8.86/3.13 YES 9.34/3.25 proof of /export/starexec/sandbox/benchmark/theBenchmark.pl 9.34/3.25 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 9.34/3.25 9.34/3.25 9.34/3.25 Left Termination of the query pattern 9.34/3.25 9.34/3.25 p(g) 9.34/3.25 9.34/3.25 w.r.t. the given Prolog program could successfully be proven: 9.34/3.25 9.34/3.25 (0) Prolog 9.34/3.25 (1) PrologToPiTRSProof [SOUND, 0 ms] 9.34/3.25 (2) PiTRS 9.34/3.25 (3) DependencyPairsProof [EQUIVALENT, 1 ms] 9.34/3.25 (4) PiDP 9.34/3.25 (5) DependencyGraphProof [EQUIVALENT, 0 ms] 9.34/3.25 (6) PiDP 9.34/3.25 (7) PiDPToQDPProof [SOUND, 0 ms] 9.34/3.25 (8) QDP 9.34/3.25 (9) QDPOrderProof [EQUIVALENT, 61 ms] 9.34/3.25 (10) QDP 9.34/3.25 (11) DependencyGraphProof [EQUIVALENT, 0 ms] 9.34/3.25 (12) QDP 9.34/3.25 (13) UsableRulesProof [EQUIVALENT, 0 ms] 9.34/3.25 (14) QDP 9.34/3.25 (15) QReductionProof [EQUIVALENT, 0 ms] 9.34/3.25 (16) QDP 9.34/3.25 (17) UsableRulesReductionPairsProof [EQUIVALENT, 6 ms] 9.34/3.25 (18) QDP 9.34/3.25 (19) PisEmptyProof [EQUIVALENT, 0 ms] 9.34/3.25 (20) YES 9.34/3.25 9.34/3.25 9.34/3.25 ---------------------------------------- 9.34/3.25 9.34/3.25 (0) 9.34/3.25 Obligation: 9.34/3.25 Clauses: 9.34/3.25 9.34/3.25 p(cons(X, nil)). 9.34/3.25 p(cons(s(s(X)), cons(Y, Xs))) :- ','(p(cons(X, cons(Y, Xs))), p(cons(s(s(s(s(Y)))), Xs))). 9.34/3.25 p(cons(0, Xs)) :- p(Xs). 9.34/3.25 9.34/3.25 9.34/3.25 Query: p(g) 9.34/3.25 ---------------------------------------- 9.34/3.25 9.34/3.25 (1) PrologToPiTRSProof (SOUND) 9.34/3.25 We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes: 9.34/3.25 9.34/3.25 p_in_1: (b) 9.34/3.25 9.34/3.25 Transforming Prolog into the following Term Rewriting System: 9.34/3.25 9.34/3.25 Pi-finite rewrite system: 9.34/3.25 The TRS R consists of the following rules: 9.34/3.25 9.34/3.25 p_in_g(cons(X, nil)) -> p_out_g(cons(X, nil)) 9.34/3.25 p_in_g(cons(s(s(X)), cons(Y, Xs))) -> U1_g(X, Y, Xs, p_in_g(cons(X, cons(Y, Xs)))) 9.34/3.25 p_in_g(cons(0, Xs)) -> U3_g(Xs, p_in_g(Xs)) 9.34/3.25 U3_g(Xs, p_out_g(Xs)) -> p_out_g(cons(0, Xs)) 9.34/3.25 U1_g(X, Y, Xs, p_out_g(cons(X, cons(Y, Xs)))) -> U2_g(X, Y, Xs, p_in_g(cons(s(s(s(s(Y)))), Xs))) 9.34/3.25 U2_g(X, Y, Xs, p_out_g(cons(s(s(s(s(Y)))), Xs))) -> p_out_g(cons(s(s(X)), cons(Y, Xs))) 9.34/3.25 9.34/3.25 The argument filtering Pi contains the following mapping: 9.34/3.25 p_in_g(x1) = p_in_g(x1) 9.34/3.25 9.34/3.25 cons(x1, x2) = cons(x1, x2) 9.34/3.25 9.34/3.25 nil = nil 9.34/3.25 9.34/3.25 p_out_g(x1) = p_out_g 9.34/3.25 9.34/3.25 s(x1) = s(x1) 9.34/3.25 9.34/3.25 U1_g(x1, x2, x3, x4) = U1_g(x2, x3, x4) 9.34/3.25 9.34/3.25 0 = 0 9.34/3.25 9.34/3.25 U3_g(x1, x2) = U3_g(x2) 9.34/3.25 9.34/3.25 U2_g(x1, x2, x3, x4) = U2_g(x4) 9.34/3.25 9.34/3.25 9.34/3.25 9.34/3.25 9.34/3.25 9.34/3.25 Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog 9.34/3.25 9.34/3.25 9.34/3.25 9.34/3.25 ---------------------------------------- 9.34/3.25 9.34/3.25 (2) 9.34/3.25 Obligation: 9.34/3.25 Pi-finite rewrite system: 9.34/3.25 The TRS R consists of the following rules: 9.34/3.25 9.34/3.25 p_in_g(cons(X, nil)) -> p_out_g(cons(X, nil)) 9.34/3.25 p_in_g(cons(s(s(X)), cons(Y, Xs))) -> U1_g(X, Y, Xs, p_in_g(cons(X, cons(Y, Xs)))) 9.34/3.25 p_in_g(cons(0, Xs)) -> U3_g(Xs, p_in_g(Xs)) 9.34/3.25 U3_g(Xs, p_out_g(Xs)) -> p_out_g(cons(0, Xs)) 9.34/3.25 U1_g(X, Y, Xs, p_out_g(cons(X, cons(Y, Xs)))) -> U2_g(X, Y, Xs, p_in_g(cons(s(s(s(s(Y)))), Xs))) 9.34/3.25 U2_g(X, Y, Xs, p_out_g(cons(s(s(s(s(Y)))), Xs))) -> p_out_g(cons(s(s(X)), cons(Y, Xs))) 9.34/3.25 9.34/3.25 The argument filtering Pi contains the following mapping: 9.34/3.25 p_in_g(x1) = p_in_g(x1) 9.34/3.25 9.34/3.25 cons(x1, x2) = cons(x1, x2) 9.34/3.25 9.34/3.25 nil = nil 9.34/3.25 9.34/3.25 p_out_g(x1) = p_out_g 9.34/3.25 9.34/3.25 s(x1) = s(x1) 9.34/3.25 9.34/3.25 U1_g(x1, x2, x3, x4) = U1_g(x2, x3, x4) 9.34/3.25 9.34/3.25 0 = 0 9.34/3.25 9.34/3.25 U3_g(x1, x2) = U3_g(x2) 9.34/3.25 9.34/3.25 U2_g(x1, x2, x3, x4) = U2_g(x4) 9.34/3.25 9.34/3.25 9.34/3.25 9.34/3.25 ---------------------------------------- 9.34/3.25 9.34/3.25 (3) DependencyPairsProof (EQUIVALENT) 9.34/3.25 Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem: 9.34/3.25 Pi DP problem: 9.34/3.25 The TRS P consists of the following rules: 9.34/3.25 9.34/3.25 P_IN_G(cons(s(s(X)), cons(Y, Xs))) -> U1_G(X, Y, Xs, p_in_g(cons(X, cons(Y, Xs)))) 9.34/3.25 P_IN_G(cons(s(s(X)), cons(Y, Xs))) -> P_IN_G(cons(X, cons(Y, Xs))) 9.34/3.25 P_IN_G(cons(0, Xs)) -> U3_G(Xs, p_in_g(Xs)) 9.34/3.25 P_IN_G(cons(0, Xs)) -> P_IN_G(Xs) 9.34/3.25 U1_G(X, Y, Xs, p_out_g(cons(X, cons(Y, Xs)))) -> U2_G(X, Y, Xs, p_in_g(cons(s(s(s(s(Y)))), Xs))) 9.34/3.25 U1_G(X, Y, Xs, p_out_g(cons(X, cons(Y, Xs)))) -> P_IN_G(cons(s(s(s(s(Y)))), Xs)) 9.34/3.25 9.34/3.25 The TRS R consists of the following rules: 9.34/3.25 9.34/3.25 p_in_g(cons(X, nil)) -> p_out_g(cons(X, nil)) 9.34/3.25 p_in_g(cons(s(s(X)), cons(Y, Xs))) -> U1_g(X, Y, Xs, p_in_g(cons(X, cons(Y, Xs)))) 9.34/3.25 p_in_g(cons(0, Xs)) -> U3_g(Xs, p_in_g(Xs)) 9.34/3.25 U3_g(Xs, p_out_g(Xs)) -> p_out_g(cons(0, Xs)) 9.34/3.25 U1_g(X, Y, Xs, p_out_g(cons(X, cons(Y, Xs)))) -> U2_g(X, Y, Xs, p_in_g(cons(s(s(s(s(Y)))), Xs))) 9.34/3.25 U2_g(X, Y, Xs, p_out_g(cons(s(s(s(s(Y)))), Xs))) -> p_out_g(cons(s(s(X)), cons(Y, Xs))) 9.34/3.25 9.34/3.25 The argument filtering Pi contains the following mapping: 9.34/3.25 p_in_g(x1) = p_in_g(x1) 9.34/3.25 9.34/3.25 cons(x1, x2) = cons(x1, x2) 9.34/3.25 9.34/3.25 nil = nil 9.34/3.25 9.34/3.25 p_out_g(x1) = p_out_g 9.34/3.25 9.34/3.25 s(x1) = s(x1) 9.34/3.25 9.34/3.25 U1_g(x1, x2, x3, x4) = U1_g(x2, x3, x4) 9.34/3.25 9.34/3.25 0 = 0 9.34/3.25 9.34/3.25 U3_g(x1, x2) = U3_g(x2) 9.34/3.25 9.34/3.25 U2_g(x1, x2, x3, x4) = U2_g(x4) 9.34/3.25 9.34/3.25 P_IN_G(x1) = P_IN_G(x1) 9.34/3.25 9.34/3.25 U1_G(x1, x2, x3, x4) = U1_G(x2, x3, x4) 9.34/3.25 9.34/3.25 U3_G(x1, x2) = U3_G(x2) 9.34/3.25 9.34/3.25 U2_G(x1, x2, x3, x4) = U2_G(x4) 9.34/3.25 9.34/3.25 9.34/3.25 We have to consider all (P,R,Pi)-chains 9.34/3.25 ---------------------------------------- 9.34/3.25 9.34/3.25 (4) 9.34/3.25 Obligation: 9.34/3.25 Pi DP problem: 9.34/3.25 The TRS P consists of the following rules: 9.34/3.25 9.34/3.25 P_IN_G(cons(s(s(X)), cons(Y, Xs))) -> U1_G(X, Y, Xs, p_in_g(cons(X, cons(Y, Xs)))) 9.34/3.25 P_IN_G(cons(s(s(X)), cons(Y, Xs))) -> P_IN_G(cons(X, cons(Y, Xs))) 9.34/3.25 P_IN_G(cons(0, Xs)) -> U3_G(Xs, p_in_g(Xs)) 9.34/3.25 P_IN_G(cons(0, Xs)) -> P_IN_G(Xs) 9.34/3.25 U1_G(X, Y, Xs, p_out_g(cons(X, cons(Y, Xs)))) -> U2_G(X, Y, Xs, p_in_g(cons(s(s(s(s(Y)))), Xs))) 9.34/3.25 U1_G(X, Y, Xs, p_out_g(cons(X, cons(Y, Xs)))) -> P_IN_G(cons(s(s(s(s(Y)))), Xs)) 9.34/3.25 9.34/3.25 The TRS R consists of the following rules: 9.34/3.25 9.34/3.25 p_in_g(cons(X, nil)) -> p_out_g(cons(X, nil)) 9.34/3.25 p_in_g(cons(s(s(X)), cons(Y, Xs))) -> U1_g(X, Y, Xs, p_in_g(cons(X, cons(Y, Xs)))) 9.34/3.25 p_in_g(cons(0, Xs)) -> U3_g(Xs, p_in_g(Xs)) 9.34/3.25 U3_g(Xs, p_out_g(Xs)) -> p_out_g(cons(0, Xs)) 9.34/3.25 U1_g(X, Y, Xs, p_out_g(cons(X, cons(Y, Xs)))) -> U2_g(X, Y, Xs, p_in_g(cons(s(s(s(s(Y)))), Xs))) 9.34/3.25 U2_g(X, Y, Xs, p_out_g(cons(s(s(s(s(Y)))), Xs))) -> p_out_g(cons(s(s(X)), cons(Y, Xs))) 9.34/3.25 9.34/3.25 The argument filtering Pi contains the following mapping: 9.34/3.25 p_in_g(x1) = p_in_g(x1) 9.34/3.25 9.34/3.25 cons(x1, x2) = cons(x1, x2) 9.34/3.25 9.34/3.25 nil = nil 9.34/3.25 9.34/3.25 p_out_g(x1) = p_out_g 9.34/3.25 9.34/3.25 s(x1) = s(x1) 9.34/3.25 9.34/3.25 U1_g(x1, x2, x3, x4) = U1_g(x2, x3, x4) 9.34/3.25 9.34/3.25 0 = 0 9.34/3.25 9.34/3.25 U3_g(x1, x2) = U3_g(x2) 9.34/3.25 9.34/3.25 U2_g(x1, x2, x3, x4) = U2_g(x4) 9.34/3.25 9.34/3.25 P_IN_G(x1) = P_IN_G(x1) 9.34/3.25 9.34/3.25 U1_G(x1, x2, x3, x4) = U1_G(x2, x3, x4) 9.34/3.25 9.34/3.25 U3_G(x1, x2) = U3_G(x2) 9.34/3.25 9.34/3.25 U2_G(x1, x2, x3, x4) = U2_G(x4) 9.34/3.25 9.34/3.25 9.34/3.25 We have to consider all (P,R,Pi)-chains 9.34/3.25 ---------------------------------------- 9.34/3.25 9.34/3.25 (5) DependencyGraphProof (EQUIVALENT) 9.34/3.25 The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 2 less nodes. 9.34/3.25 ---------------------------------------- 9.34/3.25 9.34/3.25 (6) 9.34/3.25 Obligation: 9.34/3.25 Pi DP problem: 9.34/3.25 The TRS P consists of the following rules: 9.34/3.25 9.34/3.25 U1_G(X, Y, Xs, p_out_g(cons(X, cons(Y, Xs)))) -> P_IN_G(cons(s(s(s(s(Y)))), Xs)) 9.34/3.25 P_IN_G(cons(s(s(X)), cons(Y, Xs))) -> U1_G(X, Y, Xs, p_in_g(cons(X, cons(Y, Xs)))) 9.34/3.25 P_IN_G(cons(s(s(X)), cons(Y, Xs))) -> P_IN_G(cons(X, cons(Y, Xs))) 9.34/3.25 P_IN_G(cons(0, Xs)) -> P_IN_G(Xs) 9.34/3.25 9.34/3.25 The TRS R consists of the following rules: 9.34/3.25 9.34/3.25 p_in_g(cons(X, nil)) -> p_out_g(cons(X, nil)) 9.34/3.25 p_in_g(cons(s(s(X)), cons(Y, Xs))) -> U1_g(X, Y, Xs, p_in_g(cons(X, cons(Y, Xs)))) 9.34/3.25 p_in_g(cons(0, Xs)) -> U3_g(Xs, p_in_g(Xs)) 9.34/3.25 U3_g(Xs, p_out_g(Xs)) -> p_out_g(cons(0, Xs)) 9.34/3.25 U1_g(X, Y, Xs, p_out_g(cons(X, cons(Y, Xs)))) -> U2_g(X, Y, Xs, p_in_g(cons(s(s(s(s(Y)))), Xs))) 9.34/3.25 U2_g(X, Y, Xs, p_out_g(cons(s(s(s(s(Y)))), Xs))) -> p_out_g(cons(s(s(X)), cons(Y, Xs))) 9.34/3.25 9.34/3.25 The argument filtering Pi contains the following mapping: 9.34/3.25 p_in_g(x1) = p_in_g(x1) 9.34/3.25 9.34/3.25 cons(x1, x2) = cons(x1, x2) 9.34/3.25 9.34/3.25 nil = nil 9.34/3.25 9.34/3.25 p_out_g(x1) = p_out_g 9.34/3.25 9.34/3.25 s(x1) = s(x1) 9.34/3.25 9.34/3.25 U1_g(x1, x2, x3, x4) = U1_g(x2, x3, x4) 9.34/3.25 9.34/3.25 0 = 0 9.34/3.25 9.34/3.25 U3_g(x1, x2) = U3_g(x2) 9.34/3.25 9.34/3.25 U2_g(x1, x2, x3, x4) = U2_g(x4) 9.34/3.25 9.34/3.25 P_IN_G(x1) = P_IN_G(x1) 9.34/3.25 9.34/3.25 U1_G(x1, x2, x3, x4) = U1_G(x2, x3, x4) 9.34/3.25 9.34/3.25 9.34/3.25 We have to consider all (P,R,Pi)-chains 9.34/3.25 ---------------------------------------- 9.34/3.25 9.34/3.25 (7) PiDPToQDPProof (SOUND) 9.34/3.25 Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. 9.34/3.25 ---------------------------------------- 9.34/3.25 9.34/3.25 (8) 9.34/3.25 Obligation: 9.34/3.25 Q DP problem: 9.34/3.25 The TRS P consists of the following rules: 9.34/3.25 9.34/3.25 U1_G(Y, Xs, p_out_g) -> P_IN_G(cons(s(s(s(s(Y)))), Xs)) 9.34/3.25 P_IN_G(cons(s(s(X)), cons(Y, Xs))) -> U1_G(Y, Xs, p_in_g(cons(X, cons(Y, Xs)))) 9.34/3.25 P_IN_G(cons(s(s(X)), cons(Y, Xs))) -> P_IN_G(cons(X, cons(Y, Xs))) 9.34/3.25 P_IN_G(cons(0, Xs)) -> P_IN_G(Xs) 9.34/3.25 9.34/3.25 The TRS R consists of the following rules: 9.34/3.25 9.34/3.25 p_in_g(cons(X, nil)) -> p_out_g 9.34/3.25 p_in_g(cons(s(s(X)), cons(Y, Xs))) -> U1_g(Y, Xs, p_in_g(cons(X, cons(Y, Xs)))) 9.34/3.25 p_in_g(cons(0, Xs)) -> U3_g(p_in_g(Xs)) 9.34/3.25 U3_g(p_out_g) -> p_out_g 9.34/3.25 U1_g(Y, Xs, p_out_g) -> U2_g(p_in_g(cons(s(s(s(s(Y)))), Xs))) 9.34/3.25 U2_g(p_out_g) -> p_out_g 9.34/3.25 9.34/3.25 The set Q consists of the following terms: 9.34/3.25 9.34/3.25 p_in_g(x0) 9.34/3.25 U3_g(x0) 9.34/3.25 U1_g(x0, x1, x2) 9.34/3.25 U2_g(x0) 9.34/3.25 9.34/3.25 We have to consider all (P,Q,R)-chains. 9.34/3.25 ---------------------------------------- 9.34/3.25 9.34/3.25 (9) QDPOrderProof (EQUIVALENT) 9.34/3.25 We use the reduction pair processor [LPAR04,JAR06]. 9.34/3.25 9.34/3.25 9.34/3.25 The following pairs can be oriented strictly and are deleted. 9.34/3.25 9.34/3.25 P_IN_G(cons(s(s(X)), cons(Y, Xs))) -> U1_G(Y, Xs, p_in_g(cons(X, cons(Y, Xs)))) 9.34/3.25 P_IN_G(cons(0, Xs)) -> P_IN_G(Xs) 9.34/3.25 The remaining pairs can at least be oriented weakly. 9.34/3.25 Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation: 9.34/3.25 9.34/3.25 POL( U1_G_3(x_1, ..., x_3) ) = 2x_2 + 2 9.34/3.25 POL( p_in_g_1(x_1) ) = 0 9.34/3.25 POL( cons_2(x_1, x_2) ) = x_2 + 2 9.34/3.25 POL( s_1(x_1) ) = 0 9.34/3.25 POL( U1_g_3(x_1, ..., x_3) ) = max{0, 2x_2 - 2} 9.34/3.25 POL( 0 ) = 2 9.34/3.25 POL( U3_g_1(x_1) ) = 0 9.34/3.25 POL( U2_g_1(x_1) ) = 2x_1 9.34/3.25 POL( nil ) = 0 9.34/3.25 POL( p_out_g ) = 1 9.34/3.25 POL( P_IN_G_1(x_1) ) = max{0, 2x_1 - 2} 9.34/3.25 9.34/3.25 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 9.34/3.25 none 9.34/3.25 9.34/3.25 9.34/3.25 ---------------------------------------- 9.34/3.25 9.34/3.25 (10) 9.34/3.25 Obligation: 9.34/3.25 Q DP problem: 9.34/3.25 The TRS P consists of the following rules: 9.34/3.25 9.34/3.25 U1_G(Y, Xs, p_out_g) -> P_IN_G(cons(s(s(s(s(Y)))), Xs)) 9.34/3.25 P_IN_G(cons(s(s(X)), cons(Y, Xs))) -> P_IN_G(cons(X, cons(Y, Xs))) 9.34/3.25 9.34/3.25 The TRS R consists of the following rules: 9.34/3.25 9.34/3.25 p_in_g(cons(X, nil)) -> p_out_g 9.34/3.25 p_in_g(cons(s(s(X)), cons(Y, Xs))) -> U1_g(Y, Xs, p_in_g(cons(X, cons(Y, Xs)))) 9.34/3.25 p_in_g(cons(0, Xs)) -> U3_g(p_in_g(Xs)) 9.34/3.25 U3_g(p_out_g) -> p_out_g 9.34/3.25 U1_g(Y, Xs, p_out_g) -> U2_g(p_in_g(cons(s(s(s(s(Y)))), Xs))) 9.34/3.25 U2_g(p_out_g) -> p_out_g 9.34/3.25 9.34/3.25 The set Q consists of the following terms: 9.34/3.25 9.34/3.25 p_in_g(x0) 9.34/3.25 U3_g(x0) 9.34/3.25 U1_g(x0, x1, x2) 9.34/3.25 U2_g(x0) 9.34/3.25 9.34/3.25 We have to consider all (P,Q,R)-chains. 9.34/3.25 ---------------------------------------- 9.34/3.25 9.34/3.25 (11) DependencyGraphProof (EQUIVALENT) 9.34/3.25 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. 9.34/3.25 ---------------------------------------- 9.34/3.25 9.34/3.25 (12) 9.34/3.25 Obligation: 9.34/3.25 Q DP problem: 9.34/3.25 The TRS P consists of the following rules: 9.34/3.25 9.34/3.25 P_IN_G(cons(s(s(X)), cons(Y, Xs))) -> P_IN_G(cons(X, cons(Y, Xs))) 9.34/3.25 9.34/3.25 The TRS R consists of the following rules: 9.34/3.25 9.34/3.25 p_in_g(cons(X, nil)) -> p_out_g 9.34/3.25 p_in_g(cons(s(s(X)), cons(Y, Xs))) -> U1_g(Y, Xs, p_in_g(cons(X, cons(Y, Xs)))) 9.34/3.25 p_in_g(cons(0, Xs)) -> U3_g(p_in_g(Xs)) 9.34/3.25 U3_g(p_out_g) -> p_out_g 9.34/3.25 U1_g(Y, Xs, p_out_g) -> U2_g(p_in_g(cons(s(s(s(s(Y)))), Xs))) 9.34/3.25 U2_g(p_out_g) -> p_out_g 9.34/3.25 9.34/3.25 The set Q consists of the following terms: 9.34/3.25 9.34/3.25 p_in_g(x0) 9.34/3.25 U3_g(x0) 9.34/3.25 U1_g(x0, x1, x2) 9.34/3.25 U2_g(x0) 9.34/3.25 9.34/3.25 We have to consider all (P,Q,R)-chains. 9.34/3.25 ---------------------------------------- 9.34/3.25 9.34/3.25 (13) UsableRulesProof (EQUIVALENT) 9.34/3.25 As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. 9.34/3.25 ---------------------------------------- 9.34/3.25 9.34/3.25 (14) 9.34/3.25 Obligation: 9.34/3.25 Q DP problem: 9.34/3.25 The TRS P consists of the following rules: 9.34/3.25 9.34/3.25 P_IN_G(cons(s(s(X)), cons(Y, Xs))) -> P_IN_G(cons(X, cons(Y, Xs))) 9.34/3.25 9.34/3.25 R is empty. 9.34/3.25 The set Q consists of the following terms: 9.34/3.25 9.34/3.25 p_in_g(x0) 9.34/3.25 U3_g(x0) 9.34/3.25 U1_g(x0, x1, x2) 9.34/3.25 U2_g(x0) 9.34/3.25 9.34/3.25 We have to consider all (P,Q,R)-chains. 9.34/3.25 ---------------------------------------- 9.34/3.25 9.34/3.25 (15) QReductionProof (EQUIVALENT) 9.34/3.25 We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 9.34/3.25 9.34/3.25 p_in_g(x0) 9.34/3.25 U3_g(x0) 9.34/3.25 U1_g(x0, x1, x2) 9.34/3.25 U2_g(x0) 9.34/3.25 9.34/3.25 9.34/3.25 ---------------------------------------- 9.34/3.25 9.34/3.25 (16) 9.34/3.25 Obligation: 9.34/3.25 Q DP problem: 9.34/3.25 The TRS P consists of the following rules: 9.34/3.25 9.34/3.25 P_IN_G(cons(s(s(X)), cons(Y, Xs))) -> P_IN_G(cons(X, cons(Y, Xs))) 9.34/3.25 9.34/3.25 R is empty. 9.34/3.25 Q is empty. 9.34/3.25 We have to consider all (P,Q,R)-chains. 9.34/3.25 ---------------------------------------- 9.34/3.25 9.34/3.25 (17) UsableRulesReductionPairsProof (EQUIVALENT) 9.34/3.25 By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well. 9.34/3.26 9.34/3.26 The following dependency pairs can be deleted: 9.34/3.26 9.34/3.26 P_IN_G(cons(s(s(X)), cons(Y, Xs))) -> P_IN_G(cons(X, cons(Y, Xs))) 9.34/3.26 No rules are removed from R. 9.34/3.26 9.34/3.26 Used ordering: POLO with Polynomial interpretation [POLO]: 9.34/3.26 9.34/3.26 POL(P_IN_G(x_1)) = x_1 9.34/3.26 POL(cons(x_1, x_2)) = 2*x_1 + x_2 9.34/3.26 POL(s(x_1)) = 2*x_1 9.34/3.26 9.34/3.26 9.34/3.26 ---------------------------------------- 9.34/3.26 9.34/3.26 (18) 9.34/3.26 Obligation: 9.34/3.26 Q DP problem: 9.34/3.26 P is empty. 9.34/3.26 R is empty. 9.34/3.26 Q is empty. 9.34/3.26 We have to consider all (P,Q,R)-chains. 9.34/3.26 ---------------------------------------- 9.34/3.26 9.34/3.26 (19) PisEmptyProof (EQUIVALENT) 9.34/3.26 The TRS P is empty. Hence, there is no (P,Q,R) chain. 9.34/3.26 ---------------------------------------- 9.34/3.26 9.34/3.26 (20) 9.34/3.26 YES 9.64/3.32 EOF