A Magician Looks at Gödel’s Proof

Raymond Smullyan

USA

A Magician Looks at Godel’s Proof – p. 1

Part I

An Incompleteness Theorem

A Magician Looks at Godel’s Proof – p. 2

An Incompleteness Theorem

We consider a mathematical system in which

certain expressions are calledpredicates, and

by asentenceis meant any expression of the

form HX, whereH is a predicate andX is any

expression in the language of the system.

A Magician Looks at Godel’s Proof – p. 3

Intuition:

We might think of a predicateH as being the

name of a property of expressions, andHX as

a sentence asserting thatX has the property

named byH.

A Magician Looks at Godel’s Proof – p. 4

Some definitions

The system has a negation symbolN , and

for any sentenceX, the expressionNX is

also a sentence called thenegationof X.

The system is calledconsistentiff no

sentence and its negation are both provable.

A sentence is calledundecidableiff neither

it nor its negation is provable.

A Magician Looks at Godel’s Proof – p. 5

Two conditions

The system obeys the following two conditions:

C1 There is a predicateP (called aprovability

predicate) such that for any sentenceX,

the sentenceX is provable iffPX is

provable.

C2 There is a predicateR such that for any

expressionX, the sentenceRX is provable

iff PXX is provable.A Magician Looks at Godel’s Proof – p. 6

An amazing consequence

From just these two conditions, something

quite amazing follows:

TheoremG0 If the system is consistent, then

there is an undecidable sentence.

Moreover, assuming consistency, one can

exhibit an undecidable sentence using any of

the symbolsP , N , R.

A Magician Looks at Godel’s Proof – p. 7

A problem, and the solution

Problem 1: Exhibit such a sentence.

Solution: Assuming consistency, such a

sentence isRNR.

Proof: By conditionC2, for any expressionX,

the sentenceRX is provable iffPXX is

provable, and by conditionC1, PXX is

provable iffXX is provable.

A Magician Looks at Godel’s Proof – p. 8

A problem, and the solution

ThereforeRX is provable iffXX is provable.

Since this holds for every expressionX, it

holds takingX to be the expressionNR, and

soRNR is provable iffNRNR is provable.

Thus eitherRNR and its negationNRNR are

both provable, or neither is provable. By the

assumption of consistency, they are not both

provable, hence neither one is provable, and

thusRNR is undecidable.A Magician Looks at Godel’s Proof – p. 9

Part II

Omega consistency

A Magician Looks at Godel’s Proof – p. 10

Proving at stages

We next consider a system in which the

provable sentences are proved at variousstages

(we can imagine a computer proving these

sentences in a certain order). We are given

predicatesP , R, and infinitely many predicates

P1, P2, · · ·Pk, · · · [As before,PX means that

X is provable, and nowPkX means thatX is

provable at stagek].

A Magician Looks at Godel’s Proof – p. 11

Conditions

We are now given the following conditions:

G1 If X is provable at stagek, themPkX is

provable, and ifX is not provable at stage

k, thenNPkX is provable.

G2 If for somek, the sentencePkX is

provable, thenPX is provable.

G3 [Same asC2 of Part I]RX is provable iff

PXX is provable.A Magician Looks at Godel’s Proof – p. 12

Omega consistency

The system is calledomega consistentiff it is

consistent and also it is the case that there is no

sentenceX such thatPX is provable, and at

the same time all the sentences

NP1X, NP2X, · · ·NPkX, · · · are provable.

A Magician Looks at Godel’s Proof – p. 13

Gödel’s theorem

Theorem G (After Gödel)If the system is

omega consistent, then there is an undecidable

sentence.

Problem 2: Assuming omega consistency,

exhibit such a sentence (again using any of the

symbolsP , R andN ).

A Magician Looks at Godel’s Proof – p. 14

A solution (But more interesting)

The sentence is the same as before− RNR −

but the proof now is, I believe, more interesting.

We already know that conditionC2 of Part I

holds, and we will now see that omega

consistency implies that conditionC1 also

holds.

A Magician Looks at Godel’s Proof – p. 15

A solution (But more interesting)

Well, even without the assumption of omega

consistency, ifX is provable, so isPX,

because ifX is provable, it must be proved at

same stagek, and hencePkX is provable (by

G1), and thenPX is provable (byG2). Thus if

X is provable, so isPX.

A Magician Looks at Godel’s Proof – p. 16

A solution (But more interesting)

Now for the converse: SupposePX is

provable. IfX were not provable, then it would

not be provable at any stage, hence byG1, the

sentencesNP1X, NP2X, · · · , NPkX · · ·

would all be provable, and sincePX is, the

system would not be omega consistent.

A Magician Looks at Godel’s Proof – p. 17

A solution (But more interesting)

Thus if the system is omega consistent, then the

provability ofPX implies the provability ofX,

and so thenX is provable iffPX is provable,

which is conditionC1 of Part I. The rest then

follows as seen in Part I.

A Magician Looks at Godel’s Proof – p. 18

Remarks

In the systems studied by Gödel, predicates are

not considered as names of properties of

expressions, as I have done, but rather as

properties of (natural)numbers. To each

predicateH and each numbern, is associated a

sentenceH(n) to be thought of as asserting that

n has the property denoted byH.

A Magician Looks at Godel’s Proof – p. 19

Remarks

Indeed I have found it convenient to associate

to everyexpressionX and numbern, an

expressionX(n) such that ifX is a predicate,n

is a sentence.

Now, Gödel assigned to each expression a

number, subsequently called the Gödel number

of the expression. Now, to get away from

always refering to Gödel numbers, the

following schema is quite useful.A Magician Looks at Godel’s Proof – p. 20

Remarks

For anyexpressionsX andY , I defineXY to

beX(y), wherey is the Gödel number ofY .

Also I defineXY Z to beX(Y Z). Now, using

this meta notion, in the systems to which

Gödel’s argument goes through, there really are

predicatesP, R, P1, P2, · · ·Pk, · · · satisfying

conditionsG1, G2, andG3!

A Magician Looks at Godel’s Proof – p. 21