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Rafee Kamouna said...

Proof of: “P=NP iff P!=NP” – Syntactic.

The proof is by diagonalization as follows:

NP={L_1,L_2,L_3,…}

L_1={w_11,w_12,w_13,…}

L_2={w_21,w_22,w_23,…}

L_3={w_31,w_32,w_33,…}

Each string w_ij can be written as the Turing machine describing it which can be written as two substrings, the first its input and the second is the computation configurations:

= w_iw_j

Hence we have,

L_1={w_1w_1,w_1w_2,w_1w_3,…}

L_2={w_2w_1,w_2w_2,w_2w_3,…}

L_3={w_3w_1,w_3w_2,w_3w_3,…}

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

You obtain: e(KR) in NP iff e(KR) NOT in NP, of course e(KR) irreducible to SAT iff it is reducible to SAT.

Hence, P=NP iff P!=NP.

This should not be surprising as the lambda calculus is equivalent to Turing machines. By direct encoding of the lambda calculus paradox, you obtain it on Turing machines.

Sep 3, 2010, 4:31:01 PM

Posted to Intelligent questions about the alleged P NE NP proof