YES proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 222 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) QDPOrderProof [EQUIVALENT, 22 ms] (11) QDP (12) PisEmptyProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) UsableRulesProof [EQUIVALENT, 16 ms] (16) QDP (17) QDPOrderProof [EQUIVALENT, 142 ms] (18) QDP (19) UsableRulesProof [EQUIVALENT, 0 ms] (20) QDP (21) MRRProof [EQUIVALENT, 35 ms] (22) QDP (23) PisEmptyProof [EQUIVALENT, 0 ms] (24) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(1(1(x1))) -> 1(2(1(2(0(x1))))) 0(3(1(x1))) -> 1(3(2(2(0(x1))))) 0(3(1(x1))) -> 3(2(1(2(0(x1))))) 0(3(1(x1))) -> 1(3(3(3(2(0(x1)))))) 0(4(1(x1))) -> 2(1(2(0(4(x1))))) 0(0(4(5(x1)))) -> 0(0(2(5(4(x1))))) 0(1(4(1(x1)))) -> 0(1(2(2(4(1(x1)))))) 0(1(4(5(x1)))) -> 4(0(1(2(5(4(x1)))))) 0(1(5(1(x1)))) -> 1(2(2(5(0(1(x1)))))) 0(1(5(3(x1)))) -> 0(5(3(2(1(x1))))) 0(2(4(1(x1)))) -> 1(3(3(2(0(4(x1)))))) 0(2(4(1(x1)))) -> 4(2(1(2(0(4(x1)))))) 0(2(4(5(x1)))) -> 0(2(2(5(0(4(x1)))))) 0(3(1(5(x1)))) -> 0(1(2(5(3(x1))))) 0(3(1(5(x1)))) -> 1(2(5(3(0(4(x1)))))) 0(3(5(1(x1)))) -> 1(2(5(3(0(x1))))) 0(3(5(1(x1)))) -> 0(5(2(1(2(3(x1)))))) 0(3(5(5(x1)))) -> 0(3(2(5(5(x1))))) 0(4(0(1(x1)))) -> 2(0(4(4(0(1(x1)))))) 0(4(1(5(x1)))) -> 1(2(5(0(4(x1))))) 0(4(3(5(x1)))) -> 0(4(3(2(5(4(x1)))))) 0(4(5(1(x1)))) -> 2(5(4(4(0(1(x1)))))) 3(0(1(5(x1)))) -> 3(1(4(0(5(4(x1)))))) 3(0(3(1(x1)))) -> 1(3(3(2(0(x1))))) 3(0(3(5(x1)))) -> 3(2(5(0(2(3(x1)))))) 3(3(0(1(x1)))) -> 0(1(3(2(2(3(x1)))))) 3(4(5(1(x1)))) -> 3(2(5(4(2(1(x1)))))) 4(1(3(5(x1)))) -> 1(2(5(3(4(4(x1)))))) 4(1(5(1(x1)))) -> 4(4(5(1(2(1(x1)))))) 4(4(1(5(x1)))) -> 4(1(2(5(4(x1))))) 0(1(4(5(5(x1))))) -> 0(5(1(4(2(5(x1)))))) 0(2(1(4(5(x1))))) -> 0(0(1(2(5(4(x1)))))) 0(2(1(5(5(x1))))) -> 0(1(2(2(5(5(x1)))))) 0(4(2(4(1(x1))))) -> 1(3(2(0(4(4(x1)))))) 0(4(5(4(3(x1))))) -> 2(5(0(4(4(3(x1)))))) 0(5(1(5(1(x1))))) -> 0(5(1(1(2(5(x1)))))) 0(5(2(1(5(x1))))) -> 1(2(5(5(0(4(x1)))))) 0(5(2(4(1(x1))))) -> 4(5(2(1(2(0(x1)))))) 3(0(1(4(1(x1))))) -> 0(4(4(1(3(1(x1)))))) 3(0(1(4(1(x1))))) -> 4(3(2(0(1(1(x1)))))) 3(0(3(5(5(x1))))) -> 3(3(2(5(0(5(x1)))))) 3(0(5(3(1(x1))))) -> 1(0(3(3(2(5(x1)))))) 4(0(1(4(1(x1))))) -> 4(4(0(1(3(1(x1)))))) 4(0(1(5(1(x1))))) -> 0(1(2(5(4(1(x1)))))) 4(0(2(4(5(x1))))) -> 4(0(2(5(0(4(x1)))))) 4(1(1(5(1(x1))))) -> 1(1(2(5(4(1(x1)))))) 4(5(1(4(1(x1))))) -> 4(4(1(2(1(5(x1)))))) 4(5(2(3(1(x1))))) -> 4(3(1(2(2(5(x1)))))) 4(5(4(3(1(x1))))) -> 4(1(2(5(3(4(x1)))))) 4(5(5(3(1(x1))))) -> 1(3(2(5(5(4(x1)))))) Q is empty. ---------------------------------------- (1) QTRS Reverse (EQUIVALENT) We applied the QTRS Reverse Processor [REVERSE]. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 1(1(0(x1))) -> 0(2(1(2(1(x1))))) 1(3(0(x1))) -> 0(2(2(3(1(x1))))) 1(3(0(x1))) -> 0(2(1(2(3(x1))))) 1(3(0(x1))) -> 0(2(3(3(3(1(x1)))))) 1(4(0(x1))) -> 4(0(2(1(2(x1))))) 5(4(0(0(x1)))) -> 4(5(2(0(0(x1))))) 1(4(1(0(x1)))) -> 1(4(2(2(1(0(x1)))))) 5(4(1(0(x1)))) -> 4(5(2(1(0(4(x1)))))) 1(5(1(0(x1)))) -> 1(0(5(2(2(1(x1)))))) 3(5(1(0(x1)))) -> 1(2(3(5(0(x1))))) 1(4(2(0(x1)))) -> 4(0(2(3(3(1(x1)))))) 1(4(2(0(x1)))) -> 4(0(2(1(2(4(x1)))))) 5(4(2(0(x1)))) -> 4(0(5(2(2(0(x1)))))) 5(1(3(0(x1)))) -> 3(5(2(1(0(x1))))) 5(1(3(0(x1)))) -> 4(0(3(5(2(1(x1)))))) 1(5(3(0(x1)))) -> 0(3(5(2(1(x1))))) 1(5(3(0(x1)))) -> 3(2(1(2(5(0(x1)))))) 5(5(3(0(x1)))) -> 5(5(2(3(0(x1))))) 1(0(4(0(x1)))) -> 1(0(4(4(0(2(x1)))))) 5(1(4(0(x1)))) -> 4(0(5(2(1(x1))))) 5(3(4(0(x1)))) -> 4(5(2(3(4(0(x1)))))) 1(5(4(0(x1)))) -> 1(0(4(4(5(2(x1)))))) 5(1(0(3(x1)))) -> 4(5(0(4(1(3(x1)))))) 1(3(0(3(x1)))) -> 0(2(3(3(1(x1))))) 5(3(0(3(x1)))) -> 3(2(0(5(2(3(x1)))))) 1(0(3(3(x1)))) -> 3(2(2(3(1(0(x1)))))) 1(5(4(3(x1)))) -> 1(2(4(5(2(3(x1)))))) 5(3(1(4(x1)))) -> 4(4(3(5(2(1(x1)))))) 1(5(1(4(x1)))) -> 1(2(1(5(4(4(x1)))))) 5(1(4(4(x1)))) -> 4(5(2(1(4(x1))))) 5(5(4(1(0(x1))))) -> 5(2(4(1(5(0(x1)))))) 5(4(1(2(0(x1))))) -> 4(5(2(1(0(0(x1)))))) 5(5(1(2(0(x1))))) -> 5(5(2(2(1(0(x1)))))) 1(4(2(4(0(x1))))) -> 4(4(0(2(3(1(x1)))))) 3(4(5(4(0(x1))))) -> 3(4(4(0(5(2(x1)))))) 1(5(1(5(0(x1))))) -> 5(2(1(1(5(0(x1)))))) 5(1(2(5(0(x1))))) -> 4(0(5(5(2(1(x1)))))) 1(4(2(5(0(x1))))) -> 0(2(1(2(5(4(x1)))))) 1(4(1(0(3(x1))))) -> 1(3(1(4(4(0(x1)))))) 1(4(1(0(3(x1))))) -> 1(1(0(2(3(4(x1)))))) 5(5(3(0(3(x1))))) -> 5(0(5(2(3(3(x1)))))) 1(3(5(0(3(x1))))) -> 5(2(3(3(0(1(x1)))))) 1(4(1(0(4(x1))))) -> 1(3(1(0(4(4(x1)))))) 1(5(1(0(4(x1))))) -> 1(4(5(2(1(0(x1)))))) 5(4(2(0(4(x1))))) -> 4(0(5(2(0(4(x1)))))) 1(5(1(1(4(x1))))) -> 1(4(5(2(1(1(x1)))))) 1(4(1(5(4(x1))))) -> 5(1(2(1(4(4(x1)))))) 1(3(2(5(4(x1))))) -> 5(2(2(1(3(4(x1)))))) 1(3(4(5(4(x1))))) -> 4(3(5(2(1(4(x1)))))) 1(3(5(5(4(x1))))) -> 4(5(5(2(3(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: 1^1(1(0(x1))) -> 1^1(2(1(x1))) 1^1(1(0(x1))) -> 1^1(x1) 1^1(3(0(x1))) -> 3^1(1(x1)) 1^1(3(0(x1))) -> 1^1(x1) 1^1(3(0(x1))) -> 1^1(2(3(x1))) 1^1(3(0(x1))) -> 3^1(x1) 1^1(3(0(x1))) -> 3^1(3(3(1(x1)))) 1^1(3(0(x1))) -> 3^1(3(1(x1))) 1^1(4(0(x1))) -> 1^1(2(x1)) 5^1(4(0(0(x1)))) -> 5^1(2(0(0(x1)))) 1^1(4(1(0(x1)))) -> 1^1(4(2(2(1(0(x1)))))) 5^1(4(1(0(x1)))) -> 5^1(2(1(0(4(x1))))) 5^1(4(1(0(x1)))) -> 1^1(0(4(x1))) 1^1(5(1(0(x1)))) -> 1^1(0(5(2(2(1(x1)))))) 1^1(5(1(0(x1)))) -> 5^1(2(2(1(x1)))) 1^1(5(1(0(x1)))) -> 1^1(x1) 3^1(5(1(0(x1)))) -> 1^1(2(3(5(0(x1))))) 3^1(5(1(0(x1)))) -> 3^1(5(0(x1))) 3^1(5(1(0(x1)))) -> 5^1(0(x1)) 1^1(4(2(0(x1)))) -> 3^1(3(1(x1))) 1^1(4(2(0(x1)))) -> 3^1(1(x1)) 1^1(4(2(0(x1)))) -> 1^1(x1) 1^1(4(2(0(x1)))) -> 1^1(2(4(x1))) 5^1(4(2(0(x1)))) -> 5^1(2(2(0(x1)))) 5^1(1(3(0(x1)))) -> 3^1(5(2(1(0(x1))))) 5^1(1(3(0(x1)))) -> 5^1(2(1(0(x1)))) 5^1(1(3(0(x1)))) -> 1^1(0(x1)) 5^1(1(3(0(x1)))) -> 3^1(5(2(1(x1)))) 5^1(1(3(0(x1)))) -> 5^1(2(1(x1))) 5^1(1(3(0(x1)))) -> 1^1(x1) 1^1(5(3(0(x1)))) -> 3^1(5(2(1(x1)))) 1^1(5(3(0(x1)))) -> 5^1(2(1(x1))) 1^1(5(3(0(x1)))) -> 1^1(x1) 1^1(5(3(0(x1)))) -> 3^1(2(1(2(5(0(x1)))))) 1^1(5(3(0(x1)))) -> 1^1(2(5(0(x1)))) 1^1(5(3(0(x1)))) -> 5^1(0(x1)) 5^1(5(3(0(x1)))) -> 5^1(5(2(3(0(x1))))) 5^1(5(3(0(x1)))) -> 5^1(2(3(0(x1)))) 1^1(0(4(0(x1)))) -> 1^1(0(4(4(0(2(x1)))))) 5^1(1(4(0(x1)))) -> 5^1(2(1(x1))) 5^1(1(4(0(x1)))) -> 1^1(x1) 5^1(3(4(0(x1)))) -> 5^1(2(3(4(0(x1))))) 1^1(5(4(0(x1)))) -> 1^1(0(4(4(5(2(x1)))))) 1^1(5(4(0(x1)))) -> 5^1(2(x1)) 5^1(1(0(3(x1)))) -> 5^1(0(4(1(3(x1))))) 5^1(1(0(3(x1)))) -> 1^1(3(x1)) 1^1(3(0(3(x1)))) -> 3^1(3(1(x1))) 1^1(3(0(3(x1)))) -> 3^1(1(x1)) 1^1(3(0(3(x1)))) -> 1^1(x1) 5^1(3(0(3(x1)))) -> 3^1(2(0(5(2(3(x1)))))) 5^1(3(0(3(x1)))) -> 5^1(2(3(x1))) 1^1(0(3(3(x1)))) -> 3^1(2(2(3(1(0(x1)))))) 1^1(0(3(3(x1)))) -> 3^1(1(0(x1))) 1^1(0(3(3(x1)))) -> 1^1(0(x1)) 1^1(5(4(3(x1)))) -> 1^1(2(4(5(2(3(x1)))))) 1^1(5(4(3(x1)))) -> 5^1(2(3(x1))) 5^1(3(1(4(x1)))) -> 3^1(5(2(1(x1)))) 5^1(3(1(4(x1)))) -> 5^1(2(1(x1))) 5^1(3(1(4(x1)))) -> 1^1(x1) 1^1(5(1(4(x1)))) -> 1^1(2(1(5(4(4(x1)))))) 1^1(5(1(4(x1)))) -> 1^1(5(4(4(x1)))) 1^1(5(1(4(x1)))) -> 5^1(4(4(x1))) 5^1(1(4(4(x1)))) -> 5^1(2(1(4(x1)))) 5^1(1(4(4(x1)))) -> 1^1(4(x1)) 5^1(5(4(1(0(x1))))) -> 5^1(2(4(1(5(0(x1)))))) 5^1(5(4(1(0(x1))))) -> 1^1(5(0(x1))) 5^1(5(4(1(0(x1))))) -> 5^1(0(x1)) 5^1(4(1(2(0(x1))))) -> 5^1(2(1(0(0(x1))))) 5^1(4(1(2(0(x1))))) -> 1^1(0(0(x1))) 5^1(5(1(2(0(x1))))) -> 5^1(5(2(2(1(0(x1)))))) 5^1(5(1(2(0(x1))))) -> 5^1(2(2(1(0(x1))))) 5^1(5(1(2(0(x1))))) -> 1^1(0(x1)) 1^1(4(2(4(0(x1))))) -> 3^1(1(x1)) 1^1(4(2(4(0(x1))))) -> 1^1(x1) 3^1(4(5(4(0(x1))))) -> 3^1(4(4(0(5(2(x1)))))) 3^1(4(5(4(0(x1))))) -> 5^1(2(x1)) 1^1(5(1(5(0(x1))))) -> 5^1(2(1(1(5(0(x1)))))) 1^1(5(1(5(0(x1))))) -> 1^1(1(5(0(x1)))) 5^1(1(2(5(0(x1))))) -> 5^1(5(2(1(x1)))) 5^1(1(2(5(0(x1))))) -> 5^1(2(1(x1))) 5^1(1(2(5(0(x1))))) -> 1^1(x1) 1^1(4(2(5(0(x1))))) -> 1^1(2(5(4(x1)))) 1^1(4(2(5(0(x1))))) -> 5^1(4(x1)) 1^1(4(1(0(3(x1))))) -> 1^1(3(1(4(4(0(x1)))))) 1^1(4(1(0(3(x1))))) -> 3^1(1(4(4(0(x1))))) 1^1(4(1(0(3(x1))))) -> 1^1(4(4(0(x1)))) 1^1(4(1(0(3(x1))))) -> 1^1(1(0(2(3(4(x1)))))) 1^1(4(1(0(3(x1))))) -> 1^1(0(2(3(4(x1))))) 1^1(4(1(0(3(x1))))) -> 3^1(4(x1)) 5^1(5(3(0(3(x1))))) -> 5^1(0(5(2(3(3(x1)))))) 5^1(5(3(0(3(x1))))) -> 5^1(2(3(3(x1)))) 5^1(5(3(0(3(x1))))) -> 3^1(3(x1)) 1^1(3(5(0(3(x1))))) -> 5^1(2(3(3(0(1(x1)))))) 1^1(3(5(0(3(x1))))) -> 3^1(3(0(1(x1)))) 1^1(3(5(0(3(x1))))) -> 3^1(0(1(x1))) 1^1(3(5(0(3(x1))))) -> 1^1(x1) 1^1(4(1(0(4(x1))))) -> 1^1(3(1(0(4(4(x1)))))) 1^1(4(1(0(4(x1))))) -> 3^1(1(0(4(4(x1))))) 1^1(4(1(0(4(x1))))) -> 1^1(0(4(4(x1)))) 1^1(5(1(0(4(x1))))) -> 1^1(4(5(2(1(0(x1)))))) 1^1(5(1(0(4(x1))))) -> 5^1(2(1(0(x1)))) 1^1(5(1(0(4(x1))))) -> 1^1(0(x1)) 5^1(4(2(0(4(x1))))) -> 5^1(2(0(4(x1)))) 1^1(5(1(1(4(x1))))) -> 1^1(4(5(2(1(1(x1)))))) 1^1(5(1(1(4(x1))))) -> 5^1(2(1(1(x1)))) 1^1(5(1(1(4(x1))))) -> 1^1(1(x1)) 1^1(5(1(1(4(x1))))) -> 1^1(x1) 1^1(4(1(5(4(x1))))) -> 5^1(1(2(1(4(4(x1)))))) 1^1(4(1(5(4(x1))))) -> 1^1(2(1(4(4(x1))))) 1^1(4(1(5(4(x1))))) -> 1^1(4(4(x1))) 1^1(3(2(5(4(x1))))) -> 5^1(2(2(1(3(4(x1)))))) 1^1(3(2(5(4(x1))))) -> 1^1(3(4(x1))) 1^1(3(2(5(4(x1))))) -> 3^1(4(x1)) 1^1(3(4(5(4(x1))))) -> 3^1(5(2(1(4(x1))))) 1^1(3(4(5(4(x1))))) -> 5^1(2(1(4(x1)))) 1^1(3(4(5(4(x1))))) -> 1^1(4(x1)) 1^1(3(5(5(4(x1))))) -> 5^1(5(2(3(1(x1))))) 1^1(3(5(5(4(x1))))) -> 5^1(2(3(1(x1)))) 1^1(3(5(5(4(x1))))) -> 3^1(1(x1)) 1^1(3(5(5(4(x1))))) -> 1^1(x1) The TRS R consists of the following rules: 1(1(0(x1))) -> 0(2(1(2(1(x1))))) 1(3(0(x1))) -> 0(2(2(3(1(x1))))) 1(3(0(x1))) -> 0(2(1(2(3(x1))))) 1(3(0(x1))) -> 0(2(3(3(3(1(x1)))))) 1(4(0(x1))) -> 4(0(2(1(2(x1))))) 5(4(0(0(x1)))) -> 4(5(2(0(0(x1))))) 1(4(1(0(x1)))) -> 1(4(2(2(1(0(x1)))))) 5(4(1(0(x1)))) -> 4(5(2(1(0(4(x1)))))) 1(5(1(0(x1)))) -> 1(0(5(2(2(1(x1)))))) 3(5(1(0(x1)))) -> 1(2(3(5(0(x1))))) 1(4(2(0(x1)))) -> 4(0(2(3(3(1(x1)))))) 1(4(2(0(x1)))) -> 4(0(2(1(2(4(x1)))))) 5(4(2(0(x1)))) -> 4(0(5(2(2(0(x1)))))) 5(1(3(0(x1)))) -> 3(5(2(1(0(x1))))) 5(1(3(0(x1)))) -> 4(0(3(5(2(1(x1)))))) 1(5(3(0(x1)))) -> 0(3(5(2(1(x1))))) 1(5(3(0(x1)))) -> 3(2(1(2(5(0(x1)))))) 5(5(3(0(x1)))) -> 5(5(2(3(0(x1))))) 1(0(4(0(x1)))) -> 1(0(4(4(0(2(x1)))))) 5(1(4(0(x1)))) -> 4(0(5(2(1(x1))))) 5(3(4(0(x1)))) -> 4(5(2(3(4(0(x1)))))) 1(5(4(0(x1)))) -> 1(0(4(4(5(2(x1)))))) 5(1(0(3(x1)))) -> 4(5(0(4(1(3(x1)))))) 1(3(0(3(x1)))) -> 0(2(3(3(1(x1))))) 5(3(0(3(x1)))) -> 3(2(0(5(2(3(x1)))))) 1(0(3(3(x1)))) -> 3(2(2(3(1(0(x1)))))) 1(5(4(3(x1)))) -> 1(2(4(5(2(3(x1)))))) 5(3(1(4(x1)))) -> 4(4(3(5(2(1(x1)))))) 1(5(1(4(x1)))) -> 1(2(1(5(4(4(x1)))))) 5(1(4(4(x1)))) -> 4(5(2(1(4(x1))))) 5(5(4(1(0(x1))))) -> 5(2(4(1(5(0(x1)))))) 5(4(1(2(0(x1))))) -> 4(5(2(1(0(0(x1)))))) 5(5(1(2(0(x1))))) -> 5(5(2(2(1(0(x1)))))) 1(4(2(4(0(x1))))) -> 4(4(0(2(3(1(x1)))))) 3(4(5(4(0(x1))))) -> 3(4(4(0(5(2(x1)))))) 1(5(1(5(0(x1))))) -> 5(2(1(1(5(0(x1)))))) 5(1(2(5(0(x1))))) -> 4(0(5(5(2(1(x1)))))) 1(4(2(5(0(x1))))) -> 0(2(1(2(5(4(x1)))))) 1(4(1(0(3(x1))))) -> 1(3(1(4(4(0(x1)))))) 1(4(1(0(3(x1))))) -> 1(1(0(2(3(4(x1)))))) 5(5(3(0(3(x1))))) -> 5(0(5(2(3(3(x1)))))) 1(3(5(0(3(x1))))) -> 5(2(3(3(0(1(x1)))))) 1(4(1(0(4(x1))))) -> 1(3(1(0(4(4(x1)))))) 1(5(1(0(4(x1))))) -> 1(4(5(2(1(0(x1)))))) 5(4(2(0(4(x1))))) -> 4(0(5(2(0(4(x1)))))) 1(5(1(1(4(x1))))) -> 1(4(5(2(1(1(x1)))))) 1(4(1(5(4(x1))))) -> 5(1(2(1(4(4(x1)))))) 1(3(2(5(4(x1))))) -> 5(2(2(1(3(4(x1)))))) 1(3(4(5(4(x1))))) -> 4(3(5(2(1(4(x1)))))) 1(3(5(5(4(x1))))) -> 4(5(5(2(3(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 2 SCCs with 105 less nodes. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (7) Obligation: Q DP problem: The TRS P consists of the following rules: 1^1(0(3(3(x1)))) -> 1^1(0(x1)) The TRS R consists of the following rules: 1(1(0(x1))) -> 0(2(1(2(1(x1))))) 1(3(0(x1))) -> 0(2(2(3(1(x1))))) 1(3(0(x1))) -> 0(2(1(2(3(x1))))) 1(3(0(x1))) -> 0(2(3(3(3(1(x1)))))) 1(4(0(x1))) -> 4(0(2(1(2(x1))))) 5(4(0(0(x1)))) -> 4(5(2(0(0(x1))))) 1(4(1(0(x1)))) -> 1(4(2(2(1(0(x1)))))) 5(4(1(0(x1)))) -> 4(5(2(1(0(4(x1)))))) 1(5(1(0(x1)))) -> 1(0(5(2(2(1(x1)))))) 3(5(1(0(x1)))) -> 1(2(3(5(0(x1))))) 1(4(2(0(x1)))) -> 4(0(2(3(3(1(x1)))))) 1(4(2(0(x1)))) -> 4(0(2(1(2(4(x1)))))) 5(4(2(0(x1)))) -> 4(0(5(2(2(0(x1)))))) 5(1(3(0(x1)))) -> 3(5(2(1(0(x1))))) 5(1(3(0(x1)))) -> 4(0(3(5(2(1(x1)))))) 1(5(3(0(x1)))) -> 0(3(5(2(1(x1))))) 1(5(3(0(x1)))) -> 3(2(1(2(5(0(x1)))))) 5(5(3(0(x1)))) -> 5(5(2(3(0(x1))))) 1(0(4(0(x1)))) -> 1(0(4(4(0(2(x1)))))) 5(1(4(0(x1)))) -> 4(0(5(2(1(x1))))) 5(3(4(0(x1)))) -> 4(5(2(3(4(0(x1)))))) 1(5(4(0(x1)))) -> 1(0(4(4(5(2(x1)))))) 5(1(0(3(x1)))) -> 4(5(0(4(1(3(x1)))))) 1(3(0(3(x1)))) -> 0(2(3(3(1(x1))))) 5(3(0(3(x1)))) -> 3(2(0(5(2(3(x1)))))) 1(0(3(3(x1)))) -> 3(2(2(3(1(0(x1)))))) 1(5(4(3(x1)))) -> 1(2(4(5(2(3(x1)))))) 5(3(1(4(x1)))) -> 4(4(3(5(2(1(x1)))))) 1(5(1(4(x1)))) -> 1(2(1(5(4(4(x1)))))) 5(1(4(4(x1)))) -> 4(5(2(1(4(x1))))) 5(5(4(1(0(x1))))) -> 5(2(4(1(5(0(x1)))))) 5(4(1(2(0(x1))))) -> 4(5(2(1(0(0(x1)))))) 5(5(1(2(0(x1))))) -> 5(5(2(2(1(0(x1)))))) 1(4(2(4(0(x1))))) -> 4(4(0(2(3(1(x1)))))) 3(4(5(4(0(x1))))) -> 3(4(4(0(5(2(x1)))))) 1(5(1(5(0(x1))))) -> 5(2(1(1(5(0(x1)))))) 5(1(2(5(0(x1))))) -> 4(0(5(5(2(1(x1)))))) 1(4(2(5(0(x1))))) -> 0(2(1(2(5(4(x1)))))) 1(4(1(0(3(x1))))) -> 1(3(1(4(4(0(x1)))))) 1(4(1(0(3(x1))))) -> 1(1(0(2(3(4(x1)))))) 5(5(3(0(3(x1))))) -> 5(0(5(2(3(3(x1)))))) 1(3(5(0(3(x1))))) -> 5(2(3(3(0(1(x1)))))) 1(4(1(0(4(x1))))) -> 1(3(1(0(4(4(x1)))))) 1(5(1(0(4(x1))))) -> 1(4(5(2(1(0(x1)))))) 5(4(2(0(4(x1))))) -> 4(0(5(2(0(4(x1)))))) 1(5(1(1(4(x1))))) -> 1(4(5(2(1(1(x1)))))) 1(4(1(5(4(x1))))) -> 5(1(2(1(4(4(x1)))))) 1(3(2(5(4(x1))))) -> 5(2(2(1(3(4(x1)))))) 1(3(4(5(4(x1))))) -> 4(3(5(2(1(4(x1)))))) 1(3(5(5(4(x1))))) -> 4(5(5(2(3(1(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (8) UsableRulesProof (EQUIVALENT) 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. ---------------------------------------- (9) Obligation: Q DP problem: The TRS P consists of the following rules: 1^1(0(3(3(x1)))) -> 1^1(0(x1)) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (10) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. 1^1(0(3(3(x1)))) -> 1^1(0(x1)) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial interpretation [POLO]: POL(0(x_1)) = x_1 POL(1^1(x_1)) = x_1 POL(3(x_1)) = 1 + x_1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: none ---------------------------------------- (11) Obligation: Q DP problem: P is empty. R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (12) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (13) YES ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: 1^1(3(0(x1))) -> 1^1(x1) 1^1(1(0(x1))) -> 1^1(x1) 1^1(5(1(0(x1)))) -> 1^1(x1) 1^1(4(2(0(x1)))) -> 1^1(x1) 1^1(5(3(0(x1)))) -> 1^1(x1) 1^1(3(0(3(x1)))) -> 1^1(x1) 1^1(4(2(4(0(x1))))) -> 1^1(x1) 1^1(4(1(0(3(x1))))) -> 1^1(1(0(2(3(4(x1)))))) 1^1(3(5(0(3(x1))))) -> 1^1(x1) 1^1(5(1(1(4(x1))))) -> 1^1(1(x1)) 1^1(5(1(1(4(x1))))) -> 1^1(x1) 1^1(3(2(5(4(x1))))) -> 1^1(3(4(x1))) 1^1(3(4(5(4(x1))))) -> 1^1(4(x1)) 1^1(3(5(5(4(x1))))) -> 1^1(x1) The TRS R consists of the following rules: 1(1(0(x1))) -> 0(2(1(2(1(x1))))) 1(3(0(x1))) -> 0(2(2(3(1(x1))))) 1(3(0(x1))) -> 0(2(1(2(3(x1))))) 1(3(0(x1))) -> 0(2(3(3(3(1(x1)))))) 1(4(0(x1))) -> 4(0(2(1(2(x1))))) 5(4(0(0(x1)))) -> 4(5(2(0(0(x1))))) 1(4(1(0(x1)))) -> 1(4(2(2(1(0(x1)))))) 5(4(1(0(x1)))) -> 4(5(2(1(0(4(x1)))))) 1(5(1(0(x1)))) -> 1(0(5(2(2(1(x1)))))) 3(5(1(0(x1)))) -> 1(2(3(5(0(x1))))) 1(4(2(0(x1)))) -> 4(0(2(3(3(1(x1)))))) 1(4(2(0(x1)))) -> 4(0(2(1(2(4(x1)))))) 5(4(2(0(x1)))) -> 4(0(5(2(2(0(x1)))))) 5(1(3(0(x1)))) -> 3(5(2(1(0(x1))))) 5(1(3(0(x1)))) -> 4(0(3(5(2(1(x1)))))) 1(5(3(0(x1)))) -> 0(3(5(2(1(x1))))) 1(5(3(0(x1)))) -> 3(2(1(2(5(0(x1)))))) 5(5(3(0(x1)))) -> 5(5(2(3(0(x1))))) 1(0(4(0(x1)))) -> 1(0(4(4(0(2(x1)))))) 5(1(4(0(x1)))) -> 4(0(5(2(1(x1))))) 5(3(4(0(x1)))) -> 4(5(2(3(4(0(x1)))))) 1(5(4(0(x1)))) -> 1(0(4(4(5(2(x1)))))) 5(1(0(3(x1)))) -> 4(5(0(4(1(3(x1)))))) 1(3(0(3(x1)))) -> 0(2(3(3(1(x1))))) 5(3(0(3(x1)))) -> 3(2(0(5(2(3(x1)))))) 1(0(3(3(x1)))) -> 3(2(2(3(1(0(x1)))))) 1(5(4(3(x1)))) -> 1(2(4(5(2(3(x1)))))) 5(3(1(4(x1)))) -> 4(4(3(5(2(1(x1)))))) 1(5(1(4(x1)))) -> 1(2(1(5(4(4(x1)))))) 5(1(4(4(x1)))) -> 4(5(2(1(4(x1))))) 5(5(4(1(0(x1))))) -> 5(2(4(1(5(0(x1)))))) 5(4(1(2(0(x1))))) -> 4(5(2(1(0(0(x1)))))) 5(5(1(2(0(x1))))) -> 5(5(2(2(1(0(x1)))))) 1(4(2(4(0(x1))))) -> 4(4(0(2(3(1(x1)))))) 3(4(5(4(0(x1))))) -> 3(4(4(0(5(2(x1)))))) 1(5(1(5(0(x1))))) -> 5(2(1(1(5(0(x1)))))) 5(1(2(5(0(x1))))) -> 4(0(5(5(2(1(x1)))))) 1(4(2(5(0(x1))))) -> 0(2(1(2(5(4(x1)))))) 1(4(1(0(3(x1))))) -> 1(3(1(4(4(0(x1)))))) 1(4(1(0(3(x1))))) -> 1(1(0(2(3(4(x1)))))) 5(5(3(0(3(x1))))) -> 5(0(5(2(3(3(x1)))))) 1(3(5(0(3(x1))))) -> 5(2(3(3(0(1(x1)))))) 1(4(1(0(4(x1))))) -> 1(3(1(0(4(4(x1)))))) 1(5(1(0(4(x1))))) -> 1(4(5(2(1(0(x1)))))) 5(4(2(0(4(x1))))) -> 4(0(5(2(0(4(x1)))))) 1(5(1(1(4(x1))))) -> 1(4(5(2(1(1(x1)))))) 1(4(1(5(4(x1))))) -> 5(1(2(1(4(4(x1)))))) 1(3(2(5(4(x1))))) -> 5(2(2(1(3(4(x1)))))) 1(3(4(5(4(x1))))) -> 4(3(5(2(1(4(x1)))))) 1(3(5(5(4(x1))))) -> 4(5(5(2(3(1(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (15) UsableRulesProof (EQUIVALENT) 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. ---------------------------------------- (16) Obligation: Q DP problem: The TRS P consists of the following rules: 1^1(3(0(x1))) -> 1^1(x1) 1^1(1(0(x1))) -> 1^1(x1) 1^1(5(1(0(x1)))) -> 1^1(x1) 1^1(4(2(0(x1)))) -> 1^1(x1) 1^1(5(3(0(x1)))) -> 1^1(x1) 1^1(3(0(3(x1)))) -> 1^1(x1) 1^1(4(2(4(0(x1))))) -> 1^1(x1) 1^1(4(1(0(3(x1))))) -> 1^1(1(0(2(3(4(x1)))))) 1^1(3(5(0(3(x1))))) -> 1^1(x1) 1^1(5(1(1(4(x1))))) -> 1^1(1(x1)) 1^1(5(1(1(4(x1))))) -> 1^1(x1) 1^1(3(2(5(4(x1))))) -> 1^1(3(4(x1))) 1^1(3(4(5(4(x1))))) -> 1^1(4(x1)) 1^1(3(5(5(4(x1))))) -> 1^1(x1) The TRS R consists of the following rules: 3(4(5(4(0(x1))))) -> 3(4(4(0(5(2(x1)))))) 1(1(0(x1))) -> 0(2(1(2(1(x1))))) 1(3(0(x1))) -> 0(2(2(3(1(x1))))) 1(3(0(x1))) -> 0(2(1(2(3(x1))))) 1(3(0(x1))) -> 0(2(3(3(3(1(x1)))))) 1(4(0(x1))) -> 4(0(2(1(2(x1))))) 1(4(1(0(x1)))) -> 1(4(2(2(1(0(x1)))))) 1(5(1(0(x1)))) -> 1(0(5(2(2(1(x1)))))) 1(4(2(0(x1)))) -> 4(0(2(3(3(1(x1)))))) 1(4(2(0(x1)))) -> 4(0(2(1(2(4(x1)))))) 1(5(3(0(x1)))) -> 0(3(5(2(1(x1))))) 1(5(3(0(x1)))) -> 3(2(1(2(5(0(x1)))))) 1(0(4(0(x1)))) -> 1(0(4(4(0(2(x1)))))) 1(5(4(0(x1)))) -> 1(0(4(4(5(2(x1)))))) 1(3(0(3(x1)))) -> 0(2(3(3(1(x1))))) 1(0(3(3(x1)))) -> 3(2(2(3(1(0(x1)))))) 1(5(4(3(x1)))) -> 1(2(4(5(2(3(x1)))))) 1(5(1(4(x1)))) -> 1(2(1(5(4(4(x1)))))) 1(4(2(4(0(x1))))) -> 4(4(0(2(3(1(x1)))))) 1(5(1(5(0(x1))))) -> 5(2(1(1(5(0(x1)))))) 1(4(2(5(0(x1))))) -> 0(2(1(2(5(4(x1)))))) 1(4(1(0(3(x1))))) -> 1(3(1(4(4(0(x1)))))) 1(4(1(0(3(x1))))) -> 1(1(0(2(3(4(x1)))))) 1(3(5(0(3(x1))))) -> 5(2(3(3(0(1(x1)))))) 1(4(1(0(4(x1))))) -> 1(3(1(0(4(4(x1)))))) 1(5(1(0(4(x1))))) -> 1(4(5(2(1(0(x1)))))) 1(5(1(1(4(x1))))) -> 1(4(5(2(1(1(x1)))))) 1(4(1(5(4(x1))))) -> 5(1(2(1(4(4(x1)))))) 1(3(2(5(4(x1))))) -> 5(2(2(1(3(4(x1)))))) 1(3(4(5(4(x1))))) -> 4(3(5(2(1(4(x1)))))) 1(3(5(5(4(x1))))) -> 4(5(5(2(3(1(x1)))))) 3(5(1(0(x1)))) -> 1(2(3(5(0(x1))))) 5(4(0(0(x1)))) -> 4(5(2(0(0(x1))))) 5(4(1(0(x1)))) -> 4(5(2(1(0(4(x1)))))) 5(4(2(0(x1)))) -> 4(0(5(2(2(0(x1)))))) 5(4(1(2(0(x1))))) -> 4(5(2(1(0(0(x1)))))) 5(4(2(0(4(x1))))) -> 4(0(5(2(0(4(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (17) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. 1^1(5(1(0(x1)))) -> 1^1(x1) 1^1(5(3(0(x1)))) -> 1^1(x1) 1^1(3(5(0(3(x1))))) -> 1^1(x1) 1^1(5(1(1(4(x1))))) -> 1^1(1(x1)) 1^1(5(1(1(4(x1))))) -> 1^1(x1) 1^1(3(2(5(4(x1))))) -> 1^1(3(4(x1))) 1^1(3(4(5(4(x1))))) -> 1^1(4(x1)) 1^1(3(5(5(4(x1))))) -> 1^1(x1) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial interpretation [POLO]: POL(0(x_1)) = x_1 POL(1(x_1)) = x_1 POL(1^1(x_1)) = x_1 POL(2(x_1)) = x_1 POL(3(x_1)) = x_1 POL(4(x_1)) = x_1 POL(5(x_1)) = 1 + x_1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 3(4(5(4(0(x1))))) -> 3(4(4(0(5(2(x1)))))) 1(1(0(x1))) -> 0(2(1(2(1(x1))))) 1(3(0(x1))) -> 0(2(2(3(1(x1))))) 1(3(0(x1))) -> 0(2(1(2(3(x1))))) 1(3(0(x1))) -> 0(2(3(3(3(1(x1)))))) 1(4(0(x1))) -> 4(0(2(1(2(x1))))) 1(4(1(0(x1)))) -> 1(4(2(2(1(0(x1)))))) 1(5(1(0(x1)))) -> 1(0(5(2(2(1(x1)))))) 1(4(2(0(x1)))) -> 4(0(2(3(3(1(x1)))))) 1(4(2(0(x1)))) -> 4(0(2(1(2(4(x1)))))) 1(5(3(0(x1)))) -> 0(3(5(2(1(x1))))) 1(5(3(0(x1)))) -> 3(2(1(2(5(0(x1)))))) 1(0(4(0(x1)))) -> 1(0(4(4(0(2(x1)))))) 1(5(4(0(x1)))) -> 1(0(4(4(5(2(x1)))))) 1(3(0(3(x1)))) -> 0(2(3(3(1(x1))))) 1(0(3(3(x1)))) -> 3(2(2(3(1(0(x1)))))) 1(5(4(3(x1)))) -> 1(2(4(5(2(3(x1)))))) 1(5(1(4(x1)))) -> 1(2(1(5(4(4(x1)))))) 1(4(2(4(0(x1))))) -> 4(4(0(2(3(1(x1)))))) 1(5(1(5(0(x1))))) -> 5(2(1(1(5(0(x1)))))) 1(4(2(5(0(x1))))) -> 0(2(1(2(5(4(x1)))))) 1(4(1(0(3(x1))))) -> 1(3(1(4(4(0(x1)))))) 1(4(1(0(3(x1))))) -> 1(1(0(2(3(4(x1)))))) 1(3(5(0(3(x1))))) -> 5(2(3(3(0(1(x1)))))) 1(4(1(0(4(x1))))) -> 1(3(1(0(4(4(x1)))))) 1(5(1(0(4(x1))))) -> 1(4(5(2(1(0(x1)))))) 1(5(1(1(4(x1))))) -> 1(4(5(2(1(1(x1)))))) 1(4(1(5(4(x1))))) -> 5(1(2(1(4(4(x1)))))) 1(3(2(5(4(x1))))) -> 5(2(2(1(3(4(x1)))))) 1(3(4(5(4(x1))))) -> 4(3(5(2(1(4(x1)))))) 1(3(5(5(4(x1))))) -> 4(5(5(2(3(1(x1)))))) 3(5(1(0(x1)))) -> 1(2(3(5(0(x1))))) 5(4(0(0(x1)))) -> 4(5(2(0(0(x1))))) 5(4(1(0(x1)))) -> 4(5(2(1(0(4(x1)))))) 5(4(2(0(x1)))) -> 4(0(5(2(2(0(x1)))))) 5(4(1(2(0(x1))))) -> 4(5(2(1(0(0(x1)))))) 5(4(2(0(4(x1))))) -> 4(0(5(2(0(4(x1)))))) ---------------------------------------- (18) Obligation: Q DP problem: The TRS P consists of the following rules: 1^1(3(0(x1))) -> 1^1(x1) 1^1(1(0(x1))) -> 1^1(x1) 1^1(4(2(0(x1)))) -> 1^1(x1) 1^1(3(0(3(x1)))) -> 1^1(x1) 1^1(4(2(4(0(x1))))) -> 1^1(x1) 1^1(4(1(0(3(x1))))) -> 1^1(1(0(2(3(4(x1)))))) The TRS R consists of the following rules: 3(4(5(4(0(x1))))) -> 3(4(4(0(5(2(x1)))))) 1(1(0(x1))) -> 0(2(1(2(1(x1))))) 1(3(0(x1))) -> 0(2(2(3(1(x1))))) 1(3(0(x1))) -> 0(2(1(2(3(x1))))) 1(3(0(x1))) -> 0(2(3(3(3(1(x1)))))) 1(4(0(x1))) -> 4(0(2(1(2(x1))))) 1(4(1(0(x1)))) -> 1(4(2(2(1(0(x1)))))) 1(5(1(0(x1)))) -> 1(0(5(2(2(1(x1)))))) 1(4(2(0(x1)))) -> 4(0(2(3(3(1(x1)))))) 1(4(2(0(x1)))) -> 4(0(2(1(2(4(x1)))))) 1(5(3(0(x1)))) -> 0(3(5(2(1(x1))))) 1(5(3(0(x1)))) -> 3(2(1(2(5(0(x1)))))) 1(0(4(0(x1)))) -> 1(0(4(4(0(2(x1)))))) 1(5(4(0(x1)))) -> 1(0(4(4(5(2(x1)))))) 1(3(0(3(x1)))) -> 0(2(3(3(1(x1))))) 1(0(3(3(x1)))) -> 3(2(2(3(1(0(x1)))))) 1(5(4(3(x1)))) -> 1(2(4(5(2(3(x1)))))) 1(5(1(4(x1)))) -> 1(2(1(5(4(4(x1)))))) 1(4(2(4(0(x1))))) -> 4(4(0(2(3(1(x1)))))) 1(5(1(5(0(x1))))) -> 5(2(1(1(5(0(x1)))))) 1(4(2(5(0(x1))))) -> 0(2(1(2(5(4(x1)))))) 1(4(1(0(3(x1))))) -> 1(3(1(4(4(0(x1)))))) 1(4(1(0(3(x1))))) -> 1(1(0(2(3(4(x1)))))) 1(3(5(0(3(x1))))) -> 5(2(3(3(0(1(x1)))))) 1(4(1(0(4(x1))))) -> 1(3(1(0(4(4(x1)))))) 1(5(1(0(4(x1))))) -> 1(4(5(2(1(0(x1)))))) 1(5(1(1(4(x1))))) -> 1(4(5(2(1(1(x1)))))) 1(4(1(5(4(x1))))) -> 5(1(2(1(4(4(x1)))))) 1(3(2(5(4(x1))))) -> 5(2(2(1(3(4(x1)))))) 1(3(4(5(4(x1))))) -> 4(3(5(2(1(4(x1)))))) 1(3(5(5(4(x1))))) -> 4(5(5(2(3(1(x1)))))) 3(5(1(0(x1)))) -> 1(2(3(5(0(x1))))) 5(4(0(0(x1)))) -> 4(5(2(0(0(x1))))) 5(4(1(0(x1)))) -> 4(5(2(1(0(4(x1)))))) 5(4(2(0(x1)))) -> 4(0(5(2(2(0(x1)))))) 5(4(1(2(0(x1))))) -> 4(5(2(1(0(0(x1)))))) 5(4(2(0(4(x1))))) -> 4(0(5(2(0(4(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (19) UsableRulesProof (EQUIVALENT) 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. ---------------------------------------- (20) Obligation: Q DP problem: The TRS P consists of the following rules: 1^1(3(0(x1))) -> 1^1(x1) 1^1(1(0(x1))) -> 1^1(x1) 1^1(4(2(0(x1)))) -> 1^1(x1) 1^1(3(0(3(x1)))) -> 1^1(x1) 1^1(4(2(4(0(x1))))) -> 1^1(x1) 1^1(4(1(0(3(x1))))) -> 1^1(1(0(2(3(4(x1)))))) The TRS R consists of the following rules: 3(4(5(4(0(x1))))) -> 3(4(4(0(5(2(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (21) MRRProof (EQUIVALENT) By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. Strictly oriented dependency pairs: 1^1(3(0(x1))) -> 1^1(x1) 1^1(1(0(x1))) -> 1^1(x1) 1^1(4(2(0(x1)))) -> 1^1(x1) 1^1(3(0(3(x1)))) -> 1^1(x1) 1^1(4(2(4(0(x1))))) -> 1^1(x1) 1^1(4(1(0(3(x1))))) -> 1^1(1(0(2(3(4(x1)))))) Strictly oriented rules of the TRS R: 3(4(5(4(0(x1))))) -> 3(4(4(0(5(2(x1)))))) Used ordering: Polynomial interpretation [POLO]: POL(0(x_1)) = 3 + x_1 POL(1(x_1)) = 2*x_1 POL(1^1(x_1)) = 3*x_1 POL(2(x_1)) = 1 + x_1 POL(3(x_1)) = 3 + 2*x_1 POL(4(x_1)) = 1 + 2*x_1 POL(5(x_1)) = 2 + 2*x_1 ---------------------------------------- (22) Obligation: Q DP problem: P is empty. R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (23) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (24) YES