Complete proofs of Gödel's Incompleteness Theorems.

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Oct 8, 2013 (3 years and 8 months ago)

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COMPLETE PROOFS OF G

ODEL'S INCOMPLETENESS
THEOREMS
LECTURES BY B.KIM
Step 0:Preliminary Remarks
We de ne recursive and recursively enumerable functions and relations,enumer-
ate several of their properties,prove Godel's -Function Lemma,and demonstrate
its rst applications to coding techniques.
Denition.
For R !
n
a relation,
R
:!
n
!!,the characteristic function on
R,is given by

R
(
a) =
{
1 if:R(
a),
0 if R(
a).
Denition.
A function from!
m
to!(m  0) is called recursive (or com-
putable) if it is obtained by nitely many applications of the following rules:
R1.
I
n
i
:!
n
!!,1  i  n,de ned by (x
1
;:::;x
n
) 7!x
i
is recursive;

+:!!!!and :!!!!are recursive;


<
:!!!!is recursive.
R2.
(Composition) For recursive functions G;H
1
;:::;H
k
such that H
i
:!
n
!!
and G:!
k
!!,F:!
n
!!,de ned by
F(
a) = G(H
1
(
a);:::;H
k
(
a)):
is recursive.
R3.
(Minimization) For G:!
n+1
!!recursive,such that for all
a 2!
n
there
exists some x 2!such that G(
a;x) = 0,F:!
n
!!,de ned by
F(
a) = x(G(
a;x) = 0)
is recursive.(Recall that xP(x) for a relation P is the minimal x 2!such
that x 2 P obtains.)
Denition.
R(!
k
) is called recursive,or computable (R is a recursive rela-
tion) if 
R
is a recursive function.
Proofs in this note are adaptation of those in [Sh] into the deduction system described in [E].
Many thanks to Peter Ahumada and Michael Brewer who wrote up this note.
2 LECTURES BY B.KIM
Properties of Recursive Functions and Relations:
P0.
Assume :f1;:::;kg!f1;:::;ng is given.If G:!
k
!!is recursive,then
F:!
n
!!de ned by,for
a = (a
1
;:::;a
n
),
F(
a) = G(a
(1)
;:::;a
(k)
) = G(I
n
(1)
(
a);:::;I
n
(k)
(
a));
is recursive.Similarly,if P(x
1
;:::;x
k
) is recursive,then so is
R(x
1
;:::;x
n
)  P(x
(1)
;:::;x
(k)
):
P1.
For Q !
k
a recursive relation,and H
1
;:::;H
k
:!
n
!!recursive func-
tions,
P = f
a 2!
n
j Q(H
1
(
a);:::;H
k
(
a))g
is a recursive relation.
Proof.

P
(
a) = 
Q
(H
1
(
a);:::;H
k
(
a)) is a recursive function by R2.
P2.
For P !
n+1
,a recursive relation such that for all
a 2!
n
there exists
some x 2!such that P(
a;x),then F:!
n
!!,de ned by
F(
a) = xP(
a;x)
is recursive.
Proof.
F(
a) = x(
P
(
a;x) = 0),so we may apply R3.
P3.
Constant functions,C
n;k
:!
n
!!such that C
n;k
(
a) = k,are recursive.
(Hence for recursive F:!
m+n
!!or P !
m+n
,and
b 2!
n
,both the
map (x
1
;:::;x
m
) 7!F(x
1
;:::;x
m
;
b) and P(x
1
;:::;x
m
;
b) !
m
are recur-
sive.)
Proof.
By induction:
C
n;0
(
a) = x(I
n+1
n+1
(
a;x) = 0)
C
n;k+1
(
a) = x(C
n;k
(
a) < x)
are recursive by R3 and P2,respectively.
P4.
For Q;P !
n
,recursive relations,:P,P _ Q,and P ^ Q are recursive.
Proof.
We have that

:P
(
a) = 
<
(0;
P
(
a));

P _Q
(
a) = 
P
(
a)  
Q
(
a);
P ^ Q =:(:P _:Q):
P5.
The predicates =,,>,and  are recursive.(Hence each nite set is
recursive.)
Proof.
For a;b 2!,
a = b iff:(a < b) ^:(b < a);
a  b iff:(a < b);
a > b iff (a  b) ^:(a = b);and
a  b iff:(a > b);
COMPLETE PROOFS OF G

ODEL'S INCOMPLETENESS THEOREMS 3
hence these are recursive by P4.
Notation.
We write,for
a 2!
n
,f:!
n
!!a function and P !
m+1
a relation,
x<f(
a) P(x;
b)  x(P(x;
b) _ x = f(
a)):
In particular,x<f(
a) P(x;
b) is the smallest integer less than f(
a) which satis es
P,if such exists,or f(
a),otherwise.
We also write
9x<f(
a) P(x)  (x<f(
a) P(x)) < f(
a);and
8x<f(
a) P(x) :(9x<f(
a) (:P(x))):
The rst is clearly satis ed if some x < f(
a) satis es P(x),while the second is
satisifed if all x < f(
a) satisfy P(x).
P6.
For P !
n+1
a recursive relation,F:!
n+1
!!,de ned by
F(a;
b) = x<aP(x;
b);
is recursive.
Proof.
F(a;
b) = x(P(x;
b) _ x = a),and thus F is recursive by P2,since
for all
b,a satis es P(x;
b) _ x = a.
P7.
For R !
n+1
a recursive relation,P;Q !
n+1
such that
P(a;
b)  9x<aR(x;
b);Q(a;
b)  8x<aR(x;
b)
are recursive.(Hence,with P1,it follows both
Div(y;z)( yjz) = 9x < z +1(z = x  y);
and PN,the set of all prime numbers,are recursive.)
Proof.
Note that P is de ned by composition of recursive functions and
predicates,hence recursive by P1,and Q is de ned by composition of re-
cursive functions,recursive predicates,and negation,hence recursive by P1
and P4.
P8.
_
:!!!!,de ned by
a
_
b =
{
a b if a  b,
0 otherwise,
is recursive.
Proof.
Note that
a
_
b = x(b +x = a _a < b):
4 LECTURES BY B.KIM
P9.
If G
1
;:::;G
k
:!
n
!!are recursive functions,and R
1
;:::;R
k
!
n
are
recursive relations partitioning!
n
(i.e.,for each
a 2!
n
,there exists a
unique i such that R
i
(
a)),then F:!
n
!!,de ned by
F(
a) =
8
>
>
>
>
<
>
>
>
>
:
G
1
(
a) if R
1
(
a),
G
2
(
a) if R
2
(
a),
.
.
.
.
.
.
G
k
(
a) if R
k
(
a),
is recursive.
Proof.
Note that
F = G
1

:R
1
+   +G
k

:R
k
:
P10.
If Q
1
;:::;Q
k
!
n
are recursive relations,and R
1
;:::;R
k
!
n
are recur-
sive relations partitioning!
n
,then P !
n
,de ned by
P(
a) iff
8
>
>
<
>
>
:
Q
1
(
a) if R
1
(
a),
.
.
.
.
.
.
Q
k
(
a) if R
k
(
a),
is recursive.
Proof.
Note that

P
(
a) =
8
>
>
<
>
>
:

Q
1
(
a) if R
1
(
a),
.
.
.
.
.
.

Q
k
(
a) if R
k
(
a),
is recursive by P9.
Denition.
A relation P !
n
is recursively enumerable (r.e.) if there exists
some recursive relation Q !
n+1
such that
P(
a) iff 9xQ(
a;x):
Remark If a relation R !
n
is recursive,then it is recursively enumerable,since
R(
a) iff 9x(R(
a) ^ x = x).
Negation Theorem.
A relation R !
n
is recursive if and only if R and:R are
recursively enumerable.
Proof.
If R is recursive,then:R is recursive.Hence by above remark,both are r.e.
Now,let P and Q be recursive relations such that for
a 2!
n
,R(
a) iff 9xQ(
a;x)
and:R(
a) iff 9xP(
a;x).
De ne F:!
n
!!by
F(
a) = x(Q(
a;x) _P(
a;x));
recursive by P2,since either R(
a) or:R(
a) must hold.
We show that
R(
a) iff Q(
a;F(
a)):
COMPLETE PROOFS OF G

ODEL'S INCOMPLETENESS THEOREMS 5
In particular,Q(
a;F(
a)) implies there exists x (namely,F(
a)) such that Q(
a;x),
thus R(
a) holds.Further,if:Q(
a;F(
a)),then P(
a;F(
a)),since F(
a) satis es
Q(
a;x) _ P(
a;x).Thus:R(
a) holds.
The -Function Lemma.
-Function Lemma (Godel).There is a recursive function :!
2
!!such that
(a;i)  a
_
1 for all a;i 2!,and for any a
0
;a
1
;:::;a
n 1
2!,there is an a 2!
such that (a;i) = a
i
for all i < n.
Remark 1.
Let A = fa
1
;:::a
n
g !∖f0;1g (n  2) be a set such that any two
distinct elements of A are realtively prime.Then given non-empty subset B of A,
there is y 2!such that for any a 2 A,ajy iff a 2 B.(y is a product of elements in
B.)
Lemma 2.
If kjz for z ̸= 0,then (1 +(j +k)z;1 +jz) are relatively prime for any
j 2!.
Proof.
Note that for p prime,pjz implies that p=j 1 +jz.But if pj1 +(j +k)z and
pj1 +jz,then pjkz,implying pjkjz or pjz,and thus pjz,a contradiction.
Lemma 3.
J:!
2
!!,de ned by J(a;b) = (a +b)
2
+(a +1),is one-to-one.
Proof.
If a +b < a

+b

,then
J(a;b) = (a+b)
2
+a+1  (a+b)
2
+2(a+b)+1 = (a+b+1)
2
 (a

+b

)
2
< J(a

;b

):
Thus if J(a;b) = J(a

;b

),then a +b = a

+b

,and
0 = J(a

;b

) J(a;b) = a

a;
implying that a = a

and b = b

,as desired.
Proof of -Function Lemma.
De ne
(a;i) = x<a
_
1(9y<a(9z <a(a = J(y;z) ^ Div(1 +(J(x;i) +1)  z;y))));
It is clear that is recursive,and that (a;i)  a
_
1.
Given a
1
;:::;a
n 1
2!,we want to nd a 2!such that (a;i) = a
i
for all
i < n.Let
c = max
i<n
fJ(a
i
;i) +1g;
and choose z 2!,nonzero,such that for all j < c nonzero,jjz.
By Lemma 2,for all j;l such that 1  j < l  c,(1 +jz;1 +lz) are relatively
prime,since 0 < l j < c implies that (l j)jz.By Remark 1,there exists y 2!
such that for all j < c,
1 +(j +1)z j y iff j = J(a
i
;i) for some i < n:()
Let a = J(y;z).
We note the following,for each a
i
:
(i)
a
i
< y < a and z < a;
In particular,y;z < a by the de nition of J,and that a
i
< y by ().
(ii)
Div(1 +(J(a
i
;i) +1)  z;y);
From ().
6 LECTURES BY B.KIM
(iii)
For all x < a
i
,1 +(J(x;i) +1)z=j y;
Since J is one-to-one,x < a
i
implies J(x;i) ̸= J(a
i
;i),and for j ̸= i,
J(x;i) ̸= J(a
j
;j).Thus,by (),x does not satisfy the required predicate
for y and z as chosen above.
Since for any other y

and z

,a = J(y;z) ̸= J(y

;z

),we have that a
i
is in fact
the minimal integer satisfying the predicate de ning ,and thus (a;i) = a
i
,as
desired.
The -function will be the basis for various systems of coding.Our rst use will
be in encoding sequences of numbers:
Denition.
The sequence number of a sequence of natural numbers a
1
;:::;a
n
,
is given by
<a
1
;:::;a
n
>= x( (x;0) = n ^ (x;1) = a
1
^   ^ (x;n) = a
n
):
Note that the map <> is de ned on all sequences due to the properties of
proved above.Further,since is recursive,<> is recursive,and <> is one-to-one,
since
<a
1
;:::;a
n
>=<b
1
;:::;b
m
>
implies that n = m and a
i
= b
i
for each i.Note,too,that the sequence number of
the empty sequence is
<>= x( (x;0) = 0) = 0:
An important feature of our coding is that we can recover a given sequence from
its sequence number:
Denition.
For each i 2!,we have a function ()
i
:!!!,given by
(a)
i
= (a;i):
Clearly ()
i
is recursive for each i.()
0
will be called the length and denoted lh.
As intended,it follows from these de nitions that ( < a
1
:::a
n
>)
i
= a
i
and
lh( <a
1
:::a
n
>) = n.
Note also that whenever a > 0;we have lh(a) < a and (a)
i
< a.
Denition.
The relation Seq !is given by
Seq(a) iff 8x < a(lh(x) ̸= lh(a) _ 9i < lh(a)((x)
i+1
̸= (a)
i+1
):
That Seq is recursive is evident from properties enumerated above.From our
de nition,it is clear that Seq(a) if and only if a is the sequence number for some
sequence (in particular,a =<(a)
1
;:::;(a)
lh(a)
>).Note that
:Seq(a) iff 9x < a(lh(x) = lh(a) ^ 8i < lh(a)((x)
i+1
= (a)
i+1
):
Denition.
The initial sequence function Init:!
2
!!is given by
Init(a;i) = x(lh(x) = i ^ 8j < i((x)
j+1
= (a)
j+1
):
Again,Init is evidently recursive.Note that for 1  i  n,
Init( <a
1
;:::;a
n
>;i) =<a
1
;:::;a
i
>;
as intended.
COMPLETE PROOFS OF G

ODEL'S INCOMPLETENESS THEOREMS 7
Denition.
The concatenation function :!
2
!!is given by
a  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 recursive,and that
<a
1
:::a
n
>  <b
1
:::b
m
>=<a
1
:::a
n
;b
1
:::b
m
>;
as desired.
Denition.
For F:!!
k
!!,we de ne
F:!!
k
!!by
F(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 recursive if and only
if F is recursive.
Properties of Recursive Functions and Relations (continued):
P11.
For G:!!!
n
!!a recursive function,the function F:!!
n
!!,
given by
F(a;
b) = G(
F(a;
b);a;
b);
is recursive.Because
F(a;
b) is de ned in terms of values F(x;
b),for x
strictly smaller than a,this inductive de nition of F makes sense.
Proof.
Note that
F(a;
b) = G(H(a;
b);a;
b)
where
H(a;
b) = x(Seq(x) ^ lh(x) = a ^ 8i < a((x)
i+1
= G(Init(x;i);i;
b)):
According 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 computation is cumbersome,but possible,for any particular value a.
P12.
For G:!!
n
!!and H:!!
n
!!recursive functions,F:!!
n
!!
de ned by
F(a;
b) =
{
F(G(a;
b);
b) if G(a;
b) < a,and
H(a;
b) otherwise,
is recursive.
Proof.
Note that when G(a;
b) < a,we have
F(G(a;
b);
b) = (
F(a;
b))
G(a;
b)+1
= (
F(a;
b);G(a;
b) +1) = G

(
F(a;
b);a;
b)
with recursive G

(x;y;
z) = (x;G(y;
z)+1).Thus F is recursive by P11.
8 LECTURES BY B.KIM
For most purposes,when we de ne a function F inductively by cases,we must
satisfy two requirements to guarantee that our function is well-de ned.First,if
F(x;
b) appears in a de ning case involving a,we must show that x < a whenever
this case is true.Second,we must show that our base case is not de ned in terms
of F.In particular,this means that we cannot use F in a de ning case which is
used to compute F(0; ).
P13.
Given recursive G:!
n
!!and H:!
2
!
n
!!,F:!!
n
!!given
by
F(a;
b) =
{
H(F(a 1;
b);a 1;
b) if a > 0,and
G(
b) otherwise,
is recursive.(For example,the maps
n 7!n!=
{
(n 1)! n if n > 0
1 n = 0;
(n;m) 7!m
n
=
{
m
(n 1)
 m if n > 0,
1 n = 0;
and
n 7!(n +1)
th
prime =
{
x(x > n
th
prime ^ PN(x)) if n > 0
2 n = 0
are all recursive.)
Proof.
Note that H(F(a 1;
b);a 1;
b) = H( (
F(a;
b);a);a 1;
b) has the
form of P11.
P14.
Given recursive relations Q !
n+1
and R !
n+1
and recursive H:
!!
n
!!such that H(a;
b) < a whenever Q(a;
b) holds,the relation
P !
n+1
,given by
P(a;
b) iff
{
P(H(a;
b);
b) if Q(a;
b),
R(a;
b) otherwise,
is recursive.
Proof.
De ne H

:!!
n
!!by
H

(a;
b) =
{
H(a;
b) if Q(a;
b),and
a otherwise.
H

is clearly recursive.Note

P
(a;
b) =
{

P
(H

(a;
b);
b) if H

(a;
b) < a,and

R
(a;
b) otherwise.
The following example will prove useful:
COMPLETE PROOFS OF G

ODEL'S INCOMPLETENESS THEOREMS 9
Denition.
Let A !
2
be given by
A(a;c) iff Seq(c) ^ lh(c) = a ^ 8i < a((c)
i+1
= 0 _ (c)
i+1
= 1);
and let F:!
2
!!be given by
F(a;i) =
8
>
<
>
:
x(A(a;x)) if i = 0,
x(F(a;i 1) < x ^ A(a;x)) if 0 < i < 2
a
,and
0 otherwise.
Then the function bd:!!!is given by
bd(n) = F(n;2
n
1):
Evidently,A,F,and bd are all recursive.In fact,
bd(n) = maxf< c
1
c
2
:::c
n
> j c
i
= 0 or 1g:
Step 1:Representability of Recursive Functions in Q
We de ne Q,a subtheory of the natural numbers,and prove the Representability
Theorem,stating that all recursive functions are representable in this subtheory.
Consider the language of natural numbers L
N
= f+;;S;<;0g.We specify the
theory Q with the following axioms.
Q1.
8x Sx ̸= 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.
Note that the natural numbers,N,are a model of the theory Q.If we add to
this theory the set of all generalizations of formulas of the form

x
0
^ 8x(φ!φ
x
Sx
))!φ;
providing the capability for induction,we call this theory Peano Arithmetic,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 function f:!
n
!!is representable in Q if there exists an
L
N
-formula φ(x
1
;:::;x
n
;y) such that
Q ⊢ 8y(φ(k
1
;:::;k
n
;y) !y = f(k
1
;:::;k
n
)
)
for all k
1
;:::;k
n
2!.We say φ represents f in Q.
10 LECTURES BY B.KIM
Denition.
Arelation P !
n
is representable in Qif there exists an L
N
-formula
φ(x
1
;:::;x
n
) such that for all k
1
;:::;k
n
2!,
P(k
1
;:::;k
n
)!Q ⊢ φ(k
1
;:::;k
n
)
and
:P(k
1
;:::;k
n
)!Q ⊢:φ(k
1
;:::;k
n
):
Again,we say that φ 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 m̸= n,then Q ⊢:(m
= n
).
Proof.
It is enough to demonstrate this for m > n.For n = 0,our result follows
from axiom Q1.Assume,then,that the result holds for k = n and all l > k.Then
we have that,for a given m > n + 1,Q ⊢ m 1
̸= n
.By axiom Q2 we have,
Q ⊢ m 1
̸= n
!m
̸= n +1
.Hence we conclude that Q ⊢ m
̸= n +1
,and the
result 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 result
holds 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 n
Proof.
For n = 0,our result follows from axiom Q5.Assume,then,that the
result holds for k = n.Then Q ⊢ m
 n
= mn
.Applying Q6,we have that
Q ⊢ m
 n +1
= mn
+m
,and applying the previous lemma,we have the result for
k = 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 hold
for k = n.We show both claims hold for k = n +1 as well.
First,suppose m < n + 1.Either m < n,and Q ⊢ m
< n
by the induction
hypothesis,or m= n,and Q ⊢ m
= n
by Lemma 1.In either case,by Q8 and Rule
T,we have that Q ⊢ m
< n +1
.
Second,suppose m  n + 1.Then m > n and by the induction hypothesis,
Q ⊢:(m
< n
).By Lemma 1,we also have Q ⊢:(m
= n
).Again applying Q8 and
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
is
representable.
Proof.
Assume P is representable and that φ(x
1
:::x
n
) represents P.Let
(
x;y)  (φ(
x) ^ y = 0) _ (:φ(
x) ^ y = 1
):
We claim (
x;y) represents 
P
:
Suppose P(k
1
;:::;k
n
) holds.Then Q ⊢ φ(k
1
;:::;k
n
).Now since
φ(k
1
;:::;k
n
)!(y = 0 ! (k
1
;:::;k
n
;y))
COMPLETE PROOFS OF G

ODEL'S INCOMPLETENESS THEOREMS 11
is a tautology,we have Q ⊢ y = 0 ! (k
1
;:::;k
n
;y),as required.Similarly,if
:P(k
1
;:::;k
n
) holds,then Q ⊢:φ(k
1
;:::;k
n
),and since
⊢:φ(k
1
;:::;k
n
)!(y = 1
! (k
1
;:::;k
n
;y);
we obtain that Q ⊢ y = 1
! (k
1
;:::;k
n
;y),as required.Thus, (
x;y) repre-
sents 
P
.
Assume now that (
x;y) represents 
P
.Then (
x;0) represents P.
In particular,when P(k
1
;:::;k
n
) holds,we have
Q ⊢ (k
1
;:::;k
n
;y) !y = 0:
Substitution of y by 0 yields Q ⊢ (k
1
;:::;k
n
;0);as desired.Similarly,when
:P(k
1
;:::;k
n
) holds,we have
Q ⊢ (k
1
:::k
n
;y) !y = 1
;
and because Q ⊢:(0 = 1
) we may conclude Q ⊢: (k
1
:::k
n
;0),as needed.Thus
is P representable.
Lemma 6.
For a formula φ in L
N
,
Q ⊢ φ
x
0
!  !(φ
x
k 1
!(x < k
!φ))
Proof.
The proof is by induction on k.When k is 0,we have
Q ⊢ (x < 0!φ):
This is (vacuously) true by axiom Q7.Now,assume that
Q ⊢ φ
x
0
!:::!(φ
x
k 1
!(x < k
!φ)):
We must show that
Q ⊢ φ
x
0
!  !(φ
x
k
!(x < k +1
!φ)):
Equivalently,we want to show that ⊢ φ where = Q[ fφ
x
0
;:::;φ
x
k
;x < k +1
g.
By Q8, ⊢ x < k
_ x = k
.In the rst case,the inductive hypothesis implies that
⊢ φ,while in the latter case,j= x = k
!(φ
x
k
!φ),and hence ⊢ φ.By either
route, proves φ.
Lemma 7.
If (a) Q ⊢:φ
x
k
for each k < n,and (b) Q ⊢ φ
x
n
,then for z ̸= x not
appearing in φ,
Q ⊢ (φ ^ 8z(z < x!:φ
x
z
)) !x = n
:
Proof.
We de ne
 (φ ^ 8z(z < x!:φ
x
z
)):
Now,we obtain
j= x = n
!( !(φ
x
n
^ 8z(z < n
!:φ
x
z
))):()
By (a) and Lemma 6,we get
Q ⊢ x < n
!:φ;()
and,applying substitution and generalization,we obtain
Q ⊢ 8z(z < n
!:φ
x
z
):
Combining this with (b) and (),we conclude
Q ⊢ x = n
! :
12 LECTURES BY B.KIM
For the reverse implication,we note that
j= 8z(z < x!:φ
x
z
)!(n
< x!:φ
x
n
);
and thus (b) implies Q ⊢ !:(n
< x).Now Q[f ;x < n
g ⊢ φ ^:φ by () and
the de nition of .Therefore Q ⊢ !:(x < n
) and by Axiom Q9 we conclude
Q ⊢ !x = n
.
Representability Theorem.
Every recursive function or relation is representable
in Q.
Proof.
It suffices to prove representability of functions having the forms enumerated
in the de nition of recursiveness:
R1.
I
n
i
,+,,and 
<
.
The latter three are representable by Lemmas 2,3,and 4.In particular,
for +,say,we have that φ(x
1
;x
2
;y)  y = x
1
+x
2
represents + in Q,since
for any m;n 2!,
Q ⊢ m
+n
= m+n
;
Q ⊢ y = m
+n
!y = m+n
;
Q ⊢ φ(m
;n
;y) !y = m+n
;and hence
Q ⊢ 8y(φ(m
;n
;y) !y = m+n
);
as required. and 
<
are similar (with 
<
making additional use of Lemma
5).
I
n
i
is representable by φ(x
1
;:::;x
n
;y)  x
i
= y.In particular,for any
k
1
;:::;k
n
2!,I
n
i
(k
1
;:::;k
n
) = k
i
,and hence
Q ⊢ φ(k
1
;:::;k
n
;y) !y = k
i
!y = I
n
i
(k
1
;:::;k
n
)
;
by our choice of φ.Generalization completes the result.
R2.
F(
a) = G(H
1
(
a);:::;H
k
(
a)),where Gand each of the H
i
are representable.
Assume that G is represented in Q by φ and the H
i
are represented in
Q by
i
,respectively.We show that F is represented by
(
x;y)  9z
1
;:::;z
k
(
1
(
x;z
1
) ^    ^
k
(
x;z
k
) ^ φ(z
1
;:::;z
k
;y)):
In other word we want to show,for any a
1
;:::;a
n
2!,
Q ⊢ (a
1
;:::;a
n
;y) !y = G(H
1
(
a);:::;H
k
(
a))
(y)
where
a = (a
1
:::a
n
).
Now,for = Q[ f (a
1
;:::;a
n
;y)g,since the
i
represent H
i
,we have
that ⊢ 9z
1
;:::;z
k
(z
1
= H
1
(
a)
^    ^ z
k
= H
k
(
a)
^ φ(z
1
;:::;z
k
;y)):
Hence we have
j= 9z
1
;:::;z
k
(φ(H
1
(
a)
;:::;H
k
(
a)
;y));
and since the z
i
do not appear,
j= φ(H
1
(
a)
;:::;H
k
(
a)
;y):
Since φ represents G,we have
j= y = G(H
1
(
a);:::;H
k
(
a))
;
as required.
COMPLETE PROOFS OF G

ODEL'S INCOMPLETENESS THEOREMS 13
On the other hand,for  = Q[ fy = G(H
1
(
a);:::;H
k
(
a))
g,
 ⊢ φ(H
1
(
a)
;:::;H
k
(
a)
;y)
 ⊢ 9z
1
;:::;z
k
(z
1
= H
1
(
a)
^    z
k
= H
k
(
a)
^ φ(z
1
;:::;z
k
;y))
 ⊢ 9z
1
;:::;z
k
(
1
(
a;z
i
) ^   
k
(
a;z
k
) ^ φ(z
1
;:::;z
k
;y))
 ⊢ (a
1
;:::;a
n
;y)
Thus (y) is established.
R3.
F(
a) = x(G(
a;x) = 0),where G is representable in Q and for all
a there
exists x such that G(
a;x) = 0,is representable in Q.
Assume G is represented in Q by φ(x
1
;:::;x
n
;x;y).Let
(x
1
;:::;x
n
;x)  φ
y
0
^ 8z(z < x!:φ
yx
0z
):
Let F(
a) = b and k
i
= G(
a;i) for i 2!.Then
Q ⊢ φ(a
1
;:::;a
n
;i
;y) !y = k
i
;
thus
Q ⊢ φ(a
1
;:::;a
n
;i
;0) !0 = k
i
;
.Hence now if j < b,so that k
j
̸= 0,then
Q ⊢:φ(a
1
;:::;a
n
;j
;0):
On the other hand,k
b
= 0,so
Q ⊢ φ(a
1
;:::;a
n
;b
;0):
Hence,by Lemma 7,
Q ⊢ (φ(
a;x;y)
y
0
^ 8z(z < x!:φ(
a;x;y)
y
0
x
z
)) !x = b
;
and thus,
Q ⊢ (
a;x) !x = b
:
By generalization,we have that represents F in Q,as desired.
Step 2:Axiomatizable Complete Theories are Decidable
We begin by showing that we may encode terms and formulas of a reasonable
language in such a way that important classes of formulas,e.g.,the logical axioms,
are mapped to recursive subsets of the natural numbers.We use this to derive the
main result.
Denition.
Let L be a countable language with subsets C,F,and P of constant,
function,and predicate symbols,respectively (=2 P).Let V be a set of variables
for L.L is called reasonable if the following two functions exist:

h:L[f:;!;8g[V!!injective such that V
= h(V),C
= h(C),F
= h(F),
and P
= h(P) are all recursive.

AR:!!!∖f0g recursive such that AR(h(f)) = n and AR(h(P)) = n
for n-ary function and predicate symbols f and P.
For the rest of this note,the language L is countable and reasonable.
Now we de ne a coding ⌈⌉:fL-terms and L-formulasg!!inductively,by

For x 2 V [ C,⌈x⌉ =<h(x)>.
14 LECTURES BY B.KIM
 For L-terms u
1
;:::;u
n
and n-ary f 2 F,
⌈fu
1
u
2
:::u
n
⌉ =<h(f);⌈u
1
⌉;⌈u
2
⌉;:::;⌈u
n
⌉>:

For L-terms t
1
;:::;t
n
and P 2 P,
⌈Pt
1
t
2
:::t
n
⌉ =<h(P);⌈t
1
⌉;:::;⌈t
n
⌉>:

For L-formulas φ and ,
⌈φ! ⌉ =<h(!);⌈φ⌉;⌈ ⌉>;
⌈:φ⌉ =<h(:);⌈φ⌉>;
⌈8xφ⌉ =<h(8);⌈x⌉;⌈φ⌉>:
Note that our de nition of ⌈⌉ is one-to-one.Given a term or formula ,we call
⌈⌉ the Godel number of .
We show the following predicates and functions are recursive (We follow de ni-
tions for syntax in [E].):
(1)
Vble = f⌈v⌉ j v 2 Vg !and Const = f⌈c⌉ j c 2 Cg !.
Proof.
Note
Vble(x) iff x =<(x)
1
> ^V
((x)
1
);
Const(x) iff x =<(x)
1
> ^C
((x)
1
):
(2)
Term = f⌈t⌉ j t an L-termg !.
Proof.
Note
Term(a) iff
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.
(3)
AtF = f⌈⌉ j  an atomic L-formulag !.
Proof.
Note
AtF(a) iff Seq(a) ^ P
((a)
1
) ^ (AR((a)
1
) = lh(a)
_
1)
^ 8j <(lh(a)
_
1) (Term((a)
j+2
)):
(4)
Form = f⌈φ⌉ j φ an L-formulag !.
Proof.
Note
Form(a) iff
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
!!,such that Sub(⌈t⌉;⌈x⌉;⌈u⌉) = ⌈t
x
u
⌉ and Sub(⌈φ⌉;⌈x⌉;⌈u⌉) =
⌈φ
x
u
⌉ for terms t and u,variable x,and formula φ.
COMPLETE PROOFS OF G

ODEL'S INCOMPLETENESS THEOREMS 15
Proof.
De ne
Sub(a;b;c) =
8
>
>
>
>
>
>
>
>
<
>
>
>
>
>
>
>
>
:
c if Vble(a) ^ a = b,
<(a)
1
;Sub((a)
2
;b;c);:::if lh(a) > 1 ^ (a)
1
̸= h(8)
:::;Sub((a)
lh(a)
;b;c)> ^Seq(a);
<(a)
1
;(a)
2
;Sub((a)
3
;b;c)> if a =<h(8);(a)
2
;(a)
3
>,
^(a)
2
̸= b
a otherwise.
Note that,if well-de ned,the function has the properties desired above.
We show Sub is well-de ned by induction on a:a = 0 falls into the
rst or last category since lh(0) = 0,hence Sub(0;b;c) is well-de ned for
all b;c 2!.If a ̸= 0,then (a)
i
< a for all i  lh(a),and thus we may
assume the values Sub((a)
i
;b;c) are well-de ned,showing Sub(a;b;c) to be
well-de ned in all cases.
(6)
Free !
2
,such that for formula φ,term ,and variable x,Free(⌈φ⌉;⌈x⌉)
if and only if x occurs free in φ,and Free(⌈⌉;⌈x⌉) if and only if x occurs
in 
Proof.
De ne
Free(a;b) iff
8
>
<
>
:
9j <(lh(a)
_
1) (Free((a)
j+2
;b)) if lh(a) > 1 ^ (a)
1
̸= h(8),
Free((a)
3
;b) ^ (a)
2
̸= b if lh(a) > 1 ^ (a)
1
= h(8),
a = b otherwise.
Free clearly has the desired property,and that it is well-de ned follows by
essentially the same induction on a as above.
(7)
Sent = f⌈φ⌉ j φ is an L-sentenceg !.
Proof.
Note
Sent(a) iff Form(a) ^ 8b<a(:Vble(b) _:Free(a;b)):
(8)
Subst(a;b;c) !
3
such that for a given formula φ,variable x,and term t,
Subst(⌈φ⌉;⌈x⌉;⌈t⌉) if and only if t is substitutable for x in φ.
Proof.
De ne
Subst(a;b;c) iff
8
>
>
>
>
>
>
<
>
>
>
>
>
>
:
Subst((a)
2
;b;c) if a =<h(:);(a)
2
>,
Subst((a)
2
;b;c) ^ Subst((a)
3
;b;c) if a =<h(!);(a)
2
;(a)
3
>,
:Free(a;b) _ (:Free(c;(a)
2
) if a =<h(8);(a)
2
;(a)
3
>,
^Subst((a)
3
;b;c))
0 = 0 otherwise.
Note that Subst has the desired property,and is well-de ned by essentially
the same induction used above.
16 LECTURES BY B.KIM
(9)
We de ne
False(a;b) iff
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 recursive by the same induction as applied above.We note the
signi cance of False presently.
To each b 2!,we may associate a truth assignment v
b
such that for a prime
formula (atomic or of the form 8xφ),
v
b
( ) = F iff (b)
⌈ ⌉
= 0:
Further,for any truth assignment v:A!fT;Fg,where A is a nite set of prime
formulas,there exists a b such that v = v
b
:we may write A = fφ
1
;:::;φ
n
g such
that ⌈φ
1
⌉ < ⌈φ
2
⌉ <    < ⌈φ
n
⌉.For 1  j  ⌈φ
n
⌉ de ne c
j
= 0 when j = ⌈φ
i

for some i  n and v(φ
i
) = F,and c
j
= 1 otherwise.Then b =<c
1
;:::;c
⌈φ
n

>
satis es v
b
= v on A.
Then moreover,for any formula φ built up from A,
v(φ) = F iff
v
b
(φ) = F iff False(⌈φ⌉;b):
(10)
De ne Taut = f⌈⌉ j is a tautologyg !.
Proof.
Recall bd:!!!such that bd(a) = maxf < c
1
;:::;c
a
> j c
i
2
f0;1gg,recursive,has been previously de ned.De ne
Taut(a) iff Form(a) ^ 8b<(bd(a) +1) (:False(a;b)):
(11)
AG2
= f⌈φ⌉ j φ is in axiom group 2g !.
Proof.
Recall axiomgroup 2 contains formulas of the form8x !
x
t
,with
term t substitutable for x in .Thus
AG2
(a) iff 9x;y;z <a(Vble(x) ^ Form(y) ^ Term(z) ^ Subst(y;x;z)
^ a =<h(!);<h(8);x;y>;Sub(y;x;z)>);
where 9x;y;z <aP(x;y;z) abbreviates what one would expect.
(12)
AG3
= f⌈φ⌉ j φ is in axiom group 3g !.
Proof.
Recall we take axiomgroup 3 to be the formulas having the following
form:8x( !

)!(8x !8x

).Thus
AG3
(a) iff 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
= f⌈φ⌉ j φ is in axiom group 4g !.
COMPLETE PROOFS OF G

ODEL'S INCOMPLETENESS THEOREMS 17
Proof.
Recall axiomgroup 4 contains formulas of the form !8x ,where
x does not occur free in .Thus
AG4
(a) iff 9x;y<a(Vble(x) ^ Form(y)
^:Free(y;x) ^ a =<h(!);y;<h(8);x;y>>)
(14)
AG5
= f⌈φ⌉ j φ is in axiom group 5g !.
Proof.
Recall axiom group 5 contains formulas of the form x = x,for a
variable x,hence
AG5
(a) iff 9x<a(Vble(x) ^ a =<h(=);x;x>):
(15)
AG6
= f⌈φ⌉ j φ is in axiom group 6g !.
Proof.
Recall formulas of axiom group 6 have the form x = y!( !

),
where is an atomic formula and

is obtained by from by replacing
one or more occurrences of x with y.Thus
AG6
(a) iff 9x;y;b;c<a(Vble(x) ^ Vble(y) ^ AtF(b) ^ AtF(c)
^ lh(b) = lh(c) ^ 8j < lh(b) +1((c)
j
= (b)
j
_ ((c)
j
= y ^ (b)
j
= x))
^ a =<h(!);<h(=);x;y>;<h(!);b;c>>)
(16)
Gen(a;b) !
2
,such that Gen(⌈φ⌉;⌈ ⌉) if and only if φ is a generalization
of (i.e.,φ = 8x
1
:::8x
n
for some nite fx
i
g  V).
Proof.
Note that
Gen(a;b) iff
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)

= f⌈⌉ j  2 g !,where  is the set of logical axioms.
Proof.
Note that

(a) iff 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 codings:<> on sequences of natural num-
bers,h on the language and logical symbols,and ⌈⌉ on the terms and formulas.We
presently de ne a fourth coding,of sequences of formulas:
⌈⌈⌉⌉:fsequences of L-formulasg!!;
given by
⌈⌈φ
1
;:::;φ
n
⌉⌉ =<⌈φ
1
⌉;:::;⌈φ
n
⌉>:
18 LECTURES BY B.KIM
This map is one-to-one,as it is derived from the established (injective) codings,
and in particular,we can determine,for a given number,if it lies in the image of
⌈⌈⌉⌉,and,if so,recover the associated sequence of formulas.
Denition.
Given L,let T be a theory (a collection of sentences) in L.De ne
T
= f⌈⌉ j  2 Tg:
We say that T is axiomatizable if there exists a theory S,axiomatizing T (that
is,such that CnS = CnT),such that S
is recursive.We say that T is decidable
if CnT
is recursive.
We shall make use of the following relations:

Ded
T
= f⌈⌈φ
1
;:::;φ
n
⌉⌉ j φ
1
;:::;φ
n
is a deduction from Tg !.
Note that
Ded
T
(a) iff Seq(a) ^ lh(a) ̸= 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
>))

Prf
T
!
2
,given by Prf
T
(a;b) iff Ded
T
(b) ^ a = (b)
lh(b)
.

Pf
T
!,given by Pf
T
(a) iff Sent(a) ^ 9xPrf
T
(a;x).
Note that we may read Prf
T
(a;b) as\b is a proof of a from T,"and Pf
T
(a) as
\a is a sentence provable from T."In particular
Pf
T
= CnT
= f⌈⌉ j T ⊢ g:
We use this fact to prove the following:
Theorem.
If T is axiomatizable,then Pf
T
= CnT
is recursively enumerable.
Proof.
Let S axiomatize T,where S is recursive.From the above de nitions,we
see that Ded
S
and Prf
S
are recursive relations,hence Pf
S
is an r.e.relation.But
Pf
S
= Pf
T
,since CnS = CnT.
Theorem.
If T is axiomatizable and complete in L,then T is decidable.
Proof.
By the negation theorem,it suffices to show that:Pf
T
is recursively enu-
merable.Note that since T is complete,for any sentence ,T ⊬  if and only if
T ⊢:.Hence
:Pf
T
(a) iff:Sent(a) _ 9mPrf
T
( <h(:);a>;m)
iff 9m(:Sent(a) _ Prf
T
( <h(:);a>;m)):
Thus:Pf
T
is recursively enumerable,and Pf
T
is recursive.
We can see that if we say T is axiomatizable in wider sense when S axiomatiz-
ing T is recursively enumerable,then the above two theorems still hold with this
seemingly weaker notion.In fact,two notions are equivalent,which is known as
Craig's Theorem.
Step 3:The Incompleteness Theorems and Other Results
We return now to the language of natural numbers,L
N
.Recall that we de ne,
for a natural number n,
n
 SS:::S
|
{z
}
n
0:
COMPLETE PROOFS OF G

ODEL'S INCOMPLETENESS THEOREMS 19
Denition.
The diagonalization of an L
N
formula φ is a new formula
d(φ)  9v
0
(v
0
= ⌈φ⌉
^ φ);
where 9 and ^ provide the usual abbreviations in L
N
.
In particular,we note d(φ) is satis able precisely when φ is satis able by some
truth assignment taking v
0
to the Godel number of φ,and L
N
j= d(φ) precisely
when φ is satis ed by every truth assignment taking v
0
to ⌈φ⌉.
Lemma.
There exists a recursive function dg:!!!such that for any L
N
formula,dg(⌈φ⌉) = ⌈d(φ)⌉.
Proof.
De ne num:!!!by num(0) =<0> and,for n 2!
num(n +1) =<h(S);num(n)>:
In particular,note that num(n) = ⌈n
⌉.
De ne
dg(a) =<h(:);<h(8);⌈v
0
⌉;<h(:);
<h(:);<h(!);<h(=);⌈v
0
⌉;num(a)>;<h(:);a>>>>>>
Then
dg(⌈φ⌉) =<h(:);<h(8);⌈v
0
⌉;<h(:);
<h(:);<h(!);<h(=);⌈v
0
⌉;num(⌈φ⌉)>;<h(:);⌈φ⌉>>>>>>;
=<h(:);<h(8);⌈v
0
⌉;<h(:);
<h(:);<h(!);<h(=);⌈v
0
⌉;⌈⌈φ⌉
⌉>;<h(:);⌈φ⌉>>>>>>:
However,writing out what formula this encodes and introducing our usual abbre-
viations,we have
dg(⌈φ⌉) = ⌈:8v
0
:(:(v
0
= ⌈φ⌉
!:φ))⌉
= ⌈9v
0
(v
0
= ⌈φ⌉
^ φ)⌉
= ⌈d(φ)⌉;
as desired.
Fixed Point Theorem(Godel).
For any L
N
-formula φ(x) (i.e.,either a sentence
or a formula having x as the only free variable),there is some L
N
-sentence  such
that
Q ⊢  !φ(⌈⌉
):
Proof.
Since dg is recursive,it is representable in Q by Step 1,say by (x;y).Then
Q ⊢ 8y( (n
;y) !y = dg(n)
):
Let (v
0
)  9y( (v
0
;y) ^ φ(y)),and let n = ⌈(v
0
)⌉.De ne
  d((v
0
))  9v
0
(v
0
= n
^ (v
0
)):
Then if we let k = dg(n) = ⌈⌉,we have
j=  !(n
) !9y( (n
;y) ^ φ(y)):
But
Q ⊢ (n
;y) !y = k
;
20 LECTURES BY B.KIM
and therefore
Q ⊢  !9y(y = k
^ φ(y)) !φ(k
) !φ(⌈⌉
);
as required.
Tarski Undenability Theorem.
ThN
= f⌈⌉ j N j= g is not denable.
Proof.
Suppose ThN
were de nable by (x).Then by the xed point lemma,with
φ =: ,there exists a sentence  such that
N j=  !: (⌈⌉
):
Then N j=  implies that N ̸j= (⌈⌉
),implying N ̸j= ,or N j=:,since ThN
is complete.On the other hand,N ̸j=  implies N j=:,and thus that N j=
(⌈⌉
),implying N j= .The contradictions together imply that cannot represent
ThN
.
Strong Undecidability of Q.
Let T be a theory in L  L
N
.If T [Q is consistent
in L,then T is not decidable in L (CnT
is not recursive).
Proof.
Assume that CnT
is recursive.We rst show that this implies recursiveness
of Cn(T [Q)
.Since Q is nite,it suffices to show that for any sentence  in the
language,Cn(T [ fg)
is recursive.
In particular,note that 2 Cn(T [ fg) iff ! 2 CnT.Thus
a 2 Cn(T [fg)
iff Sent(a) ^ <h(!);⌈⌉;a>2 CnT
:
Hence Cn(T [fg)
is recursive,as desired.
To prove the theorem,then,it suffices to show that Cn(T [ Q)
is not recursive.
If this were the case,then it would be representable,say by (x),in Q.By the
xed point lemma,there exists an L
N
sentence  such that
Q ⊢  !: (⌈⌉
):
If T [ Q ⊢ ,then
Q ⊢ (⌈⌉
);
by the representability of Cn(T [Q)
by (x) in Q.In particular,
Q ⊢:;
a contradiction.On the other hand,if T [ Q ⊬ ,then by representability,
Q ⊢: (⌈⌉
);
and hence
Q ⊢ ;
a contradiction,implying that Cn(T [ Q)
is not representable,and hence not re-
cursive.
Corollary.
ThN,PA,and Q are all undecidable.
Proof.
We need note only that each of these theories is consistent with Q.
Moreover,we have:
COMPLETE PROOFS OF G

ODEL'S INCOMPLETENESS THEOREMS 21
Undecidability of First Order Logic (Church).
For a reasonable countable
language L  L
N
,the set of all Godel numbers of valid sentences (f⌈⌉ j ∅ ⊢ g)
is not recursive (the set of valid sentences is not decidable).
In fact,the above corollary is true for any countable L containing a k-ary pred-
icate or function symbol,k  2,or at least two unary function symbols.
Godel-Rosser First Incompleteness Theorem.
If T is a theory in a countable
reasonable L  L
N
,with T [ Q consistent and T axiomatizable,then T is not
complete.
Proof.
By Step 2,if T is complete,then T is decidable,contradicting the strong
undecidability of Q.
Remarks.
In (N;+),0,<,and S are de nable.Hence the same result follows if we
take L

N
= f+;g instead of our usual L
N
.In particular,Th(N;+;) is undecidable,
and for any T

 Q

(where Q

is simply Q written in the language of L

N
),we have
that T

is,if consistent,undecidable,and,if axiomatizable,incomplete.
It is important to note that for an undecidable theory T,we may have T  T

,
where T

is a decidable theory.As an example,the theory of groups is undecidable,
whereas the theory of divisible torsion-free groups is decidable.
We turn our attention now to the proof of the result used in Godel's original
paper.In particular,Godel 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 consistent theory in a reasonable countable language L  L
N
,
and presume that T
is recursive.Then
T ⊢  )Q ⊢ Pf
T
(⌈⌉
):
In particular,T ⊢  implies that Prf
T
(⌈⌉
;m) for some m 2!.Since Prf
T
is
recursive,it is representable in Q,hence Q ⊢ Prf
T
(⌈⌉
;m
),and
Q ⊢ 9xPrf
T
(⌈⌉
;x);
or
Q ⊢ Pf
T
(⌈⌉
):
By the xed point lemma,there exists a sentence such that
T  Q ⊢ !:Pf
T
(⌈ ⌉
):()
If T ⊢ ,then Q ⊢ Pf
T
(⌈ ⌉
),and thus Q ⊢: ,and hence T ⊢: ,a contradiction.
Thus T ⊬ .
On the other hand,if T is!-consistent (i.e.,whenever T ⊢ 9xφ(x),then for
some n 2!,T ⊬:φ(n
)),then T ⊬: .In particular,if T ⊢: ,then
T ⊢ Pf
T
(⌈ ⌉
);
by ().That is,
T ⊢ 9xPrf
T
(⌈ ⌉
;x):
However,if Prf
T
(⌈ ⌉
;m) for some m2!,then T ⊢ ,contradicting the consis-
tency of T.Thus we must have:Prf
T
(⌈ ⌉
;m) for all m 2!.Since Q represents
Prf
T
,
T  Q ⊢:Prf
T
(⌈ ⌉
;m)
22 LECTURES BY B.KIM
for all m2!,contradicting the!-consistency of T.
Rosser generalized Godel's proof by singling out for T a sentence such that
T ⊬ and T ⊬: ,without the assumption of!-consistency.
We now begin our approach to Godel's Second Incompleteness Theorem.We x
T,a theory in a countable reasonable language L  L
N
.
We note the following fact from Hilbert and Bernays'Grundlagen der Mathe-
matik,1934.
Fact.
If T is consistent,T ⊢ PA,and T
is recursive,then for any sentences  and
 in L,
I.T ⊢  )Q ⊢ Pf
T
(⌈⌉
)
II.PA ⊢ (Pf
T
(⌈⌉
) ^ Pf
T
(⌈!⌉
))!Pf
T
(⌈⌉
)
III.PA ⊢ Pf
T
(⌈⌉
)!Pf
T
(
⌈Pf
T
(⌈⌉
)⌉
)
Notation.
We will write Con
T
:Pf
T
(⌈0 ̸= 0⌉
).Clearly Con
T
holds if and only
if T is consistent.
Lemma.
If T ⊢ !,then PA ⊢ Pf
T
(⌈⌉
)!Pf
T
(⌈⌉
).
Proof.
If T ⊢ !,then by (I) above,
PA ⊢ Pf
T
(⌈!⌉
);
and by (II),
PA ⊢ Pf
T
(⌈⌉
)!Pf
T
(⌈⌉
):
Godel's Second Incompleteness Theorem.
If T is consistent,T
is recursive,
and T ⊢ PA,then T ⊬ Con
T
.
Proof.
By the xed point lemma,there exists  such that
Q ⊢  !:Pf
T
(⌈⌉
):(y)
By (III),above,
PA ⊢ Pf
T
(⌈⌉
)!Pf
T
(
⌈Pf
T
(⌈⌉
)⌉
)
:(z)
And further,by Lemma,we have
PA ⊢ Pf
T
(
⌈Pf
T
(⌈⌉
)⌉
)
!Pf
T
(⌈:⌉
):
Combining this result with (z),we have
PA ⊢ Pf
T
(⌈⌉
)!Pf
T
(⌈:⌉
):
Now note that ⊢: !(!(0 ̸= 0)).By the lemma,
PA ⊢ Pf
T
(⌈⌉
)!Pf
T
(⌈!(0 ̸= 0)⌉
):
In particular,
PA ⊢ Pf
T
(⌈⌉
)!Pf
T
(⌈⌉
) ^ Pf
T
(⌈!(0 ̸= 0)⌉
);
hence,by (II),
PA ⊢ Pf
T
(⌈⌉
)!Pf
T
(⌈0 ̸= 0⌉
);
COMPLETE PROOFS OF G

ODEL'S INCOMPLETENESS THEOREMS 23
i.e.
PA ⊢ Pf
T
(⌈⌉
)!:Con
T
:
Thus PA ⊢ Con
T
!,by (y).
Now,suppose that T ⊢ Con
T
.Then T ⊢ ,and hence by (I),T  Q ⊢ Pf
T
(⌈⌉
).
But again,by (y),this implies that T ⊢:,a contradiction,showing that T cannot
prove its own consistency.
We remark that one may carry the proof through using only the assumption that
T
is recursively enumerable.
Lob's Theorem.
Suppose T is a consistent theory in L  L
N
,such that T
re-
cursive,and T ⊢ PA.Then for any L-sentence ,if T ⊢ Pf
T
(⌈⌉
)!,then
T ⊢ .
Proof.
By the xed point lemma,there exists  such that
Q ⊢  !(Pf
T
(⌈⌉
)!):
Since T ⊢ PA  Q,T proves the same result.From this we may deduce that
PA ⊢ Pf
T
(⌈⌉
)!Pf
T
(⌈⌉
):
In particular,by our lemma,we have
PA ⊢ Pf
T
(⌈⌉
)!Pf
T
(
⌈Pf
T
(⌈⌉
)!⌉
)
;
and,combining this with (III) from above,
PA ⊢ Pf
T
(⌈⌉
)!Pf
T
(
⌈Pf
T
(⌈⌉
)⌉
)
^ Pf
T
(
⌈Pf
T
(⌈⌉
)!⌉
)
;
and thus,by (II),
PA ⊢ Pf
T
(⌈⌉
)!Pf
T
(⌈⌉
);
as desired.
Now assume that T ⊢ Pf
T
(⌈⌉
)!.Then,by the above,
T ⊢ Pf
T
(⌈⌉
)!:
By our choice of ,this in turn implies that T ⊢ .By (I),we have that Q ⊢
Pf
T
(⌈⌉
),and hence T proves the same result,implying that T ⊢ ,as desired.
Remark.
Godel's Second Incompleteness Theorem in fact follows from Lob's The-
orem.In particular,given T as in the hypotheses of both theorems,if T ⊢ Con
T
,
then
T ⊢ Pf
T
(⌈0 ̸= 0⌉
)!0 ̸= 0:
But by Lob's Theorem,this in turn implies that T ⊢ 0 ̸= 0,showing that such a
theory,if consistent,cannot prove its own consistency.
References
[BJ]
G.S.Boolos and R.C.Jeffrey,Computability and logic.
[E]
H.Enderton,A mathematical introduction to logic.
[Sh]
J.R.Shoen eld,Mathematical logic.
[Sm]
R.M.Smullyan,Godel's incompleteness theorems.