Case Study of the Proof of Cook's theorem - Interpretation of A(w)
CCase Study of the Proof of
Cooks theorem - Interpretation of A ( w ) Yu LI
MIS, Universit´e de Picardie Jules Verne, 33 rue Saint-Leu, 80090 Amiens, France
Abstract
Cooks theorem is commonly expressed such as any polynomial time-verifiable problem can be reducedto the
SAT problem. The proof of
Cooks theorem consists in constructing a propositional formula A ( w )to simulate a computation of TM , and such A ( w ) is claimed to be CNF to represent a polynomial time-verifiable problem w . In this paper, we investigate A ( w ) through a very simple example and show that, A ( w ) has just an appearance of CNF , but not a true logical form. This case study suggests that thereexists the begging the question in Cooks theorem . Cooks theorem [1] is now expressed as any polynomial time-verifiable problem can be reduced tothe
SAT (SATisfiability) problem. The proof of
Cooks theorem consists in simulating a compu-tation of
TM (Turing Machine) by constructing a propositional formula A ( w ) that is claimed tobe CNF (Conjonctive Normal Form) to represent the polynomial time-verifiable problem [1].In this paper we investigate whether this A ( w ) is a true logical form to represent a problemthrough a very simple example. A polynomial time-verifiable problem refers to a problem w for which there exists a TuringMachine M to verify a certificat u in polynomial time, that is, check whether u is a solution to w .Let us study a very simple polynomial time-verifiable problem :Given a propositional formula w = ¬ x for which there exists a Turing Machine M to verifywhether a truth value u of x is a solution to w . Email address: [email protected] (Yu LI). a r X i v : . [ c s . CC ] A p r he transition function of M can be represented as follows: q → N q q → N q q → R q Y q → R q N where N means that the tape head does not move, and R means that the tape head moves toright; q Y refers to the state where M stops and indicates that u is a solution to w , and q N refersto the state where M stops and indicates that u is not a solution to w . A computation of M consists of a sequence of configurations: C (1) , C (2) , ..., C ( T ), where T = Q ( | w | ) and Q ( n ) is a polynomial function. A configuration C ( t ) represents the situation of M at time t where M is in a state, with some symbols on its tape, with its head scanning a square,and the next configuration is determined by the transition function of M .Fig.1 and Fig. 2 illustrate two computations of M on inputs : x = 0 and x = 1.Fig.
1: The computation on input x = 0. A ( w ) According to the proof of
Cooks theorem [1][2], the formula A ( w ) is built by simulating a com-putation of M , such as A ( w ) = B ∧ C ∧ D ∧ E ∧ F ∧ G ∧ H ∧ I . A ( w ) is claimed to represent aproblem w .We construct A ( w ) for the above example. 2ig.
2: The computation on input x = 1. The machine M possesses: • { q , q , q = q Y , q = q N } , where q is the initial state, and q , q are two finalstates. • { σ = b, σ = 0 , σ = 1 } , where σ is the blank symbol. • { s = 1 , s = 2 } . • • n is the input size, n = 2; p ( n ) is a polynomial function of n , and p (2) = 3. • t = 1 , t = 2 , t = 3) et 2 steps to verify a certificat u of w , where t = 1 correspondsto the time for the initial state of the machine. Three types of proposition symbols to represent a configuration of M : • P is,t for 1 ≤ i ≤
3, 1 ≤ s ≤
2, 1 ≤ t ≤ P is,t is true iff at step t the square number s contains the symbol σ i . • Q it for 1 ≤ i ≤
4, 1 ≤ t ≤ Q it is true iff at step t the machine is in state q i . • S s,t for 1 ≤ s ≤
2, 1 ≤ t ≤ t the tape head scans square number s .3 .3 Propositions E = E ∧ E ∧ E , where E t represents the truth values of P is,t , Q it and S s,t at time t : • E = Q ∧ S , ∧ P , ∧ P , ( x = 0( σ )); E = Q ∧ S , ∧ P , ∧ P , ( x = 1( σ )) • E and E are determined by the transition function of M B = B ∧ B ∧ B , where B t asserts that at time t one and only one square is scanned : • B = ( S , ∨ S , ) ∧ ( ¬ S , ∨ ¬ S , ) • B = ( S , ∨ S , ) ∧ ( ¬ S , ∨ ¬ S , ) • B = ( S , ∨ S , ) ∧ ( ¬ S , ∨ ¬ S , )3. C = C ∧ C ∧ C , where C t asserts that at time t there is one and only one symbol at eachsquare. C t is the conjunction of all the C i,t . C = C , ∧ C , : • C , = ( P , ∨ P , ∨ P , ) ∧ ( ¬ P , ∨ ¬ P , ) ∧ ( ¬ P , ∨ ¬ P , ) ∧ ( ¬ P , ∨ ¬ P , ) • C , = ( P , ∨ P , ∨ P , ) ∧ ( ¬ P , ∨ ¬ P , ) ∧ ( ¬ P , ∨ ¬ P , ) ∧ ( ¬ P , ∨ ¬ P , ) C = C , ∧ C , : • C , = ( P , ∨ P , ∨ P , ) ∧ ( ¬ P , ∨ ¬ P , ) ∧ ( ¬ P , ∨ ¬ P , ) ∧ ( ¬ P , ∨ ¬ P , ) • C , = ( P , ∨ P , ∨ P , ) ∧ ( ¬ P , ∨ ¬ P , ) ∧ ( ¬ P , ∨ ¬ P , ) ∧ ( ¬ P , ∨ ¬ P , ) C = C , ∧ C , : • C , = ( P , ∨ P , ∨ P , ) ∧ ( ¬ P , ∨ ¬ P , ) ∧ ( ¬ P , ∨ ¬ P , ) ∧ ( ¬ P , ∨ ¬ P , ) • C , = ( P , ∨ P , ∨ P , ) ∧ ( ¬ P , ∨ ¬ P , ) ∧ ( ¬ P , ∨ ¬ P , ) ∧ ( ¬ P , ∨ ¬ P , )4. D = D ∧ D ∧ D , where D t asserts that at time t the machine is in one and only one state. • D = ( Q ∨ Q ∨ Q ∨ Q ) ∧ ( ¬ Q ∨ ¬ Q ) ∧ ( ¬ Q ∨ ¬ Q ) ∧ ( ¬ Q ∨ ¬ Q ) ∧ ( ¬ Q ∨ ¬ Q ) ∧ ( ¬ Q ∨ ¬ Q ) ∧ ( ¬ Q ∨ ¬ Q ) • D = ( Q ∨ Q ∨ Q ∨ Q ) ∧ ( ¬ Q ∨ ¬ Q ) ∧ ( ¬ Q ∨ ¬ Q ) ∧ ( ¬ Q ∨ ¬ Q ) ∧ ( ¬ Q ∨ ¬ Q ) ∧ ( ¬ Q ∨ ¬ Q ) ∧ ( ¬ Q ∨ ¬ Q ) • D = ( Q ∨ Q ∨ Q ∨ Q ) ∧ ( ¬ Q ∨ ¬ Q ) ∧ ( ¬ Q ∨ ¬ Q ) ∧ ( ¬ Q ∨ ¬ Q ) ∧ ( ¬ Q ∨ ¬ Q ) ∧ ( ¬ Q ∨ ¬ Q ) ∧ ( ¬ Q ∨ ¬ Q ) 4. F , G , and H assert that for each time t the values of the P is,t , Q it and S s,t are updatedproperly. F = F ∧ F , where F t is the conjunction over all i and j of F ti,j , where F ti,j asserts that at time t the machine is in state q i scanning symbol σ j , then at time t + 1 σ j is changed into σ l , where σ l is the symbol given by the transition function for M . F = F , ∧ F , : • F , = ( ¬ Q ∨ ¬ S , ∨ ¬ P , ∨ P , ), with the rule ( q , → , N, q ) • F , = ( ¬ Q ∨ ¬ S , ∨ ¬ P , ∨ P , ), with the rule ( q , → , N, q ) F = F , ∧ F , : • F , = ( ¬ Q ∨ ¬ S , ∨ ¬ P , ∨ P , ), with the rule ( q , → , R, q N ) • F , = ( ¬ Q ∨ ¬ S , ∨ ¬ P , ∨ P , ), with the rule ( q , → , R, q Y ) G = G ∧ G , where G t is the conjunction over all i and j of G ti,j , where G ti,j asserts that attime t the machine is in state q i scanning symbol σ j , then at time t + 1 the machine is in state q k , where q k is the state given by the transition function for M . G = G , ∧ G , : • G , = ( ¬ Q ∨ ¬ S , ∨ ¬ P , ∨ Q ), with the rule ( q , → , N, q ) • G , = ( ¬ Q ∨ ¬ S , ∨ ¬ P , ∨ Q ), with the rule ( q , → , N, q ) G = G , ∧ G , : • G , = ( ¬ Q ∨ ¬ S , ∨ ¬ P , ∨ Q ), with the rule ( q , → , R, q N ) • G , = ( ¬ Q ∨ ¬ S , ∨ ¬ P , ∨ Q ), with the rule ( q , → , R, q Y ) H = H ∧ H , where H t is the conjunction over all i and j of G ti,j , where H ti,j asserts that attime t the machine is in state q i scanning symbol σ j , then at time t + 1 the tape head movesaccording to the transition function for M . H = H , ∧ H , : • H , = ( ¬ Q ∨ ¬ S , ∨ ¬ P , ∨ S , ), with the rule ( q , → , N, q ) • H , = ( ¬ Q ∨ ¬ S , ∨ ¬ P , ∨ S , ), with the rule ( q , → , N, q ) H = H , ∧ H , : • H , = ( ¬ Q ∨ ¬ S , ∨ ¬ P , ∨ S , ), with the rule ( q , → , R, q N ) • H , = ( ¬ Q ∨ ¬ S , ∨ ¬ P , ∨ S , ), with the rule ( q , → , R, q Y )6. I = ( Q ∨ Q ) ∧ ( Q ∨ ¬ Q ) ∧ ( ¬ Q ∨ Q ), asserts that the machine reaches the state q y or q N at time 3.Finally, A ( w ) = B ∧ C ∧ D ∧ E ∧ F ∧ G ∧ H ∧ I .5 Conjunctive form of A ( w ) We develop A ( w ) as a computation of M for x = 0 as input (see Fig. 1) in order to clarify thereal sense of A ( w ).Let us define the configuration and the transition of configurations of M : C ( t ) : the truth values of P is,t , Q it , S s,t and their constraints. C ( t ) → C ( t + 1) : C ( t ) is changed to C ( t + 1) according to the transition function of M .1. At t = 1, C (1) = E ∧ B ∧ C ∧ D : • E = Q ∧ S , ∧ P , ∧ P , , representing the initial configuration where M is in q , the tapehead scans the square of number 1, and a string 0 b is on the tape. • B = ( S , ∨ S , ) ∧ ( ¬ S , ∨ ¬ S , ). • C = C , ∧ C , : – C , = ( P , ∨ P , ∨ P , ) ∧ ( ¬ P , ∨ ¬ P , ) ∧ ( ¬ P , ∨ ¬ P , ) ∧ ( ¬ P , ∨ ¬ P , ) – C , = ( P , ∨ P , ∨ P , ) ∧ ( ¬ P , ∨ ¬ P , ) ∧ ( ¬ P , ∨ ¬ P , ) ∧ ( ¬ P , ∨ ¬ P , ) • D = ( Q ∨ Q ∨ Q ∨ Q ) ∧ ( ¬ Q ∨ ¬ Q ) ∧ ( ¬ Q ∨ ¬ Q ) ∧ ( ¬ Q ∨ ¬ Q ) ∧ ( ¬ Q ∨ ¬ Q ) ∧ ( ¬ Q ∨ ¬ Q ) ∧ ( ¬ Q ∨ ¬ Q )2. At t = 2, C (2) = E ∧ B ∧ C ∧ D is obtained from C (1) ∧ ( C (1) → C (2)). C (1) → C (2) is represented by F , G and H at t = 1 :- F , = ( ¬ Q ∨ ¬ S , ∨ ¬ P , ∨ P , ), with the rule ( q , → , N, q )- G , = ( ¬ Q ∨ ¬ S , ∨ ¬ P , ∨ Q ), with the rule ( q , → , N, q )- H , = ( ¬ Q ∨ ¬ S , ∨ ¬ P , ∨ S , ), with the rule ( q , → , N, q )6 E = Q ∧ S , ∧ P , ∧ P , , with Q = 1, S , = 1, P , = 1, P , = 1, and other propositionsymbols concerning t = 2 are assigned with 0. • B = ( S , ∨ S , ) ∧ ( ¬ S , ∨ ¬ S , ) • C = C , ∧ C , : – C , = ( P , ∨ P , ∨ P , ) ∧ ( ¬ P , ∨ ¬ P , ) ∧ ( ¬ P , ∨ ¬ P , ) ∧ ( ¬ P , ∨ ¬ P , ) – C , = ( P , ∨ P , ∨ P , ) ∧ ( ¬ P , ∨ ¬ P , ) ∧ ( ¬ P , ∨ ¬ P , ) ∧ ( ¬ P , ∨ ¬ P , ) • D = ( Q ∨ Q ∨ Q ∨ Q ) ∧ ( ¬ Q ∨ ¬ Q ) ∧ ( ¬ Q ∨ ¬ Q ) ∧ ( ¬ Q ∨ ¬ Q ) ∧ ( ¬ Q ∨ ¬ Q ) ∧ ( ¬ Q ∨ ¬ Q ) ∧ ( ¬ Q ∨ ¬ Q )3. At t = 3, C (3) = E ∧ B ∧ C ∧ D is obtained from C (2) ∧ ( C (2) → C (3)). C (2) → C (3) is represented by F , G and H at t = 2 : F , = ( ¬ Q ∨ ¬ S , ∨ ¬ P , ∨ P , ), with the rule ( q , → , R, q Y ) G , = ( ¬ Q ∨ ¬ S , ∨ ¬ P , ∨ Q ), with the rule ( q , → , R, q Y ) H , = ( ¬ Q ∨ ¬ S , ∨ ¬ P , ∨ S , ), with the rule ( q , → , R, q Y ) • E = Q ∧ S , ∧ P , ∧ P , , with Q = 1, S , = 1, P , = 1, P , = 1 , and other propositionsymbols concerning t = 3 are assigned with 0. • B = ( S , ∨ S , ) ∧ ( ¬ S , ∨ ¬ S , ) • C = C , ∧ C , : – C , = ( P , ∨ P , ∨ P , ) ∧ ( ¬ P , ∨ ¬ P , ) ∧ ( ¬ P , ∨ ¬ P , ) ∧ ( ¬ P , ∨ ¬ P , ) – C , = ( P , ∨ P , ∨ P , ) ∧ ( ¬ P , ∨ ¬ P , ) ∧ ( ¬ P , ∨ ¬ P , ) ∧ ( ¬ P , ∨ ¬ P , ) • D = ( Q ∨ Q ∨ Q ∨ Q ) ∧ ( ¬ Q ∨ ¬ Q ) ∧ ( ¬ Q ∨ ¬ Q ) ∧ ( ¬ Q ∨ ¬ Q ) ∧ ( ¬ Q ∨ ¬ Q ) ∧ ( ¬ Q ∨ ¬ Q ) ∧ ( ¬ Q ∨ ¬ Q )Therefore, the computation of M for x = 0 as input can be represented as : C (1) ∧ ( C (1) → C (2)) ∧ ( C (2) → C (3))= C (1) ∧ C (2) ∧ C (3)= ( E ∧ B ∧ C ∧ D ) ∧ ( E ∧ B ∧ C ∧ D ) ∧ ( E ∧ B ∧ C ∧ D )= E ∧ B ∧ C ∧ D = A ( w )It can be seen that A ( w ) is just the conjonction of all configurations of M to simulate a concretcomputation of M for verifying a certificat u of w . Given an input u ( x = 0 or x = 1 inthis example), whether M accepts it or not, A ( w ) is always true. Obviously, A ( w ) has just anappearance of conjunctive form, but not a true logical form.7 Conclusion
In fact, a true
CNF formula is implied in the transition function of M corresponding to F , G , H as well as C ( t ) → C ( t + 1), however the transition function of M is based on the expressiblelogical structure of a problem.Therefore, it is not that any polynomial time-verifiable problem can be reduced to the SAT prob-lem, but any polynomial time-verifiable problem itself asserts that such problem is representableby a CNF formula. In other words, there exists the begging the question in Cooks theorem . Acknowledgements
Thanks to Mr Chumin LI for his suggestion to use this simple example to study A ( w ). References [1] Stephen Cook, The complexity of theorem proving procedures. Proceedings of the Third AnnualACM Symposium on Theory of Computing. p151-158 (1971)[2] Garey Michael R., David S. Johnson, Computers and Intractability: A Guide to the Theory ofNP-Completeness. W. H. Freeman and company (1979)[1] Stephen Cook, The complexity of theorem proving procedures. Proceedings of the Third AnnualACM Symposium on Theory of Computing. p151-158 (1971)[2] Garey Michael R., David S. Johnson, Computers and Intractability: A Guide to the Theory ofNP-Completeness. W. H. Freeman and company (1979)