formalizing analytic geometries · our formalization of analytic geometry aims at establishing the...

IntroductionFormalizing Cartesian GeometryUsing Isometric Transformations

Tarski’s geometryHilbert’s geometry

Conclusions and Further Work

Formalizing Analytic Geometries

Danijela [email protected]

Department of Computer Science

Faculty of Mathematics

University of Belgrade

ADG 2012

Danijela Petrovic [email protected] Formalizing Analytic Geometries




Outline

1 Introduction

2 Formalizing Cartesian Geometry

3 Using Isometric Transformations

4 Tarski’s geometryAxioms of Betweenness

5 Hilbert’s geometryAxioms of OrderAxioms of Continuity

6 Conclusions and Further Work





Many different geometries

Euclidean (synthetic) geometry

Descartes’s coordinate system, bridged the gap betweenalgebra and geometry

Several attempts to formalize different geometries anddifferent approaches to geometry

Using the proof assistants significantly raises the level of rigour

Isabelle/HOL





Main applications:

- automated theorem proving in geometry- mathematical education and teaching ofgeometry

Automated theorem proving:algebraic methods vs. theorem provers based on syntheticaxiomatizations

Geometry in education





Goals

Formalize Cartesian geometry within a proof assistant

Different definitions of basic notions of analytic geometry allturn out to be equivalent

The standard Cartesian plane geometry represents a model ofseveral geometry axiomatizations

Formally analyze model-theoretic properties of differentaxiomatic systems

Formally analyze axiomatizations and models ofnon-Euclidean geometries and their properties

Formally establish connections of the Cartesian planegeometry with algebraic methods





Outline

1 Introduction










Our formalization of analytic geometry aims at establishingthe connection with synthetic geometries so it followsprimitive notions given in synthetic approaches

objects: points, lines

relations: incidence, betweenness, congruence





Points

pairs of real numbers (R2)

easily formalized in Isabelle/HOL by type synonym

pointag = ”real × real”





The Order of Points

Betweenness relation: B(A,B,C)

Some axiomatizations, e.g. Tarski’s, allow the case when B isequal to A or C

BagT (xa, ya) (xb, yb) (xc, yc)←→(∃(k :: real). 0 ≤ k ∧ k ≤ 1 ∧

(xb− xa) = k · (xc− xa) ∧ (yb− ya) = k · (yc− ya))

Hilbert’s axiomatizations does not allow points to be equal

BagH (xa, ya) (xb, yb) (xc, yc)←→(∃(k :: real). 0 < k ∧ k < 1 ∧

(xb− xa) = k · (xc− xa) ∧ (yb− ya) = k · (yc− ya))





Congruence

Square distance

d2ag (x1, y1) (x2, y2) = (x2 − x1) · (x2 − x1) + (y2 − y1) · (y2 − y1)

Congruence

A1B1∼=ag A2B2 ←→ d2ag A1 B1 = d2ag A2 B2





typedef

Type definition

Any non-empty subset of an existing type can be turned intoa new type

Example:typedef three = ”{0 :: nat, 1, 2}”





Constants declared:- three :: nat set- Rep three :: three −→ nat- Abs three :: nat −→ three

Some features:- Rep three surjective: Rep three x ∈ three- Rep three inverse: Abs three (Rep three x) = x- Abs three inverse:y ∈ three ⇒ Rep three (Abs three y) = y





LinesLine equations

A · x+B · y + C = 0 where A 6= 0 ∨B 6= 0

same line:

- A · x+B · y + C = 0- k ·A · x+ k ·B · y + k · C = 0, for a real k 6= 0

Lines

typedef line coeffsag =

{((A :: real), (B :: real), (C :: real)). A 6= 0 ∨B 6= 0}





LinesLine equations

Equivalent triplets

l1 ≈ag l2 ←→(∃ A1B1C1A2B2C2.(Rep line coeffs l1 = (A1, B1, C1)) ∧Rep line coeffs l2 = (A2, B2, C2) ∧(∃k. k 6= 0 ∧ A2 = k ·A1 ∧ B2 = k ·B1 ∧ C2 = k · C1))

(x, y) ∈agH l

ag in h (x, y) l←→(∃ A B C. Rep line coeffs l = (A, B, C)∧(A · x+B · y + C = 0))





LinesAffine definition

vecag = ”real × real”

typedef line point vecag = {(p :: pointag, v :: vecag). v 6= (0, 0)}

Equivalent triplets

l1 ≈ag l2 ←→ (∃ p1 v1 p2 v2.Rep line point vec l1 = (p1, v1)∧Rep line point vec l2 = (p2, v2)∧(∃km. v1 = k · v2 ∧ p2 = p1 +m · v1))

Incidence

ag in h p l ←→ (∃ p0 v0.Rep line point vec l = (p0, v0)∧(∃k. p = p0 + k · v0))





Isometries

Translation

transpag (v1, v2) (x1, x2) = (v1 + x1, v2 + x2)

Rotation

rotpag α (x, y) = ((cosα) · x− (sinα) · y, (sinα) · x+ (cosα) · y)





Isometries

All isometries are invariance properties

BagT A B C ←→BagT (transpag v A) (transpag v B) (transpag v C)

AB ∼=ag CD ←→(transpag v A)(transpag v B) ∼=ag (transpag v C)(transpag v D)

BagT A B C ←→ Bag

T (rotpag α A) (rotpag α B) (rotpag α C)

AB ∼=ag CD ←→(rotpag α A)(rotpag α B) ∼=ag (rotpag α C)(rotpag α D)





Isometries

used only to transform points to canonical position

∃v. transpag v P = (0, 0)

∃α. rotpag α P = (0, p)

BagT (0, 0) P1 P2 −→ ∃α p1 p2. rotpag α P1 = (0, p1)

∧rotpag α P2 = (0, p2)





Outline

1 Introduction










Motivation example

BagT A X B

AX

B





Motivation example

BagT A X B ∧ BagT A B Y

XA

YB





Motivation example

BagT A X B ∧ BagT A B Y −→ BagT X B Y

AX

YB





Motivation example


AX

YB

A = (xA, yA), B = (xB, yB), and X = (xX , yX)

X = B holds trivially

from BagT A X B there is a real number k1, 0 ≤ k1 ≤ 1 suchthat

- (xX − xA) = k1 · (xB − xA)- (yX − yA) = k1 · (yB − yA)





Motivation example

AX

YB

from BagT A B Y there is a real number k2, 0 ≤ k2 ≤ 1-(xB − xA) = k2 · (xY − xA) - (yB − yA) = k2 · (yY − yA)

k = (k2 − k2 · k1)/(1− k2 · k1)

if X 6= B, it is shown that- 0 ≤ k ≤ 1- (xB − xX) = k · (xY − xX) (yB − yX) = k · (yY − yX)

and thus BagT X B Y holds





Motivation example

A B YX

we can assume: A = (0, 0), B = (0, yB), and X = (0, yX),and that 0 ≤ yX ≤ yB.

yB = 0 holds trivially

Otherwise, xY = 0 and 0 ≤ yB ≤ yY . Hence yX ≤ yB ≤ yY





Invariant property

inv P t←→ (∀ A B C. P A B C ←→ P (tA) (tB) (tC))

Used to reduce the statement to any three collinear points to thepositive part of the y-axis

lemma

assumes "∀ yB yC . 0 ≤ yB ∧ yB ≤ yC −→P (0, 0) (0, yB) (0, yC)""∀ v. inv P (transpag v )""∀α. inv P (rotpag α )"

shows "∀ABC. BagT A B C −→ P A B C"

John Harrison, Without loss of generality





Axioms of Betweenness

Outline

1 Introduction











AB ∼=t BAAB ∼=t CC −→ A = BAB ∼=t CD ∧ AB ∼=t EF −→ CD ∼=t EF

Bt(A,B,A) −→ A = B

Bt(A,P,C) ∧ Bt(B,Q,C) −→ (∃X. (Bt(P,X,B) ∧ Bt(Q,X,A)))

∃ A B C. ¬ Ct(A,B,C)

(∃a. ∀x. ∀y. φ x ∧ ψ y −→ Bt(a, x, y)) −→(∃b. ∀x. ∀y. φ x ∧ ψ y −→ Bt(x, b, y))






AP ∼=t AQ ∧ BP ∼=t BQ ∧ CP ∼=t CQ ∧ P 6= Q −→ Ct(A,B,C)

∃E. Bt(A,B,E) ∧ BE ∼=t CD

AB ∼=t A′B′ ∧ BC ∼=t B

′C ′ ∧ AD ∼=t A′D′ ∧ BD ∼=t B

′D′ ∧Bt(A,B,C) ∧ Bt(A

′, B′, C ′) ∧ A 6= B −→ CD ∼=t C′D′

Bt(A,D, T ) ∧ Bt(B,D,C) ∧ A 6= D −→(∃XY. (Bt(A,B,X) ∧ Bt(A,C, Y ) ∧ Bt(X,T, Y )))






The axiom of Pasch

The axiom of Pasch

Bt(A,P,C) ∧ Bt(B,Q,C) −→ (∃X. (Bt(P,X,B) ∧ Bt(Q,X,A)))

A

B

P C

Q

X






The axiom of Pasch

Properties used in proof of Pasch axiom

BagT A A B

BagT A B C −→ BagT C B A






The axiom of Pasch


BagT A A B


BagT A X B

A X B






The axiom of Pasch


BagT A A B



XA B Y






The axiom of Pasch


BagT A A B



A X B Y






The axiom of Pasch


BagT A A B



A X B Y

BagT A X B

A X B






The axiom of Pasch


BagT A A B



A X B Y


XA B Y






The axiom of Pasch


BagT A A B



A X B Y

BagT A X B ∧ BagT A B Y −→ BagT A X Y

BXA Y






The axiom of Pasch

∃X. (Bt(P,X,B) ∧ Bt(Q,X,A))

If P = C, then Q is thepoint sought

A

B

C= P

Q






The axiom of Pasch

∃X. (Bt(P,X,B) ∧ Bt(Q,X,A))

If P = C, then Q is thepoint sought

A

B

C= P

Q

Bt(A,B,C), B is the pointsought

A P Q CB






The axiom of Pasch

use isometric transformations so that:A = (0, 0), P = (0, yP ), C = (0, yC)B = (xB, yB), Q = (xQ, yQ),X = (xX , yX)

from Bt(A,P,C), there is a real numberk3 such that

- 0 ≤ k3 ≤ 1 yP = k3 · yC

from Bt(B,Q,C), there is a real numberk4 such that

- 0 ≤ k4 ≤ 1- xQ − xB = −k4 ∗ xB- yQ − yB = k4 ∗ (yC − yB)

A P C

Q

B

X






The axiom of Pasch

define a real number k1 =k3·(1−k4)

k4+k3−k3·k4and show

- 0 ≤ k1 ≤ 1- xX = k1 · xB- yX − yP = k1 · (yB − yP )

thus Bt(P,X,B) holds

define a real number k2 =k4·(1−k3)

k4+k3−k3·k4and show

- 0 ≤ k2 ≤ 1- xX − xQ = −k2 · xQ- yX − yQ = −k2 · yQ

thus Bt(Q,X,A) holds

A P C

Q

B

X






Axiom (Schema) of Continuity

(∃a. ∀x. ∀y. φ x ∧ ψ y −→ Bt(a, x, y)) −→(∃b. ∀x. ∀y. φ x ∧ ψ y −→ Bt(x, b, y))

A x yB

Isabelle/HOL does not restrict predicate φ and ψ to be FOLpredicates

One of the sets is empty, the statement trivially hold

The sets have a point in common, that point is the pointsought







A = (0, 0)

P QB

Isometry transformations are applied so that all points fromboth sets lie on the positive part of the y-axis.

lemma

assumes

"P = {x. x ≥ 0 ∧ φ(0, x)}" "Q = {y. y ≥ 0 ∧ ψ(0, y)}""¬(∃b. b ∈ P ∧ b ∈ Q)" "∃x0. x0 ∈ P" "∃y0. y0 ∈ Q""∀x ∈ P. ∀y ∈ Q. BagT (0, 0) (0, x) (0, y)"shows

"∃b. ∀x ∈ P. ∀y ∈ Q. BagT (0, x) (0, b) (0, y)"







A = (0, 0)

P QB

Completeness of reals

(∃x. x ∈ P ) ∧ (∃y. ∀x ∈ P. x < y) −→∃S. (∀y. (∃x ∈ P. y < x)↔ y < S)

P satisfies the supremum property

∀x ∈ P. ∀y ∈ Q. x < y

there is b such that:∀x ∈ P. x ≤ b and ∀y ∈ Q. b ≤ y





Axioms of OrderAxioms of Continuity

Outline

1 Introduction











A 6= B −→ ∃! l. A ∈h l ∧ B ∈h l

∃AB. A 6= B ∧ A ∈h l ∧ B ∈h l

∃ABC. ¬ Ch(A,B,C)

Bh(A,B,C) −→ A 6= B ∧ A 6= C ∧ B 6= C ∧Ch(A,B,C) ∧ Bh(C,B,A)

A 6= C −→ ∃B. Bh(A,C,B)

A ∈h l ∧ B ∈h l ∧ C ∈h l ∧ A 6= B ∧ B 6= C ∧ A 6= C −→(Bh(A,B,C) ∧ ¬Bh(B,C,A) ∧ ¬Bh(C,A,B)) ∨(¬Bh(A,B,C) ∧ Bh(B,C,A) ∧ ¬Bh(C,A,B)) ∨(¬Bh(A,B,C) ∧ ¬Bh(B,C,A) ∧ Bh(C,A,B))

A 6= B ∧ B 6= C ∧ C 6= A ∧ Bh(A,P,B)∧P ∈h l ∧ ¬C ∈h l ∧ ¬A ∈h l ∧ ¬B ∈h lh −→∃Q. (Bh(A,Q,C) ∧ Q ∈h l) ∨ (Bh(B,Q,C) ∧ Q ∈h l)






A 6= B ∧ A ∈h l ∧ B ∈h l ∧ A′ ∈h l

′ −→∃B′ C ′. B′ ∈h l

′ ∧ C ′ ∈h l′ ∧ Bh(C

′, A′, B′)∧AB ∼=h A

′B′ ∧ AB ∼=h A′C ′

AB ∼=h A′B′ ∧ AB ∼=h A

′′B′′ −→ A′B′ ∼=h A′′B′′

Bh(A,B,C) ∧ Bh(A′, B′, C ′) ∧ AB ∼=h A

′B′ ∧ BC ∼=h B′C ′ −→

AC ∼=h A′C ′

¬P ∈h l −→ ∃! l′. P ∈h l′ ∧ ¬(∃ P1. P1 ∈h l ∧ P1 ∈h l

′)

Let A1 be any point upon a straight line between the arbitrarilychosen points A and B. Choose the points A2, A3, A4, . . . so that A1

lies between A and A2, A2 between A1 and A3, A3 between A2 andA4 etc. Moreover, let the segments AA1, A1A2, A2A3, A3A4, . . . beequal to one another. Then, among this series of points, therealways exists a point An such that B lies between A and An.






Axiom of Pasch

A 6= B ∧ B 6= C ∧ C 6= A ∧ Bh(A,P,B)∧P ∈h l ∧ ¬C ∈h l ∧ ¬A ∈h l ∧ ¬B ∈h l −→∃Q. (Bh(A,Q,C) ∧ Q ∈h l) ∨ (Bh(B,Q,C) ∧ Q ∈h l)

A

B

P

C

Q

The statement holds trivially if A, B and C are collinear






A = (0, 0), B = (xB, 0), P = (xP , 0),C = (xC , yC) andRep line coeffs l = (lA, lB, lC)

Bh(A,P,B) −→ lA · yB 6= 0

k1 =−lClA·yB

k2 =lA·yB+lClA·yB

0 < k1 < 1, Q = (xQ, yQ) is determed by- xQ = k1 · xC yQ = k1 · yC- Bh(A,Q,C)

0 < k2 < 1, Q = (xQ, yQ) is determed by- xQ = k2 · (xC − xB) + xB- yQ = k2 · yC- Bt(B,Q,C)

A

B

P

C

Q






Axiom of Archimedes

Let A1 be any point upon a straight line between the arbitrarilychosen points A and B. Choose the points A2, A3, A4, . . . so thatA1 lies between A and A2, A2 between A1 and A3, A3 betweenA2 and A4 etc. Moreover, let the segmentsAA1, A1A2, A2A3, A3A4, . . . be equal to one another. Then,among this series of points, there always exists a point An suchthat B lies between A and An.

A A1

B






Axiom of Archimedes


B

A A1 A2 A3 A4 A5






Axiom of Archimedes


A1 A2 A3 A5

B

A A4






Axiom of Archimedes

definition

congruentl l −→ length l ≥ 3 ∧∀i. 0 ≤ i ∧ i+ 2 < length l −→(l ! i)(l ! (i+ 1)) ∼=h (l ! (i+ 1))(l ! (i+ 2)) ∧Bh((l ! i), (l ! (i+ 1)), (l ! (i+ 2)))

Axiom of Archimedes

Bh(A,A1, B) −→(∃l. congruentl(A # A1 # l) ∧ (∃i. Bh(A,B, (l ! i))))






Axiom of Archimedes

d2ag A A′ > d2ag A B and d2ag A A′ = t · d2ag A A1

t · d2ag A A1 > d2ag A B (applying Archimedes’ rule for realnumbers)

inductively proved that there exists a list l such that:

- congruentl l,- longer then t,- first two elements are A and A1





Outline

1 Introduction










Conclusions

Developed a formalization of Cartesian plane geometry withinIsabelle/HOL

Formally shown that the Cartesian plane satisfies all Tarski’saxioms and most of the Hilbert’s axioms

Proof that analytic geometry models geometric axioms arerather complex

Applying isometry transformations

Our formalization of the analytic geometry relies on theaxioms of real numbers and properties of reals are usedthroughout our proofs





Further Work

Formalizing analytic models of non-Euclidean geometries

Connect our formal developments to the implementation ofalgebraic methods for automated deduction in geometry


formalizing analytic geometries · our formalization of analytic geometry aims at establishing the...

Documents