17.87/5.50 YES 18.11/5.52 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 18.11/5.52 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 18.11/5.52 18.11/5.52 18.11/5.52 Termination w.r.t. Q of the given QTRS could be proven: 18.11/5.52 18.11/5.52 (0) QTRS 18.11/5.52 (1) QTRS Reverse [EQUIVALENT, 0 ms] 18.11/5.52 (2) QTRS 18.11/5.52 (3) FlatCCProof [EQUIVALENT, 0 ms] 18.11/5.52 (4) QTRS 18.11/5.52 (5) RootLabelingProof [EQUIVALENT, 0 ms] 18.11/5.52 (6) QTRS 18.11/5.52 (7) QTRSRRRProof [EQUIVALENT, 36 ms] 18.11/5.52 (8) QTRS 18.11/5.52 (9) DependencyPairsProof [EQUIVALENT, 0 ms] 18.11/5.52 (10) QDP 18.11/5.52 (11) DependencyGraphProof [EQUIVALENT, 0 ms] 18.11/5.52 (12) AND 18.11/5.52 (13) QDP 18.11/5.52 (14) UsableRulesProof [EQUIVALENT, 2 ms] 18.11/5.52 (15) QDP 18.11/5.52 (16) QDPSizeChangeProof [EQUIVALENT, 0 ms] 18.11/5.52 (17) YES 18.11/5.52 (18) QDP 18.11/5.52 (19) UsableRulesProof [EQUIVALENT, 0 ms] 18.11/5.52 (20) QDP 18.11/5.52 (21) QDPSizeChangeProof [EQUIVALENT, 0 ms] 18.11/5.52 (22) YES 18.11/5.52 (23) QDP 18.11/5.52 (24) UsableRulesProof [EQUIVALENT, 15 ms] 18.11/5.52 (25) QDP 18.11/5.52 (26) QDPOrderProof [EQUIVALENT, 44 ms] 18.11/5.52 (27) QDP 18.11/5.52 (28) PisEmptyProof [EQUIVALENT, 0 ms] 18.11/5.52 (29) YES 18.11/5.52 18.11/5.52 18.11/5.52 ---------------------------------------- 18.11/5.52 18.11/5.52 (0) 18.11/5.52 Obligation: 18.11/5.52 Q restricted rewrite system: 18.11/5.52 The TRS R consists of the following rules: 18.11/5.52 18.11/5.52 a(x1) -> x1 18.11/5.52 a(b(x1)) -> c(a(x1)) 18.11/5.52 c(c(x1)) -> c(b(c(b(a(x1))))) 18.11/5.52 18.11/5.52 Q is empty. 18.11/5.52 18.11/5.52 ---------------------------------------- 18.11/5.52 18.11/5.52 (1) QTRS Reverse (EQUIVALENT) 18.11/5.52 We applied the QTRS Reverse Processor [REVERSE]. 18.11/5.52 ---------------------------------------- 18.11/5.52 18.11/5.52 (2) 18.11/5.52 Obligation: 18.11/5.52 Q restricted rewrite system: 18.11/5.52 The TRS R consists of the following rules: 18.11/5.52 18.11/5.52 a(x1) -> x1 18.11/5.52 b(a(x1)) -> a(c(x1)) 18.11/5.52 c(c(x1)) -> a(b(c(b(c(x1))))) 18.11/5.52 18.11/5.52 Q is empty. 18.11/5.52 18.11/5.52 ---------------------------------------- 18.11/5.52 18.11/5.52 (3) FlatCCProof (EQUIVALENT) 18.11/5.52 We used flat context closure [ROOTLAB] 18.11/5.52 As Q is empty the flat context closure was sound AND complete. 18.11/5.52 18.11/5.52 ---------------------------------------- 18.11/5.52 18.11/5.52 (4) 18.11/5.52 Obligation: 18.11/5.52 Q restricted rewrite system: 18.11/5.52 The TRS R consists of the following rules: 18.11/5.52 18.11/5.52 a(a(x1)) -> a(x1) 18.11/5.52 b(a(x1)) -> b(x1) 18.11/5.52 c(a(x1)) -> c(x1) 18.11/5.52 a(b(a(x1))) -> a(a(c(x1))) 18.11/5.52 b(b(a(x1))) -> b(a(c(x1))) 18.11/5.52 c(b(a(x1))) -> c(a(c(x1))) 18.11/5.52 a(c(c(x1))) -> a(a(b(c(b(c(x1)))))) 18.11/5.52 b(c(c(x1))) -> b(a(b(c(b(c(x1)))))) 18.11/5.52 c(c(c(x1))) -> c(a(b(c(b(c(x1)))))) 18.11/5.52 18.11/5.52 Q is empty. 18.11/5.52 18.11/5.52 ---------------------------------------- 18.11/5.52 18.11/5.52 (5) RootLabelingProof (EQUIVALENT) 18.11/5.52 We used plain root labeling [ROOTLAB] with the following heuristic: 18.11/5.52 LabelAll: All function symbols get labeled 18.11/5.52 18.11/5.52 As Q is empty the root labeling was sound AND complete. 18.11/5.52 18.11/5.52 ---------------------------------------- 18.11/5.52 18.11/5.52 (6) 18.11/5.52 Obligation: 18.11/5.52 Q restricted rewrite system: 18.11/5.52 The TRS R consists of the following rules: 18.11/5.52 18.11/5.52 a_{a_1}(a_{a_1}(x1)) -> a_{a_1}(x1) 18.11/5.52 a_{a_1}(a_{b_1}(x1)) -> a_{b_1}(x1) 18.11/5.52 a_{a_1}(a_{c_1}(x1)) -> a_{c_1}(x1) 18.11/5.52 b_{a_1}(a_{a_1}(x1)) -> b_{a_1}(x1) 18.11/5.52 b_{a_1}(a_{b_1}(x1)) -> b_{b_1}(x1) 18.11/5.52 b_{a_1}(a_{c_1}(x1)) -> b_{c_1}(x1) 18.11/5.52 c_{a_1}(a_{a_1}(x1)) -> c_{a_1}(x1) 18.11/5.52 c_{a_1}(a_{b_1}(x1)) -> c_{b_1}(x1) 18.11/5.52 c_{a_1}(a_{c_1}(x1)) -> c_{c_1}(x1) 18.11/5.52 a_{b_1}(b_{a_1}(a_{a_1}(x1))) -> a_{a_1}(a_{c_1}(c_{a_1}(x1))) 18.11/5.52 a_{b_1}(b_{a_1}(a_{b_1}(x1))) -> a_{a_1}(a_{c_1}(c_{b_1}(x1))) 18.11/5.52 a_{b_1}(b_{a_1}(a_{c_1}(x1))) -> a_{a_1}(a_{c_1}(c_{c_1}(x1))) 18.11/5.52 b_{b_1}(b_{a_1}(a_{a_1}(x1))) -> b_{a_1}(a_{c_1}(c_{a_1}(x1))) 18.11/5.52 b_{b_1}(b_{a_1}(a_{b_1}(x1))) -> b_{a_1}(a_{c_1}(c_{b_1}(x1))) 18.11/5.52 b_{b_1}(b_{a_1}(a_{c_1}(x1))) -> b_{a_1}(a_{c_1}(c_{c_1}(x1))) 18.11/5.52 c_{b_1}(b_{a_1}(a_{a_1}(x1))) -> c_{a_1}(a_{c_1}(c_{a_1}(x1))) 18.11/5.52 c_{b_1}(b_{a_1}(a_{b_1}(x1))) -> c_{a_1}(a_{c_1}(c_{b_1}(x1))) 18.11/5.52 c_{b_1}(b_{a_1}(a_{c_1}(x1))) -> c_{a_1}(a_{c_1}(c_{c_1}(x1))) 18.11/5.52 a_{c_1}(c_{c_1}(c_{a_1}(x1))) -> a_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{a_1}(x1)))))) 18.11/5.52 a_{c_1}(c_{c_1}(c_{b_1}(x1))) -> a_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{b_1}(x1)))))) 18.11/5.52 a_{c_1}(c_{c_1}(c_{c_1}(x1))) -> a_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{c_1}(x1)))))) 18.11/5.52 b_{c_1}(c_{c_1}(c_{a_1}(x1))) -> b_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{a_1}(x1)))))) 18.11/5.52 b_{c_1}(c_{c_1}(c_{b_1}(x1))) -> b_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{b_1}(x1)))))) 18.11/5.52 b_{c_1}(c_{c_1}(c_{c_1}(x1))) -> b_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{c_1}(x1)))))) 18.11/5.52 c_{c_1}(c_{c_1}(c_{a_1}(x1))) -> c_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{a_1}(x1)))))) 18.11/5.52 c_{c_1}(c_{c_1}(c_{b_1}(x1))) -> c_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{b_1}(x1)))))) 18.11/5.52 c_{c_1}(c_{c_1}(c_{c_1}(x1))) -> c_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{c_1}(x1)))))) 18.11/5.52 18.11/5.52 Q is empty. 18.11/5.52 18.11/5.52 ---------------------------------------- 18.11/5.52 18.11/5.52 (7) QTRSRRRProof (EQUIVALENT) 18.11/5.52 Used ordering: 18.11/5.52 Polynomial interpretation [POLO]: 18.11/5.52 18.11/5.52 POL(a_{a_1}(x_1)) = x_1 18.11/5.52 POL(a_{b_1}(x_1)) = x_1 18.11/5.52 POL(a_{c_1}(x_1)) = 1 + x_1 18.11/5.52 POL(b_{a_1}(x_1)) = 1 + x_1 18.11/5.52 POL(b_{b_1}(x_1)) = 1 + x_1 18.11/5.52 POL(b_{c_1}(x_1)) = x_1 18.11/5.52 POL(c_{a_1}(x_1)) = x_1 18.11/5.52 POL(c_{b_1}(x_1)) = x_1 18.11/5.52 POL(c_{c_1}(x_1)) = 1 + x_1 18.11/5.52 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 18.11/5.52 18.11/5.52 b_{a_1}(a_{c_1}(x1)) -> b_{c_1}(x1) 18.11/5.52 a_{c_1}(c_{c_1}(c_{a_1}(x1))) -> a_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{a_1}(x1)))))) 18.11/5.52 a_{c_1}(c_{c_1}(c_{b_1}(x1))) -> a_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{b_1}(x1)))))) 18.11/5.52 a_{c_1}(c_{c_1}(c_{c_1}(x1))) -> a_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{c_1}(x1)))))) 18.11/5.52 c_{c_1}(c_{c_1}(c_{a_1}(x1))) -> c_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{a_1}(x1)))))) 18.11/5.52 c_{c_1}(c_{c_1}(c_{b_1}(x1))) -> c_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{b_1}(x1)))))) 18.11/5.52 c_{c_1}(c_{c_1}(c_{c_1}(x1))) -> c_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{c_1}(x1)))))) 18.11/5.52 18.11/5.52 18.11/5.52 18.11/5.52 18.11/5.52 ---------------------------------------- 18.11/5.52 18.11/5.52 (8) 18.11/5.52 Obligation: 18.11/5.52 Q restricted rewrite system: 18.11/5.52 The TRS R consists of the following rules: 18.11/5.52 18.11/5.52 a_{a_1}(a_{a_1}(x1)) -> a_{a_1}(x1) 18.11/5.52 a_{a_1}(a_{b_1}(x1)) -> a_{b_1}(x1) 18.11/5.52 a_{a_1}(a_{c_1}(x1)) -> a_{c_1}(x1) 18.11/5.52 b_{a_1}(a_{a_1}(x1)) -> b_{a_1}(x1) 18.11/5.52 b_{a_1}(a_{b_1}(x1)) -> b_{b_1}(x1) 18.11/5.52 c_{a_1}(a_{a_1}(x1)) -> c_{a_1}(x1) 18.11/5.52 c_{a_1}(a_{b_1}(x1)) -> c_{b_1}(x1) 18.11/5.52 c_{a_1}(a_{c_1}(x1)) -> c_{c_1}(x1) 18.11/5.52 a_{b_1}(b_{a_1}(a_{a_1}(x1))) -> a_{a_1}(a_{c_1}(c_{a_1}(x1))) 18.11/5.52 a_{b_1}(b_{a_1}(a_{b_1}(x1))) -> a_{a_1}(a_{c_1}(c_{b_1}(x1))) 18.11/5.52 a_{b_1}(b_{a_1}(a_{c_1}(x1))) -> a_{a_1}(a_{c_1}(c_{c_1}(x1))) 18.11/5.52 b_{b_1}(b_{a_1}(a_{a_1}(x1))) -> b_{a_1}(a_{c_1}(c_{a_1}(x1))) 18.11/5.52 b_{b_1}(b_{a_1}(a_{b_1}(x1))) -> b_{a_1}(a_{c_1}(c_{b_1}(x1))) 18.11/5.52 b_{b_1}(b_{a_1}(a_{c_1}(x1))) -> b_{a_1}(a_{c_1}(c_{c_1}(x1))) 18.11/5.52 c_{b_1}(b_{a_1}(a_{a_1}(x1))) -> c_{a_1}(a_{c_1}(c_{a_1}(x1))) 18.11/5.52 c_{b_1}(b_{a_1}(a_{b_1}(x1))) -> c_{a_1}(a_{c_1}(c_{b_1}(x1))) 18.11/5.52 c_{b_1}(b_{a_1}(a_{c_1}(x1))) -> c_{a_1}(a_{c_1}(c_{c_1}(x1))) 18.11/5.52 b_{c_1}(c_{c_1}(c_{a_1}(x1))) -> b_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{a_1}(x1)))))) 18.11/5.52 b_{c_1}(c_{c_1}(c_{b_1}(x1))) -> b_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{b_1}(x1)))))) 18.11/5.52 b_{c_1}(c_{c_1}(c_{c_1}(x1))) -> b_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{c_1}(x1)))))) 18.11/5.52 18.11/5.52 Q is empty. 18.11/5.52 18.11/5.52 ---------------------------------------- 18.11/5.52 18.11/5.52 (9) DependencyPairsProof (EQUIVALENT) 18.11/5.52 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 18.11/5.52 ---------------------------------------- 18.11/5.52 18.11/5.52 (10) 18.11/5.52 Obligation: 18.11/5.52 Q DP problem: 18.11/5.52 The TRS P consists of the following rules: 18.11/5.52 18.11/5.52 B_{A_1}(a_{a_1}(x1)) -> B_{A_1}(x1) 18.11/5.52 B_{A_1}(a_{b_1}(x1)) -> B_{B_1}(x1) 18.11/5.52 C_{A_1}(a_{a_1}(x1)) -> C_{A_1}(x1) 18.11/5.52 C_{A_1}(a_{b_1}(x1)) -> C_{B_1}(x1) 18.11/5.52 A_{B_1}(b_{a_1}(a_{a_1}(x1))) -> A_{A_1}(a_{c_1}(c_{a_1}(x1))) 18.11/5.52 A_{B_1}(b_{a_1}(a_{a_1}(x1))) -> C_{A_1}(x1) 18.11/5.52 A_{B_1}(b_{a_1}(a_{b_1}(x1))) -> A_{A_1}(a_{c_1}(c_{b_1}(x1))) 18.11/5.52 A_{B_1}(b_{a_1}(a_{b_1}(x1))) -> C_{B_1}(x1) 18.11/5.52 A_{B_1}(b_{a_1}(a_{c_1}(x1))) -> A_{A_1}(a_{c_1}(c_{c_1}(x1))) 18.11/5.52 B_{B_1}(b_{a_1}(a_{a_1}(x1))) -> B_{A_1}(a_{c_1}(c_{a_1}(x1))) 18.11/5.52 B_{B_1}(b_{a_1}(a_{a_1}(x1))) -> C_{A_1}(x1) 18.11/5.52 B_{B_1}(b_{a_1}(a_{b_1}(x1))) -> B_{A_1}(a_{c_1}(c_{b_1}(x1))) 18.11/5.52 B_{B_1}(b_{a_1}(a_{b_1}(x1))) -> C_{B_1}(x1) 18.11/5.52 B_{B_1}(b_{a_1}(a_{c_1}(x1))) -> B_{A_1}(a_{c_1}(c_{c_1}(x1))) 18.11/5.52 C_{B_1}(b_{a_1}(a_{a_1}(x1))) -> C_{A_1}(a_{c_1}(c_{a_1}(x1))) 18.11/5.52 C_{B_1}(b_{a_1}(a_{a_1}(x1))) -> C_{A_1}(x1) 18.11/5.52 C_{B_1}(b_{a_1}(a_{b_1}(x1))) -> C_{A_1}(a_{c_1}(c_{b_1}(x1))) 18.11/5.52 C_{B_1}(b_{a_1}(a_{b_1}(x1))) -> C_{B_1}(x1) 18.11/5.52 C_{B_1}(b_{a_1}(a_{c_1}(x1))) -> C_{A_1}(a_{c_1}(c_{c_1}(x1))) 18.11/5.52 B_{C_1}(c_{c_1}(c_{a_1}(x1))) -> B_{A_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{a_1}(x1)))))) 18.11/5.52 B_{C_1}(c_{c_1}(c_{a_1}(x1))) -> A_{B_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{a_1}(x1))))) 18.11/5.52 B_{C_1}(c_{c_1}(c_{a_1}(x1))) -> B_{C_1}(c_{b_1}(b_{c_1}(c_{a_1}(x1)))) 18.11/5.52 B_{C_1}(c_{c_1}(c_{a_1}(x1))) -> C_{B_1}(b_{c_1}(c_{a_1}(x1))) 18.11/5.52 B_{C_1}(c_{c_1}(c_{a_1}(x1))) -> B_{C_1}(c_{a_1}(x1)) 18.11/5.52 B_{C_1}(c_{c_1}(c_{b_1}(x1))) -> B_{A_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{b_1}(x1)))))) 18.11/5.52 B_{C_1}(c_{c_1}(c_{b_1}(x1))) -> A_{B_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{b_1}(x1))))) 18.11/5.52 B_{C_1}(c_{c_1}(c_{b_1}(x1))) -> B_{C_1}(c_{b_1}(b_{c_1}(c_{b_1}(x1)))) 18.11/5.52 B_{C_1}(c_{c_1}(c_{b_1}(x1))) -> C_{B_1}(b_{c_1}(c_{b_1}(x1))) 18.11/5.52 B_{C_1}(c_{c_1}(c_{b_1}(x1))) -> B_{C_1}(c_{b_1}(x1)) 18.11/5.52 B_{C_1}(c_{c_1}(c_{c_1}(x1))) -> B_{A_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{c_1}(x1)))))) 18.11/5.52 B_{C_1}(c_{c_1}(c_{c_1}(x1))) -> A_{B_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{c_1}(x1))))) 18.11/5.52 B_{C_1}(c_{c_1}(c_{c_1}(x1))) -> B_{C_1}(c_{b_1}(b_{c_1}(c_{c_1}(x1)))) 18.11/5.52 B_{C_1}(c_{c_1}(c_{c_1}(x1))) -> C_{B_1}(b_{c_1}(c_{c_1}(x1))) 18.11/5.52 B_{C_1}(c_{c_1}(c_{c_1}(x1))) -> B_{C_1}(c_{c_1}(x1)) 18.11/5.52 18.11/5.52 The TRS R consists of the following rules: 18.11/5.52 18.11/5.52 a_{a_1}(a_{a_1}(x1)) -> a_{a_1}(x1) 18.11/5.52 a_{a_1}(a_{b_1}(x1)) -> a_{b_1}(x1) 18.11/5.52 a_{a_1}(a_{c_1}(x1)) -> a_{c_1}(x1) 18.11/5.52 b_{a_1}(a_{a_1}(x1)) -> b_{a_1}(x1) 18.11/5.52 b_{a_1}(a_{b_1}(x1)) -> b_{b_1}(x1) 18.11/5.52 c_{a_1}(a_{a_1}(x1)) -> c_{a_1}(x1) 18.11/5.52 c_{a_1}(a_{b_1}(x1)) -> c_{b_1}(x1) 18.11/5.52 c_{a_1}(a_{c_1}(x1)) -> c_{c_1}(x1) 18.11/5.52 a_{b_1}(b_{a_1}(a_{a_1}(x1))) -> a_{a_1}(a_{c_1}(c_{a_1}(x1))) 18.11/5.52 a_{b_1}(b_{a_1}(a_{b_1}(x1))) -> a_{a_1}(a_{c_1}(c_{b_1}(x1))) 18.11/5.52 a_{b_1}(b_{a_1}(a_{c_1}(x1))) -> a_{a_1}(a_{c_1}(c_{c_1}(x1))) 18.11/5.52 b_{b_1}(b_{a_1}(a_{a_1}(x1))) -> b_{a_1}(a_{c_1}(c_{a_1}(x1))) 18.11/5.52 b_{b_1}(b_{a_1}(a_{b_1}(x1))) -> b_{a_1}(a_{c_1}(c_{b_1}(x1))) 18.11/5.52 b_{b_1}(b_{a_1}(a_{c_1}(x1))) -> b_{a_1}(a_{c_1}(c_{c_1}(x1))) 18.11/5.52 c_{b_1}(b_{a_1}(a_{a_1}(x1))) -> c_{a_1}(a_{c_1}(c_{a_1}(x1))) 18.11/5.52 c_{b_1}(b_{a_1}(a_{b_1}(x1))) -> c_{a_1}(a_{c_1}(c_{b_1}(x1))) 18.11/5.52 c_{b_1}(b_{a_1}(a_{c_1}(x1))) -> c_{a_1}(a_{c_1}(c_{c_1}(x1))) 18.11/5.52 b_{c_1}(c_{c_1}(c_{a_1}(x1))) -> b_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{a_1}(x1)))))) 18.11/5.52 b_{c_1}(c_{c_1}(c_{b_1}(x1))) -> b_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{b_1}(x1)))))) 18.11/5.52 b_{c_1}(c_{c_1}(c_{c_1}(x1))) -> b_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{c_1}(x1)))))) 18.11/5.52 18.11/5.52 Q is empty. 18.11/5.52 We have to consider all minimal (P,Q,R)-chains. 18.11/5.52 ---------------------------------------- 18.11/5.52 18.11/5.52 (11) DependencyGraphProof (EQUIVALENT) 18.11/5.52 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 23 less nodes. 18.11/5.52 ---------------------------------------- 18.11/5.52 18.11/5.52 (12) 18.11/5.52 Complex Obligation (AND) 18.11/5.52 18.11/5.52 ---------------------------------------- 18.11/5.52 18.11/5.52 (13) 18.11/5.52 Obligation: 18.11/5.52 Q DP problem: 18.11/5.52 The TRS P consists of the following rules: 18.11/5.52 18.11/5.52 C_{A_1}(a_{b_1}(x1)) -> C_{B_1}(x1) 18.11/5.52 C_{B_1}(b_{a_1}(a_{a_1}(x1))) -> C_{A_1}(x1) 18.11/5.52 C_{A_1}(a_{a_1}(x1)) -> C_{A_1}(x1) 18.11/5.52 C_{B_1}(b_{a_1}(a_{b_1}(x1))) -> C_{B_1}(x1) 18.11/5.52 18.11/5.52 The TRS R consists of the following rules: 18.11/5.52 18.11/5.52 a_{a_1}(a_{a_1}(x1)) -> a_{a_1}(x1) 18.11/5.52 a_{a_1}(a_{b_1}(x1)) -> a_{b_1}(x1) 18.11/5.52 a_{a_1}(a_{c_1}(x1)) -> a_{c_1}(x1) 18.11/5.52 b_{a_1}(a_{a_1}(x1)) -> b_{a_1}(x1) 18.11/5.52 b_{a_1}(a_{b_1}(x1)) -> b_{b_1}(x1) 18.11/5.52 c_{a_1}(a_{a_1}(x1)) -> c_{a_1}(x1) 18.11/5.52 c_{a_1}(a_{b_1}(x1)) -> c_{b_1}(x1) 18.11/5.52 c_{a_1}(a_{c_1}(x1)) -> c_{c_1}(x1) 18.11/5.52 a_{b_1}(b_{a_1}(a_{a_1}(x1))) -> a_{a_1}(a_{c_1}(c_{a_1}(x1))) 18.11/5.52 a_{b_1}(b_{a_1}(a_{b_1}(x1))) -> a_{a_1}(a_{c_1}(c_{b_1}(x1))) 18.11/5.52 a_{b_1}(b_{a_1}(a_{c_1}(x1))) -> a_{a_1}(a_{c_1}(c_{c_1}(x1))) 18.11/5.52 b_{b_1}(b_{a_1}(a_{a_1}(x1))) -> b_{a_1}(a_{c_1}(c_{a_1}(x1))) 18.11/5.52 b_{b_1}(b_{a_1}(a_{b_1}(x1))) -> b_{a_1}(a_{c_1}(c_{b_1}(x1))) 18.11/5.52 b_{b_1}(b_{a_1}(a_{c_1}(x1))) -> b_{a_1}(a_{c_1}(c_{c_1}(x1))) 18.11/5.52 c_{b_1}(b_{a_1}(a_{a_1}(x1))) -> c_{a_1}(a_{c_1}(c_{a_1}(x1))) 18.11/5.52 c_{b_1}(b_{a_1}(a_{b_1}(x1))) -> c_{a_1}(a_{c_1}(c_{b_1}(x1))) 18.11/5.52 c_{b_1}(b_{a_1}(a_{c_1}(x1))) -> c_{a_1}(a_{c_1}(c_{c_1}(x1))) 18.11/5.52 b_{c_1}(c_{c_1}(c_{a_1}(x1))) -> b_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{a_1}(x1)))))) 18.11/5.52 b_{c_1}(c_{c_1}(c_{b_1}(x1))) -> b_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{b_1}(x1)))))) 18.11/5.52 b_{c_1}(c_{c_1}(c_{c_1}(x1))) -> b_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{c_1}(x1)))))) 18.11/5.52 18.11/5.52 Q is empty. 18.11/5.52 We have to consider all minimal (P,Q,R)-chains. 18.11/5.52 ---------------------------------------- 18.11/5.52 18.11/5.52 (14) UsableRulesProof (EQUIVALENT) 18.11/5.52 We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. 18.11/5.52 ---------------------------------------- 18.11/5.52 18.11/5.52 (15) 18.11/5.52 Obligation: 18.11/5.52 Q DP problem: 18.11/5.52 The TRS P consists of the following rules: 18.11/5.52 18.11/5.52 C_{A_1}(a_{b_1}(x1)) -> C_{B_1}(x1) 18.11/5.52 C_{B_1}(b_{a_1}(a_{a_1}(x1))) -> C_{A_1}(x1) 18.11/5.52 C_{A_1}(a_{a_1}(x1)) -> C_{A_1}(x1) 18.11/5.52 C_{B_1}(b_{a_1}(a_{b_1}(x1))) -> C_{B_1}(x1) 18.11/5.52 18.11/5.52 R is empty. 18.11/5.52 Q is empty. 18.11/5.52 We have to consider all minimal (P,Q,R)-chains. 18.11/5.52 ---------------------------------------- 18.11/5.52 18.11/5.52 (16) QDPSizeChangeProof (EQUIVALENT) 18.11/5.52 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. 18.11/5.52 18.11/5.52 From the DPs we obtained the following set of size-change graphs: 18.11/5.52 *C_{B_1}(b_{a_1}(a_{a_1}(x1))) -> C_{A_1}(x1) 18.11/5.52 The graph contains the following edges 1 > 1 18.11/5.52 18.11/5.52 18.11/5.52 *C_{B_1}(b_{a_1}(a_{b_1}(x1))) -> C_{B_1}(x1) 18.11/5.52 The graph contains the following edges 1 > 1 18.11/5.52 18.11/5.52 18.11/5.52 *C_{A_1}(a_{a_1}(x1)) -> C_{A_1}(x1) 18.11/5.52 The graph contains the following edges 1 > 1 18.11/5.52 18.11/5.52 18.11/5.52 *C_{A_1}(a_{b_1}(x1)) -> C_{B_1}(x1) 18.11/5.52 The graph contains the following edges 1 > 1 18.11/5.52 18.11/5.52 18.11/5.52 ---------------------------------------- 18.11/5.52 18.11/5.52 (17) 18.11/5.52 YES 18.11/5.52 18.11/5.52 ---------------------------------------- 18.11/5.52 18.11/5.52 (18) 18.11/5.52 Obligation: 18.11/5.52 Q DP problem: 18.11/5.52 The TRS P consists of the following rules: 18.11/5.52 18.11/5.52 B_{A_1}(a_{a_1}(x1)) -> B_{A_1}(x1) 18.11/5.52 18.11/5.52 The TRS R consists of the following rules: 18.11/5.52 18.11/5.52 a_{a_1}(a_{a_1}(x1)) -> a_{a_1}(x1) 18.11/5.52 a_{a_1}(a_{b_1}(x1)) -> a_{b_1}(x1) 18.11/5.52 a_{a_1}(a_{c_1}(x1)) -> a_{c_1}(x1) 18.11/5.52 b_{a_1}(a_{a_1}(x1)) -> b_{a_1}(x1) 18.11/5.52 b_{a_1}(a_{b_1}(x1)) -> b_{b_1}(x1) 18.11/5.52 c_{a_1}(a_{a_1}(x1)) -> c_{a_1}(x1) 18.11/5.52 c_{a_1}(a_{b_1}(x1)) -> c_{b_1}(x1) 18.11/5.52 c_{a_1}(a_{c_1}(x1)) -> c_{c_1}(x1) 18.11/5.52 a_{b_1}(b_{a_1}(a_{a_1}(x1))) -> a_{a_1}(a_{c_1}(c_{a_1}(x1))) 18.11/5.52 a_{b_1}(b_{a_1}(a_{b_1}(x1))) -> a_{a_1}(a_{c_1}(c_{b_1}(x1))) 18.11/5.52 a_{b_1}(b_{a_1}(a_{c_1}(x1))) -> a_{a_1}(a_{c_1}(c_{c_1}(x1))) 18.11/5.52 b_{b_1}(b_{a_1}(a_{a_1}(x1))) -> b_{a_1}(a_{c_1}(c_{a_1}(x1))) 18.11/5.52 b_{b_1}(b_{a_1}(a_{b_1}(x1))) -> b_{a_1}(a_{c_1}(c_{b_1}(x1))) 18.11/5.52 b_{b_1}(b_{a_1}(a_{c_1}(x1))) -> b_{a_1}(a_{c_1}(c_{c_1}(x1))) 18.11/5.52 c_{b_1}(b_{a_1}(a_{a_1}(x1))) -> c_{a_1}(a_{c_1}(c_{a_1}(x1))) 18.11/5.52 c_{b_1}(b_{a_1}(a_{b_1}(x1))) -> c_{a_1}(a_{c_1}(c_{b_1}(x1))) 18.11/5.52 c_{b_1}(b_{a_1}(a_{c_1}(x1))) -> c_{a_1}(a_{c_1}(c_{c_1}(x1))) 18.11/5.52 b_{c_1}(c_{c_1}(c_{a_1}(x1))) -> b_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{a_1}(x1)))))) 18.11/5.52 b_{c_1}(c_{c_1}(c_{b_1}(x1))) -> b_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{b_1}(x1)))))) 18.11/5.52 b_{c_1}(c_{c_1}(c_{c_1}(x1))) -> b_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{c_1}(x1)))))) 18.11/5.52 18.11/5.52 Q is empty. 18.11/5.52 We have to consider all minimal (P,Q,R)-chains. 18.11/5.52 ---------------------------------------- 18.11/5.52 18.11/5.52 (19) UsableRulesProof (EQUIVALENT) 18.11/5.52 We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. 18.11/5.52 ---------------------------------------- 18.11/5.52 18.11/5.52 (20) 18.11/5.52 Obligation: 18.11/5.52 Q DP problem: 18.11/5.52 The TRS P consists of the following rules: 18.11/5.52 18.11/5.52 B_{A_1}(a_{a_1}(x1)) -> B_{A_1}(x1) 18.11/5.52 18.11/5.52 R is empty. 18.11/5.52 Q is empty. 18.11/5.52 We have to consider all minimal (P,Q,R)-chains. 18.11/5.52 ---------------------------------------- 18.11/5.52 18.11/5.52 (21) QDPSizeChangeProof (EQUIVALENT) 18.11/5.52 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. 18.11/5.52 18.11/5.52 From the DPs we obtained the following set of size-change graphs: 18.11/5.52 *B_{A_1}(a_{a_1}(x1)) -> B_{A_1}(x1) 18.11/5.52 The graph contains the following edges 1 > 1 18.11/5.52 18.11/5.52 18.11/5.52 ---------------------------------------- 18.11/5.52 18.11/5.52 (22) 18.11/5.52 YES 18.11/5.52 18.11/5.52 ---------------------------------------- 18.11/5.52 18.11/5.52 (23) 18.11/5.52 Obligation: 18.11/5.52 Q DP problem: 18.11/5.52 The TRS P consists of the following rules: 18.11/5.52 18.11/5.52 B_{C_1}(c_{c_1}(c_{a_1}(x1))) -> B_{C_1}(c_{a_1}(x1)) 18.11/5.52 B_{C_1}(c_{c_1}(c_{a_1}(x1))) -> B_{C_1}(c_{b_1}(b_{c_1}(c_{a_1}(x1)))) 18.11/5.52 B_{C_1}(c_{c_1}(c_{b_1}(x1))) -> B_{C_1}(c_{b_1}(b_{c_1}(c_{b_1}(x1)))) 18.11/5.52 B_{C_1}(c_{c_1}(c_{b_1}(x1))) -> B_{C_1}(c_{b_1}(x1)) 18.11/5.52 B_{C_1}(c_{c_1}(c_{c_1}(x1))) -> B_{C_1}(c_{b_1}(b_{c_1}(c_{c_1}(x1)))) 18.11/5.52 B_{C_1}(c_{c_1}(c_{c_1}(x1))) -> B_{C_1}(c_{c_1}(x1)) 18.11/5.52 18.11/5.52 The TRS R consists of the following rules: 18.11/5.52 18.11/5.52 a_{a_1}(a_{a_1}(x1)) -> a_{a_1}(x1) 18.11/5.52 a_{a_1}(a_{b_1}(x1)) -> a_{b_1}(x1) 18.11/5.52 a_{a_1}(a_{c_1}(x1)) -> a_{c_1}(x1) 18.11/5.52 b_{a_1}(a_{a_1}(x1)) -> b_{a_1}(x1) 18.11/5.52 b_{a_1}(a_{b_1}(x1)) -> b_{b_1}(x1) 18.11/5.52 c_{a_1}(a_{a_1}(x1)) -> c_{a_1}(x1) 18.11/5.52 c_{a_1}(a_{b_1}(x1)) -> c_{b_1}(x1) 18.11/5.52 c_{a_1}(a_{c_1}(x1)) -> c_{c_1}(x1) 18.11/5.52 a_{b_1}(b_{a_1}(a_{a_1}(x1))) -> a_{a_1}(a_{c_1}(c_{a_1}(x1))) 18.11/5.52 a_{b_1}(b_{a_1}(a_{b_1}(x1))) -> a_{a_1}(a_{c_1}(c_{b_1}(x1))) 18.11/5.52 a_{b_1}(b_{a_1}(a_{c_1}(x1))) -> a_{a_1}(a_{c_1}(c_{c_1}(x1))) 18.11/5.52 b_{b_1}(b_{a_1}(a_{a_1}(x1))) -> b_{a_1}(a_{c_1}(c_{a_1}(x1))) 18.11/5.52 b_{b_1}(b_{a_1}(a_{b_1}(x1))) -> b_{a_1}(a_{c_1}(c_{b_1}(x1))) 18.11/5.52 b_{b_1}(b_{a_1}(a_{c_1}(x1))) -> b_{a_1}(a_{c_1}(c_{c_1}(x1))) 18.11/5.52 c_{b_1}(b_{a_1}(a_{a_1}(x1))) -> c_{a_1}(a_{c_1}(c_{a_1}(x1))) 18.11/5.52 c_{b_1}(b_{a_1}(a_{b_1}(x1))) -> c_{a_1}(a_{c_1}(c_{b_1}(x1))) 18.11/5.52 c_{b_1}(b_{a_1}(a_{c_1}(x1))) -> c_{a_1}(a_{c_1}(c_{c_1}(x1))) 18.11/5.52 b_{c_1}(c_{c_1}(c_{a_1}(x1))) -> b_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{a_1}(x1)))))) 18.11/5.52 b_{c_1}(c_{c_1}(c_{b_1}(x1))) -> b_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{b_1}(x1)))))) 18.11/5.52 b_{c_1}(c_{c_1}(c_{c_1}(x1))) -> b_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{c_1}(x1)))))) 18.11/5.52 18.11/5.52 Q is empty. 18.11/5.52 We have to consider all minimal (P,Q,R)-chains. 18.11/5.52 ---------------------------------------- 18.11/5.52 18.11/5.52 (24) UsableRulesProof (EQUIVALENT) 18.11/5.52 We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. 18.11/5.52 ---------------------------------------- 18.11/5.52 18.11/5.52 (25) 18.11/5.52 Obligation: 18.11/5.52 Q DP problem: 18.11/5.52 The TRS P consists of the following rules: 18.11/5.52 18.11/5.52 B_{C_1}(c_{c_1}(c_{a_1}(x1))) -> B_{C_1}(c_{a_1}(x1)) 18.11/5.52 B_{C_1}(c_{c_1}(c_{a_1}(x1))) -> B_{C_1}(c_{b_1}(b_{c_1}(c_{a_1}(x1)))) 18.11/5.52 B_{C_1}(c_{c_1}(c_{b_1}(x1))) -> B_{C_1}(c_{b_1}(b_{c_1}(c_{b_1}(x1)))) 18.11/5.52 B_{C_1}(c_{c_1}(c_{b_1}(x1))) -> B_{C_1}(c_{b_1}(x1)) 18.11/5.52 B_{C_1}(c_{c_1}(c_{c_1}(x1))) -> B_{C_1}(c_{b_1}(b_{c_1}(c_{c_1}(x1)))) 18.11/5.52 B_{C_1}(c_{c_1}(c_{c_1}(x1))) -> B_{C_1}(c_{c_1}(x1)) 18.11/5.52 18.11/5.52 The TRS R consists of the following rules: 18.11/5.52 18.11/5.52 b_{c_1}(c_{c_1}(c_{a_1}(x1))) -> b_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{a_1}(x1)))))) 18.11/5.52 b_{c_1}(c_{c_1}(c_{b_1}(x1))) -> b_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{b_1}(x1)))))) 18.11/5.52 b_{c_1}(c_{c_1}(c_{c_1}(x1))) -> b_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{c_1}(x1)))))) 18.11/5.52 c_{b_1}(b_{a_1}(a_{a_1}(x1))) -> c_{a_1}(a_{c_1}(c_{a_1}(x1))) 18.11/5.52 c_{b_1}(b_{a_1}(a_{b_1}(x1))) -> c_{a_1}(a_{c_1}(c_{b_1}(x1))) 18.11/5.52 c_{b_1}(b_{a_1}(a_{c_1}(x1))) -> c_{a_1}(a_{c_1}(c_{c_1}(x1))) 18.11/5.52 c_{a_1}(a_{c_1}(x1)) -> c_{c_1}(x1) 18.11/5.52 c_{a_1}(a_{a_1}(x1)) -> c_{a_1}(x1) 18.11/5.52 c_{a_1}(a_{b_1}(x1)) -> c_{b_1}(x1) 18.11/5.52 a_{b_1}(b_{a_1}(a_{a_1}(x1))) -> a_{a_1}(a_{c_1}(c_{a_1}(x1))) 18.11/5.52 a_{b_1}(b_{a_1}(a_{b_1}(x1))) -> a_{a_1}(a_{c_1}(c_{b_1}(x1))) 18.11/5.52 a_{b_1}(b_{a_1}(a_{c_1}(x1))) -> a_{a_1}(a_{c_1}(c_{c_1}(x1))) 18.11/5.52 b_{a_1}(a_{a_1}(x1)) -> b_{a_1}(x1) 18.11/5.52 b_{a_1}(a_{b_1}(x1)) -> b_{b_1}(x1) 18.11/5.52 b_{b_1}(b_{a_1}(a_{a_1}(x1))) -> b_{a_1}(a_{c_1}(c_{a_1}(x1))) 18.11/5.52 b_{b_1}(b_{a_1}(a_{b_1}(x1))) -> b_{a_1}(a_{c_1}(c_{b_1}(x1))) 18.11/5.52 b_{b_1}(b_{a_1}(a_{c_1}(x1))) -> b_{a_1}(a_{c_1}(c_{c_1}(x1))) 18.11/5.52 a_{a_1}(a_{c_1}(x1)) -> a_{c_1}(x1) 18.11/5.52 18.11/5.52 Q is empty. 18.11/5.52 We have to consider all minimal (P,Q,R)-chains. 18.11/5.52 ---------------------------------------- 18.11/5.52 18.11/5.52 (26) QDPOrderProof (EQUIVALENT) 18.11/5.52 We use the reduction pair processor [LPAR04,JAR06]. 18.11/5.52 18.11/5.52 18.11/5.52 The following pairs can be oriented strictly and are deleted. 18.11/5.52 18.11/5.52 B_{C_1}(c_{c_1}(c_{a_1}(x1))) -> B_{C_1}(c_{a_1}(x1)) 18.11/5.52 B_{C_1}(c_{c_1}(c_{a_1}(x1))) -> B_{C_1}(c_{b_1}(b_{c_1}(c_{a_1}(x1)))) 18.11/5.52 B_{C_1}(c_{c_1}(c_{b_1}(x1))) -> B_{C_1}(c_{b_1}(b_{c_1}(c_{b_1}(x1)))) 18.11/5.52 B_{C_1}(c_{c_1}(c_{b_1}(x1))) -> B_{C_1}(c_{b_1}(x1)) 18.11/5.52 B_{C_1}(c_{c_1}(c_{c_1}(x1))) -> B_{C_1}(c_{b_1}(b_{c_1}(c_{c_1}(x1)))) 18.11/5.52 B_{C_1}(c_{c_1}(c_{c_1}(x1))) -> B_{C_1}(c_{c_1}(x1)) 18.11/5.52 The remaining pairs can at least be oriented weakly. 18.11/5.52 Used ordering: Polynomial interpretation [POLO]: 18.11/5.52 18.11/5.52 POL(B_{C_1}(x_1)) = x_1 18.11/5.52 POL(a_{a_1}(x_1)) = x_1 18.11/5.52 POL(a_{b_1}(x_1)) = x_1 18.11/5.52 POL(a_{c_1}(x_1)) = 1 + x_1 18.11/5.52 POL(b_{a_1}(x_1)) = 1 + x_1 18.11/5.52 POL(b_{b_1}(x_1)) = 1 + x_1 18.11/5.52 POL(b_{c_1}(x_1)) = x_1 18.11/5.52 POL(c_{a_1}(x_1)) = x_1 18.11/5.52 POL(c_{b_1}(x_1)) = x_1 18.11/5.52 POL(c_{c_1}(x_1)) = 1 + x_1 18.11/5.52 18.11/5.52 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 18.11/5.52 18.11/5.52 c_{a_1}(a_{c_1}(x1)) -> c_{c_1}(x1) 18.11/5.52 c_{a_1}(a_{a_1}(x1)) -> c_{a_1}(x1) 18.11/5.52 c_{a_1}(a_{b_1}(x1)) -> c_{b_1}(x1) 18.11/5.52 b_{c_1}(c_{c_1}(c_{a_1}(x1))) -> b_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{a_1}(x1)))))) 18.11/5.52 b_{c_1}(c_{c_1}(c_{b_1}(x1))) -> b_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{b_1}(x1)))))) 18.11/5.52 b_{c_1}(c_{c_1}(c_{c_1}(x1))) -> b_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{c_1}(x1)))))) 18.11/5.52 c_{b_1}(b_{a_1}(a_{a_1}(x1))) -> c_{a_1}(a_{c_1}(c_{a_1}(x1))) 18.11/5.52 c_{b_1}(b_{a_1}(a_{b_1}(x1))) -> c_{a_1}(a_{c_1}(c_{b_1}(x1))) 18.11/5.52 c_{b_1}(b_{a_1}(a_{c_1}(x1))) -> c_{a_1}(a_{c_1}(c_{c_1}(x1))) 18.11/5.52 a_{b_1}(b_{a_1}(a_{a_1}(x1))) -> a_{a_1}(a_{c_1}(c_{a_1}(x1))) 18.11/5.52 a_{b_1}(b_{a_1}(a_{b_1}(x1))) -> a_{a_1}(a_{c_1}(c_{b_1}(x1))) 18.11/5.52 a_{b_1}(b_{a_1}(a_{c_1}(x1))) -> a_{a_1}(a_{c_1}(c_{c_1}(x1))) 18.11/5.52 b_{a_1}(a_{a_1}(x1)) -> b_{a_1}(x1) 18.11/5.52 b_{a_1}(a_{b_1}(x1)) -> b_{b_1}(x1) 18.11/5.52 a_{a_1}(a_{c_1}(x1)) -> a_{c_1}(x1) 18.11/5.52 b_{b_1}(b_{a_1}(a_{a_1}(x1))) -> b_{a_1}(a_{c_1}(c_{a_1}(x1))) 18.11/5.52 b_{b_1}(b_{a_1}(a_{b_1}(x1))) -> b_{a_1}(a_{c_1}(c_{b_1}(x1))) 18.11/5.52 b_{b_1}(b_{a_1}(a_{c_1}(x1))) -> b_{a_1}(a_{c_1}(c_{c_1}(x1))) 18.11/5.52 18.11/5.52 18.11/5.52 ---------------------------------------- 18.11/5.52 18.11/5.52 (27) 18.11/5.52 Obligation: 18.11/5.52 Q DP problem: 18.11/5.52 P is empty. 18.11/5.52 The TRS R consists of the following rules: 18.11/5.52 18.11/5.52 b_{c_1}(c_{c_1}(c_{a_1}(x1))) -> b_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{a_1}(x1)))))) 18.11/5.52 b_{c_1}(c_{c_1}(c_{b_1}(x1))) -> b_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{b_1}(x1)))))) 18.11/5.52 b_{c_1}(c_{c_1}(c_{c_1}(x1))) -> b_{a_1}(a_{b_1}(b_{c_1}(c_{b_1}(b_{c_1}(c_{c_1}(x1)))))) 18.11/5.52 c_{b_1}(b_{a_1}(a_{a_1}(x1))) -> c_{a_1}(a_{c_1}(c_{a_1}(x1))) 18.11/5.52 c_{b_1}(b_{a_1}(a_{b_1}(x1))) -> c_{a_1}(a_{c_1}(c_{b_1}(x1))) 18.11/5.52 c_{b_1}(b_{a_1}(a_{c_1}(x1))) -> c_{a_1}(a_{c_1}(c_{c_1}(x1))) 18.11/5.52 c_{a_1}(a_{c_1}(x1)) -> c_{c_1}(x1) 18.11/5.52 c_{a_1}(a_{a_1}(x1)) -> c_{a_1}(x1) 18.11/5.52 c_{a_1}(a_{b_1}(x1)) -> c_{b_1}(x1) 18.11/5.52 a_{b_1}(b_{a_1}(a_{a_1}(x1))) -> a_{a_1}(a_{c_1}(c_{a_1}(x1))) 18.11/5.52 a_{b_1}(b_{a_1}(a_{b_1}(x1))) -> a_{a_1}(a_{c_1}(c_{b_1}(x1))) 18.11/5.52 a_{b_1}(b_{a_1}(a_{c_1}(x1))) -> a_{a_1}(a_{c_1}(c_{c_1}(x1))) 18.11/5.52 b_{a_1}(a_{a_1}(x1)) -> b_{a_1}(x1) 18.11/5.52 b_{a_1}(a_{b_1}(x1)) -> b_{b_1}(x1) 18.11/5.52 b_{b_1}(b_{a_1}(a_{a_1}(x1))) -> b_{a_1}(a_{c_1}(c_{a_1}(x1))) 18.11/5.52 b_{b_1}(b_{a_1}(a_{b_1}(x1))) -> b_{a_1}(a_{c_1}(c_{b_1}(x1))) 18.11/5.52 b_{b_1}(b_{a_1}(a_{c_1}(x1))) -> b_{a_1}(a_{c_1}(c_{c_1}(x1))) 18.11/5.52 a_{a_1}(a_{c_1}(x1)) -> a_{c_1}(x1) 18.11/5.52 18.11/5.52 Q is empty. 18.11/5.52 We have to consider all minimal (P,Q,R)-chains. 18.11/5.52 ---------------------------------------- 18.11/5.52 18.11/5.52 (28) PisEmptyProof (EQUIVALENT) 18.11/5.52 The TRS P is empty. Hence, there is no (P,Q,R) chain. 18.11/5.52 ---------------------------------------- 18.11/5.52 18.11/5.52 (29) 18.11/5.52 YES 18.23/5.58 EOF