We will now make precise what is meant by a syntactic characterization of the set of arithmetically true sentences.

Transcription

1 2.4 Incompleteness We will now make precise what is meant by a syntactic characterization of the set of arithmetically true sentences. Definition A theory T is called axiomatisable if there is a recursive set Γ T s.t. T ClpΓq. The above recursive set Γ can best be thought of as a set of axioms for T which motivates the choice of the term axiomatisable. Being axiomatisable is a property which we require in order to accept a characterisation of a theory as a syntactic one. Lemma 2.4. If T is an axiomatisable theory then it is recursively enumerable. Proof. Let T ClpΓq and Γ be recursive. Let A 0,A 1,A 2,... be an enumeration of all valid sentences in the language of arithmetic which exists by the abstract completeness theorem (Theorem2.2). ThefollowingalgorithmenumeratesT: Input: i P N if A i is of the form B 1 ^... ^B k Ñ C and B j P Γ for j 1,...,k then print C else print D end if where D is any sentence provable in T. Lemma 2.5. If T is an axiomatisable and complete theory then it is recursive. Proof. If T is inconsistent, it is the set of all sentences which is recursive. So assume T is consistent. Then, as T is also complete, T & A iff T $ A. Let A 0,A 1,A 2,... be a recursive enumeration of T which exists by Lemma 2.4. The following algorithm is a decision procedure for T: Input: formula A k Ð 0 loop if A k A then return 1 end if if A k A then return 0 end if end loop By completeness of T the above algorithm terminates for every input A. The kernel of the proof of the first incompleteness theorem is a diagonal argument which we first formulate in a general, abstract form. Let R N 2, for m P N we write R m for tn P N pm,nq P Ru. Lemma 2.6 (Diagonal Lemma). If R N 2 and P tn P N pn,nq R Ru then P R m for all m. 30

2 Proof. Suppose P R m, then pm,mq P R iff m P R m iff m P P iff pm,mq R R. Lemma 2.7. If T Q is a consistent theory, then T is not recursive. Proof. We fix a variable z. For a formula A define EpAq : tn P N T $ Arzzsnsu ForeveryrecursivesetS NthereisaformulaAwhichrepresentsS int bytherepresentability theorem, i.e. 1. n P S implies T $ Arzzsns hence n P EpAq and 2. n R S implies T $ Arzzsns which, by consistency of T, implies T & Arzzsns hence n R EpAq. Therefore S EpAq. Let now R : tpm,nq P N 2 m #F,T $ F rzzsnsu and observe that R #A tn P N T $ Arzzsnsu EpAq. Furthermore, let P : tn P N pn,nq R Ru tn P N n #F or n #F,T & F rzzsnsu. By the diagonal lemma P R m for all m so, in particular, P R #A for all A so P EpAq for all A. But every recursive set is representable in Q hence in T therefore P is not recursive. Then also T is not recursive for suppose it would be, then also P would be recursive. Theorem 2.3 (First Incompleteness Theorem). If T Q is a consistent and axiomatisable theory, then T is not complete. Proof. SupposeT Qisconsistent,axiomatisableandcomplete. ByconsistencyandLemma2.7 it is not recursive. But by axiomatizability and completeness and Lemma 2.5 it is recursive. Contradiction. Corollary 2.1. ThpN q is not axiomatisable. Proof. ThpN q Q because the axioms of Q are true in N. Also ThpN q is (as every theory of a structure) consistent and complete. Corollary 2.2 (UndecidabilityofFirst-OrderLogic). The set of valid sentences is not recursive. Proof. Let A be the conjunction of the axioms of Q. Suppose the set of valid sentences would be recursive then also Q would be recursive because Q $ B iff A Ñ B is valid. But Q is not recursive by the above Lemma

3 The set of valid sentences is therefore an example for a set which is recursively enumerable but not recursive. There are more direct proofs of the undecidability of first-order logic which do not use the first incompleteness theorem. A sentence A is called undecidable in a theory T if T & A and T & A. Be careful to not confuse this notion with that of a decidable set (which in our terminology is a recursive set). The first incompleteness theorem shows that every consistent and axiomatisable extension of Q has an undecidable sentence. However, the above proof does not directly give an example of such a sentence. We will now carry out an alternative proof by constructing such a sentence explicitely. This other proof will also be useful for the second incompleteness theorem. Before that we need some preparatory steps: Definition A bounded quantifier is a quantifier of the form Dx px t Ñ Bq. t ^Bq px AformulaAiscalledΣ 1 -formulaifitislogicallyequivalenttoaformulaoftheform Dx 1 Dx n B where all quantifiers in B are bounded. AformulaAiscalledΠ 1 n B where all quantifiers in B are bounded. Lemma 2.8 (Σ 1 -completeness of Q). If A is a Σ 1 -sentence with N ( A, then Q $ A. Without Proof. Lemma 2.9. There is a Σ 1 -formula Numpx,yq s.t. for all k,n P N with n #s k: Q pnumps k,yq Ø y snq There is a Σ 1 -formula Subpx 1,x 2,x 3,x 4 q s.t. for all a,v,t,b P N where a #A for a formula A, v #z for a variable z, t #s for a term s and b #Arzzts: Q psubpsa,sv,s,yq Ø y s bq If T is an axiomatisable theory, then there is a Σ 1 -formula Prov T pxq s.t. for all n P N Q $ Prov T psnq iff n #A and T $ A for some formula A. Proof Sketch. The corresponding functions and predicate are recursive. : We will use Gödel-numerals, these are the numerals of Gödel-numbers. More precisely, for an expression (i.e. a term, a formula,...) e, the Gödel-numeral of e is defined as xey #e. Example P N ands 0 0 is also an LN -term. x0y is again an L N -term, but x0y 0 because x0y #0 #p2,0,0q and #p2,0,0q P N and #p2,0,0q 0. For any expression e: Q pnumpxey,yq Ø y xxeyyq. We can now give an alternative proof of a slightly modified version of the first incompleteness theorem. Theorem 2.4. If T is a consistent and axiomatisable theory with ThpN q T Q then T is incomplete. 32

4 In the proof of Lemma 2.7 we have used sets R P N 2 and P P N s.t. #A P P iff p#a,#aq R R iff T & ArzzxAys Proof. Now we will formalise the above condition inside the language of arithmetic. To that aim define the formula Bpzq : DuDv pnumpz,uq ^Subpz, xzy,u,vq ^Prov T pvqq. which exists as T is axiomatisable. Let us define the so-called Gödel-sentence G : BpxByq which expresses that G is not provable in T because: N ( G iff N * DuDv pnumpxby,uq ^SubpxBy, xzy,u,vq ^Prov T pvqq iff T & BrzzxBys iff T & G Suppose N * G, then by the above T $ G. Furthermore, G is a Π 1 -sentence, so G is a Σ 1 -sentence and by Σ 1 -completeness we know that N ( G implies T $ G and T would be inconsistent, contradiction, so N ( G. Suppose now T $ G, then N ( G, contradiction, so T & G. Suppose T $ G then N * G by the above so N ( G and by Σ 1 -completeness T $ G which contradicts consistency, so T & G. The above sentence G of a theory T is therefore true but undecidable in T. The second incompleteness theorem will give another example of a sentence which is undecidable in T: the sentence which expresses the consistency of T Con T : Prov T pxkyq. The second incompleteness theorem applies to arithmetical theories which are stronger than minimal arithmetic. The reason for this is that the proof of the second incompleteness theorem is essentially a formalisation in T of the above alternative proof of the first incompleteness theorem about T. Definition The arithmetical theory of Peano Arithmetic (PA) is defined as the deductive closure of Q and all sentences of the form the induction axioms. Ap0q papxq Ñ Apspxqqq Theorem 2.5 (Second Incompleteness Theorem). If T PA is consistent and axiomatisable, then T & Con T. 33

5 Proof. Let G be the Gödel-sentence for T. Then T $ G Ñ Prov T px Gyq because PA proves the Σ 1 -completeness of Q. On the other hand G DuDv pnumpxby,uq ^SubpxBy, xzy,u,vq ^Prov T pvqq and by Σ 1 -completeness T $ NumpxBy, xxbyyq and T $ SubpxBy, xzy, xxbyy, xgyq, hence using representability of the Num- and Sub-functions T $ G Ñ Prov T pxgyq. Furthermore, T proves that Prov T is closed under the inference rules of NK so T $ G Ñ Prov T pxkyq, i.e. T $ Con T Ñ G but as we have seen in the alternative proof of the first incompleteness theorem T & G and therefore T & Con T. In order to fully appreciate the importance of the incompleteness theorems of Gödel it is necessary to say a few words about the historical context. Gödel s work is a reaction to Hilbert s programme, which called for a formalisation of mathematical reasoning and for a proof of the consistency of this formalism by finitary means. It has never been made completely precise what should be understood as finitary but the intention of providing such a proof was to obtain an absolute justification of mathematical reasoning. So the methods employed in this consistencyproofshouldthemselvesbeundoubtable. Thesecondincompletenesstheorem shows that this program cannot be fulfilled: for proving the consistency of a theory the theory must be transcended. Finally one should discuss the mathematical significance of the incompleteness theorems. The examples of undecidable sentences we have seen so far, the Gödel-sentence of a theory as well as the consistency of a theory, are purely logical statements. One might think at first sight that this incompleteness phenomenon only manifests itself on such self-referential statements which are outside of logic hardly used in mathematics. However, this impression is not justified. There are sentences of purely mathematical (as opposed to logical) nature that have been shown to be independent of strong theories. The primary example is the continuum hypothesis, the first of Hilbert s famous 23 problems. We again work in the language of set theory; the linear ordering of sets by comparing their cardinality is given by x y : Dϕ pϕ : x Ñ y ^injectivepϕqq 34

6 and Then the continuum hypothesis is the statement x y : x y ^ y x CH Dx pn x Rq which has been shown to be independent of Zermelo-Fraenkel set theory ZFC by Gödel and Cohen. 35