Proof2: ℙ≠ℕℙ is released

This ℙ≠ℕℙ proof is the shortest one and easy to understand

https://sourceforge.net/projects/cscall/files/MisFiles/PNP-proof-en.txt/download
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| Proof2: ℙ≠ℕℙ |
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STM::= Turing machine represented by string {0,1}* string
Num(x)::= Function that maps {0,1}* to natural numbers

 Let s∈STM, s:STM->{0,1}. s is a function that computes a Turing machine
 decision function in Ptime and is defined as follows:

   s(f)=1 if ∃c∈<=Num(f), f(c)=1.

 Therefore, Problem(s)∈ℕℙ (the problem solved by s is ℕℙ problem)
 The proof of this assertion is quite straightforward and trivial, so it is
 omitted. Doubts about whether certain natural numbers are legal Turing
 Machines are not a problem, because the definition of ℕℙ does not consider
 (and can handle) this situation.

 Assume that Problem(s)∈ℙ holds, then by definition
   s(s)=1 if ∃c∈<=Num(s), s(c)=1.
 But when c=Num(s), there will be a self-referential situation where
   s(s)=1 if s(s)=1 (similar to the undecidable situation of the halting
 problem). Therefore, the assumption that ℙ=ℕℙ does not hold, that is, ℙ≠ℕℙ.