the suprise examination paradox and godel

8/13/2019 The Suprise Examination Paradox and Godel

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The Surprise Examination

Paradox and the SecondIncompleteness TheoremShira Kritchman and Ran Raz

Few theorems in the history of mathematicshave inspired mathematicians and philosophersas much as Godels incompleteness theorems.The first incompleteness theorem states that, forany rich enough1 consistent mathematical theory,2

there exists a statement that cannot be proved ordisproved within the theory. The second incom-pleteness theorem states that for any rich enoughconsistent mathematical theory, the consistencyof the theory itself cannot be proved (or disproved)within the theory.

In this paper we give a new proof for Godelssecond incompleteness theorem, based on Kol-mogorov complexity, Chaitins incompletenesstheorem, and an argument that resembles thesurprise examination paradox.

We then go the other way around and suggestthat the second incompleteness theorem gives apossible resolution of the surprise examinationparadox. Roughly speaking, we argue that theflaw in the derivation of the paradox is that itcontains a hidden assumption that one can provethe consistency of the mathematical theory in

Shira Kritchman did her M.Sc in applied mathematics at

the Weizmann Institute of Science and is currently work-

ing in logic design at ECI Telecom. Her email address is

[email protected] .

Ran Raz is professor on the faculty of mathematics and

computer science at the Weizmann Institute of Science.

His email address is [email protected].

Research supported by the Israel Science Foundation (ISF),

the Binational Science Foundation (BSF), and the Minerva

Foundation. Article2010 Shira Kritchman, Ran Raz.1We require that the theory can express and prove ba-

sic arithmetical truths. In particular, ZFC and Peano

arithmetic (PA) are rich enough.2Here and below, we consider only first-order theories

with recursively enumerable sets of axioms. For simplic-

ity, let us assume that the set of axioms is computable.

which the derivation is done; which is impossibleby the second incompleteness theorem.

The First Incompleteness TheoremGodels original proof for the first incompletenesstheorem [Gdel31] is based on the liar paradox.

The liar paradox: consider thestatement this statement is false.The statement can be neither truenor false.

Godel considered the related statement this state-ment has no proof. He showed that this statementcan be expressed in any theory that is capable of

expressing elementary arithmetic. If the statementhas a proof, then it is false; but since in a consistenttheory any statement that has a proof must betrue, we conclude that if the theory is consistent,the statement has no proof. Since the statementhas no proof, it is true (over N). Thus, if the the-ory is consistent, we have an example for a truestatement (over N) that has no proof.

The main conceptual difficulty in Godels orig-inal proof is the self-reference of the statementthis statement has no proof. A conceptuallysimpler proof of the first incompleteness the-orem, based on Berrys paradox, was given by

Chaitin [Chaitin71].Berrys paradox: consider the ex-pression the smallest positive in-teger not definable in under elevenwords. This expression definesthat integer in under eleven words.

To formalize Berrys paradox, Chaitin uses thenotion ofKolmogorov complexity. The Kolmogorovcomplexity K(x) of an integerx is defined to be thelength (in bits) of the shortest computer programthat outputs x (and stops). Formally, to define K(x)one has to fix a programming language, such as

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LISP, Pascal or C++. Alternatively, one can defineK(x)by considering any universal Turing machine.

Chaitins incompleteness theorem states thatfor any rich enough consistent mathematicaltheory, there exists a (large enough) integer L(depending on the theory and on the programminglanguage that is used to define Kolmogorov com-plexity) such that, for any integer x, the statementK(x) > L cannot be proved within the theory.

The proof given by Chaitin is as follows. Let Lbea large enough integer. Assume for a contradictionthat, for some integer x, there is a proof for thestatement K(x) > L. Let w be the first proof(say, according to the lexicographic order) for astatement of the form K(x) > L. Let z be theintegerx such thatwproves K(x) > L. It is easyto give a computer program that outputs z: theprogram enumerates all possible proofsw, one byone, and for the first wthat proves a statement ofthe form K(x) > L, the program outputs x andstops. The length of this program is a constant

+ log L. Thus, ifLis large enough, the Kolmogorovcomplexity ofz is less than L. Since w is a prooffor K(z) > L (which is a false statement), weconclude that the theory is inconsistent.

Note that the number of computer programs oflengthLbits is at most 2L+1. Hence, for any integerL, there exists an integer 0 x 2L+1, such thatK(x) > L. Thus, for some integer x, the statementK(x) > L is a true statement (over N) that has noproof.

A different proof for Godels first incomplete-ness theorem, also based on Berrys paradox, wasgiven by Boolos [Boolos89] (see also [Vopenka66,Kikuchi94]). Other proofs for the first incomplete-ness theorem are also known (for a recent survey,see [Kotlarski04]).

The Second Incompleteness TheoremThe second incompleteness theorem follows di-rectly from Godels original proof for the first in-completeness theorem. As described above, Godelexpressed the statement this statement has noproof and showed that, if the theory is consistent,this is a true statement (over N) that has no proof.Informally, because the proof that this is a truestatement can be obtained within any rich enoughtheory, such as Peano arithmetic (PA) or ZFC, if

the consistency of the theory itself can also beproved within the theory, then the statement can

be proved within the theory, which is a contra-diction. Hence, if the theory is rich enough, theconsistency of the theory cannot be proved withinthe theory.

Thus the second incompleteness theorem fol-lows directly from Godels original proof for thefirst incompleteness theorem. However, the sec-ond incompleteness theorem doesnt follow fromChaitins and Booloss simpler proofs for the firstincompleteness theorem. The problem is that these

proofs only show the existence of a true statement(over N) that has no proof, without giving anexplicit example of such a statement.

A different proof for the second incomplete-ness theorem, based on Berrys paradox, wasgiven by Kikuchi [Kikuchi97]. This proof is model-theoretic and seems to us somewhat less intuitivefor people who are less familiar with model theory.For previous model-theoretic proofs for the sec-ond incompleteness theorem see [Kreisel50] (seealso [Smorynski77]).

Our ApproachWe give a new proof for the second incomplete-ness theorem, based on Chaitins incompletenesstheorem and an argument that resembles thesurprise examination paradox (also known as theunexpected hanging paradox).

The surprise examination para-dox: the teacher announces inclass: next week you are goingto have an exam, but you will not

be able to know on which day ofthe week the exam is held untilthat day. The exam cannot beheld on Friday, because otherwise,the night before the students willknow that the exam is going to beheld the next day. Hence, in thesame way, the exam cannot be heldon Thursday. In the same way, theexam cannot be held on any of thedays of the week.

Let Tbe a (rich enough) mathematical theory,

such as PA or ZFC. For simplicity, the reader canassume thatTis ZFC,the theory of allmathematics;thus any mathematical proof, and in particular anyproof in this paper, is obtained withinT.

Let L be the integer guaranteed by Chaitinsincompleteness theorem. Thus, for any integer x,the statement K(x) > L cannot be proved (in thetheory T) unless the theory is inconsistent. Note,however, that for any integerxsuch thatK(x) L,there is a proof (in T) for the statement K(x) L,simply by giving thecomputerprogram of length atmostLthat outputsxand stops and by describingthe running of that computer program until it

stops.Let m be the number of integers 0 x 2L+1

such that K(x) > L. (The number m is analogousto the day of the week on which the exam isheld in the surprise examination paradox.) Recallthat because the number of computer programsof length L bits is at most 2L+1, there exists atleast one integer 0 x 2L+1 such that K(x) > L.Hence,m 1.

Assume that m = 1. Thus there exists a sin-gle integer x {0, . . . , 2L+1} such that K(x) > L,and every other integer y {0, . . . , 2L+1} satisfies

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K(y) L. In this case, one can prove that x satis-fies K(x) > L by proving that every other integery {0, . . . , 2L+1} satisfies K(y) L (and recall thatthere is a proof for every such statement). Becausewe proved that m 1, the only x for which wedidnt proveK(x) Lmust satisfy K(x) > L.

Thus if m = 1, then for some integer x thestatement K(x) > L can be proved (in T). Butwe know that for any integer x the statementK(x) > L cannot be proved (in T) unless thetheory is inconsistent. Hence, if the theory isconsistent, m 2. As we assume that T is a richenough theory, we can prove the last conclusion inT. That is, we can prove inTthat ifT is consistent,thenm 2.

Assume for a contradiction that the consistencyofTcan be proved within T. Thus we can provein Tthe statement m 2. In the same way, wecan work our way up and prove that m i + 1,for every i 2L+1 + 1. In particular, m >2L+1 + 1,which is a contradiction, as m 2L+1 + 1 (by the

definition ofm).

The Formal ProofTo present the proof formally, one needs to beable to express provability within T, in the lan-guage of T. The standard way of doing that is

by assuming that the language of T containsthe language of arithmetics and by encoding ev-ery formula and every proof in T by an integer,usually referred to as the Godel number of thatformula or proof. For a formula A, let A beits Godel number. Let PrT(A) be the followingformula:there existswthat is the Godel number of

aT-proof for the formulaA. Intuitively, PrT(A)expresses the provability of the formula A. For-mally, the formulas PrT(A) satisfy the so-calledHilbert-Bernays derivability conditions (see, forexample, [Mendelson97]):

1. IfTprovesA, thenTproves PrT(A).2. Tproves: PrT(A) PrT(PrT(A)).3. T proves: PrT(A B) (PrT(A)

PrT(B)).

The consistency ofT is usually expressed as theformula Con(T ) PrT(0 = 1). Inall that comes

below,T Adenotes TprovesA. We will provethatT Con(T ), unless Tis inconsistent.

For our proof, we will need two facts aboutprovability of claims concerning Kolmogorov com-plexity. First, we need to know that Con(T ) PrT(K(x) > L). We will use the followingform of Chaitins incompleteness theorem (see,for example, [Kikuchi97], Theorem 3.3).

T Con(T ) x {0, . . . , 2L+1}(1)

PrT(K(x) > L).

Second, we need to know that (K(y) L) PrT(K(y) L). We will use the following form(formally, this follows, as K(y) L is a 1formula;

see, for example, [Kikuchi97], Theorem 1.2 andSection 2).

T y {0, . . . , 2L+1} ((K(y) L)(2)

PrT(K(y) L)) .

Assume for a contradiction that Tis consistentandT Con(T ). Then, by Equation 1,

(3) T x {0, . . . , 2L+

1} PrT(K(x) > L).We will derive a contradiction by proving by

induction that, for every i 2L+1 + 1, T (m i + 1), where m is defined as in the previoussection. Because, obviously, T (m 2L+1 + 1),this is a contradiction to the assumption that Tis consistent andT Con(T ). Because we alreadyknow that T (m 1), we already have the

base case of the induction. Assume (the inductionhypothesis) that for some 1 i 2L+1 + 1,

T (m i).

We will show that T (m i + 1) as follows. Let

r=

2

L+1 +1

i.1. By the definition of m, T (m = i)

different y1, . . . , y r {0, . . . , 2L+1}r

j=1

(K(yj) L).2. Hence, by Equation 2, T (m = i)

different y1, . . . , y r {0, . . . , 2L+1}r

j=1

PrT(K(yj) L).3. Foreverydifferenty1, . . . , y r {0, . . . , 2L+1},

and every x {0, . . . , 2L+1} \ {y1, . . . , y r},

T (m i) r

j=1(K(yj)L) (K(x) >

L)

, (by the definition of m), andhence by Hilbert-Bernays derivabil-ity conditions, T PrT(m i)

rj=1 PrT(K(yj) L)PrT(K(x)>L) .4. By the previous two items, T ((m =

i) PrT(m i)) x {0, . . . , 2L+1}PrT(K(x) > L).

5. As T (m i ) (by the induction hypoth-esis), T PrT(m i). Hence, T (m =i) x {0, . . . , 2L+1} PrT(K(x) > L).

6. Hence, by Equation 3,T (m = i).7. Hence, asT (m i),T (m i + 1).

A Possible Resolution of the SurpriseExamination Paradox

In the previous sections we gave a proof forGodelssecond incompleteness theorem by an argumentthat resembles the surprise examination paradox.In this section we go the other way around andsuggest that the second incompleteness theoremgives a possible resolution of the surprise exami-nation paradox. Roughly speaking, we argue thatthe flaw in the derivation of the paradox is that itcontains a hidden assumption that one can provethe consistency of the mathematical theory inwhich the derivation is done, which is impossible

by the second incompleteness theorem.



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The important step in analyzing the paradoxis the translation of the teachers announcementinto a mathematical language. The key point liesin the formalization of the notions of surprise andknowledge.

As before, let Tbe a rich enough mathematicaltheory (say, ZFC). Let {1, . . . , 5} be the days ofthe week and let m denote the day of the weekon which the exam is held. Recall the teachersannouncement: next week you are going to havean exam, but you will not be able to know onwhich day of the week the exam is held untilthat day. The first part of the announcement isformalized asm {1, . . . , 5}. A standard way thatappears in the literature to formalize the secondpart is by replacing the notion ofknowledge bythe notion ofprovability[Shaw58, Fitch64] (for arecent survey see [Chow98]). The second part isrephrased as on the night before the exam youwill not be able to prove, using this statement,that the exam is tomorrow, or, equivalently, for

every 1 i 5, if you are able to prove, using thisstatement, that (m i) (m = i), then m = i.This can be formalized as the following statementthat we denote byS(the statementScontains bothparts of the teachers announcement):

S [m {1, . . . , 5}]

1i5PrT ,S(m i m = i) (m i)

,

where PrT ,S(A) expresses the provability of aformulaAfrom the formulaSin the theoryT (for-mally, PrT ,S(A) is the formula: there existsw thatis the Godel number of aT-proof for the formulaA

from the formulaS). Note that the formula S isself-referential. Nevertheless, it is well known thatthis is not a real problem and that such a formulaS can be formulated (see [Shaw58, Chow98]; formore about this issue, see below).

Let us try to analyze the paradox when theteachers announcement is formalized as the abovestatement S. We will start from the last day. Thestatement m 5 together with m {1, . . . , 5}implym = 5. Hence, PrT ,S(m 5 m = 5), and

by S we can conclude m = 5. Thus S impliesm {1, . . . , 4}. In the same way, working our waydown, we can prove PrT ,S(m 4 m = 4), and

by S we can conclude m = 4. In the same way,

m = 3, m = 2, and m = 1. In other words, Simpliesm {1, . . . , 5}. ThusScontradicts itself.

The fact that Scontradicts itself gives a certainexplanation for the paradox; the teachers an-nouncement is just a contradiction. On the otherhand, we feel that this formulation doesnt fullyexplain the paradox: Note that, because Sis a con-tradiction, it can be used to prove any statement.So, for example, on Tuesday night the studentscan use S to prove that the exam will be heldon Wednesday. Is it fair to say that this meansthat they know that the exam will be held on

Wednesday? No, because they can also use S toprove that the exam will be held on Thursday. Thuswe conclude that, since Sis a contradiction, prov-ability from S doesnt imply knowledge. Recall,however, that the very intuition behind the formal-ization of the teachers announcement as S wasthat the notion ofknowledge can be replaced bythe notion ofprovability. But if provability from Sdoesnt imply knowledge, the statement Sdoesntseem to be an accurate translation of the teachersannouncement into a mathematical language.

Is there a better way to formalize the teachersannouncement? To answer this question, let us an-alyze the situation from the students point of viewon Tuesday night. There are three possibilities:

1. On Tuesday night, the students are notable to prove that the exam will be held onWednesday.

2. On Tuesday night, the students are ableto prove that the exam will be held onWednesday, but they are also able to prove

for some other day that the exam will beheld on that day.(Note that this possibility can only occurif the system is inconsistent and is infact equivalent to the inconsistency of thesystem).

3. On Tuesday night, the students are ableto prove that the exam will be held onWednesday, and they are not able to provefor any other day that the exam will beheld on that day.

We feel that only in the third case is it fair to saythat the students know that the exam will be held

on Wednesday. They know that the exam will beheld on Wednesday only if they are able to provethat the exam will be held on Wednesday, and theyare not able to prove for any other day that theexam will be held on that day.

We hence rephrase the second part of theteachers announcement as for every 1 i 5, ifone canprove(usingthis statement) that (m i) (m = i), and there is no j = i for which one canprove (using this statement) (m i) (m = j),then m = i. Thus the teachers announcement isthe following statement:3

S [m {1, . . . , 5}]

1i5

PrT ,S(m i m = i)

1j5,ji

PrT,Sm i m = j

(m i) .

Let us try to analyze the paradox whenthe teachers announcement is formalized as

3This statement is equivalent to one of the suggestions

(the statement I5) made by Halpern and Moses [HM86].However, the analysis of the paradox there is different

from the one shown here and makes no use of Godels

second incompleteness theorem.

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the new statement S. As before, m 5 to-gether with m {1, . . . , 5} imply m = 5. Hence,PrT ,S(m 5 m = 5). However, this timeone cannot use S to conclude m = 5, as itis possible that for some j = 5 we also havePrT ,S

m 5 m = j

. This happens iff the

system T + S is inconsistent. Formally, this timeone cannot use Sto deduce m = 5, but rather theformula

Con(T,S) (m = 5),

where Con(T,S) PrT ,S(0 =1) expresses theconsistency of T + S. Because by the second in-completeness theorem one cannot prove Con(T,S)within T + S, we cannot conclude that S impliesm = 5 and hence cannot continue the argument.

More precisely, because S doesnt imply m {1, . . . , 4}, but rather Con(T,S) m {1, . . . , 4},when we try to work our way down we do not getthe desired formula PrT ,S(m 4 m = 4) butrather the formula

PrT ,S(Con(T,S) (m 4) m = 4) ,which is not enough to continue the argument.

Thus our conclusion is that if the studentsbelieve in the consistency ofT+S, the exam cannotbe held on Friday, because on Thursday night thestudents will know that ifT+ Sis consistent theexam will be held on Friday. However, the examcan be held on any other day of the week becausethe students cannot prove the consistency ofT+S.

Finally, for completeness, let us address theissue of the self-reference of the statement S. Theissue of self-referentiality of a statement goes backto Godels original proof for the first incomplete-

ness theorem. The self-reference is what makesGodels original proof conceptually difficult andwhat makes the teachers announcement in thesurprise examination paradox paradoxical.

To solve this issue, Godel introduced the tech-nique of diagonalization. The same technique can

be used here. To formalizeS, we will use the nota-tion a b to indicate implication between Godelnumbers a and b. That is, a b is a statementindicating thatais a Godel number of a statementA and b is a Godel number of a statement B,such that A B. We will also need the functionSub(a,b) that represents substitution ofb in theformula with Godel number a. That is, if a is aGodel number of a formula A(x)with free variablex and b is a number, then Sub(a,b) is the Godelnumber of the statement A(b).

Letvij m i m = j. Denote by Q(x)theformula

Q(x)[m {1, . . . , 5}]

1i5

PrT(Sub(x,x) vii )

1j5,ji

PrT

Sub(x,x) vij (m i)

.

Let q be the Godel number of the formula Q(x).The statement S is formalized as S Q(q). Tosee that this statement is the one that we areinterested in, denote by s the Godel number ofSand note that s = Sub(q,q). Thus

S [m {1, . . . , 5}]

1i5

PrT(s vii )

1j5,ji

PrT

s vij (m i)

.

References[Boolos89] G. Boolos, A new proof of the Gdel

incompleteness theorem, Notices Amer.Math. Soc. 36 (1989), 388390.

[Chaitin71] G. J. Chaitin, Computational complex-ity and Gdels incompleteness theorem,ACM SIGACT News9 (1971), 1112.

[Chow98] T. Y. Chow, The surprise examinationor unexpected hanging paradox, Amer.

Math. Monthly 105(1998), 4151.[Fitch64] F. Fitch, A Gdelized formulation of theprediction paradox,Amer. Phil. Quart. 1(1964), 161164.

[Gdel31] K. Gdel, ber formal unentscheid-bare Stze der Principia Mathematicaund verwandter Systeme I, Monatshehtefr Mathematik und Physik 38 (1931),173198.

[HM86] J. Halpernand Y. Moses, Taken by sur-prise: The paradox of the surprise testrevisited, Journal of Philosophical Logic15(1986), 281304.

[Kikuchi94] M. Kikuchi, A note on Booloss proof ofthe incompleteness theorem, Math. Logic

Quart.40 (1994), 528532.[Kikuchi97] , Kolmogorov complexity and thesecond incompleteness theorem, Arch.Math. Logic 36(1997), 437443.

[Kotlarski04] H. Kotlarski, The incompleteness the-orems after 70 years, Ann. Pure Appl.Logic 126(2004), 125138.

[Kreisel50] G. Kreisel, Notes on arithmetical modelsfor consistent formulae of the predi-cate calculus, Fund. Math. 37 (1950),265285.

[Mendelson97] E. Mendelson, Introduction to Mathe-matical Logic, CRC Press, 1997.

[Shaw58] R. Shaw, The unexpected examination,Mind 67(1958), 382384.

[Smorynski77] C. A. Smorynski, The incompletenesstheorem. In: Introduction to Mathemati-cal Logic(J. Barwise, ed.), North-Holland,1977, 821865.

[Vopenka66] P. Vopenka, A new proof on the Gdelsresult of nonprovability of consistency,Bull. Acad. Polon. Sci. 14 (1966), 111116.


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