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.
Denition.
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).
Denition.
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.)
Denition.
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.
Denition.
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 nonempty 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 onetoone.
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 onetoone,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:
Denition.
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 onetoone,
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:
Denition.
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.
Denition.
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
):
Denition.
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
Denition.
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.
Denition.
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 wellde 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
Denition.
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
(φ
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:
Denition.
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
Denition.
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.
Denition.
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 nary 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 ⌈⌉:fLterms and Lformulasg!!inductively,by
For x 2 V [ C,⌈x⌉ =<h(x)>.
14 LECTURES BY B.KIM
For Lterms u
1
;:::;u
n
and nary f 2 F,
⌈fu
1
u
2
:::u
n
⌉ =<h(f);⌈u
1
⌉;⌈u
2
⌉;:::;⌈u
n
⌉>:
For Lterms t
1
;:::;t
n
and P 2 P,
⌈Pt
1
t
2
:::t
n
⌉ =<h(P);⌈t
1
⌉;:::;⌈t
n
⌉>:
For Lformulas φ and ,
⌈φ! ⌉ =<h(!);⌈φ⌉;⌈ ⌉>;
⌈:φ⌉ =<h(:);⌈φ⌉>;
⌈8xφ⌉ =<h(8);⌈x⌉;⌈φ⌉>:
Note that our de nition of ⌈⌉ is onetoone.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 Ltermg !.
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 Lformulag !.
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 Lformulag !.
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 wellde ned,the function has the properties desired above.
We show Sub is wellde ned by induction on a:a = 0 falls into the
rst or last category since lh(0) = 0,hence Sub(0;b;c) is wellde 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 wellde ned,showing Sub(a;b;c) to be
wellde 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 wellde ned follows by
essentially the same induction on a as above.
(7)
Sent = f⌈φ⌉ j φ is an Lsentenceg !.
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 wellde 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 Lformulasg!!;
given by
⌈⌈φ
1
;:::;φ
n
⌉⌉ =<⌈φ
1
⌉;:::;⌈φ
n
⌉>:
18 LECTURES BY B.KIM
This map is onetoone,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.
Denition.
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
Denition.
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 Undenability Theorem.
ThN
= f⌈⌉ j N j= g is not denable.
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 Godel 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 kary pred
icate or function symbol,k 2,or at least two unary function symbols.
GodelRosser 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 torsionfree 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
(⌈⌉
):
Godel'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.
Lob's Theorem.
Suppose T is a consistent theory in L L
N
,such that T
re
cursive,and T ⊢ PA.Then for any Lsentence ,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,Godel's incompleteness theorems.
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