the author’s version of the work. It is posted here by permission of

ACM for you personal use. Not for redistribution. The definitive version

was published in the proceedings of SAC 2009.

SAC’09 March 8-12, 2009, Honolulu, Hawaii, U.S.A.

Copyright 2009 ACM 978-1-60558-166-8/09/03 ...$5.00.

mustbe handled properly by the solver as shown in [20℄. The stru tural ontradi tions are easy to dete t, whereas thoseresulting from geometri theoremsare more di ult todealwith.

We fo us on proje tive geometry whi h is a simplebut powerfulenoughsettingtoexpressarbitrarily omplex prob-lems as shownin[16℄. Moreover, in3D (orhigher) proofs be ome mu hmore di ult than in 2D: rst, Desargues' propertybe omesatheorem and onsequentlyall the pro-je tive spa es arisefrom a division eld; se ond in iden e geometry has to deal not only with points and lines, but also with planes or, more generally, ats; third there is a ombinatori explosionof ases.

We propose to use the on ept of rank, whi h pro-vides a generi way to des ribe in iden e relations. In-formally, ranks allow to distinguish between equal/non-equal points, ollinear/non- ollinear points, oplanar/non- oplanarpoints,et . Tovalidatethisapproa h,we arryout ame hanizedproof(usingonlyranks)ofoneofthe funda-mentaltheoremsoftheproje tivespa e,namelyDesargues' theorem.

Related work Proof assistantshavealready beenusedin the ontextofgeometry. Numerouspapershaveemphasized theimportan eoftheproblemofdegenerate asesinformal geometry[8,15,10,19℄. BrandtandS hneiderstudiedhow tohandledegenerate asesfortheorientationpredi atesin omputationalgeometryusingthreevaluedlogi [3℄. Bezem andHendri ksformalizedHessenberg'stheoreminCoq[2℄. GuilhothasformalizedinCoqaproofofDesargues'theorem inanegeometry[10℄. NarbouxhasformalizedinCoqthe areamethodofChou,GaoandZhang[5,18℄andappliedit toobtainaproofofDesargues'theoreminanegeometry. Kusak has formalized inMizar Desargues' theorem in the Fanoian proje tive at least 3-dimensional spa e [13℄. The assumption that the spa e is Fanoian makes the theorem more spe ialized thanours. We have performed a formal-ization ofproje tiveplane geometryinCoq [14℄. Theidea ofprovingproje tivespa etheoremswithranksissuggested byMi helu iandS hre kin[17℄.

OutlineThepaperisorganizedasfollows. Inse tion2,we presenttheaxiomsfortheproje tiveatleast3-dimensional spa e. In se tion 3,weintrodu e the on ept of rank and anotheraxiomsystemto aptureproje tivegeometryusing ranks. Weshowhownon-degenera y onditions anbe ex-pressed ni ely usingthe notion of rank. In se tion 4, we proposeaproofofDesargues'theorempurelybasedonthe on eptofrank. Inse tion5,wepresenttheme hanization oftheproofusingtheCoqproofassistant[1,6℄.

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2. AXIOM SYSTEM

Proje tivegeometryisageneralsettinginthehierar hyof geometrieswhi hassumesthattwolinesinaplane always meet [7, 4℄. We rst assume that we have two kinds of obje ts(pointsandlines)

1

.Wethen onsiderarelation(

∈

) between elements of thesetwo sets. Thefollowing axioms des ribeproje tiveatleast3-dimensionalspa e:

Line-Existen e

∀A B : P oint, ∃l : Line, A ∈ l ∧ B ∈ l

Pas h

∀A B C D : P oint, ∀l

AB

l

CD

l

AC

l

BD

: Line,

A

6= B ∧ A 6= C ∧ A 6= D ∧ B 6= C ∧ B 6= D ∧ C 6= D∧

A

∈ l

AB

∧ B ∈ l

AB

∧ C ∈ l

CD

∧ D ∈ l

CD

∧

A

∈ l

AC

∧ C ∈ l

AC

∧ B ∈ l

BD

∧ D ∈ l

BD

∧

(∃I : P oint, I ∈ l

AB

∧ I ∈ l

CD

)

⇒

(∃J : P oint, J ∈ l

AC

∧ J ∈ l

BD

)

Three-Points

∀l : Line, ∃A B C : P oint,

A

6= B ∧ B 6= C ∧ A 6= C ∧ A ∈ l ∧ B ∈ l ∧ C ∈ l

Uniqueness

∀A B : P oint, ∀l m : Line,

A

∈ l ∧ B ∈ l ∧ A ∈ m ∧ B ∈ m ⇒ A = B ∨ l = m

Lower-Dimension

∃l m : Line, ∀p : P oint, p 6∈ l ∨ p 6∈ m

Pas h's axiom ensures that two o-planar lines always meet. We presented this standard axiomsystem as a ref-eren e,buttoeasetheformalizationinCoq,weproposean alternativeaxiomsystembasedonthenotionofrank.

3. RANKS

The on eptofrankisageneralnotionofmatroidtheory. Anintegerfun tion

rk

on

E

istherankfun tionofamatroid ifandonlyifthefollowing onditions aresatised:

R1

∀X ⊆ E, 0 ≤ rk(X) ≤ |X|

(nonnegativeandsub ardinal) R2

∀XY ⊆ E, X ⊆ Y ⇒ rk(X) ≤ rk(Y )

(nonde reasing) R3

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

(submodular)

Inproje tivegeometry,we andenearankfun tiononsets ofpointswhi hverifytheaxioms above: aatbeingaset of points losed by the ollinearity relation, the rank of a setofpoints

A

isthe ardinalofasmallestsetgenerating

A

(seegure1forexamples).

rk{A, B} = 1

A

= B

rk{A, B} = 2

A

6= B

rk{A, B, C} = 2

A, B, C

are ollinear

withatleasttwoofthemdistin t

rk{A, B, C} ≤ 2

A, B, C

are ollinear

rk{A, B, C} = 3

A, B, C

arenot ollinear

rk{A, B, C, D} = 3

A, B, C, D

are o-planar, notall ollinear

rk{A, B, C, D} = 4

A, B, C, D

arenot o-planar

rk{A, B, C, D, E} ≤ 2 A, B, C, D, E

areall ollinear

Figure 1: Rank statements and their geometri in-terpretations

1

Planesarenotbasi obje ts inthisaxiomsystem,butare

Usingthisdenition,one anshowthateveryproje tive spa ehasamatroidstru ture,butthe onverseisnottrue. Inthenextse tion,weintrodu eadditionalaxiomsto ap-ture3Dorhigherproje tivegeometry.

3.1 A rank-based axiom system

Contrarytotheaxiomsystemshowninse tion2,we as-sumethatwehaveonlyonekindofobje ts,namelypoints. To apturethewholeproje tivespa e,weneedtoaddsome newaxiomstothematroid'sones:

Rk-Singleton

∀P : P oint, rk{P } ≥ 1

Rk-Couple

∀P Q : P oint, P 6= Q ⇒ rk{P, Q} ≥ 2

Rk-Pas h

∀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

The rsttwo ones ensure that the rank fun tionis not degenerate. Rk-Pas histhetranslationofPas h'saxiom:

rk{A, B, C, D} ≤ 3

meansthese points are oplanar, thus thatthetwolines

AB

and

CD

interse t.

Usingthisaxiomsystemweformallyprovedalltheaxioms of se tion 2. Inparti ular the following Rk-Uniqueness lemmais derivable and anbeusedtoprovethe Unique-nessaxiom: Lemma1 (Rk-Uniqueness).

∀ABCDM P,

0 B

B

@

rk{A, B} = 2

rk{C, D} = 2

rk{A, B, M } ≤ 2

rk{C, D, M } ≤ 2

rk{A, B, P } ≤ 2

rk{C, D, P } ≤ 2

rk{A, B, C, D} ≥ 3

1 C

C

A

⇒ rk{M, P } = 1

Proof. Seese tion 6.

We analsoderivealemmawhi hexpresses on iselythat foreverypointthereexistsonewhi hisdierent,forevery linethereexistsapointnotonthislineandforeveryplane thereexistsapointnotonthisplane:

Lemma2 (Constru tion).

∀E, rk(E) ≤ 3 ⇒ ∃P, rk(E ∪ {P }) = rk(E) + 1

Overall, this axiomsystem is onvenient as rst it only deals with pointsand hen e the theory is dimension-inde-pendent,se ondranksallowtosummarizebothpositiveand negativeassumptionsaboutsetsofpointshomogeneously.

3.2 Proof techniques using ranks

Inthis se tion we des ribe two proofs te hniques whi h aresimplebutimportanttosimplifyformalproofs.

First, all equalities about ranks(say

rk(a) = rk(b)

) are usuallyprovedintwosteps: rstshowingthat

rk(a)

≤ rk(b)

andthenthat

rk(a)

≥ rk(b)

. Consequently,whenstatinga lemma,itisworthbeing autiousaboutwhetherthea tual equalityisrequiredorifoneofthetwoinequalitiesisenough

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proofsintheCoqproofassistant.

Se ond,intheprovingpro ess,wemakeoftenuseof ax-iomR3. Forinstan e,ifweneedtoproveastatementlike:

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

we ould be tempted to instantiate axiomR3 with

X

:=

{A, B, I}

and

Y

:=

{C, D, I}

. Butunfortunately,this state-ment is not a dire t onsequen e of axiom R3. For in-stan e

A

maybeequalto

C

and onsequently

{A, B, I} ∩

{C, D, I} = {A, I}

. Determiningtheinterse tionoftwo -nite sets of points requires to distinguish ases about the equality of these points. This leads to intri ate proofs in Coq. Therefore, inthe rest of this paper, we shall never onsidertherealsettheoreti alinterse tionbutalower ap-proximationoftheinterse tion(noted

⊓

).

Definition1 (Literalinterse tion). Let

L

1

and

L

2

be two sets of points. By denition

L

1 ⊓ L

2

is the in-terse tion ofthetwosetsofpoints onsideredsynta ti ally.

Usingliteralinterse tionwe anderiveamore onvenient versionofaxiomR3whi hleadstofewer ase distin tions:

Lemma3 (R3-lit).

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

InCoq,itis notpossibletodene thelitteralinterse tion. To apturetheintentofthislemmaweratheruse:

Lemma4 (R3-alt).

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

Thislemmawillbeusedheavilyinthenextse tions.

4. DESARGUES’ THEOREM

Desargues'theoremstatesthat:

let E be a 3D or higher proje tive spa e and

A

,

B

,

C

,

A

′

,

B

′

,

C

′

bepointsin

E

,ifthethreelines joining the orresponding verti es of triangles

ABC

and

A

′

_B

′

_C

′

all meet in a point

O

, then the three interse tions of pairsof orresponding sides

α

,

β

and

γ

lieonaline.

b A b B b C bO b C' b A' b B' b b b

β

γ

,

B

′

and

C

′

lieonasame

ranktheproofofthe3Dversionisstraightforwardand spe- ial ases anbe handledsmoothly. Inse tion 4.3,wewill showhowwea tuallybuildthe2Dversionand on ludethe proofoftheoriginaltheorem.

4.2 A 3D version of Desargues’ Theorem

Inthisse tion,weproveDesargues' 3Dtheorem.

Theorem 1 (Desargues3D). Let's onsider two (nondegenerate) triangles

ABC

and

abc

su hthat theyare perspe tivefrom agivenpoint

O

:

rk{A, B, C} = rk{a, b, c} = 3

rk{a, A, O} = rk{b, B, O} = rk{c, C, O} = 2

Weassume thisformsanonplanargure:

rk{A, B, C, a, b, c} ≥ 4

anddenethree points

α

,

β

,

γ

su hthat:

rk{A, B, γ} = rk{a, b, γ} = 2

rk{A, C, β} = rk{a, c, β} = 2

rk{B, C, α} = rk{b, c, α} = 2

Undertheseassumptions,

rk{α, β, γ} ≤ 2

holds. Proof ofDesargues3D.

Lemma 5.

rk{A, B, C, α} = 3

Proof. Byassumption

rk{A, B, C} = 3

,hen eusing ax-iom R2,

rk{A, B, C, α} ≥ 3

. Moreover using R3-alt we have:

rk{A, B, C, α} + rk{B, C} ≤ rk{A, B, C} + rk{α, B, C}

rk{A, B, C, α} + 2 ≤ 3 + 2

Hen e, we an on lude that

rk{A, B, C, α} = 3

. Similar proofs anbedonewith

β

and

γ

.

Lemma 6.

rk{A, B, C, α, β} = 3

Proof. First, using axiom R2 and lemma 5 we have

rk{A, B, C, α, β} ≥ 3

. Se ond,usingR3-altwehave:

rk{A, B, C, α, β} + rk{A, B, C}

≤ rk{A, B, C, α} + rk{A, B, C, β}

rk{A, B, C, α, β} + 3 ≤ 3 + 3

Hen e,we an on ludethat

rk{A, B, C, α, β} = 3

. Lemma 7.

rk{A, B, C, α, β, γ} = rk{a, b, c, α, β, γ} = 3

Proof. Theproofissimilartolemma6.

Lemma 8.

rk{A, B, C, a, b, c, α, β, γ} ≥ 4

Proof. Byassumption

rk{A, B, C, a, b, c} ≥ 4

,hen e us-ingaxiomR2,

rk{A, B, C, a, b, c, α, β, γ} ≥ 4

.

Usingtheselemmaswe an on ludetheproof: FromR3-altweknowthat:

rk{A, B, C, a, b, c, α, β, γ} + rk{α, β, γ}

≤ rk{A, B, C, α, β, γ} + rk{a, b, c, α, β, γ}

Hen e,usinglemmas7and8

rk{α, β, γ} ≤ 2

holds.

4.3 Lifting from 2D to 3D

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4.3.1 Statement

Mostassumptionsarethesameasinthe3Dversion. Let's onsidertwotriangles

ABC

and

A

′

_B

′

_C

′

su hthattheyare perspe tivefromagivenpoint

O

:

rk{A, B, C} = rk{A

′

_{, B}

′

_{, C}

′

_{} = 3}

rk{A

′

_{, A, O} = rk{B}

′

_{, B, O} = rk{C}

′

_{, C, O} = 2}

Wedenethreepoints

α

,

β

,

γ

su hthat:

rk{A, B, γ} = rk{A

′

_{, B}

′

_{, γ} = 2}

rk{A, C, β} = rk{A

′

_{, C}

′

_{, β} = 2}

rk{B, C, α} = rk{B

′

_{, C}

′

_{, α} = 2}

Contrary to the 3D ase, we assume this forms a planar gure:

rk{A, B, C, A

′

_{, B}

′

_{, C}

′

_{, O} = 3}

Inadditionto these assumptionswhi h are losely related to those of Desargues' 3D theorem, the following non-degenera y onditionsarerequired:

rk{A, B, O} = rk{A, C, O} = rk{B, C, O} = 3

rk{A, A

′

_{} = rk{B, B}

′

_{} = rk{C, C}

′

A

′

,

B

′

,

C

′

,

O

,

α

,

β

and

γ

are o-planar. If

P

isthe sun,triangle

abc

then asts itsshadowin

A

′

_B

′

_C

′

.

4.3.2 Informal Proof

We haveto lift triangle

A

′

_B

′

_C

′

into anew triangle

abc

whi his not oplanar with triangle

ABC

inorderto have a onguration of pointsin whi hDesargues' 3D theorem anbeapplied. The onstru tionisshowningure2.Here arethe mainsteps: werst onstru t apoint

P

whi hlies outsidethe planeformedby

A, B, C, A

′

_{, B}

′

_{, C}

A

′

_B

′

_C

′

are perspe tive) and onsider athird point

o

onthis line (axiomRk-Three-Points en-suressu hapoint exists and is dierent fromboth

O

and

P

). We onstru tanewpoint

a

astheinterse tion oflines

P A

′

and

oA

. Weknowthesetwolinesinterse tbe auseof

Pas h'saxiomandthefa tthatlines

de-ned asthe interse tionof lines

BC

and

B

′

_C

′

isthe same asthe

α

ofDesargues'3Dtheoremwhi histheinterse tion of

BC

and

bc

. Thisrequirement anbesatised bysimply provingthat

α

isin identto

bc

. Thesamerequirement ap-plies for

β

and

γ

. Overall,we haveto provethe following statementswhi hare requirementstoapply Desargues'3D version (proofs are given inappendix B). Note thatwhen applyingthetheorem,thepoint

o

playstheroleofpoint

O

.

rk{A, B, C} = rk{a, b, c} = 3

rk{A, B, C, a, b, c} ≥ 4

rk{a, b, γ} = rk{a, c, β} = rk{b, c, α} = 2

rk{A, B, γ} = rk{A, C, β} = rk{B, C, α} = 2

4.3.3 General Lemmas

Most proofs are fairly te hni al, simply using the ma-troid axiomsofrank. However,somelemmas anbe high-lighted,espe iallyfortheirgeneri ityandtheirpervasiveuse throughouttheproofs. Amongthem,somestabilitylemmas statethatoneofthepointsofaset hara terizingaat(e.g aplaneorthewholespa e) anberepla edbyanotherone belongingtothisat.

Lemma9 (Planerepresentation hange) .

0 B

B

@

rk{A, B, C} = 3

rk{A, B, C, M } = 3

rk{B, C, M } = 3

rk{A, B, C, P } = 4

1 C

C

A

⇒ rk{M, B, C, P } = 4

This lemma is heavily used to prove all possible state-ments expressing that

P

lies outside the plane formed by

A, B, C, A

′

_{, B, C}

′

_{, O}

.

Other lemmas about oplanarity and also upper bound onrankswhenmergingaplaneandalineare onvenientas well. They ouldformthebasisofanautomationpro edure whendoing omputer- he kedformalproofs.

5. FORMALIZATION IN COQ

Weme hanizetheproofofDesargues' theoremusingthe Coq proof assistant whi himplementsahigherorder intu-itionisti logi basedontypetheory. Insu haproof assis-tant, every step of reasoning is proposed by the user but he kedby the system. It dramati ally in reases the reli-ability of the proofs omparedto paper-and-pen ilproofs. Inaddition,duringthe developmentpro ess,being able to hange theaxiomsystem easilyis very onvenient. Proofs anbeautomati ally re he kedbythesystemand hanges onlyrequireminorrewritingoftheproofs.

The formalization in Coq of our axiom system is quite straightforward

2

. To in rease reusability of the proofs we deneitasamoduletypeofCoq'smodulesystem(see g-ure 3). Thismodule dependson

DecP oints

whi hdenes thetypeofpointswithade idableequality.

InCoq,Desargues'theorem anbestatedasfollows: TheoremDesargues xD' :

∀

≥

2. Axiompas h:

∀

AB C D,rk (quadrupleABC D)

≤

3

→

∃

J,rk (tripleAB J)=2

∧

rk (triple C D J)=2. Axiomthree points :

∀

AB,

∃

C,

rk

isamorphism withrespe ttosetequality.

Overall,ourdevelopment onsistsofmorethan6000lines and200lemmasandtheirformalproofsorganizedasshown inthe gures below. Theproof of Desargues' theorem in itself is relatively on ise be ause we proved some generi lemmasandreusedtheme ientlytakingadvantageofthe symmetriesinthestatementofDesargues'theorem.

3D 2D Framework Total linesofCoqspe s 215 450 777 1442 linesofCoqproofs 963 976 2331 4670

6. CONCLUSIONS

We proposed a new way to express ni ely in iden e re-lations in a 3D setting thanks to ranks and designed an axiomsystem to apture proje tive geometry usingranks. Wesu essfullyappliedittoproveDesargues'theorem. We presentedproofengineeringte hniqueswhi hallowtohave proofs of reasonable size. In the future, we plan to study howthenotionofrank anbeusedtoautomati allyderive in iden eproperties. Webelieve that thegeneri ity ofthe notationwillhelptheautomationpro ess.

Availability ThefullCoq development isavailableatthe

A knowledgmentsWewishtothankNegarGholamizadeh Behbahaniwhostudiedsomelemmasduringaninternship.

7. REFERENCES

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theProofofHessenberg'sTheoreminCoherentLogi . J.ofAutomated Reasoning,40(1):6185,2008. [3℄ J.BrandtandK.S hneider.Usingthree-valuedlogi

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APPENDIX

A. PROOFS RELEVANT TO THE FIRST

SECTION

Lemma10 (Rk-Uniqueness).

∀ABCDM P,

0 B

B

@

rk{A, B} = 2

rk{C, D} = 2

rk{A, B, M } ≤ 2

rk{C, D, M } ≤ 2

rk{A, B, P } ≤ 2

rk{C, D, P } ≤ 2

rk{A, B, C, D} ≥ 3

1 C

C

A

⇒ rk{M, P } = 1

Proof. UsingR3-altwehave:

rk{A, B, M, P } + rk{A, B} ≤ rk{A, B, M } + rk{A, B, P }

Hen e

rk{A, B, M, P } = 2

, similarly we an show that

rk{C, D, M, P } = 2

. Moreover

rk{A, B, C, D, M, P } ≥ 3

as

rk{A, B, C, D} ≥ 3

and

{A, B, C, D} ⊆ {A, B, C, D, M, P }

. Finally,usingR3-altwehave:

rk{A, B, C, D, M, P } + rk{M, P }

≤ rk{A, B, M, P } + rk{C, D, M, P }

Hen e:

3 + rk{M, P } ≤ 2 + 2

.

OurdevelopmentisbasedonmatroidaxiomsR1,R2and R3. Butone anprovethat theyareequivalenttothe fol-lowingsetofaxioms:

R1'

rk(∅) = 0

R2'

rk(X)

≤ rk(X ∪ {x}) ≤ rk(X) + 1

R3'

rk(X

∪ {y}) = rk(X ∪ {z}) = rk(X) ⇒ rk(X) =

rk(X

∪ {y, z})

Lemma11 (LowerDimension).

∃ABCD, ∀M, rk{A, B, M } 6= 2 ∨ rk{C, D, M } 6= 2

Proof. UsingaxiomRk-Lower-Dimension,weobtain

A

,

B

,

C

and

D

su hthat

rk{A, B, C, D} = 4

.

Supposethat

rk{A, B, M } = 2

and

rk{C, D, M } = 2

. UsingR3-altwehavethat:

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

Hen e

rk{A, B, C, D, M } ≤ 3

,whi hisin ontradi tionwith

rk{A, B, C, D} = 4

.

Lemma12 (Constru tion).

∀E, rk(E) ≤ 3 ⇒ ∃P, rk(E ∪ {P }) = rk(E) + 1

Proof. Consider

E

su hthat

rk(E)

≤ 3

. Using axiom Rk-Lower-Dimensionweobtain

A

,

B

,

C

and

D

su hthat

rk{A, B, C, D} = 4

. Using R2' we know that

rk(E)

≤

rk(E

∪ {A}) ≤ rk(E) + 1

and similarly for

B

,

C

and

D

. Suppose that

rk(E

∪ {A}) = rk(E ∪ {B}) = rk(E ∪

{C}) = rk(E ∪ {D}) = rk(E)

, then we would obtain

rk(E

∪ {A, B, C, D}) = rk(E)

by repeated appli ations of R3'. Thisisin ontradi tionwith

rk{A, B, C, D} = 4

sin e

rk(E)

≤ 3

. Hen ethereexistsa

P

su hthat

rk(E

∪ {P }) =

rk(E) + 1

.

B. PROVING DESARGUES’ 2D THEOREM

Weremindthereaderofthepropertiesweneedtoprove inordertoapplyDesargues'3D theorem,namely: