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Comment by scscsc | Hacker News Reader

Your comment shows you don't know much about logic.

No it doesn't! Justify your claim. I just stated the definition of the P?=NP problem. You can find it in any good book on complexity theory. For example:

http://www.amazon.com/Introduction-Theory-Computation-Michae...

I was not refering to your definition of P and NP (which I've just reread and I think it's not good; you want "accept", not "recognize"). I was refering to the fact that you do not make the difference between "true" and "provable", which can be found in any standard introductory book on logic; or, if you prefer a novel, Hofstadter's Goedel, Escher, Bach: an Eternal Golden Braid.

First of all- the definition is not mine. It is what is in the books. The terms "recognize" and "recognizable" is defined in the context of complexity theory. So are the terms "decide" and "decidable". "Accept" is not. So when you say you don't think some definition is good please read the exact definition first. There are many serious and great mathematicians and computer scientist involved in that theory and it is irrespectable to them to comment it without being familiar with it.

Second - What I meant to say in my first comment is that the way P?=NP problem is defined it is provable. The fact that it is provable can be proved easily and I will leave this as exercise to you.

Dude, you have no idea what you're talking about. So please stop with the bullshit about great mathematicians and computer scientists when this has nothing to do with them -- only with you -- you cannot accept being wrong. Do your homework first.

Recognizable is not the same thing as acceptable. For example, some undecidable languages are recognizable by Turing machines.

You can leave it as an exercise for anyone you wish, but your assertion (existance of a simple metaproof that there is [a proof of either P=NP or P!=NP]) is false. I base my assertion about your assertion on the fact that about 40 years of research has not produced such a metaproof.

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