theoremizing paradoxes

7/27/2019 Theoremizing Paradoxes

1/15

ROSSER PHENOMENON: Applications in Recursion Theory IPM Weekly Seminars on Math. Logic

ROSSER PHENOMENON:

Applications in Recursion Theory

Saeed Salehi

University of Tabriz

http://SaeedSalehi.ir/

April 25, 2013

Logic Group, School of Mathematics, IPM

Saeed Salehi http://SaeedSalehi.ir/

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2/15


Rossers Trickhttp://en.wikipedia.org/wiki/Rossers_trick

In mathematical logic, Rossers trick is a method for proving Godels

incompleteness theorems without the assumption that the theory

being considered is consistent (...). This method was introduced by

J. Barkley Rosser in 1936, as an improvement of Godels original

proof of the incompleteness theorems that was published in 1931.

While Godels original proof uses a sentence that says (informally)This sentence is not provable, Rossers trick uses a formula that says

If this sentence is provable, there is a shorter proof of its negation.


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http://en.wikipedia.org/wiki/Rosser's_trickhttp://en.wikipedia.org/wiki/Rosser's_trick


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J. Barkley Rosser

http://en.wikipedia.org/wiki/J._Barkley_Rosser

John Barkley Rosser Sr. (December 6, 1907 September 5, 1989) was an

American logician, a student of Alonzo Church, and known for his part in

the ChurchRosser theorem, in lambda calculus. He also developed what is

now called the Rosser sieve, in number theory. He was later director of the

Army Mathematics Research Center at the University ofWisconsinMadison. Rosser wrote mathematical textbooks as well.

In 1936, he proved Rossers trick, a stronger version of Godels first

incompleteness theorem which shows that the requirement for

consistency may be weakened to consistency. Rather than using the liar

paradox sentence equivalent to I am not provable, he used a sentence that

stated For every proof of me, there is a shorter proof of my negation.

In prime number theory, he proved Rossers theorem. The KleeneRosser

paradox showed that the original lambda calculus was inconsistent.


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http://en.wikipedia.org/wiki/J._Barkley_Rosserhttp://en.wikipedia.org/wiki/J._Barkley_Rosser


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The Trick

For every proof of me, there is a shorter proof of my negation.

Number x is a Rosser proof of the formula y in theory T:

ProofRT(x, y) ProofT(x, y) zx ProofT(z, neg(y))

The Rosser Sentence: x ProofRT(x, )

x [ProofT(x, ) zx ProofT(z, )]

Rosser Consistency:

ConR(T) x [ProofT(x, ) zx ProofT(z, )]but then T ConR(T)! Contradictory with Godels 2nd Thm.


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5/15


Rossers ProofIf T is 1Complete and Consistent then T ,

T x [ProofT(x, ) zx ProofT(z, )]

IfT then for some n N, ProofT(n, ) and so by

1completeness, T ProofT(n, ). By the definition of,T zn ProofT(z, ). So, for some mn, ProofT(m, )whence T ; contradiction!

IfT then T x[ProofT(x, ) zxProofT(z, )],

also T ProofT(n, ) for some n N.So xn (T nx xn). But then T

in ProofT(i, ),

and so T ; contradiction!


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R A li i i R i Th IPM W kl S i M th L i


6/15


Reduction Principle

Every Two Sets (RE sets) Can Be Separated

Let A = {u | x (u, x)} and B = {u | x (u, x)}. Then for

A = {u | x[(u, x) zx(u, z)]}

B = {u | x[(u, x) z < x(u, z)]}

we have A A, B B, A B = , A B = A B.

For RE sets We = {u | z T(e,u,z)} let

Wi Wj = {u | z[T(i,u,z) yzT(j, u, y)]},Wi Wj = {u | z[T(i,u,z) y < zT(j, u, y)]}.Whence, A = Wi Wj and B

= Wj Wi will do.

IfWi Wj = N Then Wi Wj and Wj Wi are Recursive!


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ROSSER PHENOMENON A li ti i R i Th IPM Weekl Seminars on Math Logic


7/15


Can All RE Sets Be Separated Recursively?Recursively Inseparable Sets

A and B are said to be recursively inseparablewhen there

exists no recursive set C such that A C and B C = .

Example: A = {u | u(u) = 0} and B = {u | u(u) = 0}.If A C and B C = and C = e, then

e C = e(e) = 1 = e B = e C

e C = e(e) = 0 = e A = e C

So, e C e C; contradiction!


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ROSSER PHENOMENON: Applications in Recursion Theory IPM Weekly Seminars on Math Logic


8/15


RE and Undecidable SetsRE and Recursively Inseparable Sets

Resembles K = {u | u(u) } = {u | u Wu}.If K = W then K K.

Which resembles DF = {a A | a F(a)} for F : A P(A).If DF = F() then DF DF.

Similar to Russels Paradox:

R = {x | x x} implies R R R R.

More Examples of Rec. Insep. RE Sets:Ai = {u | u(u) = i} and Aj = {u | u(u) = j} for i = j.Ai = {u | u(0) = i} and Aj = {u | u(0) = j} for i = j.


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9/15

ROSSER PHENOMENON: Applications in Recursion Theory IPM Weekly Seminars on Math Logic


10/15


Many More Examples of Recursively Inseparable Sets

Rices Theorem

Any Two Non-Empty Disjoint Index Sets A and B

Are Recursively Inseparable.

Proof: For a A and b B, if A C B and C = e, then let

n(z) =

a(z) if e(n) = 0

b(z) if e(n) = 0

otherwise (if e(n) )

n C e(n) = 1 n = b n B n Cn C e(n) = 0 n = a n A n C

So, n C n C, contradiction!


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Connections with Incompleteness

Godels First Theorem Existence of a 1 1 Set

For a 1 1 Set A and a 1 Sound Theory T,1 {a | T a A} A 1.

So, there exists some A such that T A!

This argument shows the incompleteness of any 1completeand consistent theory.

Another example of a 1 1 set:

{ | T } for an RE and sound theory Tlike ZFC or PA or In or Iopen or Q ...

Every 1 Theory Can Be Completed To a 1 Theory.


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12/15

pp y y g


Incompleteness Theorem of GodelRosser

Rossers Trick shows that the sets of

TDerivable Sentences A = { | T }and TRefutable Sentences B = { | T }

are Recursively Inseparable.H.B. ENDERTON, A Mathematical Introduction to Logic, 2nd ed. Acad. Pres. 2001. ex. 1 p. 245

If C 1 satisfies A C B andC = {u | y (y, u)}, C = {u | y (y, u)} then let

T y[(y, ) zy (z, )]

One Can Show C C, Contradiction!


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13/15

pp y y g


GodelRosser Theorem = Recursively Inseparable RE SetsA = { | T } C = {u | y (y, u)}

B = { | T } C = {u | y (y, u)}T y[(y, ) zy (z, )]

(1) C

n N

(n, )

m N

(m, )

T y[= n](y, ) zy(z, ) T

B

C

(2)

C

n N (n, )

m N(m, )

T yn((y, )) [(y, ) zy (z, )]

T yn(zy (z, ))

T yn[(y, ) zy (z, )]

T

A

C

.

Thus, C C; Contradiction!


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14/15


Recursively Inseparable RE Sets = GodelRosser Theorem

D. VAN DALEN, Logic and Structure, 5nd ed. Springer (1980)2013. Th. 8.7.10 (Undec. PA)

If A and B are REC. INSEP. RE & T is 1complete such thatT x B x A or T x A x B

Then A {u | T u A} {u | T u B} = TB 1and B {u | T u B} {u | T u A} = TA 1

satisfy A TA = = B TB.If TA TB = N then TB TA separates A and B recursively!

A

TB T

A( 1

) =TA

TB(

1

) B

So, TA TB = N. Put k TA TB.T k A, and T k B so T k A.

Whence, T k A and T k A; so T is incomplete.


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15/15

Thank You!

Thanks To

The Participants

for Listening and for Your Patience!

and thanks to The Organizers.


irROSSER PHENOMENON: Applications in Recursion Theory IPM Weekly Seminars on Math. Logic

theoremizing paradoxes

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