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Proof-checking Euclid

Abstract

We used computer proof-checking methods to verify the correctness of our proofs of the propositions in Euclid Book I. We used axioms as close as possible to those of Euclid, in a language closely related to that used in Tarski’s formal geometry. We used proofs as close as possible to those given by Euclid, but filling Euclid’s gaps and correcting errors. Euclid Book I has 48 propositions; we proved 235 theorems. The extras were partly “Book Zero”, preliminaries of a very fundamental nature, partly propositions that Euclid omitted but were used implicitly, partly advanced theorems that we found necessary to fill Euclid’s gaps, and partly just variants of Euclid’s propositions. We wrote these proofs in a simple fragment of first-order logic corresponding to Euclid’s logic, debugged them using a custom software tool, and then checked them in the well-known and trusted proof checkers HOL Light and Coq.

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Appendices

Appendix A: Formal proof of Prop. I.1

The reader may compare the following proof to Euclid’s. The conclusion ELABC, that ABC is equilateral, is reached about halfway through, and that corresponds to the end of Euclid’s proof. The firsts half of our proof corresponds fairly naturally to Euclid’s, except for quoting the circle-circle axiom, and verifying that is hypotheses are satisfied.

The last half of our proof is devoted to proving that ABC is a triangle, that is, the three points are not collinear. Note the use of lemma partnotequalwhole. If Euclid had noticed the need to prove that ABC actually is a triangle, he would have justified it using the common notions, applied to equality (congruence) of lines. This version of “the part is not equal to the whole” is not an axiom for us, but a theorem.

At the request of the referee we present this proof in a typeset form rather than in its native Polish form. Obviously further mechanical processing can increase its superficial resemblance to Euclid’s style, but the point of our present work is simply its mechanically-checked correctness.

Proposition A.1 (Prop. I.1)

On a given finite straight line to construct an equilateral triangle.

$$\forall A B\qquad A \neq B \implies \exists X\qquad ABX is equilateral~ \wedge ~ABX is a triangle $$

Proof

Let J be such that J is the circle of center A and radius AB by postulate Euclid3.

BA by lemma inequalitysymmetric.

Let K be such that K is the circle of center B and radius BA by postulate Euclid3.

Let D be such that A is strictly between B and DADAB by lemma localextension.

ADBA by lemma congruenceflip.

BABA by common notion congruencereflexive.

D is outside circle K by definition of outside.

B = B by common notion equalityreflexive.

B is inside circle K by definition of inside.

ABAB by common notion congruencereflexive.

B is on circle J by definition of on.

D is on circle J by definition of on.

A = A by common notion equalityreflexive.

A is inside circle J by definition of inside.

LetC be such that C is on circle KC is on circle J by postulate circle-circle.

ACAB by axiom circle-center-radius.

ABAC by lemma congruencesymmetric.

BCBA by axiom circle-center-radius.

BCAB by lemma congruenceflip.

BCAC by lemma congruencetransitive.

ABBC by lemma congruencesymmetric.

ACCA by common notion equalityreverse.

BCCA by lemma congruencetransitive.

ABC is equilateral by definition of equilateral.

BC by axiom nocollapse.

CA by axiom nocollapse.

Let show that C is strictly between A and B does not hold by contradiction:

{

\({A}{C} \ncong {A}{B}\) by lemma partnotequalwhole.

CAAC by common notion equalityreverse.

CAAB by lemma congruencetransitive.

ACCA by common notion equalityreverse.

ACAB by lemma congruencetransitive.

We have a contradiction.

}

Let show that B is strictly between A and C does not hold by contradiction:

{

\({A}{B} \ncong {A}{C}\) by lemma partnotequalwhole.

ABCA by lemma congruencetransitive.

CAAC by common notion equalityreverse.

ABAC by lemma congruencetransitive.

We have a contradiction.

}

Let show that A is strictly between B and C does not hold by contradiction:

{

\({B}{A} \ncong {B}{C}\) by lemma partnotequalwhole.

BAAB by common notion equalityreverse.

BABC by lemma congruencetransitive.

We have a contradiction.

}

Let show that ABC are collinear does not hold by contradiction:

{

AC by lemma inequalitysymmetric.

A = BA = CB = CA is strictly between B and CB is strictly between B and CC is strictly between A and B by definition of collinear.

We have a contradiction.

}

ABC is a triangle by definition of triangle. □

Appendix B: Axioms and definitions

The following formulas are presented in a format that can be cut and pasted, even from a pdf file.

Definitions

A and B are distinct points

figure a

A, B, and C are collinear

figure b

A, B, and C are not collinear

figure c

P is inside (some) circle J of center C and radius AB

figure d

P is outside (some) circle J of center U and radius VW

figure e

B is on (some) circle J of center U and radius XY

figure f

ABC is equilateral

figure g

ABC is a triangle

figure h

C lies on ray AB

figure i

AB is less than CD

figure j

B is the midpoint of AC

figure k

Angle ABC is equal to angle abc

figure l

DBF is a supplement of ABC

figure m

ABC is a right angle

figure n

PQ is perpendicular to AB at C and NCABP

figure o

PQ is perpendicular to AB

figure p

P is in the interior of angle ABC

figure q

P and Q are on opposite sides of AB

figure r

P and Q are on the same side of AB

figure s

ABC is isosceles with base BC

figure t

AB cuts CD in E

figure u

Triangle ABC is congruent to abc

figure v

Angle ABC is less than angle DEF

figure w

AB and CD are together greater than EF

figure x

AB, CD are together greater than EF,GH

figure y

ABC and DEF make together two right angles

figure z

AB meets CD

figure aa

AB crosses CD

figure ab

AB and CD are Tarski parallel

figure ac

AB and CD are parallel

figure ad

ABC and DEF are together equal to PQR

figure ae

ABCD is a parallelogram

figure af

ABCD is a square

figure ag

ABCD is a rectangle

figure ah

ABCD and abcd are congruent rectangles

figure ai

ABCD and abcd are equal rectangles

figure aj

ABCD is a base rectangle of triangle BCE

figure ak

ABC and abc are equal triangles

figure al

ABCD and abcd are equal quadrilaterals

figure am

Common notions

figure an
figure ao
figure ap
figure aq
figure ar
figure as
figure at
figure au

Axioms of betweenness and congruence

figure av
figure aw
figure ax
figure ay
figure az
figure ba

Postulates

figure bb
figure bc
figure bd
figure be
figure bf
figure bg
figure bh
figure bi

Axioms for equal figures

figure bj
figure bk
figure bl
figure bm
figure bn
figure bo
figure bp
figure bq
figure br
figure bs
figure bt
figure bu
figure bv
figure bw
figure bx
figure by

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Beeson, M., Narboux, J. & Wiedijk, F. Proof-checking Euclid. Ann Math Artif Intell 85, 213–257 (2019). https://doi.org/10.1007/s10472-018-9606-x

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Keywords

  • Euclid
  • Proof-checking
  • Euclidean geometry
  • HOL light
  • Coq

Mathematics Subject Classification (2010)

  • 03B30
  • 51M05