YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) RootLabelingProof [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 26 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) QDP (8) QDPSizeChangeProof [EQUIVALENT, 0 ms] (9) YES (10) QDP (11) QDPOrderProof [EQUIVALENT, 7687 ms] (12) QDP (13) QDPOrderProof [EQUIVALENT, 7906 ms] (14) QDP (15) PisEmptyProof [EQUIVALENT, 0 ms] (16) YES (17) QDP (18) SplitQDPProof [EQUIVALENT, 0 ms] (19) AND (20) QDP (21) SemLabProof [SOUND, 0 ms] (22) QDP (23) DependencyGraphProof [EQUIVALENT, 0 ms] (24) QDP (25) QDPOrderProof [EQUIVALENT, 33 ms] (26) QDP (27) QDPOrderProof [EQUIVALENT, 22 ms] (28) QDP (29) PisEmptyProof [SOUND, 0 ms] (30) TRUE (31) QDP (32) QDPOrderProof [EQUIVALENT, 8061 ms] (33) QDP (34) PisEmptyProof [EQUIVALENT, 0 ms] (35) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(b(a(a(b(x1))))) -> b(a(a(b(b(x1))))) b(a(a(a(b(b(b(x1))))))) -> b(b(b(a(a(a(b(x1))))))) b(a(b(a(a(b(x1)))))) -> b(a(b(b(a(a(a(b(x1)))))))) b(a(a(a(b(a(b(x1))))))) -> b(a(a(b(a(b(x1)))))) Q is empty. ---------------------------------------- (1) RootLabelingProof (EQUIVALENT) We used plain root labeling [ROOTLAB] with the following heuristic: LabelAll: All function symbols get labeled As Q is empty the root labeling was sound AND complete. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(x1))))) b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(x1))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(x1))))))) -> b_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))))) -> b_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))))) b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))))) b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1))))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))))) Q is empty. ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: B_{B_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))) -> B_{A_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(x1))))) B_{B_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))) -> B_{B_1}(b_{b_1}(x1)) B_{B_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))) -> B_{A_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(x1))))) B_{B_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))) -> B_{B_1}(b_{a_1}(x1)) B_{A_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(x1))))))) -> B_{B_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))))) B_{A_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(x1))))))) -> B_{B_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))) B_{A_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(x1))))))) -> B_{A_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))) B_{A_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))))) -> B_{B_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))))) B_{A_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))))) -> B_{B_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))))) B_{A_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))))) -> B_{A_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))) B_{A_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))) -> B_{A_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))))) B_{A_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))) -> B_{B_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))) B_{A_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))) -> B_{A_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))) B_{A_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))))) -> B_{A_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))))))) B_{A_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))))) -> B_{B_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))))) B_{A_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))))) -> B_{A_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))) B_{A_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))))) -> B_{A_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))))) B_{A_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1))))))) -> B_{A_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))))) The TRS R consists of the following rules: b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(x1))))) b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(x1))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(x1))))))) -> b_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))))) -> b_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))))) b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))))) b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1))))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 12 less nodes. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (7) Obligation: Q DP problem: The TRS P consists of the following rules: B_{B_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))) -> B_{B_1}(b_{a_1}(x1)) B_{B_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))) -> B_{B_1}(b_{b_1}(x1)) The TRS R consists of the following rules: b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(x1))))) b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(x1))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(x1))))))) -> b_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))))) -> b_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))))) b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))))) b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1))))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (8) QDPSizeChangeProof (EQUIVALENT) By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs: *B_{B_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))) -> B_{B_1}(b_{a_1}(x1)) The graph contains the following edges 1 > 1 *B_{B_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))) -> B_{B_1}(b_{b_1}(x1)) The graph contains the following edges 1 > 1 ---------------------------------------- (9) YES ---------------------------------------- (10) Obligation: Q DP problem: The TRS P consists of the following rules: B_{A_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))))) -> B_{A_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))) B_{A_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(x1))))))) -> B_{A_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))) The TRS R consists of the following rules: b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(x1))))) b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(x1))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(x1))))))) -> b_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))))) -> b_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))))) b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))))) b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1))))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (11) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. B_{A_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(x1))))))) -> B_{A_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))) The remaining pairs can at least be oriented weakly. Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: <<< POL(B_{A_1}(x_1)) = [[0A]] + [[0A, 0A, -I]] * x_1 >>> <<< POL(a_{a_1}(x_1)) = [[-I], [-I], [-I]] + [[0A, -I, -I], [0A, -I, 0A], [-I, 0A, -I]] * x_1 >>> <<< POL(a_{b_1}(x_1)) = [[-I], [-I], [-I]] + [[0A, -I, -I], [-I, -I, 0A], [0A, -I, -I]] * x_1 >>> <<< POL(b_{b_1}(x_1)) = [[-I], [0A], [0A]] + [[0A, -I, 0A], [0A, 0A, 0A], [1A, -I, -I]] * x_1 >>> <<< POL(b_{a_1}(x_1)) = [[-I], [0A], [-I]] + [[0A, -I, -I], [0A, -I, -I], [-I, -I, 0A]] * x_1 >>> The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(x1))))))) -> b_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))))) -> b_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))))) b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))))))) b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1))))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))))) b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(x1))))) b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(x1))))) ---------------------------------------- (12) Obligation: Q DP problem: The TRS P consists of the following rules: B_{A_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))))) -> B_{A_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))) The TRS R consists of the following rules: b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(x1))))) b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(x1))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(x1))))))) -> b_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))))) -> b_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))))) b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))))) b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1))))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (13) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. B_{A_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))))) -> B_{A_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))) The remaining pairs can at least be oriented weakly. Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: <<< POL(B_{A_1}(x_1)) = [[-I]] + [[0A, 0A, -I]] * x_1 >>> <<< POL(a_{a_1}(x_1)) = [[0A], [0A], [0A]] + [[0A, -I, 0A], [0A, 0A, 0A], [0A, 0A, 0A]] * x_1 >>> <<< POL(a_{b_1}(x_1)) = [[0A], [0A], [-I]] + [[-I, -I, -I], [0A, 0A, 0A], [-I, -I, -I]] * x_1 >>> <<< POL(b_{b_1}(x_1)) = [[0A], [0A], [-I]] + [[0A, 0A, 0A], [0A, -I, 1A], [0A, 1A, 0A]] * x_1 >>> <<< POL(b_{a_1}(x_1)) = [[0A], [-I], [0A]] + [[0A, -I, -I], [-I, -I, 0A], [0A, -I, 0A]] * x_1 >>> The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(x1))))))) -> b_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))))) -> b_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))))) b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))))))) b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1))))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))))) b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(x1))))) b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(x1))))) ---------------------------------------- (14) Obligation: Q DP problem: P is empty. The TRS R consists of the following rules: b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(x1))))) b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(x1))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(x1))))))) -> b_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))))) -> b_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))))) b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))))) b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1))))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (15) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (16) YES ---------------------------------------- (17) Obligation: Q DP problem: The TRS P consists of the following rules: B_{A_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))))) -> B_{A_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))))))) B_{A_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))) -> B_{A_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))))) The TRS R consists of the following rules: b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(x1))))) b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(x1))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(x1))))))) -> b_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))))) -> b_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))))) b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))))) b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1))))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (18) SplitQDPProof (EQUIVALENT) We show in the first subproof that some pairs and rules can be removed, afterwards, we continue with the remaining DP-Problem ---------------------------------------- (19) Complex Obligation (AND) ---------------------------------------- (20) Obligation: Q DP problem: The TRS P consists of the following rules: B_{A_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))))) -> B_{A_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))))))) B_{A_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))) -> B_{A_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))))) The TRS R consists of the following rules: b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(x1))))) b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(x1))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(x1))))))) -> b_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))))) -> b_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))))) b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))))) b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1))))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (21) SemLabProof (SOUND) We found the following model for the rules of the TRSs R and P. Interpretation over the domain with elements from 0 to 1. b_{b_1}: 0 b_{a_1}: 0 B_{A_1}: 0 a_{b_1}: 1 a_{a_1}: 0 By semantic labelling [SEMLAB] we obtain the following labelled QDP problem. ---------------------------------------- (22) Obligation: Q DP problem: The TRS P consists of the following rules: B_{A_1}.1(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1)))))) -> B_{A_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1)))))))) B_{A_1}.1(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1)))))) -> B_{A_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1)))))))) B_{A_1}.1(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.1(x1)))))) -> B_{A_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.1(x1)))))))) B_{A_1}.1(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1)))))) -> B_{A_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1)))))))) The TRS R consists of the following rules: b_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{b_1}.0(x1))))) b_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.1(x1))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{b_1}.1(x1))))) b_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(x1))))) b_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.1(x1))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{b_1}.0(b_{b_1}.0(x1))))))) -> b_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1))))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{b_1}.0(b_{b_1}.1(x1))))))) -> b_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.1(x1))))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{b_1}.0(b_{a_1}.0(x1))))))) -> b_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1))))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{b_1}.0(b_{a_1}.1(x1))))))) -> b_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1))))))) b_{a_1}.1(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1)))))) -> b_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1)))))))) b_{a_1}.1(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.1(x1)))))) -> b_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.1(x1)))))))) b_{a_1}.1(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1)))))) -> b_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1)))))))) b_{a_1}.1(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1)))))) -> b_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1)))))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1))))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1)))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{b_1}.1(x1))))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{b_1}.1(x1)))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1))))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1)))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1))))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (23) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (24) Obligation: Q DP problem: The TRS P consists of the following rules: B_{A_1}.1(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1)))))) -> B_{A_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1)))))))) B_{A_1}.1(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1)))))) -> B_{A_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1)))))))) B_{A_1}.1(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1)))))) -> B_{A_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1)))))))) The TRS R consists of the following rules: b_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{b_1}.0(x1))))) b_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.1(x1))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{b_1}.1(x1))))) b_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(x1))))) b_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.1(x1))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{b_1}.0(b_{b_1}.0(x1))))))) -> b_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1))))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{b_1}.0(b_{b_1}.1(x1))))))) -> b_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.1(x1))))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{b_1}.0(b_{a_1}.0(x1))))))) -> b_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1))))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{b_1}.0(b_{a_1}.1(x1))))))) -> b_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1))))))) b_{a_1}.1(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1)))))) -> b_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1)))))))) b_{a_1}.1(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.1(x1)))))) -> b_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.1(x1)))))))) b_{a_1}.1(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1)))))) -> b_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1)))))))) b_{a_1}.1(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1)))))) -> b_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1)))))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1))))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1)))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{b_1}.1(x1))))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{b_1}.1(x1)))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1))))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1)))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1))))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (25) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. B_{A_1}.1(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1)))))) -> B_{A_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1)))))))) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial interpretation [POLO]: POL(B_{A_1}.1(x_1)) = x_1 POL(a_{a_1}.0(x_1)) = 1 POL(a_{a_1}.1(x_1)) = 1 + x_1 POL(a_{b_1}.0(x_1)) = x_1 POL(b_{a_1}.0(x_1)) = 1 + x_1 POL(b_{a_1}.1(x_1)) = 0 POL(b_{b_1}.0(x_1)) = x_1 POL(b_{b_1}.1(x_1)) = x_1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: b_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{b_1}.0(x1))))) b_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.1(x1))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{b_1}.1(x1))))) b_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(x1))))) b_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.1(x1))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{b_1}.0(b_{b_1}.0(x1))))))) -> b_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1))))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{b_1}.0(b_{b_1}.1(x1))))))) -> b_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.1(x1))))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{b_1}.0(b_{a_1}.0(x1))))))) -> b_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1))))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{b_1}.0(b_{a_1}.1(x1))))))) -> b_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1))))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1))))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1)))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{b_1}.1(x1))))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{b_1}.1(x1)))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1))))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1)))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1))))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1)))))) b_{a_1}.1(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1)))))) -> b_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1)))))))) b_{a_1}.1(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1)))))) -> b_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1)))))))) b_{a_1}.1(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1)))))) -> b_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1)))))))) b_{a_1}.1(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.1(x1)))))) -> b_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.1(x1)))))))) ---------------------------------------- (26) Obligation: Q DP problem: The TRS P consists of the following rules: B_{A_1}.1(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1)))))) -> B_{A_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1)))))))) B_{A_1}.1(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1)))))) -> B_{A_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1)))))))) The TRS R consists of the following rules: b_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{b_1}.0(x1))))) b_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.1(x1))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{b_1}.1(x1))))) b_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(x1))))) b_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.1(x1))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{b_1}.0(b_{b_1}.0(x1))))))) -> b_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1))))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{b_1}.0(b_{b_1}.1(x1))))))) -> b_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.1(x1))))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{b_1}.0(b_{a_1}.0(x1))))))) -> b_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1))))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{b_1}.0(b_{a_1}.1(x1))))))) -> b_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1))))))) b_{a_1}.1(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1)))))) -> b_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1)))))))) b_{a_1}.1(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.1(x1)))))) -> b_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.1(x1)))))))) b_{a_1}.1(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1)))))) -> b_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1)))))))) b_{a_1}.1(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1)))))) -> b_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1)))))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1))))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1)))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{b_1}.1(x1))))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{b_1}.1(x1)))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1))))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1)))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1))))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (27) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. B_{A_1}.1(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1)))))) -> B_{A_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1)))))))) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial interpretation [POLO]: POL(B_{A_1}.1(x_1)) = x_1 POL(a_{a_1}.0(x_1)) = 1 POL(a_{a_1}.1(x_1)) = x_1 POL(a_{b_1}.0(x_1)) = x_1 POL(b_{a_1}.0(x_1)) = x_1 POL(b_{a_1}.1(x_1)) = 1 POL(b_{b_1}.0(x_1)) = 0 POL(b_{b_1}.1(x_1)) = x_1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: b_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{b_1}.0(x1))))) b_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.1(x1))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{b_1}.1(x1))))) b_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(x1))))) b_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.1(x1))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{b_1}.0(b_{b_1}.0(x1))))))) -> b_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1))))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{b_1}.0(b_{a_1}.0(x1))))))) -> b_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1))))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{b_1}.0(b_{a_1}.1(x1))))))) -> b_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1))))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1))))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1)))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1))))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1)))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1))))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1)))))) b_{a_1}.1(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1)))))) -> b_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1)))))))) b_{a_1}.1(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1)))))) -> b_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1)))))))) b_{a_1}.1(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1)))))) -> b_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1)))))))) b_{a_1}.1(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.1(x1)))))) -> b_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.1(x1)))))))) ---------------------------------------- (28) Obligation: Q DP problem: The TRS P consists of the following rules: B_{A_1}.1(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1)))))) -> B_{A_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1)))))))) The TRS R consists of the following rules: b_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{b_1}.0(x1))))) b_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.1(x1))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{b_1}.1(x1))))) b_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(x1))))) b_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.1(x1))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{b_1}.0(b_{b_1}.0(x1))))))) -> b_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1))))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{b_1}.0(b_{b_1}.1(x1))))))) -> b_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.1(x1))))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{b_1}.0(b_{a_1}.0(x1))))))) -> b_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1))))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{b_1}.0(b_{a_1}.1(x1))))))) -> b_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1))))))) b_{a_1}.1(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1)))))) -> b_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1)))))))) b_{a_1}.1(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.1(x1)))))) -> b_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{b_1}.1(x1)))))))) b_{a_1}.1(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1)))))) -> b_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1)))))))) b_{a_1}.1(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1)))))) -> b_{a_1}.1(a_{b_1}.0(b_{b_1}.0(b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1)))))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1))))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{b_1}.0(x1)))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{b_1}.1(x1))))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{b_1}.1(x1)))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1))))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{a_1}.0(x1)))))) b_{a_1}.0(a_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1))))))) -> b_{a_1}.0(a_{a_1}.1(a_{b_1}.0(b_{a_1}.1(a_{b_1}.0(b_{a_1}.1(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (29) PisEmptyProof (SOUND) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (30) TRUE ---------------------------------------- (31) Obligation: Q DP problem: The TRS P consists of the following rules: B_{A_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))) -> B_{A_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))))) The TRS R consists of the following rules: b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(x1))))) b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(x1))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(x1))))))) -> b_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))))) -> b_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))))) b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))))) b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1))))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (32) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. B_{A_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))) -> B_{A_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))))) The remaining pairs can at least be oriented weakly. Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: <<< POL(B_{A_1}(x_1)) = [[0A]] + [[-I, 0A, -I]] * x_1 >>> <<< POL(a_{b_1}(x_1)) = [[0A], [0A], [1A]] + [[1A, 0A, 0A], [0A, -I, -I], [0A, 0A, 1A]] * x_1 >>> <<< POL(b_{a_1}(x_1)) = [[0A], [0A], [0A]] + [[0A, -I, 0A], [0A, 0A, 0A], [-I, 0A, -I]] * x_1 >>> <<< POL(a_{a_1}(x_1)) = [[0A], [0A], [0A]] + [[0A, 0A, -I], [-I, -I, 0A], [-I, 0A, -I]] * x_1 >>> <<< POL(b_{b_1}(x_1)) = [[0A], [-I], [-I]] + [[-I, -I, 0A], [0A, 0A, -I], [0A, -I, -I]] * x_1 >>> The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(x1))))) b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(x1))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(x1))))))) -> b_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))))) -> b_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1))))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))))) b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))))))) b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))))) ---------------------------------------- (33) Obligation: Q DP problem: P is empty. The TRS R consists of the following rules: b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(x1))))) b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(x1))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(x1))))))) -> b_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))))) -> b_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))))) b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))))))) b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1))))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (34) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (35) YES