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Mar. 20, 2001
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Subject: Notes For Puma Grass lecture (Mar. 2001 )
@:= Speak
#:=Write on board
@- First, I would like to thank William for his talk last week in which he explained some of the many ways
in which people have misinterpreted Godels first incompleteness theorem
-In this talk I hope, among other things, to give a proof of this theorem
-I will then leave you to make your own misinterpretations
@-What we are going to prove is
#-For any consistent formal system Z with computable axioms which is powerful enough to contain (N, 0,
1, +,*) there is a sentence G such that neither G nor ~G can be proved.
@-Now, the proof I will present here is not the one which Godel first came up with. In fact, when Godel
originally proved his theorem, he could only prove it for \omega-consistent theories.
-The sentence we will try and construct is a formalized version of the liars paradox. We want the
statement to say that I am not provable.
-Before we start, we have to determine which system we are working in.
-We will let our language be:
#- L:
1
0
2
S
3
~
11
x_1
12
x_2
13
x_3
…
…
4
v
5

6
=
7
+
8
*
9
(
10
)
#Term0, x_I are terms
if a, b are terms then so are Sa, (a+b), and (a*b)
#Formulasif s, t are terms then s=t is a formula
if H,G are formula's and v is a variable then ~(F), (F) v (G) and v(F) are formula's
@-Our axioms will be all logical axioms and equality axioms as well as the following
#-Number Theoretic Axioms
If F is a formula, x,y are terms and v,w is a variable x_i
1) Number Theoretic Axioms
a) ~(Sx = 0)
b) Sx=Sy <-> x=y
c) x+0 = x
d) x+Sy=S(x+y)
e) x*0=0
f) x*Sy=(x*y)+x
g) F(0) and w(F(w) -> F(Sw)) -> F(v)
@-It is important to note that while there are infinitely many axioms, given a formula, it is possible to
determine absolutely if it is an axiom. Godels theorem only applies to systems with this property.
-Now that we have defined our system, we have to come up with a way to talk about sentences of our
system. This is done by Godel numbering the language.
#-Put the Godel numbers above each symbol.
@-Now that we have encoded our language by numbers, we want a way to encode a formula by a
number in an effective manner.
-We will do this by looking at the exponents of the number we are interested in.
-For example
#-x_1+3 = 4: is encoded by 2^11 * 3^7 * 5^2 * 7^2 * 11^2 * 13^1 * 17^6 * 19^2 * 23^2 * 29^2 *
31^2 * 37^1
#-We will let [\phi] be the Godel number of the formula \phi, and we will let \gamma(x) [\g(x)] be the
formula representing the number x.
@-Now this encoding isn't very useful unless we can mirror the manipulations done in the system in the
Godel numbers.
-In order to do this, we first need the notion of a computable function and a computable relation.
-For simplicity of notation, we will only give the definition for one variable. It can easily be extended to
more than one variable.
#-We say that + is a computable function
-x<=y is a computable relation
-If f, g are computable functions, Q,R are computable relations then so are
1) h_1(x)=f(g(x)) (recursion)
2) Recursion:
h_2(0)= c
h_2(k+1):= f(k,h_2(k))
3) Bounded leastness:
h(n) = least x<= f(n) & R(x)
4)~Q, Q v R
5) Bounded Quantification:
T(x)<=> v (v< f(n) -> R(v))
6) R(x,y)<=> f(x)=y
@-We will not prove it here, but if R is a computable relation then there is a formula \phi_R of our
system such that:
#|-\phi_R(x,y) <=> R(x,y)
|-~\phi_R(x,y) <=> ~R(x,y)
@- Further, if R is a computable function we can say have
#-Description Lemma
If R is a computable function, with formula \phi_R and q=\phi(m) then
|-F(Z(q)) <=>|- F(\phi_R(Z(m)))
@-Basically what this says, is that if we have a computable function which has a value q at m then we
can prove that for any formula F if we apply F to q we get the same truth value as we would if we first
composed F with \phi_R and then took the value at
-With this in hand, we can begin to construct the machinery we will need.
-We will first need several functions (Explain what they are):
#-Neg(x) =[~( \g(x) ) ]
-Dis(x,y) =[( \g(x) ) v ( \g(y) )) ]
-Gen(x,y) =[\g(x) ( \g(y) )]
Axiom(x) <=>x is an axiom
UG(x,y) <=> \g(y) is a variable, and \g(x) = \g(y) ( \g(x) )
MP(x,y,z) <=> \g(x) = a & \g(y) = a->b & \g(z)= b
-Our goal now is to create a computable function and a computable relation such that
#-Sub(x,v,y)=[\g(x) with \g(y) substituted for all free occurances of variable \g(y)]
-Proof(x,y)<->y is a proof of \g(x) <->
(y= 2^a_1 3^a_2… Prime(n)^a_n ) & (a_n=x) &
( i<=n (Axiom(a_i) v (j,k<i MP(a_j,a_k,a_i)) v (j<i ,v MP(a_j,v,a_i)))
@-In Godel's original paper, he gave very explicit definitions of all the functions he needed. But as we
don’t have that much time, I will just hope you believe the they are computable
@-Now that we have finally established the machinery we just need one more fact before we can prove
the Godel’s theorem of the lecture: The fixed point theorem.
#-The Fixed Point Theorem
Let F(x_1) be any formula with only one free variable. Then there is a formula G such that
|-G<=>F( [G] )
Proof:
Let q=[F(s(x_1))] and let G = F(s( Z(q) ))
Now, [G] = Sub ([F(s(x_1))], [x_1], q) = Sub (q, [x_1], q) = s(q)
So, |-G <=> F(s(q))
<=> F([G])
QED.
@-Now we are finally ready to prove Godel's incompleteness Theorem.
-#Define Prov(x)<=> (y Proof(x,y)) & (z<y ~Proof(Neg(x), z)
#Let G be the fixed point of the formula ~Prov(x). So we have
|-G <=> ~Prov([G])
#Assume |-G (and so we have Proof([G],p) for a specific p we can find)
|-~Prov([G])
|-y~Proof([G],y)
|-~Proof([G],p) =><=
#Assume |-~G (and so we have Proof([~G],p) for a specific p we can find)
|-Prov([G])
|-(y Proof([G],y)) & (y<p)
So, we can check each Proof([G], y) at each value less than p and find the value such that
|-Proof([G],p) and so we can therefore convert this number to an actual proof and get
|-G =><=
-If we have time.
-Now that we have the fixed point theorem, it is easy to prove Tarski's Undefinability of Truth Theorem
which says
-#There is no formula which defines truth.