gödel.kurt..(by hao kim) - complete proofs of godel incompleteness theorems

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Page 1: Gödel.Kurt..(by Hao Kim) - Complete Proofs of Godel Incompleteness Theorems

COMPLETE PROOFS OF G�ODEL'S INCOMPLETENESSTHEOREMSLECTURES BY B. KIMStep 0: Preliminary RemarksWe de�ne re ursive and re ursively enumerable fun tions and relations, enumer-ate several of their properties, prove G�odel's �-Fun tion Lemma, and demonstrateits �rst appli ations to oding te hniques.De�nition. For R � !n a relation, �R : !n ! !, the hara teristi fun tion onR, is given by �R(a) = (1 if :R(a),0 if R(a).De�nition. A fun tion from !m to ! (m � 0) is alled re ursive (or om-putable) if it is obtained by �nitely many appli ations of the following rules:R1. � Ini : !n ! !, 1 � i � n, de�ned by (x1; : : : ; xn) 7! xi is re ursive;� + : ! � ! ! ! and � : ! � ! ! ! are re ursive;� �< : ! � ! ! ! is re ursive.R2. (Composition) For re ursive fun tions G;H1; : : : ; Hk su h thatHi : !n ! !and G : !k ! !, F : !n ! !, de�ned byF (a) = G(H1(a); : : : ; Hk(a)):is re ursive.R3. (Minimization) For G : !n+1 ! ! re ursive, su h that for all a 2 !n thereexists some x 2 ! su h that G(a; x) = 0, F : !n ! !, de�ned byF (a) = �x(G(a; x) = 0)is re ursive. (Re all that �xP (x) for a relation P is the minimal x 2 ! su hthat x 2 P obtains.)De�nition. R(� !k) is alled re ursive, or omputable (R is a re ursive rela-tion) if �R is a re ursive fun tion.Proofs in this note are adaptation of those in [Sh℄ into the dedu tion system des ribed in [E℄.Many thanks to Peter Ahumada and Mi hael Brewer who wrote up this note.

Page 2: Gödel.Kurt..(by Hao Kim) - Complete Proofs of Godel Incompleteness Theorems

2 LECTURES BY B. KIMProperties of Re ursive Fun tions and Relations:P1. For Q � !k a re ursive relation, and H1; : : : ; Hk : !n ! ! re ursive fun -tions, P = fa 2 !n j Q(H1(a); : : : ; Hk(a))gis a re ursive relation.Proof. �P (a) = �Q(H1(a); : : : ; Hk(a)) is a re ursive fun tion by R2.P2. For P � !n+1, a re ursive relation su h that for all a 2 !n there existssome x 2 ! su h that P (a; x), then F : !n ! !, de�ned byF (a) = �xP (a; x)is re ursive.Proof. F (a) = �x(�P (a; x) = 0), so we may apply R3.P3. Constant fun tions, Cn;k : !n ! ! su h that Cn;k(a) = k, are re ursive.Proof. By indu tion:Cn;0(a) = �x(In+1n+1 (a; x) = 0)Cn;k+1(a) = �x(Cn;k(a) < x)are re ursive by R3 and P2, respe tively.P4. For Q;P � !n, re ursive relations, :P , P _ Q, and P ^ Q are re ursive.Proof. We have that �:P (a) = �<(0; �P (a));�P _Q(a) = �P (a) � �Q(a);P ^ Q = :(:P _ :Q):P5. The predi ates =, �, >, and � are re ursive.Proof. For a; b 2 !,a = b i� :(a < b) ^ :(b < a);a � b i� :(a < b);a > b i� (a � b) ^ :(a = b); anda � b i� :(a > b);hen e these are re ursive by P4.Notation. We write, for a 2 !n, f : !n ! ! a fun tion and P � !m+1 a relation,�x<f(a)P (x; b) � �x(P (x; b) _ x = f(a)):In parti ular, �x<f(a)P (x; b) is the smallest integer less than f(a) whi h satis�esP , if su h exists, or f(a), otherwise.We also write 9x<f(a)P (x) � (�x<f(a)P (x)) < f(a); and8x<f(a)P (x) � :(9x<f(a) (:P (x))):

Page 3: Gödel.Kurt..(by Hao Kim) - Complete Proofs of Godel Incompleteness Theorems

COMPLETE PROOFS OF G�ODEL'S INCOMPLETENESS THEOREMS 3The �rst is learly satis�ed if some x < f(a) satis�es P (x), while the se ond issatisifed if all x < f(a) satisfy P (x).P6. For P � !n+1 a re ursive relation, F : !n+1 ! !, de�ned byF (a; b) = �x<aP (x; b);is re ursive.Proof. F (a; b) = �x(P (x; b) _ x = a), and thus F is re ursive by P2, sin efor all b, a satis�es P (x; b) _ x = a.P7. For R � !n a re ursive relation, P;Q � !n+1 su h thatP (a; b) � 9x<aR(x; b)Q(a; b) � 8x<aR(x; b)are re ursive.Proof. Note that P is de�ned by omposition of re ursive fun tions andpredi ates, hen e re ursive by P1, and Q is de�ned by omposition of re- ursive fun tions, re ursive predi ates, and negation, hen e re ursive by P1and P4.P8. _� : ! � ! ! !, de�ned bya _�b = (a� b if a � b,0 otherwise,is re ursive.Proof. Note that a _�b = �x(b+ x = a _ a < b):P9. If G1; : : : ; Gk : !n ! ! are re ursive fun tions, and R1; : : : ; Rk � !n arere ursive relations partitioning !n (i.e., for ea h a 2 !n, there exists aunique i su h that Ri(a)), then F : !n ! !, de�ned byF (a) =8>>>><>>>>:G1(a) if R1(a),G2(a) if R2(a),... ...Gk(a) if Gk(a),is re ursive.Proof. Note that F = G1�:R1 + � � �+Gk�:Rk :

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4 LECTURES BY B. KIMP10. If Q1; : : : ; Qk � !n are re ursive relations, and R1; : : : ; Rk � !n are re ur-sive relations partitioning !n, then P � !n, de�ned byP (a) i� 8>><>>:Q1(a) if R1(a),... ...Qk(a) if Rk(a),is re ursive.Proof. Note that �P (a) = 8>><>>:�Q1(a) if R1(a),... ...�Qk (a) if Rk(a),is re ursive by P9.De�nition. A relation P � !n is re ursively enumerable (r.e.) if there existssome re ursive relation Q � !n+1 su h thatP (a) i� 9xQ(a; x):Remark If a relation R � !n is re ursive, then it is re ursively enumerable, sin eR(a) i� 9x(R(a) ^ x = x).Negation Theorem. A relation R � !n is re ursive if and only if R and :R arere ursively enumerable.Proof. If R is re ursive, then :R is re ursive. Hen e by above remark, both are r.e.Now, let P and Q be re ursive relations su h that for a 2 !n, R(a) i� 9xQ(a; x)and :R(a) i� 9xP (a; x).De�ne F : !n ! ! by F (a) = �x(Q(a; x) _ P (a; x));re ursive by P2, sin e either R(a) or :R(a) must hold.We show that R(a) i� Q(a; F (a)):In parti ular, Q(a; F (a)) implies there exists x (namely, F (a)) su h that Q(a; x),thus R(a) holds. Further, if :Q(a; F (a)), then P (a; F (a)), sin e F (a) satis�esQ(a; x) _ P (a; x). Thus :R(a) holds.The �-Fun tion Lemma.�-Fun tion Lemma (G�odel). There is a re ursive fun tion � : !2 ! ! su h that�(a; i) � a _�1 for all a; i 2 !, and for any a0; a1; : : : ; an�1 2 !, there is an a 2 !su h that �(a; i) = ai for all i < n.Remark 1. Let A = fa1; :::ang � ! r f0; 1g (n � 2) be a set su h that any twodistin t elements of A are realtively prime. Then given non-empty subset B of A,there is y 2 ! su h that for any a 2 A, ajy i� a 2 B. (y is a produ t of elements inB.)Lemma 2. If kjz for z 6= 0, then (1 + (j + k)z; 1+ jz) are relatively prime for anyj 2 !.

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COMPLETE PROOFS OF G�ODEL'S INCOMPLETENESS THEOREMS 5Proof. Note that for p prime, pjz implies that p=j1 + jz. But if pj1 + (j + k)z andpj1 + jz, then pjkz, implying pjkjz or pjz, and thus pjz, a ontradi tion.Lemma 3. J : !2 ! !, de�ned by J(a; b) = (a+ b)2 + (a+ 1), is one-to-one.Proof. If a+ b < a0 + b0, thenJ(a; b) = (a+b)2+a+1 � (a+b)2+2(a+b)+1 = (a+b+1)2 � (a0+b0)2 < J(a0; b0):Thus if J(a; b) = J(a0; b0), then a+ b = a0 + b0, and0 = J(a0; b0)� J(a; b) = a0 � a;implying that a = a0 and b = b0, as desired.Proof of �-Fun tion Lemma. De�ne�(a; i) = �x<a _�1 (9y<a (9z<a (a = J(y; z) ^ Div(1 + (J(x; i) + 1) � z; y))));where Div(x; y) � 9z < y + 1 (y = z � x) (satis�ed i� xjy) is re ursive. It is learthat � is re ursive, and that �(a; i) � a _�1.Given a1; : : : ; an�1 2 !, we want to �nd a 2 ! su h that �(a; i) = ai for alli < n. Let = maxi<n fJ(ai; i) + 1g;and hoose z 2 !, nonzero, su h that for all j < nonzero, jjz.By Lemma 2, for all j; l su h that 1 � j < l � , (1 + jz; 1 + lz) are relativelyprime, sin e 0 < l � j < implies that (l � j)jz. By Remark 1, there exists y 2 !su h that for all j < ,1 + (j + 1)z j y i� j = J(ai; i) for some i < n: (�)Let a = J(y; z).We note the following, for ea h ai:(i) ai < y < a and z < a.In parti ular, y; z < a by the de�nition of J , and that ai < y by (�).(ii) Div(1 + (J(ai; i) + 1) � z; y).From (�).(iii) For all x < ai, 1 + (J(x; i) + 1)z=jy.Sin e J is one-to-one, x < ai implies J(x; i) 6= J(ai; i), and for j 6= i,J(x; i) 6= J(aj ; j). Thus, by (�), x does not satisfy the required predi atefor y and z as hosen above.Sin e for any other y0 and z0, a = J(y; z) 6= J(y0; z0), we have that ai is in fa tthe minimal integer satisfying the predi ate de�ning �, and thus �(a; i) = ai, asdesired.The �-fun tion will be the basis for various systems of oding. Our �rst use willbe in en oding sequen es of numbers:De�nition. The sequen e number of a sequen e of natural numbers a1; : : : ; an,is given by<a1; : : : an>= �x(�(x; 0) = n ^ �(x; 1) = a1 ^ � � � ^ �(x; n) = an):

Page 6: Gödel.Kurt..(by Hao Kim) - Complete Proofs of Godel Incompleteness Theorems

6 LECTURES BY B. KIMNote that the map <> is de�ned on all sequen es due to the properties of �proved above. Further, sin e � is re ursive, <> is re ursive, and <> is one-to-one,sin e <a1; : : : ; an>=<b1; : : : ; bm>implies that n = m and ai = bi for ea h i. Note, too, that the sequen e number ofthe empty sequen e is <>= �x(�(x; 0) = 0) = 0:An important feature of our oding is that we an re over a given sequen e fromits sequen e number:De�nition. For ea h i 2 !, we have a fun tion ()i : ! ! !, given by(a)i = �(a; i):Clearly ()i is re ursive for ea h i. ()0 will be alled the length and denoted lh.As intended, it follows from these de�nitions that ( < a1 : : : an >)i = ai andlh(<a1 : : : an>) = n.Note also that whenever a > 0; we have lh(a) < a and (a)i < a.De�nition. The relation Seq � ! is given bySeq(a) i� 8x < a(lh(x) 6= lh(a) _ 9i < lh(a)((x)i+1 6= (a)i+1):That Seq is re ursive is evident from properties enumerated above. From ourde�nition, it is lear that Seq(a) if and only if a is the sequen e number for somesequen e (in parti ular, a =<(a)1; : : : ; (a)lh(a)>). Note that:Seq(a) i� 9x < a(lh(x) = lh(a) ^ 8i < lh(a)((x)i+1 = (a)i+1):De�nition. The initial sequen e fun tion Init : !2 ! ! is given byInit(a; i) = �x(lh(x) = i ^ 8j < i((x)j+1 = (a)j+1):Again, Init is evidently re ursive. Note that for 1 � i � n,Init(<a1; : : : ; an>) =<a1; : : : ; ai>;as intended.De�nition. The on atenation fun tion � : !2 ! ! is given bya � b = �x(lh(x) = lh(a) + lh(b)^ 8i < lh(a)((x)i+1 = (a)i+1) ^ 8j < lh(b)((x)lh(a)+j+1 = (b)j+1):Note that � is re ursive, and that<a1 : : : an> � <b1 : : : bm>=<a1 : : : an; b1 : : : bm>;as desired.De�nition. For F : ! � !k ! !, we de�ne F : ! � !k ! ! byF (a; b) =<F (0; b); : : : ; F (a� 1; b)>;or, equivalently, �x(lh(x) = a ^ 8i < a((x)i+1 = F (i; b))):Note that F (a; b) = (F (a + 1; b))a+1, thus we have that F is re ursive if andonly if F is re ursive. Be ause F (a; �) is de�ned in terms of values F (x; �), for xstri tly smaller than a, this onstru tion will enable us to de�ne F indu tively.

Page 7: Gödel.Kurt..(by Hao Kim) - Complete Proofs of Godel Incompleteness Theorems

COMPLETE PROOFS OF G�ODEL'S INCOMPLETENESS THEOREMS 7Properties of Re ursive Fun tions and Relations ( ontinued):P11. For G : !�!�!n ! ! a re ursive fun tion, the fun tion F : !�!n ! !,given by F (a; b) = G(F (a; b); a; b);is re ursive.Proof. Note that F (a; b) = G(H(a; b); a; b)whereH(a; b) = �x(Seq(x) ^ lh(x) = a ^ 8i < a((x)i+1 = G(Init(x; i); i; a)):A ording to this de�nition, F (0; b) = G(<>; 0; b) = G(0; 0; b),F (1; b) = G(<G(0; 0; b)>; 1; b);and F (2; b) = G(<G(0; 0; b); G(<G(0; 0; b)>; 1; b)>; 2; b);showing that omputation is umbersome, but possible, for any parti ular value a.P12. For G : ! � !n ! ! and H : ! � !n ! !, F : ! � !n ! !, de�ned byF (a; b) = (F (G(a; b); b) if G(a; b) < a, andH(a; b) otherwise,is re ursive.Proof. Note that when G(a; b) < a, we haveF (G(a; b); b) = (F (a; b))G(a;b)+1;whi h is re ursive by P11.For most purposes, when we de�ne a fun tion F indu tively by ases, we mustsatisfy two requirements to guarantee that our fun tion is well-de�ned. First, ifF (x; b) appears in a de�ning ase involving a, we must show that x < a wheneverthis ase is true. Se ond, we must show that our base ase is not de�ned in termsof F . In parti ular, this means that we annot use F in a de�ning ase whi h isused to ompute F (0; �).P13. Given re ursive G : !n ! ! and H : !2 � !n ! !, F : ! � !n ! !, givenby F (a; b) = (H(F (a� 1; b); a� 1; b) if a > 0, andG(b) otherwise,is re ursive.Proof. Note that F has the form of P12.

Page 8: Gödel.Kurt..(by Hao Kim) - Complete Proofs of Godel Incompleteness Theorems

8 LECTURES BY B. KIMP14. Given re ursive relations Q � !n+1 and R � !n+1 and re ursive H :! � !n ! ! su h that H(a; b) < a whenever Q(a; b) holds, the relationP � !n+1, given byP (a; b) i� (P (H(a; b); b) if Q(a; b),R(a; b) otherwise,is re ursive.Proof. De�ne H 0 : ! � !n ! ! byH 0(a; b) = (H(a; b) if Q(a; b), anda otherwise.H 0 is learly re ursive. Note�P (a; b) = (�P (H 0(a; b); b) if H 0(a; b) < a, and�R(a; b) otherwise.The following example will prove useful:De�nition. Let A � !2 be given byA(a; ) i� Seq( ) ^ lh( ) = a ^ 8i < a(( )i+1 = 0 _ ( )i+1 = 1);and let F : !2 ! ! be given byF (a; i) = 8><>:�x(A(a; x)) if i = 0,�x(F (a; i� 1) < x ^ A(a; x) if 0 < i < 2a, and0 otherwise.Then the fun tion bd : ! ! ! is given bybd(n) = F (n; 2n � 1):Evidently, A, F , and bd are all re ursive. In fa t,bd(n) = maxf< 1 2::: n > j 1 = 0 or 1g:Step 1: Representability of Re ursive Fun tions in QWe de�ne Q, a subtheory of the natural numbers, and prove the RepresentabilityTheorem, stating that all re ursive fun tions are representable in this subtheory.Consider the language of natural numbers LN = fN;+; �; S;<; 0g. We spe ifythe theory Q with the following axioms.Q1. 8x Sx 6= 0.Q2. 8x8y Sx = Sy ! x = y.Q3. 8x x+ 0 = x.Q4. 8x8y x+ Sy = S(x+ y).Q5. 8x x � 0 = 0.Q6. 8x8y x � Sy = x � y + x.Q7. 8x :(x < 0).Q8. 8x8y x < Sy ! x < y _ x = y.Q9. 8x8y x < y _ x = y _ y < x.

Page 9: Gödel.Kurt..(by Hao Kim) - Complete Proofs of Godel Incompleteness Theorems

COMPLETE PROOFS OF G�ODEL'S INCOMPLETENESS THEOREMS 9Note that the natural numbers, N, are a model of the theory Q. If we add tothis theory the set of all generalizations of formulas of the form('x0 ^ 8x('! 'xSx))! ';providing the apability for indu tion, we all this theory Peano Arithmeti , or PA.Thus Q � PA, and PA ` Q.Notation. We de�ne, for a natural number n,n � SS : : : S| {z }n 0:De�nition. A fun tion f : !n ! ! is representable in Q if there exists anLN-formula '(x1; : : : ; xn; y) su h thatQ ` 8y('(k1; : : : ; kn; y) ! y = f(k1; : : : ; kn))for all k1; : : : ; kn 2 !. We say ' represents f in Q.De�nition. A relation P � !n is representable inQ if there exists an LN-formula'(x1; : : : ; xn) su h that for all k1; : : : ; kn 2 !,P (k1; : : : ; kn)! Q ` '(k1; : : : ; kn)and :P (k1; : : : ; kn)! Q ` :'(k1; : : : ; kn):Again, we say that ' represents P in Q.To prove the Representability Theorem, we will require the following:Lemma 1. If m = n, then Q ` m = n, and if m 6= n, then Q ` :(m = n).Proof. It is enough to demonstrate this for m > n. For n = 0, our result followsfrom axiom Q1. Assume, then, that the result holds for k = n and all l > k. Thenwe have that, for a given m > n + 1, Q ` m� 1 6= n. By axiom Q2 we have,Q ` m� 1 6= n ! m 6= n+ 1. Hen e we on lude that Q ` m 6= n+ 1, and theresult holds for k = n+ 1, as required.Lemma 2. Q ` m+ n = m+ n:Proof. For n = 0, our result follows from axiom Q3. Assume, then, that the resultholds for k = n. We must show it holds for k = n + 1 as well. But Q ` m + n =m+ n, and we obtain Q ` m+ n+ 1 = m+ n+ 1 by Q4.Lemma 3. Q ` m � n = m � nProof. For n = 0, our result follows from axiom Q5. Assume, then, that theresult holds for k = n. Then Q ` m � n = mn. Applying Q6, we have thatQ ` m � n+ 1 = mn+m, and applying the previous lemma, we have the result fork = n+ 1, as required.Lemma 4. If m < n, then Q ` m < n. Further, if m � n, we have Q ` :(m < n).Proof. For n = 0, the result follows from Q7. Assume, then, that the results holdfor k = n. We show both laims hold for k = n+ 1 as well.First, suppose m < n + 1. Either m < n, and Q ` m < n by the indu tionhypothesis, or m = n, and Q ` m = n by Lemma 1. In either ase, by Q8, we havethat Q ` m < n+ 1.

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10 LECTURES BY B. KIMSe ond, suppose m � n + 1. Then m > n and by the indu tion hypothesis,Q ` :(m < n). By Lemma 1, we also have Q ` :(m = n). Applying Q8 and RuleT, we have Q ` m > n. Again applying Rule T, we have that Q ` :(m < n+ 1);as desired.Lemma 5. For any relation P � !n, P is representable in Q if and only if �P isrepresentable.Proof. Assume P is representable and that '(x1 : : : xn) represents P . Let (x; y) � ('(x) ^ y = 0) _ (:'(x) ^ y = 1):Then (x; y) represents �P .In parti ular, suppose P (k1; : : : ; kn) holds. Then Q ` '(k1; : : : ; kn). But by ourde�nition of and Rule T,Q ` '(k1; : : : ; kn)! (y = 0 ! (k1; : : : ; kn; y));and thus Q ` y = 0 ! (k1; : : : ; kn; y), as required. Similarly, if :P (k1; : : : ; kn)holds, then Q ` :'(k1; : : : ; kn), and sin eQ ` :'(k1; : : : ; kn)! (y = 1 ! : (k1; : : : ; kn; y);we have thatQ ` y = 1 ! : (k1; : : : ; kn; y), as required. Thus, (x; y) represents�P .Assume now that (x; y) represents �P . Then (x; 0) represents P .In parti ular, when P (k1; : : : ; kn) holds, we haveQ ` (k1; : : : ; kn; y) ! y = 0:Substitution of y by 0 yields Q ` (k1; : : : ; kn; 0); as desired. Similarly, when:P (k1; : : : ; kn) holds, we haveQ ` (k1 : : : kn; y) ! y = 1;and be ause Q ` :(0 = 1) we may on lude Q ` : (k1 : : : kn; 0), as needed. Thusis P representable.Lemma 6. For a formula ' in LN,Q ` 'x0 ! � � � ! ('xk�1 ! (x < k ! '))Proof. The proof is by indu tion on k. When k is 0, we haveQ ` (x < 0! '):This is (va uously) true by axiom Q7. Now, assume thatQ ` 'x0 ! : : :! ('xk�1 ! (x < k ! ')):We must show that Q ` 'x0 ! � � � ! ('xk ! (x < k + 1! ')):Equivalently, we want to show that � ` ' where � = Q [ f'x0 ; :::; 'xk ; x < k + 1g.By Q8, � ` x < k _ x = k. In the �rst ase, the indu tive hypothesis implies that� ` ', while in the latter ase, j= x = k ! ('xk ! '), and hen e � ` '. By eitherroute, � proves '.

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COMPLETE PROOFS OF G�ODEL'S INCOMPLETENESS THEOREMS 11Lemma 7. If (a) Q ` :'xk for k < n, and (b) Q ` 'xn, then for z 6= x not appearingin ', Q ` (' ^ 8z(z < x! :'xz )) ! x = n:Proof. We de�ne � (' ^ 8z(z < x! :'xz )):Now, we obtain j= x = n! ( ! ('xn ^ 8z(z < n! :'xz ))): (�)By (a) and Lemma 6, we get Q ` x < n! :'; (��)and, applying substitution and generalization, we obtainQ ` 8z(z < n! :'xz ):Combining this with (b) and (�), we on ludeQ ` x = n! :For the reverse impli ation, we note thatj= 8z(z < x! :'xz )! (n < x! :'xn);and thus (b) implies Q ` ! :(n < x). Now Q[f ; x < ng ` ' ^ :' by (��) andthe de�nition of . Therefore Q ` ! :(x < n) and by Axiom Q9 we on ludeQ ` ! x = n.Representability Theorem. Every re ursive fun tion or relation is representablein Q.Proof. It suÆ es to prove representability of fun tions having the forms enumeratedin the de�nition of re ursiveness:R1. Ini , +, �, and �<.The latter three are representable by Lemmas 2, 3, and 4. In parti ular,for +, say, we have that '(x1; x2; y) � y = x1+x2 represents + in Q, sin efor any m;n 2 !,Q ` m+ n = m+ n;Q ` y = m+ n ! y = m+ n;Q ` '(m;n; y) ! y = m+ n; and hen eQ ` 8y('(m;n; y) ! y = m+ n);as required. � and �< are similar (with �< making additional use of Lemma5).Ini is representable by '(x1; : : : ; xn; y) � xi = y. In parti ular, for anyk1; : : : ; kn 2 !, Ini (k1; : : : ; kn) = ki, and hen eQ ` '(k1; : : : ; kn; y) ! y = ki ! y = Ini (k1; : : : ; kn);by our hoi e of '. Generalization ompletes the result.

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12 LECTURES BY B. KIMR2. F (a) = G(H1(a); : : : ; Hk(a)), whereG and ea h of the Hi are representable.Assume that G is represented in Q by ' and the Hi are represented inQ by i, respe tively. We show that F is represented by�(x; y) � 9z1; : : : ; zk( 1(x; z1) ^ � � � ^ k(x; zk) ^ '(z1; : : : ; zk; y)):In other word we want to show, for any a1; :::; an 2 !,Q ` �(a1; : : : ; an; y) ! y = G(H1(a); : : : ; Hk(a)) (y)where a = (a1:::an).Now, for � = Q [ f�(a1; : : : ; an; y)g, sin e the i represent Hi, we havethat � ` 9z1; : : : ; zk(z1 = H1(a) ^ � � � ^ zk = Hk(a) ^ '(z1; : : : ; zk; y)):Hen e we have� j= 9z1; : : : ; zk('(H1(a); : : : ; Hk(a); y));and sin e the zi do not appear,� j= '(H1(a); : : : ; Hk(a); y):Sin e ' represents G, we have� j= y = G(H1(a); : : : ; Hk(a));as required.On the other hand, for � = Q [ fy = G(H1(a); : : : ; Hk(a))g,� ` '(H1(a); : : : ; Hk(a); y)� ` 9z1; : : : ; zk(z1 = H1(a) ^ � � � zk = Hk(a) ^ '(z1; : : : ; zk; y))� ` 9z1; : : : ; zk( 1(a; zi) ^ � � � k(a; zk) ^ '(z1; : : : ; zk; y))� ` �(a1; : : : ; an; y)Thus (y) is established.R3. F (a) = �x(G(a; x) = 0), where G is representable in Q and for all a thereexists x su h that G(a; x) = 0, is representable in Q.Assume G is represented in Q by '(x1; : : : ; xn; x; y). Let (x1; : : : ; xn; x) � 'y0 ^ 8z(z < x! :'yx0z ):Let F (a) = b and ki = G(a; i) for i 2 !. ThenQ ` '(a1; : : : ; an; i; y) ! y = ki;thus Q ` '(a1; : : : ; an; i; 0) ! 0 = ki;and if j < b, so that kj 6= 0, thenQ ` :'(a1; : : : ; an; j; 0):On the other hand, kb = 0, soQ ` '(a1; : : : ; an; b; 0):Hen e, by Lemma 7,Q ` ('(a; x; y)y0 ^ 8z(z < x! :'(a; x; y)y0xz )) ! x = b;and thus, Q ` (a; x) ! x = b:By generalization, we have that represents F in Q, as desired.

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COMPLETE PROOFS OF G�ODEL'S INCOMPLETENESS THEOREMS 13Step 2: Axiomatizable Complete Theories are De idableWe begin by showing that we may en ode terms and formulas of a reasonablelanguage in su h a way that important lasses of formulas, e.g., the logi al axioms,are mapped to re ursive subsets of the natural numbers. We use this to derive themain result.De�nition. Let L be a ountable language with subsets C, F, and P of onstant,fun tion, and predi ate symbols, respe tively (=2 P). Let V be a set of variablesfor L. L is alled reasonable if the following two fun tions exist:� h : L[f:;!;8g[V! ! inje tive su h that V = h(V), C = h(C), F = h(F),and P = h(P) are all re ursive.� AR : ! ! ! r f0g re ursive su h that AR(h(f)) = n and AR(h(P )) = nfor n-ary fun tion and predi ate symbols f and P .For the rest of this note, the language L is ountable and reasonable.Now we de�ne a oding de : fL-terms and L-formulasg ! ! indu tively, by� For x 2 V [ C, dxe =<h(x)>.� For u1; : : : ; un 2 V [ C and f 2 F,dfu1u2 : : : une =<h(f); du1e; du2e; : : : ; dune> :� For L-terms t1; : : : ; tn and P 2 P,dPt1t2 : : : tne =<h(P ); dt1e; : : : ; dtne> :� For L-formulas ' and ,d'! e =<h(!); d'e; d e>;d:'e =<h(:); d'e>;d8x'e =<h(8); dxe; d'e> :Note that our de�nition of de is one-to-one. Given a term or formula �, we alld�e the G�odel number of �.We show the following predi ates and fun tions are re ursive (We follow de�ni-tions for syntax in [E℄.):(1) Vble = fdve j v 2 Vg � ! and Const = fd e j 2 Cg � !.Proof. Note Vble(x) i� x =<(x)1> ^V((x)1);Const(x) i� x =<(x)1> ^C((x)1):(2) Term = fdte j t an L-termg � !.Proof. NoteTerm(a) i� 8><>:8j<(lh(a) _�1)Term((a)j+2) if Seq(a) ^ F((a)1)^ AR((a)1) = lh(a) _�1;Vble(a) _ Const(a) otherwise.

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14 LECTURES BY B. KIM(3) AtF = fd�e j � an atomi L-formulag � !.Proof. NoteAtF(a) i� Seq(a) ^ P((a)1) ^ (AR((a)1) = lh(a)� 1)^ 8j<(lh(a)� 1) (Term((a)j+2)):(4) Form = fd'e j ' an L-formulag � !.Proof. NoteForm(a) i� 8>>><>>>:Form((a)2) if a =<h(:); (a)2>,Form((a)2) ^ Form((a)3) if a =<h(!); (a)2; (a)3>,Vble((a)2) ^ Form((a)3) if a =<h(8); (a)2; (a)3>,AtF (a) otherwise.(5) Sub : !3 ! !, su h that Sub(dte; dxe; due) = dtxue and Sub(d'e; dxe; due) =d'xue for terms t and u, variable x, and formula '.Proof. De�neSub(a; b; ) = 8>>>>>>>><>>>>>>>>: if Vble(a) ^ a = b,<(a)1;Sub((a)2; b; ); : : : if lh(a) > 1 ^ (a)1 6= h(8): : : ;Sub((a)lh(a); b; )> ^Seq(a);<(a)1; (a)2;Sub((a)3; b; )> if a =<h(8); (a)2; (a)3>,^ (a)2 6= ba otherwise.Note that, if well-de�ned, the fun tion has the properties desired above.We show Sub is well-de�ned by indu tion on a: a = 0 falls into the�rst or last ategory sin e lh(0) = 0, hen e Sub(0; b; ) is well-de�ned forall b; 2 !. If a 6= 0, then (a)i < a for all i � lh(a), and thus we mayassume the values Sub((a)i; b; ) are well-de�ned, showing Sub(a; b; ) to bewell-de�ned in all ases.(6) Free � !2, su h that for formula ', term � , and variable x, Free(d'e; dxe)if and only if x o urs free in ', and Free(d�e; dxe) if and only if x o ursin �Proof. De�neFree(a; b) i� 8><>:9j<(lh(a)� 1) (Free((a)j+2; b)) if lh(a) > 1 ^ (a)1 6= h(8),Free((a)3; b) ^ (a)2 6= b if lh(a) > 1 ^ (a)1 = h(8),a = b otherwise.Free learly has the desired property, and that it is well-de�ned follows byessentially the same indu tion on a as above.(7) Sent = fd'e j ' is an L-senten eg � !.

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COMPLETE PROOFS OF G�ODEL'S INCOMPLETENESS THEOREMS 15Proof. NoteSent(a) i� Form(a) ^ 8b<a (:Vble(b) _ :Free(a; b)):(8) Subst(a; b; ) � !3 su h that for a given formula ', variable x, and term t,Subst(d'e; dxe; dte) if and only if t is substitutable for x in '.Proof. De�neSubst(a; b; ) i� 8>>>>>><>>>>>>:Subst((a)2; b; ) if a =<h(:); (a)2>,Subst((a)2; b; ) ^ Subst((a)3; b; ) if a =<h(!); (a)2; (a)3>,:Free(a; b) _ (:Free( ; (a)2) if a =<h(8); (a)2; (a)3>,^Subst((a)3; b; ))0 = 0 otherwise.Note that Subst has the desired property, and is well-de�ned by essentiallythe same indu tion used above.(9) We de�neFalse(a; b) i� 8>>><>>>::False((a)2; b) ^ False((a)3; b) if a =<h(!); (a)2; (a)3>^Form((a)2) ^ Form((a)3);:False((a)2; b) if a =<h(:); (a)2> ^Form((a)2),Form(a) ^ (b)a = 0 otherwise.False is re ursive by the same indu tion as applied above. We note thesigni� an e of False presently.To ea h b 2 !, we may asso iate a truth assignment vb su h that for a primeformula (atomi or of the form 8x'),vb( ) = F i� (b)d e = 0:Further, for any truth assignment v : A ! fT;Fg, where A is a �nite set of primeformulas, there exists a b su h that v = vb: we may write A = f'1; : : : ; 'ng su hthat d'1e < d'2e < � � � < d'ne. For 1 � j � d'ne de�ne j = 0 when j = d'iefor some i � n and v('i) = F , and j = 1 otherwise. Then b = < 1; : : : ; d'ne >satis�es vb = v on A.Then moreover, for any formula ' built up from A,v(') = F i� vb(') = F i� False(d'e; b):(10) Taut � ! su h that for a formula �, Taut(d�e) if and only if � is a tautology.Proof. Re all bd : ! ! ! su h that bd(a) = maxf < 1; : : : ; a > j i 2f0; 1gg, re ursive, has been previously de�ned. De�neTaut(a) i� Form(a) ^ 8b<(bd(a) + 1) (:False(a; b)):(11) AG2 = fd'e j ' is in axiom group 2g � !.

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16 LECTURES BY B. KIMProof. Re all axiom group 2 ontains formulas of the form 8x ! xt , withterm t substitutable for x in . ThusAG2(a) i� 9x; y; z<a (Vble(x) ^ Form(y) ^ Term(z) ^ Sub(y; x; z)^ a =<h(!); <h(8); x; y>;Subst(y; x; z)>);where 9x; y; z<aP (x; y; z) abbreviates what one would expe t.(12) AG3 = fd'e j ' is in axiom group 3g � !.Proof. Re all we take axiom group 3 to be the formulas having the followingform: 8x( ! 0)! (8x ! 8x 0). ThusAG3(a) i� 9x; y; z<a (Vble(x) ^ Form(y) ^ Form(z)^ a =<h(!); <h(8); x; <h(!); y; z>>;<h(!); <h(8); x; y>; <h(8); x; z>>>)(13) AG4 = fd'e j ' is in axiom group 4g � !.Proof. Re all axiom group 4 ontains formulas of the form ! 8x , wherex does not o ur free in . ThusAG4(a) i� 9x; y<a (Vble(x) ^ Form(y)^ :Free(y; x) ^ a =<h(!); y; <h(8); x; y>>)(14) AG5 = fd'e j ' is in axiom group 5g � !.Proof. Re all axiom group 5 ontains formulas of the form x = x, for avariable x, hen eAG5(a) i� 9x<a (Vble(x) ^ a =<h(=); x; x>):(15) AG6 = fd'e j ' is in axiom group 6g � !.Proof. Re all formulas of axiom group 6 have the form x = y ! ( ! 0),where is an atomi formula and 0 is obtained by from by repla ingone or more o urren es of x with y. ThusAG6(a) i� 9x; y; b; <a (Vble(x) ^ Vble(y) ^ AtF(b) ^ AtF( )^ lh(b) = lh( ) ^ 8j < lh(b) + 1(( )j = (b)j _ (( )j = y ^ (b)j = x))^ a =<h(!); <h(=); x; y>; <h(!); b; >>)(16) Gen(a; b) � !2, su h that Gen(d'e; d e) if and only if ' is a generalizationof (i.e., ' = 8x1 : : :8xn for some �nite fxig � V).

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COMPLETE PROOFS OF G�ODEL'S INCOMPLETENESS THEOREMS 17Proof. Note thatGen(a; b) i� 8><>:a =<h(8); (a)2; (a)3> ^Vble((a)2) ^ Gen((a)3; b) if a > b,0 = 0 if a = b,0 = 1 if a < b.(17) � = fd�e j � 2 �g � !, where � is the set of logi al axioms.Proof. Note that�(a) i� 9b<a+ 1 (Form(a) ^ Gen(a; b)^ (Taut(b) _ AG2(b) _ AG3(b) _ AG4(b) _ AG5(b) _ AG6(b)))We have, to this point, de�ned three odings: <> on sequen es of natural num-bers, h on the language and logi al symbols, and de on the terms and formulas. Wepresently de�ne a fourth oding, of sequen es of formulas:ddee : fsequen es of L-formulasg ! !;given by dd'1; : : : ; 'nee =<d'1e; : : : ; d'ne> :This map is one-to-one, as it is derived from the established (inje tive) odings,and in parti ular, we an determine, for a given number, if it lies in the image ofddee, and, if so, re over the asso iated sequen e of formulas.De�nition. Given L, let T be a theory (a olle tion of senten es) in L. De�neT = fd�e j � 2 Tg:We say that T is axiomatizable if there exists a theory S, axiomatizing T (thatis, su h that CnS = CnT ), su h that S is re ursive. We say that T is de idableif CnT is re ursive.We shall make use of the following relations:� DedT = fdd'1; : : : ; 'nee j '1; : : : ; 'n is a dedu tion from Tg � !.Note thatDedT (a) i� Seq(a) ^ lh(a) 6= 0^8j< lh(a) (�((a)j+1)_T ((a)j+1)_9i; k<j+1 ((a)k+1 =<h(!); (a)i+1; (a)j+1>))� PrfT � !2, given by PrfT (a; b) i� DedT (b) ^ a = (b)lh(b).� PfT � !, given by PfT (a) i� Sent(a) ^ 9xPrfT (a; x).Note that we may read PrfT (a; b) as \b is a proof of a from T ," and PfT (a) as\a is a senten e provable from T ." In parti ularPfT = CnT = fd�e j T ` �g:We use this fa t to prove the following:Theorem. If T is axiomatizable, then PfT = CnT is re ursively enumerable.Proof. Let S axiomatize T , where S is re ursive. From the above de�nitions, wesee that DedS and PrfS are re ursive relations, hen e PfS is an r.e. relation. ButPfS = PfT , sin e CnS = CnT .

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18 LECTURES BY B. KIMTheorem. If T is axiomatizable and omplete in L, then T is de idable.Proof. By the negation theorem, it suÆ es to show that :PfT is re ursively enu-merable. Note that sin e T is omplete, for any senten e �, T 0 � if and only ifT ` :�. Hen e :PfT (a) i� :Sent(a) _ 9mPrfT (<h(:); a>;m)i� 9m(:Sent(a) _ PrfT (<h(:); a>;m)):Thus :PfT is re ursively enumerable, and PfT is re ursive.We an see that if we say T is axiomatizable in wider sense when S axiomatiz-ing T is re ursively enumerable, then the above two theorems still hold with thisseemingly weaker notion. In fa t, two notions are equivalent, whi h is known asCraig's Theorem.Step 3: The In ompleteness Theorems and Other ResultsWe return now to the language of natural numbers, LN. Re all that we de�ne,for a natural number n, n � SS : : : S| {z }n 0:De�nition. The diagonalization of an LN formula ' is a new formulad(') � 9v0(v0 = d'e ^ ');where 9 and ^ provide the usual abbreviations in LN.In parti ular, we note d(') is satis�able pre isely when ' is satis�able by sometruth assignment taking v0 to the G�odel number of ', and LN j= d(') pre iselywhen ' is satis�ed by every truth assignment taking v0 to d'e.Lemma. There exists a re ursive fun tion dg : ! ! ! su h that for any LNformula, dg(d'e) = dd(')e.Proof. De�ne num : ! ! ! by num(0) =<0> and, for n 2 !num(n+ 1) =<h(S);num(n)> :In parti ular, note that num(n) = dne.De�nedg(a) =<h(:); <h(8); dv0e; <h(:);<h(:); <h(!); <h(=); dv0e; num(a)>; <h(:); a>>>>>>Thendg(d'e) =<h(:); <h(8); dv0e; <h(:);<h(:); <h(!); <h(=); dv0e;num(d'e)>; <h(:); d'e>>>>>>;=<h(:); <h(8); dv0e; <h(:);<h(:); <h(!); <h(=); dv0e; dd'ee>; <h(:); d'e>>>>>> :

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COMPLETE PROOFS OF G�ODEL'S INCOMPLETENESS THEOREMS 19However, writing out what formula this en odes and introdu ing our usual abbre-viations, we have dg(d'e) = d:8v0:(:(v0 = d'e ! :'))e= d9v0(v0 = d'e ^ ')e= dd(')e;as desired.Fixed Point Theorem (G�odel). For any LN-formula '(x) (i.e., either a senten eor a formula having x as the only free variable), there is some LN-senten e � su hthat Q ` � ! '(d�e):Proof. Sin e dg is re ursive, it is representable in Q by Step 1, say by (x; y). ThenQ ` 8y( (n; y) ! y = dg(n)):Let Æ(v0) � 9y( (v0; y) ^ '(y)), and let n = dÆ(v0)e. De�ne� � d(Æ(v0)) � 9v0(v0 = n ^ Æ(v0)):Then if we let k = dg(n) = d�e, we havej= � ! Æ(n) ! 9y( (n; y) ^ '(y)):But Q ` (n; y) ! y = k;and therefore Q ` � ! 9y(y = k ^ '(y)) ! '(k) ! '(d�e);as required.Tarski Unde�nability Theorem. ThN = fd�e j N j= �g is not de�nable.Proof. Suppose ThN were de�nable by �(x). Then by the �xed point lemma, with' = :�, there exists a senten e � su h thatN j= � ! :�(d�e):Then N j= � implies that N 6j= �(d�e), implying N 6j= �, or N j= :�, sin e ThNis omplete. On the other hand, N 6j= � implies N j= :�, and thus that N j=�(d�e), implying N j= �. The ontradi tions together imply that � annot representThN.Strong Unde idability of Q. Let T be a theory in L � LN. If T [Q is onsistentin L, then T is not de idable in L (CnT is not re ursive).Proof. Assume that CnT is re ursive. We �rst show that this implies re ursivenessof CnT [Q. Sin e Q is �nite, it suÆ es to show that for any senten e � in thelanguage, CnT [ f�g is re ursive.In parti ular, note that if � 2 CnT [ f�g, then � ! � 2 CnT . Thusa 2 CnT [ f�g i� Sent(a)^ <h(!); d�e; a>2 CnT :Hen e CnT [ f�g is re ursive, as desired.

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20 LECTURES BY B. KIMTo prove the theorem, then, it suÆ es to show that CnT [Q is not re ursive. Ifthis were the ase, then it would be representable, say by �(x), in Q. By the �xedpoint lemma, there exists an LN senten e � su h thatQ ` � ! :�(d�e):If T [Q ` �, then Q ` �(d�e);by the representability of CnT [Q by �(x) in Q. In parti ular,Q ` :�;a ontradi tion. On the other hand, if T [Q 0 �, then by representability,Q ` :�(d�e);and hen e Q ` �;a ontradi tion, implying that CnT [Q is not representable, and hen e not re ur-sive.Corollary. ThN, PA, and Q are all unde idable.Proof. We need note only that ea h of these theories is onsistent with Q.Moreover, we have:Unde idability of First Order Logi (Chur h). For a reasonable ountablelanguage L � LN, the set of all G�odel numbers of valid senten es (fd�e j ; ` �g)is not re ursive (the set of valid senten es is not de idable).In fa t, the above orollary is true for any ountable L ontaining a k-ary pred-i ate or fun tion symbol, k � 2, or at least two unary fun tion symbols.G�odel-Rosser First In ompleteness Theorem. If T is a theory in a ountablereasonable L � LN, with T [ Q onsistent and T axiomatizable, then T is not omplete.Proof. By Step 2, if T is omplete, then T is de idable, ontradi ting the strongunde idability of Q.Remarks. In (N;+), 0, <, and S are de�nable. Hen e the same result follows if wetake L0N = f+; �g instead of our usual LN. In parti ular, Th(N;+; �) is unde idable,and for any T 0 � Q0 (where Q0 is simply Q written in the language of L0N), we havethat T 0 is, if onsistent, unde idable, and, if axiomatizable, in omplete.It is important to note that for an unde idable theory T , we may have T � T 0,where T 0 is a de idable theory. As an example, the theory of groups is unde idable,whereas the theory of divisible torsion-free groups is de idable.We turn our attention now to the proof of the result used in G�odel's originalpaper. In parti ular, G�odel worked in the model (N;+; �; 0; <;E). (Note that E,exponentiation, is de�nable in (N;+; �; 0; <), or, equivalently, (N;+; �)).Let T � Q be a onsistent theory in a reasonable ountable language L � LN,and presume that T is re ursive. ThenT ` � ) Q ` PfT (d�e):

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COMPLETE PROOFS OF G�ODEL'S INCOMPLETENESS THEOREMS 21In parti ular, T ` � implies that PrfT (d�e;m) for some m 2 !. Sin e PrfT isre ursive, it is representable in Q, hen e Q ` PrfT (d�e;m), andQ ` 9xPrfT (d�e; x);or Q ` PfT (d�e):By the �xed point lemma, there exists a senten e � su h thatT � Q ` � ! :PfT (d�e): (�)If T ` �, then Q ` PfT (d�e), and thus Q ` :�, and hen e T ` :�, a ontradi tion.Thus T 0 �.On the other hand, if T is !- onsistent (i.e., whenever T ` 9x'(x), then forsome n 2 !, T 0 :'(n)), then T 0 :�. In parti ular, if T ` :�, thenT ` PfT (d�e);by (�). That is, T ` 9xPrfT (d�e; x):However, if PrfT (d�e;m) for some m 2 !, then T ` �, ontradi ting the onsis-ten y of T . Thus we must have :PrfT (d�e;m) for all m 2 !. Sin e Q representsPrfT , T � Q ` :PrfT (d�e;m)for all m 2 !, ontradi ting the !- onsisten y of T .Rosser generalized G�odel's proof by singling out for T a senten e � su h thatT 0 � and T 0 :�, without the assumption of !- onsisten y.We now begin our approa h to G�odel's Se ond In ompleteness Theorem. We �xT , a theory in a ountable reasonable language L � LN.We note the following fa t from Hilbert and Bernays' Grundlagen der Mathe-matik, 1934.Fa t. If T is onsistent, T ` PA, and T is re ursive, then for any senten es � andÆ in L, I. T ` � ) Q ` PfT (d�e)II. PA ` (PfT (d�e) ^ PfT (d� ! Æe))! PfT (dÆe)III. PA ` PfT (d�e)! PfT �dPfT (d�e)e�Notation. We will write ConT � :PfT (d0 6= 0e). Clearly ConT holds if and onlyif T is onsistent.Lemma. If T ` � ! Æ, then PA ` PfT (d�e)! PfT (dÆe).Proof. If T ` � ! Æ, then by (I) above,PA ` PfT (d� ! Æe);and by (II), PA ` PfT (d�e)! PfT (dÆe):

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22 LECTURES BY B. KIMG�odel's Se ond In ompleteness Theorem. If T is onsistent, T is re ursive,and T ` PA, then T 0 ConT .Proof. By the �xed point lemma, there exists � su h thatQ ` � ! :PfT (d�e): (y)By (III), above, PA ` PfT (d�e)! PfT �dPfT (d�e)e� : (z)And further, by Lemma, we havePA ` PfT �dPfT (d�e)e�! PfT (d:�e):Combining this result with (z), we havePA ` PfT (d�e)! PfT (d:�e):Now note that ` :� ! (� ! (0 6= 0)). By the lemma,PA ` PfT (d�e)! PfT (d� ! (0 6= 0)e):In parti ular, PA ` PfT (d�e)! PfT (d�e) ^ PfT (d� ! (0 6= 0)e);hen e, by (II), PA ` PfT (d�e)! PfT (d0 6= 0e);i.e. PA ` PfT (d�e)! :ConT :Thus PA ` ConT ! �, by (y).Now, suppose that T ` ConT . Then T ` �, and hen e by (I), T � Q ` PfT (d�e).But again, by (y), this implies that T ` :�, a ontradi tion, showing that T annotprove its own onsisten y.We remark that one may arry the proof through using only the assumption thatT is re ursively enumerable.L�ob's Theorem. Suppose T is a onsistent theory in L � LN, su h that T re- ursive, and T ` PA. Then for any L-senten e �, if T ` PfT (d�e) ! �, thenT ` �.Proof. By the �xed point lemma, there exists Æ su h thatQ ` Æ ! (PfT (dÆe)! �):Sin e T ` PA � Q, T proves the same result. From this we may dedu e thatPA ` PfT (dÆe)! PfT (d�e):In parti ular, by our lemma, we havePA ` PfT (dÆe)! PfT �dPfT (dÆe)! �e� ;and, ombining this with (III) from above,PA ` PfT (dÆe)! PfT �dPfT (dÆe)e� ^ PfT �dPfT (dÆe)! �e� ;and thus, by (II), PA ` PfT (dÆe)! PfT (d�e);

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COMPLETE PROOFS OF G�ODEL'S INCOMPLETENESS THEOREMS 23as desired.Now assume that T ` PfT (d�e)! �. Then, by the above,T ` PfT (dÆe)! �:By our hoi e of Æ, this in turn implies that T ` Æ. By (I), we have that Q `PfT (dÆe), and hen e T proves the same result, implying that T ` �, as desired.Remark. G�odel's Se ond In ompleteness Theorem in fa t follows from L�ob's The-orem. In parti ular, given T as in the hypotheses of both theorems, if T ` ConT ,then T ` PfT (d0 6= 0e)! 0 6= 0:But by L�ob's Theorem, this in turn implies that T ` 0 6= 0, showing that su h atheory, if onsistent, annot prove its own onsisten y.Referen es[BJ℄ G. S. Boolos and R. C. Je�rey, Computability and logi .[E℄ H. Enderton, A mathemati al introdu tion to logi .[Sh℄ J. R. Shoen�eld, Mathemati al logi .[Sm℄ R. M. Smullyan, G�odel's in ompleteness theorems.