The Surprise Examination Paradox and the Second Incompleteness Theorem Shira Kritchman and Ran Raz Few theorems in the history of mathematics which the derivation is done; which is impossible have inspired mathematicians and philosophers by the second incompleteness theorem. as much as G¨odel’s incompleteness theorems. The first incompleteness theorem states that, for The First Incompleteness Theorem 1 2 any rich enough consistent mathematical theory, G¨odel’s original proof for the first incompleteness there exists a statement that cannot be proved or theorem [Gödel31] is based on the liar paradox. disproved within the theory. The second incom- The liar paradox: consider the pleteness theorem states that for any rich enough statement “this statement is false.” consistent mathematical theory, the consistency The statement can be neither true of the theory itself cannot be proved (or disproved) nor false. within the theory. G¨odel considered the related statement “this state- In this paper we give a new proof for G¨odel’s second incompleteness theorem, based on Kol- ment has no proof.” He showed that this statement mogorov complexity, Chaitin’s incompleteness can be expressed in any theory that is capable of theorem, and an argument that resembles the expressing elementary arithmetic. If the statement surprise examination paradox. has a proof, then it is false; but since in a consistent We then go the other way around and suggest theory any statement that has a proof must be that the second incompleteness theorem gives a true, we conclude that if the theory is consistent, possible resolution of the surprise examination the statement has no proof. Since the statement N paradox. Roughly speaking, we argue that the has no proof, it is true (over ). Thus, if the the- flaw in the derivation of the paradox is that it ory is consistent, we have an example for a true N contains a hidden assumption that one can prove statement (over ) that has no proof. the consistency of the mathematical theory in The main conceptual difficulty in G¨odel’s orig- inal proof is the self-reference of the statement Shira Kritchman did her M.Sc in applied mathematics at “this statement has no proof.” A conceptually the Weizmann Institute of Science and is currently work- simpler proof of the first incompleteness the- ing in logic design at ECI Telecom. Her email address is orem, based on Berry’s paradox, was given by [email protected]. Chaitin [Chaitin71]. Ran Raz is professor on the faculty of mathematics and Berry’s paradox: consider the ex- computer science at the Weizmann Institute of Science. pression “the smallest positive in- His email address is [email protected]. teger not definable in under eleven Research supported by the Israel Science Foundation (ISF), words.” This expression defines the Binational Science Foundation (BSF), and the Minerva that integer in under eleven words. Foundation. Article©2010 Shira Kritchman, Ran Raz. To formalize Berry’s paradox, Chaitin uses the 1We require that the theory can express and prove ba- sic arithmetical truths. In particular, ZFC and Peano notion of Kolmogorov complexity. The Kolmogorov arithmetic (PA) are rich enough. complexity K(x) of an integer x is defined to be the 2Here and below, we consider only first-order theories length (in bits) of the shortest computer program with recursively enumerable sets of axioms. For simplic- that outputs x (and stops). Formally, to define K(x) ity, let us assume that the set of axioms is computable. one has to fix a programming language, such as 1454 Notices of the AMS Volume 57, Number 11 LISP, Pascal or C++. Alternatively, one can define proofs only show the existence of a true statement K(x) by considering any universal Turing machine. (over N) that has no proof, without giving an Chaitin’s incompleteness theorem states that explicit example of such a statement. for any rich enough consistent mathematical A different proof for the second incomplete- theory, there exists a (large enough) integer L ness theorem, based on Berry’s paradox, was (depending on the theory and on the programming given by Kikuchi [Kikuchi97]. This proof is model- language that is used to define Kolmogorov com- theoretic and seems to us somewhat less intuitive plexity) such that, for any integer x, the statement for people who are less familiar with model theory. “K(x) > L” cannot be proved within the theory. For previous model-theoretic proofs for the sec- The proof given by Chaitin is as follows. Let L be ond incompleteness theorem see [Kreisel50] (see a large enough integer. Assume for a contradiction also [Smory´nski77]). that, for some integer x, there is a proof for the statement “K(x) > L”. Let w be the first proof Our Approach (say, according to the lexicographic order) for a We give a new proof for the second incomplete- statement of the form “K(x) > L”. Let z be the ness theorem, based on Chaitin’s incompleteness integer x such that w proves “K(x) > L”. It is easy theorem and an argument that resembles the to give a computer program that outputs z: the surprise examination paradox (also known as the program enumerates all possible proofs w, one by unexpected hanging paradox). one, and for the first w that proves a statement of The surprise examination para- the form “K(x) > L”, the program outputs x and dox: the teacher announces in stops. The length of this program is a constant + log L. Thus, if L is large enough, the Kolmogorov class: “next week you are going complexity of z is less than L. Since w is a proof to have an exam, but you will not for “K(z) > L” (which is a false statement), we be able to know on which day of conclude that the theory is inconsistent. the week the exam is held until Note that the number of computer programs of that day.” The exam cannot be length L bits is at most 2L+1. Hence, for any integer held on Friday, because otherwise, L, there exists an integer 0 ≤ x ≤ 2L+1, such that the night before the students will K(x) > L. Thus, for some integer x, the statement know that the exam is going to be “K(x) > L” is a true statement (over N) that has no held the next day. Hence, in the proof. same way, the exam cannot be held A different proof for G¨odel’s first incomplete- on Thursday. In the same way, the ness theorem, also based on Berry’s paradox, was exam cannot be held on any of the given by Boolos [Boolos89] (see also [Vopenka66, days of the week. Kikuchi94]). Other proofs for the first incomplete- Let T be a (rich enough) mathematical theory, ness theorem are also known (for a recent survey, such as PA or ZFC. For simplicity, the reader can see [Kotlarski04]). assume that T is ZFC, the theory of all mathematics; thus any mathematical proof, and in particular any The Second Incompleteness Theorem proof in this paper, is obtained within T . The second incompleteness theorem follows di- Let L be the integer guaranteed by Chaitin’s rectly from G¨odel’s original proof for the first in- incompleteness theorem. Thus, for any integer x, completeness theorem. As described above, G¨odel the statement “K(x) > L” cannot be proved (in the expressed the statement “this statement has no theory T ) unless the theory is inconsistent. Note, proof” and showed that, if the theory is consistent, however, that for any integer x such that K(x) ≤ L, this is a true statement (over N) that has no proof. there is a proof (in T ) for the statement “K(x) ≤ L”, Informally, because the proof that this is a true simply by giving the computer program of length at statement can be obtained within any rich enough most L that outputs x and stops and by describing theory, such as Peano arithmetic (PA) or ZFC, if the running of that computer program until it the consistency of the theory itself can also be stops. proved within the theory, then the statement can Let m be the number of integers 0 ≤ x ≤ 2L+1 be proved within the theory, which is a contra- such that K(x) > L. (The number m is analogous diction. Hence, if the theory is rich enough, the to the day of the week on which the exam is consistency of the theory cannot be proved within held in the surprise examination paradox.) Recall the theory. that because the number of computer programs Thus the second incompleteness theorem fol- of length L bits is at most 2L+1, there exists at lows directly from G¨odel’s original proof for the least one integer 0 ≤ x ≤ 2L+1 such that K(x) > L. first incompleteness theorem. However, the sec- Hence, m ≥ 1. ond incompleteness theorem doesn’t follow from Assume that m = 1. Thus there exists a sin- Chaitin’s and Boolos’s simpler proofs for the first gle integer x ∈ {0,..., 2L+1} such that K(x) > L, incompleteness theorem. The problem is that these and every other integer y ∈ {0,..., 2L+1} satisfies December 2010 Notices of the AMS 1455 K(y) ≤ L. In this case, one can prove that x satis- see, for example, [Kikuchi97], Theorem 1.2 and fies K(x) > L by proving that every other integer Section 2). + ∈ { L 1} ≤ + y 0,..., 2 satisfies K(y) L (and recall that (2) T ⊢ ∀y ∈ {0,..., 2L 1} ((K(y) ≤ L) there is a proof for every such statement).