## P = NP iff P != NP

The P vs. NP Problem, see: http://www.claymath.org/millennium.

The Kleene-Rosser Paradox ==> P = NP  iff  P != NP

P = NP  iff  P != NP, A Turing Machine that Diagonalizes Against Itself

Mathematics is Inconsistent: at http://www.4shared.com/office/OpzjDoKb/3_Papers.html

As the lambda calculus is equivalent to Turing machines. By direct encoding of the lambda calculus paradox, you obtain it on Turing machines.

This proof was syntactic, the same result can be reached independently via semantics.

The line of argumentation of Cook’s proof of CNF SAT being NP-complete is as follows:

“Let A be a language in NP accepted by a non-deterministic Turing machine M. Fix an input x, we will create a 3CNF formula \phi that will be satisfiable if and only if there is a proper tableau for M and x”

1. Let w be the encoding of the Kleene-Rosser paradox, thus e^-1(w)=Not e^-1(w), where e is an encoding function.
2. M accepts w.
3. SAT is NP-complete, Cook’s theorem.
4. There must exists \phi(w):\phi(w) is satisfiable.
5. Then \phi(w) is satisfiable iff M accepts w iff e^-1(w)=Not e^-1(w)
6. M accepts w is “True”.
7. Then: \phi(w) is satisfiable iff e^-1(w)=Not e^-1(w) – Logical Contradiction.
8. Then There must be no \phi(w) which is satisfiable.
9. Then SAT is (NOT) NP-complete.

### 179 responses

1. Anonymous

You’ve been peddling this for at least two years. People tell you where your errors are and you ignore them. When are you going to stop?

September 6, 2010 at 5:54 pm

2. kamouna

I would be grateful if you point me to any single error in my work and make me stop. Even top journals were not able to do this. I hope you might be able to.

Best,

Rafee Kamouna.

September 6, 2010 at 6:07 pm

• Anonymous

I read several valid posts to you already online which try to show you how you are wrong and you keep ignoring them. The top journals are probably ignoring you at this point if you are this way with them as well. You were politely rejected from STOC. You have gone through 8 versions on arXiv over the course of a year so perhaps that community has given up on reasoning with you as well. Good day.

September 7, 2010 at 12:29 am

• Miracle Healer

To quote Kamouna “I would be grateful if you point me to any single error in my work and make me stop.”

It seems that many people are able to point the many errors in Kamouna’s logic, but they are not able to make him stop, no one is.

January 22, 2011 at 1:05 am

3. kamouna

You are just dreaming. If so, you would have included them here. STOC was never able to discover sny mistake, they said not for this forum and needs more time for vetting. In addition the above papers were released in public just few days.

Have you realized the level of your dreaming, or not yet?

Best,

Rafee Kamouna.

September 7, 2010 at 1:06 am

4. Anonymous

This is brilliant work. Please do not let the conspiracy of complexity “scientists” from stopping you from promoting it further. From the “global warming” hoax, we know full well how far “scientists” are willing to go to secure research grants.

September 9, 2010 at 4:08 pm

5. Anonymous

Yes, the world of math and science must be filled with people trying to prevent simple truths from escaping into the world.

Consider the following proof that 1=2 which many professors have called wrong.

Let y be a generic particular integer.
Let x=y
Multiply each side by x to get x^2=xy
Add x^2 to each side to get 2x^2 = x^2 + xy
Subtract 2xy from both sides to get
2x^2 – 2xy = x^2 – xy
Factor this out as
2(x^2-xy) = 1(x^2-xy)
Simplify this to 2z = 1z and divide both sides by z and you get the fact that 1=2.

September 10, 2010 at 12:30 am

• Rafee Kamouna

This is a clear divide by zero fallacy. Please be serious here.

September 10, 2010 at 5:02 am

6. Soufiane

@Anonymous: I’m not sure if you are being sarcastic (you probably are) but the proof is clearly wrong, x=y which means xy=x^2 then x^2 – xy = 0 so you can’t divide by it.

September 11, 2010 at 1:28 am

7. Rafee Kamouna

You are dividing by zero. Here we are discussing P vs. NP. Please be serious.

September 13, 2010 at 4:50 pm

8. Indra

test

September 30, 2010 at 3:51 am

9. Indra

testing again you creep

December 3, 2010 at 6:08 am

10. Indra

they have rejected your journal submission I think it is a govt conspiracy?

December 3, 2010 at 6:08 am

11. Rafee Kamouna

No journal submissiom has bee rejected. The papers had been submitted to the “Theory of Computing Journal” on 7th Feb. 2010 and no decision has been made yet.

December 9, 2010 at 12:26 am

• A. S.

You should ask them to expedite the reviewing. No journal should take close to a year, even more so for path-breaking results.

January 10, 2011 at 6:59 am

• kamouna

Thank you for your advice. I have already expdited them few months ago when the papers were only 7-months old, the Editor-in-Chief reply was:

“I will understand if you withdraw your papers due to the slow pace of our review process.”

Should I expdite him again?

January 10, 2011 at 8:30 am

• A. S.

No I don’t think you should expedite, it looks like he wants you to withdraw the paper. Perhaps you could try sending it to a less rigorous but more widely read magazine like CACM Communications of the ACM as a correspondence letter. If you are saying the proof is simple and conclusive then even better. Good luck I can understand the frustration with journals they are not for rapid publishing of results.

January 14, 2011 at 5:29 am

12. A. S.

Hi, Prof. Lipton has a new blog post on P vs. NP that might interest you:

In particular, note that he says:

“At least one side’s proofs must be wrong, else we are all in trouble, since ZF would then be inconsistent.”

But I thought you had already showed ZF to be inconsistent. If I am correct, you might want to point that result to Prof. Lipton.

January 10, 2011 at 6:58 am

• kamouna

Your note is very interesting, but Prof Lipton is aware of my results years before. I have already posted 2 posts in the above link. I hope he would guide the community to the only provable conjecture: P=NP iff P!=NP.

A simple and conclusive proof.

Best,

Rafee Kamouna.

January 10, 2011 at 8:35 am

13. Anonymous

The first statement that makes no sense: “Each string w_ij can be written as the Turing machine describing it”.

January 13, 2011 at 7:16 pm

• Anonymous

To be more explicit about what needs to be fixed in the first half of the first statement:
1. The meaning of “can be written as” is unclear. If it means “is equal to”, write “is equal to” or “=”. If it means something else, specify.
2. It’s unclear what it means for a Turing Machine to “describe” a string.
3. The meaning of the triangle brackets in “” is undefined.

January 13, 2011 at 7:42 pm

• Rafee Kamouna

Please review basic and elementary computation theory where they will tell you any string can be regarded as data as well as a program, i.e. a Turing machine.

January 14, 2011 at 7:32 am

• Anonymous

It’s true that a string can be treated as a representation of a Turing Machine, but that’s not what the first statement says.

January 16, 2011 at 6:59 am

14. Rafee Kamouna

Dear A.S.,

The papers are in review with them for 12 months now. So, if they want me to withdraw them, they would have returned them to me instead. What anyone would say is that they didn’t discover any flaw in my papers yet. Otherwise, they would have been rejected.

How can I submit to CACM in 2011 papers that JACM went wrong on in 2008? JACM wrote to me:”A Turing machine cannot diagonalize against itself as the author claims he can do…”

The papers here prove this fact to JACM since 1997. Check the reviews at Luca’s blog.

Best,

Rafee Kamouna.

January 14, 2011 at 7:30 am

• A. S.

Here is the concern I have. There is an apparent contradiction. On one hand you say your proof is a simple one-step argument. But on the other hand you are saying the journal is taking more than a year to review it, still has not found any flaw in it but is also not willing to accept. How can that be possible? You should ask this to the editor, and if you don’t get a response then you should file a grievance with the journal. Maybe the community is not prepared yet to accept the possibility that ZFC is inconsistent?

January 18, 2011 at 6:26 am

• Rafee Kamouna

I agree with you, maybe the community is not prepared to accept that ZFC is inconsistent.

I suppose there is nothing like filing a grievance against any journal.

Rafee Kamouna.

January 18, 2011 at 8:17 am

• A. S.

Mr. Kamouna I don’t know, but most formal journals have a publisher which might provide some way to complain. You could at least try and see if it works.

January 19, 2011 at 6:39 am

15. Rafee Kamouna

I received the email below some time back:

==========================================

Rafee,

I’m reviewing your PNP proof for a reputed journal, and it is
fascinating. I cannot reveal my identity right now, due to obvious
anonymity reasons, but I’m a TCS researcher at a top university on the
west coast of the United States and you’ve written on my blog in the
past. I and other reviewers have almost finished verifying your proof
and its correctness. I suggest you annonce it more widely at this
point and claim the Clay Millennium Prize:
http://www.claymath.org/millennium/

I am happy for you. Congrats on your achievement and I look forward
for more groundbreaking results from you. It is truly amazing, but
also sad at the same time, given the implications of your results.

Best,
L.

January 14, 2011 at 12:12 pm

16. Rafee Kamouna

It’s true that a string can be treated as a representation of a Turing Machine, but that’s not what the first statement says.

Rafee Kamouna.

January 16, 2011 at 8:32 am

• Anonymous

The first statement of the “P=NP iff P!=NP” makes no sense:
“An instance of a paradox is satisfiable if and only if it is unsatisfiable.”
A scientific paper requires definitions and proofs (or references to proofs), not statements out of the blue.

January 16, 2011 at 10:46 am

17. Rafee Kamouna

This statement is ther meaning of any paradox.

The semantics paper walks you through Cook’s proof of SAT being NP-complete, you are at his “IFF” with the L.H.S. as Kleene-Rosser paradox and the R.H.S., i.e. his proof is contradictory. That’s one of the reasons Theory of Computimng journal is now a complete year reviewing.

The syntactic paper proves by diagonalization that there exists a language in NP iff it is not in NP. This is an encoding of the KR paradox. Keep it between you and me, a KILLER proof.

Best,

Rafee Kamouna.

January 16, 2011 at 1:11 pm

• Anonymous

My point is, you have not defined the word “paradox” in the paper. You should provide a definition or reference a definition first before you talk about properties of its “instances”.

January 16, 2011 at 11:02 pm

18. Rafee Kamouna

Rafee Kamouna.

January 17, 2011 at 8:45 am

19. Anonymous professor

I ask you to please keep these results secret. Your theoretical results, if published, would destroy years of research and cost a lot of jobs. Please refrain from pushing this, it does not help the theory community.

Best regards,
MIT Computational Complexity professor

January 18, 2011 at 9:06 am

• A. S.

I find it hard to believe it will cost a lot of jobs. People could be employed to formulate new consistent axioms, or extend the inconsistency result to other fields. Mr. Kamouna, I agree with you honesty is more important. I wonder if you are really an MIT professor. You have already lost it if you have to bring your (prestigious) affiliation in (implicit) defense of your claim.

January 19, 2011 at 6:32 am

• Rafee Kamouna

It is not only ZFC, but also any arithmetic, with or without the induction axiom. Remember the contradiction is in the set \Sigma*. You have:

e(KR) in NP iff e(KR) not in NP

and e(KR)\subseteq \Sigma*

I agree with you, there will be no loss of jobs. Even with the fact that:”All STOC/FOCS results that are correct, so are their negations”.

Best,

Rafee Kamouna.

January 19, 2011 at 6:58 am

20. Rafee Kamouna

Dear Professor,

It seems that you prefer being dishonest rather than having my results published. You assume all the theory community being so as well as accepting jobs and money founded on a dishonest enterprise. Further, you want me to be dishonest, clarify!

Will you (or others) lose their jobs?, how??, please clarify!

Best,

Rafee Kamouna.

January 18, 2011 at 1:36 pm

21. Miracle Healer

Dear Rafee Kamouna

I am a reviewer for a prestigious university in Nigeria and to me it is very clear that your proof is working correct, but I think I have noticed a ground breaking addition to your proof that will make it more definite.

I am interested in promoting your proof using my new found evidence to all prestigious journales and showing them that they are correct so that you may win the millenium award. Unfortunately, I cannot travel by air because of lack of money.

If you would send me 2000 Dollars, I would be able to flight to western countries and show you my evidence and we can prove P=NP iif P!=NP together.

When I put my info to post this comment, I have given you my email address.

January 22, 2011 at 1:10 am

• Rafee Kamouna

Dear Miraculous Healer,

Yes, absolutely nothing can stop me. There are no reviewers for universities neither in Nigeria nor elsewhere.

Best,

Rafee Kamouna.

January 22, 2011 at 5:24 am

22. Prof from Standford

This is the best paper I have read in years and believe me, I read a lot. Wish you the best.

January 25, 2011 at 1:26 am

23. Rafee Kamouna

Dear Stanford Professor,

Thanks a lot. Why don’t you reveal your identity?

January 25, 2011 at 5:10 am

• anonymouse

There is yet another consistency conspiracy brewing on Dr. Lipton’s blog post about BQP factoring. To quote two of the comments:

“one could assert that *nothing* in theoretical computer science has “really” been proved, so long as the authors have merely described their reductions in English, and not given explicit state transition diagrams for the appropriate Turing machines!”

“even well-known computer scientists doubt the mathematical correctness of Shor’s algorithm”

I urge you to do whatever in your power to publish your proof. Submit it in parallel to multiple venues which are not in conflict. Or send it to a less reputable venue. Once accepted, you can always build more credibility for the proof later incrementally. Ultimately JACM will have no choice but to accept.

Best wishes.

January 30, 2011 at 12:00 am

24. kamouna

Thanks a lot for your comments and I shall do the best.

Rafee Kamouna.

February 2, 2011 at 11:34 am

25. Rafee Kamouna

You can write a state trasition diagram for reducing a language to another. Have they done so? How can they do this for the reduction of infinite number of languages to a single language.

I introduced a paradoxical language which obstructs this reduction not only in complexity theory but in computability theory as well. SAT is (NOT) NP-complete and the halting problem is not r.e.-complete.

As somebody wrote in slashdot:”A proof that shoved aside established belief in theoretical computer science”.

February 2, 2011 at 9:30 pm

26. Berkeley Prof

Mr. Kamouna,

Please stop this nonsense immediately, shut down this blog, go into a rehabilitation facility. It is only inviting people to come to hashemite university of Jordan and break your back. Look at the Nigeria scam person for instance. You need to stop this madness from spreading NOW.

Berkeley Prof.

February 4, 2011 at 5:46 am

27. Berkeley Prof

KAMOUNA STOP THE FUCKING MADNESS YOU FUCKING TWIT!

February 4, 2011 at 7:26 am

28. Rafee Kamouna

I appreciate all the feelings of the TCS community because of the inevitable fate of P vs. NP. Honestly, I don’t want any of them to be harmed by my results at all. But this is the objectivity of mathematics and science. However, all my proofs are logical proofs and none of them are mathematical.

Russell’s paradox is a PARADOX in naive set theory; in ZFC it is a proof that a class is a proper class. We all seek proofs to derive TRUTHS. How can you derive truth from contradictory premises. Aristotle would get angry at this. ZFC is obviously defective. Godel believed in this when he proved the consistency of the continuum hypothesis and the axiom of choice.

I wish you all the best at Berkeley, MIT & Stanford and everywhere. Nobody can stop the evolution of science. It is a divine plan for the human race.

Best,

Rafee Kamouna.

February 4, 2011 at 9:54 am

29. Rafee Kamouna

Everybody knows that Lance wrote this for me (at least I’m the only one on the web who made the matter of the appeal public). Have you read the JACM reviewer (June, 2008) for my ArXiv paper (not in this blog):

“A Turing machine cannot diagonalize against itself”

A huge intellectual failure for JACM. The FLP system in that paper has this property. I gave them more than 200 papers references only to be ignored. I gave Lance another opportuniy at ToCT (July, 2009) only to repeat the same argument. So, I had to write a complete paper with the proof of a Turing machine diagonalizing against itself.

Your question is not the right one. You should have asked why they are holding a one page proof for over a year without finding a flaw in it, yet not willing to accept. I suppose you have to ask them and not me for this.

If you are sure of yourself, reveal your identity as I do in all posts over the web. If there is a trouble that NSF or others like changing TCS departments in the faculty of arts not science, we can handle this matter together. We can strike a deal together.

Best,

Rafee Kamouna.

February 7, 2011 at 4:39 pm

30. Andrew Wakefield

Dear gentlemen,

I have closely read Kamouna’s papers and I can claim that the proof is correct. It is unfortunate that the mainstream computer scientists do not intend to allow real progress to be made in this topic.

De. Andrew Wakefield.

March 29, 2011 at 7:45 pm

31. Rafee Kamouna

I think we all should consider the Godel sentence that Godel forgot:

“The algorithm that recognizes all algorithms that do not recognize themselves.”

Does it recognize itself?!?!

Some say that the above sentence is worth 7 million dollars. Others say that this is the single true statement in all mathematics. Anything else is both true and false.

Best,

Rafee Kamouna.

April 16, 2011 at 10:41 am

32. kidhuvig

Hello Mr. Kamouna,

I am not an expert in mathematics or computer science but very much interested in your work. I am a graduate of an engineering school. I think your work is truly ground breaking and would truly appreciate if you write an article for the layman explaining the implications of your findings. I strongly believe that wrong axioms can hold advances in science back for years. I also believe that advances in science depend very much on correct foundations (you can’t build a tower with out foundations).

We hear of many great mathematicians but curiously enough not roman mathematicians and the reason is not because they are less smart than other civilizations or anything like that, its rather because of their awkward number system that hindered their abilities to do any proper mathematics.

So again I would truly appreciate if you write a short article explaining your findings for the not so technical person. What does it imply if ZFC is inconsistent? What theories will be disproved? How can we remedy this problem? What will be the future of mathematics?

July 9, 2011 at 10:15 pm

• I completely agree … It is a shame how people have misunderstood Dr. Kamouna. Even today you see on all the theory blogs, researchers questioning consistency of ZFC (e.g., http://blog.computationalcomplexity.org/2011/07/why-did-112-take-russell-and-whitehead.html). It is only a matter of time when it is accepted to be inconsistent.

Regards,
L.

August 4, 2011 at 7:55 pm

• Thanks a lot for your message. Why don’t you reveal your identity?

September 28, 2011 at 7:23 am

33. Craig Feinstein

Hi Rafee,

I’m trying to understand your paper “P = NP iff P != NP, A Turing Machine that Diagonalizes Against Itself”. It looks like you have an interesting idea, so I think it is worth my time to try to understand what you are trying to say. But I think your paper needs to be written clearer, so I’d like to make some stylistic and grammar comments which might help you improve the communication of your result and also help me understand it.

Theorem 1.1 is not obvious to me, because I don’t understand what Beautiful(Mary) is. You need to explain this better. Or perhaps you might want to consider eliminating it from the paper altogether as it doesn’t seem to be important in the rest of the paper.

It took a while for me to figure out why you said (NOT) instead of NOT. You might want to explain this better.

Finally, I’m up to Definition 2.2 but can’t seem to figure this out. You are saying that Turing machine M_wi on input wj concatenates wi with wj. But then you say M_wi accepts wiwj. How is this possible if the input is wj? So I’m not sure what this Turing machine M_wi is doing.

Also, you need to explain the diagram on the next page better. I see that if I look at row w2 and column M_w1, I get w2w1. But shouldn’t I get w1w2 instead?

Also, what is L_lambda(M_wi): w1w1w2w2w3w3w4w4….. ?

I just don’t understand this.

October 19, 2011 at 2:09 am

34. Thanks Craig,

1. Beuatiful(Mary) means Mary is beautiful.

2. There is no difference between (NOT) and NOT and not.

3. Follow the diagram not the definition: the input first.

Best,

Rafee Kamouna.

October 19, 2011 at 5:41 am

35. Craig

I followed the diagram. I’m lost. Also, I don’t understand the logical steps in your proof. For instance, what is K_lambda={k_1,k_2,…}. Why does KR_lambda={k_1k_1,k_2k_2,…}={not k_1k_1, not k_2k_2,…}. How do you know that these two sets are equal? And from above, how do you obtain e(KR) in NP iff e(KR) NOT in NP?

I understand what a diagonal argument is, but I don’t understand how you are applying your diagonal argument in this proof. Also, the argument above doesn’t seem to use the concept of polynomial-time anywhere so it would have to apply to many classes of languages, not just NP. So why didn’t you state your proof claim in a more general manner?

October 19, 2011 at 4:57 pm

36. 1. You should read the paradox definition more carefully: kk_1 = not kk_1. This is a single instance of the paradox. That’s why both sets are equivaleny.

2. Either the diagonal is accepted or its complement and not both. This yields e(KR) in NP iff e(KR) not in NP. Because both the diagonal and its complement represent e(KR).

3. The result can be stated for all bigger classes that contain NP. You can definitely say e(KR) in EXP iff e(KR) not in EXP. It doesn’t matter. But here we are concerned by the P vs. NP problem.

Best,

Rafee Kamouna.

October 19, 2011 at 5:15 pm

37. Craig

But what is K_lambda={k_1,k_2,…}?

October 19, 2011 at 5:37 pm

38. The set of all lambda terms that form an instance of the Kleene-Rosser paradox: KR_lambda={kk_1,kk_1,…}, the set K_lambda the set of all lambda terms when combined with itslef the KR paradox is formed. The decision problem studied is that of KR_lambda and K_lambda is not used.

Best,

Rafee Kamouna

October 20, 2011 at 6:46 am

39. Craig,

Have you understood the proofs?

October 26, 2011 at 7:34 am

40. Craig

Rafee,

I still don’t understand the proofs. You are skipping some steps in your logic and I am unfortunately unable to fill in the details by myself.

October 27, 2011 at 12:09 am

41. Which steps did I skip?

October 27, 2011 at 5:24 am

42. Craig

I’m just looking at the above argument. You said, “Note that both the diagonal and its complement can easily encode the Kleene-Rosser paradox as:

KR={kk_1,kk_2,kk_3,…}={Not kk_1, Not kk_2, Not kk_3…}

Use the encoding e(kk_i)=w_iw_i and e(Not kk_i) = \bar w_i\bar w_i”

I don’t see this.

October 28, 2011 at 2:25 am

43. What is your problem with this straightforward encoding?

October 28, 2011 at 3:06 am

44. Craig

The problem is I have no clue what you are talking about. I’ve tried to figure it out, but it’s not so straightforward to me. I read both of your papers too. It looks to me like you might have something interesting, but I have given up trying to figure out what, unless you provide me with more details.

October 28, 2011 at 4:20 am

• The semantics paper assumes a paradox as an input to a Turing machine. The Cook-Levin theorem would derive a SAT from it. This is a semantic contradiction. Is that clear to you?

The proof by diagonalization establishes a missing element in an infinite sequence. You argue the element is missing because if it is not missing you derive a contradiction. The complement of the diagonal is the missing element. Is that clear to you?

Best,

Rafee Kamouna.

October 28, 2011 at 6:17 am

45. Craig

I cannot understand the semantics argument either. I’d like to, but more detail needs to be provided for me to understand it. I’m guessing that a lot of readers have the same trouble too. I’m not a professional computer scientist or mathematician, but I do have graduate education in mathematics and enjoy it as a hobby.

October 28, 2011 at 1:38 pm

46. The Cook-Levin theorem states for all input w (for a non-deterministic Turing machine), w is accepted iff a SAT formula is derived.

So, this is an iff statement whose L.H.S. is a paradox and the R.H.S. is SAT

No surprise if you know the ToC journal is still reviewing it for 21 months now.

Is this contradiction clear for you?

October 28, 2011 at 2:14 pm

47. Craig

I don’t see how the LHS is a paradox. This is the part I don’t understand.

October 28, 2011 at 6:21 pm

48. M accepts w iff e^-1(w)= Not e^-1(w).

w_i = e(k_ik_i), i.e. e is an encoding of a paradox.

What is not clear.

October 28, 2011 at 7:07 pm

49. Craig

How do you know this?

October 30, 2011 at 2:05 am

50. The paper assumes a Paradox Recognizer machine.

October 30, 2011 at 4:59 am

51. Does this make sense for you?

November 1, 2011 at 1:26 pm

52. Craig

No, it still doesn’t make sense. As I said before, there are steps missing that I don’t understand. I would recommend you rewriting your paper to explain things more. I’m positive that I’m not the only one who doesn’t understand what you are saying.

November 1, 2011 at 2:48 pm

53. If you were right then the journal should have rejected my papers instead of reviewing them for 21 months by now. However, you should have mentioned steps i and j and tell me where is the missing step k in between.

November 2, 2011 at 7:27 am

54. Craig

Rafee, I’m not saying that what you have is wrong. I’m just saying that I don’t understand it. There is a big difference.

November 2, 2011 at 1:58 pm

55. I appreciate. Thanks a lot. But you said there are missing steps, so try to demark it. Where are the missing steps, Rewrite the proof as: 1,2,3,4,5,6, then tell me there is a missing step between 4 and 5.

November 2, 2011 at 2:14 pm

56. Craig

It’s this part that needs to be explained more:

“Note that both the diagonal and its complement can easily encode the Kleene-Rosser paradox as:

KR={kk_1,kk_2,kk_3,…}={Not kk_1, Not kk_2, Not kk_3…}

Use the encoding e(kk_i)=w_iw_i and e(Not kk_i) = \bar w_i\bar w_i”

November 2, 2011 at 4:53 pm

57. The diagonalization happens on strings not on formulae. Formulae are encoded into string given the encoding function. This encoding function encodes both sides of the paradox into complementary strings. That’s to say iff kk_1 is encoded into w_1w_1, then Not kk_1 is encoded into \bar w_1\bar w_1, where w_i_w_j are binary strings.

November 3, 2011 at 8:16 am

58. Craig

How does the diagonalization work?

November 4, 2011 at 4:20 pm

59. Consider the Turing machine that appens itself to its input, you get the diagonalization table shown. M_w1 on input w_1, you get the entry w_1w_1, and so on for all the table entries. Now the L.H.S. of the KR paradox encodes the diagonal, while the R.H.S. encodes the complement of the diagonal. Both are in NP, but under the condition if either is accepted the other is rejected while both encode the same logical formula.

Thus e(KR) in NP iff e(KR) not in NP

If this is not clear, ask me for more details.

November 4, 2011 at 4:34 pm

60. Craig

“but under the condition if either is accepted the other is rejected while both encode the same logical formula.”

Why is this?

November 7, 2011 at 12:50 am

61. It is a paradox before being encoded. The encoding maintains the paradoxical situation. Any encoding that tries to conceal the paradoxical situation would be the wrong encoding. However, my proof is that there exists an encoding such that (this condition) and not for all encodings…

November 7, 2011 at 4:18 am

62. It seems that you gave up.

November 9, 2011 at 11:26 am

63. MrAnderson

Your proof is correct. However, it is vacuously correct, because the hypothesis is always false. Hence, the proof is worthless.

A implies B is true if and only if (A is false OR B is true). This is elementary logic taught in the first year of college, or even to young children.

This situation is the case where A is false. We are uncertain whether B is true or not.

Do you understand? I will give you another example.

If I am the Lord of the Universe, then I am talking dog.

This is a true statement, because I am NOT the Lord of the Universe.

February 21, 2012 at 6:02 am

64. kamouna

You are correct. You are referring to Cook’s proof of SAT being NP-complete. Hence, SAT is not NP-complete. Then, my argument is correct. Yet, another indenpdent diagonalization argument confirms my results that you did not review.

Best,

Rafee Kamouna.

February 21, 2012 at 6:27 am

65. MrAnderson

I have now thoroughly reviewed the Kleene-Rosser paradox. It seems you have not.

The Kleene-Rosser paradox only applies to Church and Curry theories from before 1934. It was then discovered by Curry in 1941 the original cause of the paradox (see here: http://plato.stanford.edu/entries/paradoxes-contemporary-logic/#ParBetMetTypFreFou193, section 5.3).

Read the first few sections of here as well: http://en.wikipedia.org/wiki/Lambda_calculus . You are applying the Kleene-Rosser paradox to the FIXED versions of these theories. The fixed versions do not have this paradox occur.

I would suggest you follow up with what I am saying, find all the relevant papers, and read them for yourself, so that you may finally understand.

Alex Anderson

February 23, 2012 at 9:28 pm

66. kamouna

So, you admit the failure of your previous argument and now you have another one where you confirm that the Kleene-Rosser paradox MAY NOT be input to a Turing machine.

I feel sorry for you being so bankrupt.

Best,

Rafee Kamouna.

February 24, 2012 at 7:05 am

67. MrAnderson

I do not understand your response. You are not refuting my argument, but merely stating it?

It seems that you did not read the links. Welp, I tried. You have waste many man-hours of work with your insistence on your incorrect application of the Kleene-Rosser paradox. I suggest you email those publishers and ask them if you have applied it (the Kleene-Rosser paradox) correctly. They will of course say the same thing.

Good day, I will not be speaking with you again. It seems you don’t understand enough English to understand that you are wrong.

February 24, 2012 at 7:11 am

68. kamouna

You are saying that the Kleene-Rosser paradox MAY NOT be input to a Turing machine. So stupid even for an under-graduate, not somebody dealing with P vs. NP.

You are disgustingly stupid, not just stupid.

February 24, 2012 at 7:24 am

69. MrAnderson

Damn, you got me to respond again.

>> “You are saying that the Kleene-Rosser paradox MAY NOT be input to a Turing machine.”

Correct, it cannot be input into a Turing machine. Nothing you “claim” will make this anything else. If you want to be believed, get OTHER PEOPLE to verify this. Which of course, they will respond in the negative.

>> “So stupid even for an under-graduate, not somebody dealing with P vs. NP.
You are disgustingly stupid, not just stupid.”

This is not an argument. You are trying to deflect my argument by attacking my qualifications, which while lacking, do not affect the validity of my argument. You are wrong. Perhaps you even know it, but have been living this lie for too long to give it up.

February 24, 2012 at 7:40 am

70. kamouna

You fail and fail. My work has been reviewed by top journals now for the fifth year. Its current status is that is being reviewed by the Theory of Computing Journal for more than 2 years now. Do you think they are so stupid to tell me that Kleene-Rosser paradox cannot be input to a Turing machine. You STUPID!!!

February 24, 2012 at 8:04 am

71. MrAnderson

Well, you still don’t get it. It is clear that you have no regular contact with any contemporaries.

Again – “My work has been reviewed by top journals now for the fifth year. Its current status is that is being reviewed by the Theory of Computing Journal for more than 2 years now.”
Here you give an argument “Because this has occurred for a long time, I must be right.”
That is not a proper argument. You don’t understand the normal flow of arguments, it seems.

Here, I, an undergraduate, shall educate you. Arguments need to be based on LOGIC and they need to be based off of EVIDENCE. Doing something for a long time indicates experience. So you have experience…waiting. And perhaps a lot of experience being incompetent.

I’ll quote you from the top of this: “As the lambda calculus is equivalent to Turing machines. By direct encoding of the lambda calculus paradox, you obtain it on Turing machines.”

This is your argument, your line of logic. The second part is where you are wrong : “By direct encoding of the lambda calculus paradox…”. Because the paradox doesn’t exist anymore, not in the lambda calculus that has been around since the 1940s.

Here is EVIDENCE. I am labeling it so you understand what it is, because you don’t seem to have any: “A. Church, “An unsolvable problem of elementary number theory”, American Journal of Mathematics, Volume 58, No. 2. (April, 1936), pp. 345-363.” This is where Church publishes MODERN lambda calculus, where the Kleene-Rosser paradox DOES NOT APPLY. It is clear that all you know of the Kleene-Rosser paradox is a result of reading only what you want to read about it. “It said lambda calculus in there, so it must apply to modern lambda calculus” A very wrong line of reasoning.

Of course, this will not be enough for you. You probably won’t even read it. You’ll just type up another response that has absolutely nothing to do with the beauty of mathematics, and think that it is enough. You can’t prove something by saying “It is true” enough times.

February 24, 2012 at 2:06 pm

72. Do you think that this is new information to me. Of course I know that there are variations of the lambda calculus with the Kr paradox. My work establishes the Kleene-Rosser paradox on Turing machines. I proved by diagonalization that there is a language L in NP iff L not in NP. This applies to Turing machines regardless of your favourite version of the lambda calculus with or without a paradox. The paradox does exist on Turing machines proving P vs NP contradictory and dazzling reviewers for years.

Best,

Rafee Kamouna

February 24, 2012 at 2:32 pm

73. I mean variations without the KR paradox, in the above post.

February 24, 2012 at 2:37 pm

74. MrAnderson

Thank you for a response in which you provide evidence and reasoning against my claims.

Here is a site that I think would interest you: http://www.win.tue.nl/~gwoegi/P-versus-NP.htm
At the top, the site author refers to a journal which “guarantees a timely and competent handling of such submissions. ” So I suggest that you send them your submission, since wherever you have currently sent it does not want to reply?

Anyway, good luck, I do not wish to spend any more time on this.

February 24, 2012 at 11:15 pm

75. They replied several times since submission. However, they are absorbing the huge surprise. Of course, I know the page you are citing and that journal.

best,

Rafee Kamouna.

February 25, 2012 at 9:24 am

76. MrAnderson

Back again, because I’m from topcoder.com. I think the reason that people have trouble believing your argument relies on your use of what you call paradoxes. Could you explain them to me more clearly? Try to use analogies and examples when possible, because the high level theorems and abstractions are part of what is obscuring your proof.

May 20, 2012 at 4:26 am

77. kamouna

A paradox is a statment that leads to contradiction. In my papers I use a famous, formal, well-defined and easy-to-understand paradox which is the Kleene-Rosser paradox explained both formally and informally in pure English. What do you not understand about the Kleene-Rosser paradox?

Best,

Rafee Kamouna.

May 20, 2012 at 8:17 am

78. MrAnderson

From what I learned in my low-level classes, contradictions are statements that are always false. Doesn’t this mean that you cannot use them to prove things in the realm of mathematics, because in mathematics things need to be true? If not, then what am I misunderstanding about contradictions?

MrAnderson

May 20, 2012 at 4:49 pm

79. kamouna

A contradiction is always false, but a paradox is always both true and false, like:”This sentence is false”, which is known as the liar’s paradox. I proved that the Kleene-Rosser paradox of the lambda calculus has a form on Turing machines. Another proof that a paradox would be a counter-example to NP-completeness. Hence, SAT is both NP-complete and not NP-complete.

May 20, 2012 at 5:08 pm

80. MrAnderson

I think of true and false as two sides of a coin. Only one side can be face up at a time.

With regards to paradoxes, can’t one argue in a similar way to say a paradox is neither true nor false?

Also, paradoxes tend to show that certain formal parts of mathematics are poorly defined. How come you haven’t tried redefining the part of mathematics which is causing this paradox?

MrAnderson

May 20, 2012 at 10:46 pm

81. kamouna

One cannot say a paradox is neither true nor false, simply because the sentence:”I’m lying”, if it is true, then it is false. And if it is false, then it is true. The part of mathematics affected by paradoxes cannot be re-defined because it is the role of language (mathematics is also a language). A self-referential, self negating sentence leads to a paradoxical situation.

May 21, 2012 at 7:50 am

82. MrAnderson

I don’t understand what you mean by “it is the role of language”.

I am working with true and false as a function. A statement can be input into the function, and it outputs true or false. However, since it is a function, a single input can only be mapped to one unique output. Since paradoxes cannot be mapped to one unique output, this means that they are not in the domain of the truth function. So it doesn’t make any sense to think about whether a paradox is true or false, or to make use of it being true or false.

MrAnderson

May 21, 2012 at 7:57 am

83. kamouna

Why don’t regard it as a relation since this will solve your problem of understanding. The role of language is the type of sentence I told you about.

May 21, 2012 at 8:06 am

84. MrAnderson

But by extending the truth function to a relation, how are you not taking nearly all currently existing theorems of mathematics out of context? They were built on the assumption of using a truth function, not a truth relation. If by extending the function to a relation we break mathematics, then it is probably a good idea to not do that.

May 21, 2012 at 8:17 am

85. kamouna

May 21, 2012 at 8:21 am

86. MrAnderson

Right. So I think what your paper is really saying is not that “P == NP iff P != NP” but that you’ve discovered a paradox in current mathematics. For from any paradox, you can prove anything, and further, all paradoxes are equivalent.

And with what you said about how paradoxes are a role of language (which I still don’t understand), could you instead properly, and formally define what a language is? I can’t tell what you’re talking about. And I can’t tell when you switch from one language to another.

Quoting from your recent discussion with a journal – ” This def-

inition is rather confused (as is the entire narrative of both papers), but

let us give the author the benet of the doubt that by \equal” the author

means -equivalent, by \valid,” he means well-formed as a lambda expres-

sion (i.e., taking a syntactic meaning of the world \valid,” rather than the

logical meaning of “true in a specied context”). Then he indeed defnes a

language; it is a simple exercise to show that this language is r.e.-complete”

It is clear that people far more intelligent than I am are also confused by your ambiguity on certain terms.

May 21, 2012 at 8:43 am

87. kamouna

I discovered a version of the Kleene-Rosser paradox on Turing machines. This referee is confused but knows how to sort out his confusion. The set KR is a proper subset of LAMBDA, the set of all lambda expressions which maybe called a language in a sense. The encoding of KR, i.e. e(KR) is a subset of Sigma*, the set of all finite strings or subset of {0,1}*, the set of all binary string, the definition of a language in complexity theory.

May 21, 2012 at 9:16 am

88. MrAnderson

Now again we are back at the impasse of function versus relation.

Quoting from the first line of the post at the top of the page – “Let w be the encoding of the Kleene-Rosser paradox, thus e^-1(w)=Not e^-1(w), where e is an encoding function.”

Since e is an encoding function, it maps inputs to unique outputs. I am assuming that here you mean Not(0) = 1 and Not(1) = 0.

Then there is an error in this line, since the first bits will not match. Simplistically, you are saying “1 = 0” and then going on to use that to prove things.

May 21, 2012 at 9:34 am

89. kamouna

e^-1(w) = kk, not o and not 1.

May 21, 2012 at 10:38 am

90. MrAnderson

Yes, but all encodings must be able to be reduced to 0 and 1, since they are representations in the language of Sigma* as you stated above (Quote: “e(KR) is a subset of Sigma*, the set of all finite strings or subset of {0,1}* “)

So my point still applies, you are still essentially saying “I’ve proved 1 = 0, hence “

May 21, 2012 at 3:56 pm

91. kamouna

No. Encodings are never reduced to neither 0 nor 1. An encoding function takes an input formula, outputs a string (e.g. 110000110011, 11101010111,001100111001). 1 and 0 are strings of length 1 only. Formulae are encoded into strings of arbitrary length.

May 21, 2012 at 4:18 pm

92. MrAnderson

Sorry, you misunderstood me. I am saying that by saying e^-1(w) = NOT e^-1(w), when you look at their encodings for strings, the very first bit will be a mismatch, and so the statement is false.

Quoting myself: “Yes, but all encodings must be able to be reduced to 0 and 1, since they are representations in the language of Sigma* as you stated above (Quote: “e(KR) is a subset of Sigma*, the set of all finite strings or subset of {0,1}* “)

So my point still applies, you are still essentially saying “I’ve proved 1 = 0, hence [your proof of P == NP iff P != NP] ”

I meant to say, ‘all encodings must be able to be reduced to 0s and 1s’.

Then we simply observe the first bit by itself. Suppose the first bit of e^-1(w) is B. Then the first bit of NOT e^-1(w) is surely NOT(B). But then you say they are equal, which means you are saying that 1 = 0 is a true statement.

May 22, 2012 at 3:29 am

93. kamouna

e^-1(w)=kk, not, it is a formula and not a string, you are very much confused.

May 22, 2012 at 7:19 am

94. MrAnderson

Yes, you are right, I was confused. Could you reference me to your proof about the Kleene-Rosser paradox on Turing machines? I can’t seem to find it.

May 23, 2012 at 3:57 am

95. kamouna

In the paper:”A Turing Machine that Diagonalizes Against Itself”

May 23, 2012 at 6:52 am

96. MrAnderson

Hello, and thank you for telling me of this paper.

I am already confused by theorem 1.1

The language L is a finite Language, and hence, L is in P. And anything in P can be reduced to SAT in polynomial time.

Could you explain my mistake?

MrAnderson

May 23, 2012 at 3:25 pm

97. kamouna

It can be reduced, yes. But what is the meaning of the reduction when a paradox recognizer machine constitutes a counter-example to NP-completeness, check the other paper and the refutation of Cook’s proof. Check the diagonalization argument in this new paper.

May 23, 2012 at 3:51 pm

98. MrAnderson

I’m going to keep reading the paper, but it isn’t beneficial to your argument if you start off proving something obviously false (or where you are assuming later results to have the proof by “obvious”)

May 23, 2012 at 4:07 pm

99. kamouna

Previous results tell you that it is obvious. The first paper proves this. To consider the reduction, you have to consider the decision problem which is a counter-example to NP-completeness. Hence, the reduction is not possible.

May 23, 2012 at 4:22 pm

100. MrAnderson

I feel like you are being vague again when you say ‘first paper’ and ‘previous results’. I thought the one you referenced to me was the first one.

Also, I am wondering, did you say somewhere that there is a Turing Machine which decides the language PARADOX? If so, I am wondering what the inputs to that Turing Machine are.

Lastly, I am still reading the paper, but I don’t have enough information, since everything seems to be ‘obvious’ or ‘clear’. Could you tell me where to find the other papers that are necessary to understand the one mentioned a few posts above?

– MrAnderson

May 23, 2012 at 11:40 pm

101. MrAnderson

Continuum Hypothesis = “False”
Axiom of Choice = “False”

ZFC is Inconsistent”

You seem in many other papers to rely on the result from Theorem 1. But in the prelude to Theorem 1, you use a few undefined terms. While these terms are well-known, who is to know if you are assuming some small difference in each one, without you to tell us exactly what you mean by them.

Some terms which should have been defined 4+ years ago:
From Definition 1:::
“Let L be a language over an alphabet Σ containing at least one
constant symbol”
What kind of alphabet are you talking about? From what I have learned, alphabets consist of only constant symbols. You seem to have
taken the union of several types of elements and called it an alphabet, but I am not certain, because you don’t define alphabet.

“The set UL of all ground terms constructed from functions and
constants in L is called the Herbrand universe of L”
What is a ground term to you? Maybe you exclude a specific type or an edge case that I normally include.
How are they constructed? You should be explicit. Spell out exactly how the recursive definition is applied.

“The set BL of all ground
atomic formulas over L is called the Herbrand base of L.”
Again, what is a ground atomic formula to you?

I have several more questions, but I’ll just leave you with these few for now.

MrAnderson

P.S. Please respond to all the questions I have raised, not just one.

May 24, 2012 at 3:48 am

102. kamouna

Please concentrate on the two lambda-calculus papers here on this blog only. They have the P vs. NP results.

May 24, 2012 at 6:46 am

103. MrAnderson

Why are you so focused on P vs. NP? What you’ve proved is a contradiction. So using that why don’t you also go on to prove the rest of the millenium problems? No need to wait for the journal to finish reviewing this paper.

May 24, 2012 at 7:40 am

104. MrAnderson

Quoting one of the papers: “An instance of a paradox is satisfiable if and only if it is unsatisfiable.”

Letting P be an instance of a paradox, do we agree that this formalizes to P if and only if NOT(P) ?
Or if not, then how does it formalize into notation?

May 24, 2012 at 7:48 am

105. kamouna

A single contradiction is sufficient to collapse all mathematics and not just the millennium problems. This happenened in P vs. NP where it has been proved that SAT is both NP-complete and not NP-complete, and there is L in NP iff L is not in NP. You can formalize the above as you wish.

May 24, 2012 at 3:30 pm

106. MrAnderson

Right, so I don’t understand why you are pushing so hard your result on P vs. NP, when the result is really that you’ve found a contradiction in mathematics. You should be talking about that instead, and you also should be devising a way to fix the contradiction, since no one wants math to collapse.

May 24, 2012 at 9:16 pm

107. kamouna

As you might have seen, the mathematics community is not prepared to publish an inconsistency result. It seems to be against their interests.

May 24, 2012 at 9:44 pm

108. MrAnderson

No, they did publish an inconsistency result back when Kleene-Rosser first discovered the paradox. There must be another reason why. You should find the crux of the problem, and then there will be no doubt.

May 25, 2012 at 12:07 am

109. kamouna

Mathematicians never use lambda calculus as foundations because of this paradox.

May 25, 2012 at 12:51 am

110. MrAnderson

Now I am lost again. So you proved P vs. NP, but only in a model of mathematics that no one uses?

May 25, 2012 at 2:03 am

111. I proved the paradox on Turing machine, then ZFC is inconsistent.

May 25, 2012 at 3:06 am

112. MrAnderson

I understand. But your proof has a lot of hand-waving. By hand-waving, I mean there are a lot of places where you simply assert things are true without proving them, or you are vague in your use of words.

For example, you assert that paradoxes exist in modern day lambda calculus, but you fail to give an example within modern day lambda calculus.

Similarly you assert that it is possible to encode paradoxes, and to have a Turing machine which can identify a paradox (which is an ambiguous definition in and of itself).

May 25, 2012 at 3:33 am

113. There is nothing called modern day lambda calculus. Check the 4 referee reports, none of them denied the Kleene-Rosser paradox, check again. In paper 1, I assumed a paradox recognizer Turing machine, one referee denied the existence of such a machine. In paper 3, I proved that paradox recognition is:
1. Decidable.
2. Undecidable.
3. Decidable iff undecidable.

May 25, 2012 at 3:51 am

114. MrAnderson

I have read the discussion between you and referee 1. He makes several valid points which you ignore or hand-wave away as insignificant, when in actuality, they are very significant issues.

Firstly, while this referee does not explicitly deny the existence of the KR paradox, he says “assuming we take the K-R paradox as
a paradox in the foundations of mathematics”. But this is not the case, as he later explains, “The Kleene-Rosser paradox shows that if you try to use the lambda calculus as a foundation for logic (as opposed to as a model of computation), an inconsistency arises”.

Here, you are ignoring the fact that he is saying that when we take lambda calculus as a model of computation, the KR paradox does not apply.

I shall post some more examples of your hand-waving and disingenuous methods in a few minutes.

May 25, 2012 at 4:03 am

115. MrAnderson

Next, you respond with “The referee proved that the Paradox Recognizer Turing machine does not exist based that the language I defined is r.e.-complete, hence undecidable. Any kid student can build such a machine which is a program to recognize paradoxes”

You have stated many times that ‘any kid student can build such a machine which is a program to recognize paradoxes.’ However, if this were true, it would obviously come up in programming classes as an example or a challenge question, but it never comes up in any classes anywhere.

Furthermore, you yourself have failed to build such a program. This is not a situation where you can give pseudo-code. You need to give a legitimate implementation in a standard Turing-complete language, because the nuances you see as obvious are not obvious to almost everyone who looks at your papers.

May 25, 2012 at 4:07 am

116. MrAnderson

When you next respond the referees argument, you say “Find out about my advertised conferences and hundreds of comments on the web.” This statement is a worthless and meaningless contribution to the discussion. Comments and conferences do not say anything about the validity or invalidity of an argument. They are simply a means of discussion. Even having a few other people agree with you is equally insufficient for “correctness”, just as having everyone against you is insufficient for “incorrectness”.

May 25, 2012 at 4:10 am

117. The claim that the KR paradox does not exist in the model of computation is just a claim with no proof. As I told you above, it is possible to prove that paradox recognition is undecidable. I proved it and disproved it as above. I has been considered undecidable since 1930s, that’s why no programming class has it, but you can attempt and tell me why it is impossible for you to write such a program?

May 25, 2012 at 4:13 am

118. MrAnderson

Is it possible to write a program to solve the halting problem? For me, it is equally impossible to write a program to recognize paradoxes.

However, you seem to have missed out on a few of the things I said, and also you failed to learn a proper lesson in mathematics. The burden of proof lies on the claimant. I’ll reiterate: you say it is simple for any kid programmer to write such a program, but instead of writing the program, you deflect with statements like “you try for yourself and tell me yourself”

May 25, 2012 at 4:19 am

119. The mention of the conference is because the referees say that I’m not aware that I’m claiming mathematics is inconsistent. An entire report is based on this. Check their level! I’m not aware of this, heheheh!

May 25, 2012 at 4:20 am

120. MrAnderson

Before you go off and attempt to create a program in a standard Turing-complete language which can recognize paradoxes, can you first explain what it means for this program to recognize a paradox? It truly isn’t well defined, and yet you continue to parade it around as the foundation of all your proofs when you never define it formally.

But don’t forget to write that program in a standard Turing-complete programming language which can recognize paradoxes. You seem to have a tendency to just ignore things. Perhaps you are just forgetful, but others would say that you are ignoring the arguments for which you do not have a reasonable counterargument.

May 25, 2012 at 4:40 am

121. Can you write a program to recognize a lambda calculus expression, a C expression, a C++ expression or a Pascal one?

May 25, 2012 at 4:40 am

122. MrAnderson

Yes, yes I can.

But this isn’t about me. Stop trying to redirect the true issue that I’m pressing here, which I mentioned in my last post.

The issue here is that you have claimed many times that this program is easy to write, yet you have never done it and instead continue to tell others to do it.

May 25, 2012 at 4:46 am

123. When you bring the lambda expression recognizer, I shall tell you how to use to recognize the KR-paradox, becuase you will never know yourself.

May 25, 2012 at 4:48 am

124. MrAnderson

Do realize that simply recognizing the KR-paradox is trivial. That is not what you propose with your Paradox Turing Machine. That is a machine which recognizes ALL paradoxes.

And again, you deflect. My abilities mean nothing, because I am not the one trying to prove something here. You are the one with the paper that can’t seem to get accepted no matter which journal you send it to.

May 25, 2012 at 4:51 am

125. What is the difference between ALL paradoxes and KR paradox? Bring another paradox in formal programming context.

May 25, 2012 at 6:21 am

126. Only one journal reviewed my lambda calculus papers!

May 25, 2012 at 6:52 am

127. MrAnderson

Mathematically, all paradoxes are equivalent, this is obvious. What you ignore is that they are not all written the same way, which a TM will need to recognize all such ways to write a paradox.

And again you deflect. I am the Devil’s Advocate. But perhaps you don’t know what that is.

May 25, 2012 at 4:42 pm

128. P vs. NP is concerned with a programming paradox only and my papers are all concerned with the KR-paradox.

May 25, 2012 at 4:52 pm

129. MrAnderson

A Turing machine capable of recognizing only the KR-paradox is not sufficient for you to prove that the KR-paradox applies to Turing machines.

Furthermore, you again refuse to create the program. Yet again you change the conversation.

May 25, 2012 at 9:56 pm

130. I proved A Turing machine can recognize the KR-paradox iff it does not recognize it. A C expression recognizer would do the job.

May 26, 2012 at 6:31 am

131. MrAnderson

If you create the program you describe, the C expression recognizer would not halt…

Furthermore, you base all of your proofs on the definition of a paradox, which is not well defined. All you have done is create a new term for contradiction, and define it to be true.

May 26, 2012 at 6:40 am

132. MrAnderson

See http://mathworld.wolfram.com/Well-Defined.html for what well-defined means. A basic term.

May 26, 2012 at 6:41 am

133. A program that takes an input string and checks whether it is two substrings where the substring is self-applied to itself MUST terminate because the input is finite!

May 26, 2012 at 7:14 am

134. MrAnderson

The input may be finite, but you are ignoring recursion. The recursion is what will go on forever.

May 26, 2012 at 7:20 am

135. There is no recursion. You can do it by iteration on the input string length.

May 26, 2012 at 10:36 am

136. MrAnderson

Look.

The real issue here is in fact that your definition of paradox is not well-defined. Just from the way you defined paradox, the result is that you can prove everything whether it is provable or not.

May 26, 2012 at 2:00 pm

137. A paradox is: kk=not kk. This is a well-defined paradox as defined by Kleene and Rosser.

May 26, 2012 at 3:29 pm

138. MrAnderson

May 26, 2012 at 3:32 pm

139. My definition is that of Kleene and Rosser and none of the reviewers claimed this.

May 26, 2012 at 3:53 pm

140. MrAnderson

Why when you are asked a question do you not answer directly? You are always saying “if you do this, or look here, you will see”

May 26, 2012 at 4:04 pm

141. I never do this.

May 26, 2012 at 4:09 pm

142. MrAnderson

In Kleene and Rosser’s paper, they do not define the word paradox. How are you then using their definition in a formal context?

May 26, 2012 at 4:14 pm

143. So, Kleene and Rosser is not formal as you insist on being stupid.

May 26, 2012 at 4:15 pm

144. MrAnderson

Since they don’t define the word paradox, I’ll define it for you. http://mathworld.wolfram.com/Paradox.html
That is, it is something which APPEARS false, but is actually NOT false, but true.

This is the definition which Kleene and Rosser use, because this is the standard English definition.

May 26, 2012 at 4:19 pm

145. In logic not English.

May 26, 2012 at 4:25 pm

146. nit professor

Oh, I thought Kamouna was legit, I take it back all I have said 😦

March 12, 2013 at 1:10 am

• It is not clear what do you want to say.

Rafee Kamouna.

March 12, 2013 at 4:37 am

147. Osama Salama

What people call the ‘Liar’s Paradox’ is in fact a logical fallacy which I call the logical fallacy of ‘Incalculability due to Infinite Recursion’ . So lets go about logically evaluating the statement of the famous so called Liar’s Paradox:

(1) This sentence is false.

Now logically evaluating:

(2) [This] sentence is false

Evaluating [This] we see that it refers to the whole statement in clause (1). So we do a substitution and hence get:

(3) [This sentence is false] sentence is false

Now we go about evaluating the statement in clause (3)

(4) [{This} sentence is false] sentence is false

Evaluating {This} we see again that it refers to the whole statement in clause (1). So we do a substitution and hence get:

(5) [{This sentence is false} sentence is false] sentence is false.

And so on…

And so to reach a final logical evaluation of the original statement in clause (1) we have to follow a never ending series of substitutions and hence will never reach a final logical evaluation of the original statement. Hence it is not True nor False, rather it is incalculable due to infinite recursion.

So it is not a paradox but rather a fallacy, and that is why it is logically incorrect to use it to prove theorems.

February 9, 2014 at 8:27 pm

148. kamouna

I didn’t use the liar’s paradox. i used the KR paradox. It was never used as a definition from which I reason to prove a theorem. Instead, assuming a paradox over the lambda-calculus, a paradox was derived over Turing machines. Review latest paper.

Best,

Rafee Kamouna.

February 10, 2014 at 2:28 am

149. Rishav Kundu

Sir, has any conclusion been reached? I was riveted reading through this blog although I don’t understand anything of it. Your manner of responding to criticism has impressed me 😮 I’d like to know if there have been any developments.

May 16, 2015 at 11:09 am

150. kamouna

Thanks for your interest. Still, my results are in peer-review.

Best,

Rafee Kamouna.

May 16, 2015 at 11:17 am

• Rishav Kundu

That is sad to hear. I wish that people would either accept your results (as they should) or directly refute them giving clear reasons. This ambiguity is helping no one.

May 16, 2015 at 11:28 am

151. kamouna

I have no options, let’s wait and see.

Rafee Kamouna.

May 16, 2015 at 11:42 am

• Rishav Kundu

I wish you all the best. Someday I hope I’ll be able to understand your work. ^_^ Aspiring computer scientist.

May 16, 2015 at 11:45 am

152. kamouna

Thanks a lot.

Rafee Kamouna.

May 16, 2015 at 11:47 am

153. Jeffrey Shallit

Waiting for you to pay up, Rafee.

August 21, 2015 at 12:25 am