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If vi (p)= ⊤ then q = p ∧ (¬i) else if vi (p)= ⊥ then q = p ∨ (i) That is, V \{vi} cannot compute SAT problems. Therefore V is basis of SAT .  From descriptive complexity, P = F O + LF P [1, 2, 3]. This means that every P problem have at most a finite number of LFP operators in finite first-order logic model. Therefore P problem have at most a finite number of P-Complete basis. Theorem 4. Any p ∈ P have at most a finite number of P-Complete basis. Proof. To prove it by using reduction to absurdity. We assume that p ∈ P have infinite number of basis of P-Complete. These basis independent of each other and have independent LFP operators. But P = F O + LF P have at most finite number of LFP operators. Therefore we cannot describe p in finite length F O + LF P .  Theorem 5. P 6= NP Proof. Mentioned above 3, SAT have infinite P-Complete basis. But mentioned above 4, any p ∈ P have finite P-Complete basis. Therefore SAT is not any p ∈ P . 

3. From view of countable and continuum We show another proof from the view of completeness. Theorem 6. P 6= NP

Proof. Let hv,ii be a code number of vi. To assign this number after the dec- imal point, 0. hv,ii correspond to number within [0, 1], and [0, 0. hv,ii]+ S V = ∞ [0, 0. hv,ii]+ W vi correspond to Dedekind cut of P . i=0 ∞ If [0, 0. hv,ii]+ W vi also P then P become isomorphic as real number and i=0 contradict that P is countable. Therefore NP ∋ S V/∈ P and P 6= NP .  References

[1] Michael Sipser, Introduction to the Theory of Computation 2nd ed., 2005 [2] Neil Immerman, Descriptive and Computational Complexity, 1995 [3] Neil Immerman, Relational Queries Computable in Polynomial Time, 1982 [4] M. Vardi, Complexity of Relational Query Languages, 1982