/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- 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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 234 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 3382 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 2823 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 2543 ms] (10) QDP (11) DependencyGraphProof [EQUIVALENT, 0 ms] (12) QDP (13) QDPOrderProof [EQUIVALENT, 3435 ms] (14) QDP (15) QDPOrderProof [EQUIVALENT, 4259 ms] (16) QDP (17) QDPOrderProof [EQUIVALENT, 3693 ms] (18) QDP (19) QDPOrderProof [EQUIVALENT, 3171 ms] (20) QDP (21) QDPOrderProof [EQUIVALENT, 2737 ms] (22) QDP (23) DependencyGraphProof [EQUIVALENT, 0 ms] (24) QDP (25) UsableRulesProof [EQUIVALENT, 0 ms] (26) QDP (27) QDPSizeChangeProof [EQUIVALENT, 0 ms] (28) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(1(2(x1))) -> 0(3(1(2(x1)))) 0(1(2(x1))) -> 1(0(2(1(x1)))) 0(1(2(x1))) -> 2(1(4(0(x1)))) 0(1(2(x1))) -> 0(2(1(3(3(x1))))) 0(1(2(x1))) -> 0(2(4(1(3(x1))))) 0(1(2(x1))) -> 1(0(2(5(4(x1))))) 0(1(2(x1))) -> 2(1(1(0(3(x1))))) 0(1(2(x1))) -> 2(1(3(4(0(x1))))) 0(1(2(x1))) -> 2(1(4(3(0(x1))))) 0(1(2(x1))) -> 4(0(3(1(2(x1))))) 0(1(2(x1))) -> 4(1(0(3(2(x1))))) 0(1(2(x1))) -> 4(4(2(1(0(x1))))) 0(1(2(x1))) -> 0(0(4(1(2(5(x1)))))) 0(1(2(x1))) -> 0(4(5(1(2(5(x1)))))) 0(1(2(x1))) -> 0(5(2(3(1(4(x1)))))) 0(1(2(x1))) -> 1(0(2(2(3(4(x1)))))) 0(1(2(x1))) -> 1(1(4(0(3(2(x1)))))) 0(1(2(x1))) -> 2(0(0(3(3(1(x1)))))) 0(1(2(x1))) -> 2(0(3(4(1(5(x1)))))) 0(1(2(x1))) -> 2(3(2(0(4(1(x1)))))) 0(1(2(x1))) -> 2(5(3(0(3(1(x1)))))) 0(1(2(x1))) -> 3(2(5(3(1(0(x1)))))) 0(1(2(x1))) -> 3(4(0(5(1(2(x1)))))) 0(1(2(x1))) -> 5(5(4(2(1(0(x1)))))) 0(0(1(2(x1)))) -> 0(0(2(4(1(x1))))) 0(0(1(2(x1)))) -> 0(2(1(5(0(4(x1)))))) 0(0(1(2(x1)))) -> 1(0(2(3(5(0(x1)))))) 0(0(1(2(x1)))) -> 5(1(2(0(4(0(x1)))))) 0(0(5(2(x1)))) -> 5(3(0(3(0(2(x1)))))) 0(1(2(0(x1)))) -> 1(2(4(0(0(x1))))) 0(1(2(0(x1)))) -> 0(4(1(0(0(2(x1)))))) 0(1(2(2(x1)))) -> 3(3(0(2(1(2(x1)))))) 0(1(5(2(x1)))) -> 2(5(1(0(3(x1))))) 0(1(5(2(x1)))) -> 0(5(5(3(1(2(x1)))))) 0(1(5(2(x1)))) -> 1(3(5(0(2(2(x1)))))) 0(5(2(2(x1)))) -> 3(0(2(1(5(2(x1)))))) 1(0(1(2(x1)))) -> 0(4(1(2(1(x1))))) 1(0(1(2(x1)))) -> 1(2(1(3(0(x1))))) 1(0(1(2(x1)))) -> 0(3(1(4(1(2(x1)))))) 1(0(1(2(x1)))) -> 4(1(4(1(0(2(x1)))))) 5(0(1(2(x1)))) -> 0(2(1(3(5(x1))))) 5(0(1(2(x1)))) -> 2(0(3(1(5(x1))))) 5(0(1(2(x1)))) -> 5(3(1(0(3(2(x1)))))) 5(0(1(2(x1)))) -> 5(5(1(0(2(4(x1)))))) 0(1(0(2(2(x1))))) -> 3(0(2(1(0(2(x1)))))) 0(1(2(5(2(x1))))) -> 3(5(1(2(2(0(x1)))))) 0(1(3(0(0(x1))))) -> 0(0(4(1(3(0(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(1(5(2(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(2(1(1(x1)))))) 1(3(0(1(2(x1))))) -> 1(3(4(1(0(2(x1)))))) Q is empty. ---------------------------------------- (1) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: 0^1(1(2(x1))) -> 0^1(3(1(2(x1)))) 0^1(1(2(x1))) -> 1^1(0(2(1(x1)))) 0^1(1(2(x1))) -> 0^1(2(1(x1))) 0^1(1(2(x1))) -> 1^1(x1) 0^1(1(2(x1))) -> 1^1(4(0(x1))) 0^1(1(2(x1))) -> 0^1(x1) 0^1(1(2(x1))) -> 0^1(2(1(3(3(x1))))) 0^1(1(2(x1))) -> 1^1(3(3(x1))) 0^1(1(2(x1))) -> 0^1(2(4(1(3(x1))))) 0^1(1(2(x1))) -> 1^1(3(x1)) 0^1(1(2(x1))) -> 1^1(0(2(5(4(x1))))) 0^1(1(2(x1))) -> 0^1(2(5(4(x1)))) 0^1(1(2(x1))) -> 5^1(4(x1)) 0^1(1(2(x1))) -> 1^1(1(0(3(x1)))) 0^1(1(2(x1))) -> 1^1(0(3(x1))) 0^1(1(2(x1))) -> 0^1(3(x1)) 0^1(1(2(x1))) -> 1^1(3(4(0(x1)))) 0^1(1(2(x1))) -> 1^1(4(3(0(x1)))) 0^1(1(2(x1))) -> 1^1(0(3(2(x1)))) 0^1(1(2(x1))) -> 0^1(3(2(x1))) 0^1(1(2(x1))) -> 1^1(0(x1)) 0^1(1(2(x1))) -> 0^1(0(4(1(2(5(x1)))))) 0^1(1(2(x1))) -> 0^1(4(1(2(5(x1))))) 0^1(1(2(x1))) -> 1^1(2(5(x1))) 0^1(1(2(x1))) -> 5^1(x1) 0^1(1(2(x1))) -> 0^1(4(5(1(2(5(x1)))))) 0^1(1(2(x1))) -> 5^1(1(2(5(x1)))) 0^1(1(2(x1))) -> 0^1(5(2(3(1(4(x1)))))) 0^1(1(2(x1))) -> 5^1(2(3(1(4(x1))))) 0^1(1(2(x1))) -> 1^1(4(x1)) 0^1(1(2(x1))) -> 1^1(0(2(2(3(4(x1)))))) 0^1(1(2(x1))) -> 0^1(2(2(3(4(x1))))) 0^1(1(2(x1))) -> 1^1(1(4(0(3(2(x1)))))) 0^1(1(2(x1))) -> 1^1(4(0(3(2(x1))))) 0^1(1(2(x1))) -> 0^1(0(3(3(1(x1))))) 0^1(1(2(x1))) -> 0^1(3(3(1(x1)))) 0^1(1(2(x1))) -> 0^1(3(4(1(5(x1))))) 0^1(1(2(x1))) -> 1^1(5(x1)) 0^1(1(2(x1))) -> 0^1(4(1(x1))) 0^1(1(2(x1))) -> 5^1(3(0(3(1(x1))))) 0^1(1(2(x1))) -> 0^1(3(1(x1))) 0^1(1(2(x1))) -> 5^1(3(1(0(x1)))) 0^1(1(2(x1))) -> 0^1(5(1(2(x1)))) 0^1(1(2(x1))) -> 5^1(1(2(x1))) 0^1(1(2(x1))) -> 5^1(5(4(2(1(0(x1)))))) 0^1(1(2(x1))) -> 5^1(4(2(1(0(x1))))) 0^1(0(1(2(x1)))) -> 0^1(0(2(4(1(x1))))) 0^1(0(1(2(x1)))) -> 0^1(2(4(1(x1)))) 0^1(0(1(2(x1)))) -> 1^1(x1) 0^1(0(1(2(x1)))) -> 0^1(2(1(5(0(4(x1)))))) 0^1(0(1(2(x1)))) -> 1^1(5(0(4(x1)))) 0^1(0(1(2(x1)))) -> 5^1(0(4(x1))) 0^1(0(1(2(x1)))) -> 0^1(4(x1)) 0^1(0(1(2(x1)))) -> 1^1(0(2(3(5(0(x1)))))) 0^1(0(1(2(x1)))) -> 0^1(2(3(5(0(x1))))) 0^1(0(1(2(x1)))) -> 5^1(0(x1)) 0^1(0(1(2(x1)))) -> 0^1(x1) 0^1(0(1(2(x1)))) -> 5^1(1(2(0(4(0(x1)))))) 0^1(0(1(2(x1)))) -> 1^1(2(0(4(0(x1))))) 0^1(0(1(2(x1)))) -> 0^1(4(0(x1))) 0^1(0(5(2(x1)))) -> 5^1(3(0(3(0(2(x1)))))) 0^1(0(5(2(x1)))) -> 0^1(3(0(2(x1)))) 0^1(0(5(2(x1)))) -> 0^1(2(x1)) 0^1(1(2(0(x1)))) -> 1^1(2(4(0(0(x1))))) 0^1(1(2(0(x1)))) -> 0^1(0(x1)) 0^1(1(2(0(x1)))) -> 0^1(4(1(0(0(2(x1)))))) 0^1(1(2(0(x1)))) -> 1^1(0(0(2(x1)))) 0^1(1(2(0(x1)))) -> 0^1(0(2(x1))) 0^1(1(2(0(x1)))) -> 0^1(2(x1)) 0^1(1(2(2(x1)))) -> 0^1(2(1(2(x1)))) 0^1(1(2(2(x1)))) -> 1^1(2(x1)) 0^1(1(5(2(x1)))) -> 5^1(1(0(3(x1)))) 0^1(1(5(2(x1)))) -> 1^1(0(3(x1))) 0^1(1(5(2(x1)))) -> 0^1(3(x1)) 0^1(1(5(2(x1)))) -> 0^1(5(5(3(1(2(x1)))))) 0^1(1(5(2(x1)))) -> 5^1(5(3(1(2(x1))))) 0^1(1(5(2(x1)))) -> 5^1(3(1(2(x1)))) 0^1(1(5(2(x1)))) -> 1^1(2(x1)) 0^1(1(5(2(x1)))) -> 1^1(3(5(0(2(2(x1)))))) 0^1(1(5(2(x1)))) -> 5^1(0(2(2(x1)))) 0^1(1(5(2(x1)))) -> 0^1(2(2(x1))) 0^1(5(2(2(x1)))) -> 0^1(2(1(5(2(x1))))) 0^1(5(2(2(x1)))) -> 1^1(5(2(x1))) 0^1(5(2(2(x1)))) -> 5^1(2(x1)) 1^1(0(1(2(x1)))) -> 0^1(4(1(2(1(x1))))) 1^1(0(1(2(x1)))) -> 1^1(2(1(x1))) 1^1(0(1(2(x1)))) -> 1^1(x1) 1^1(0(1(2(x1)))) -> 1^1(2(1(3(0(x1))))) 1^1(0(1(2(x1)))) -> 1^1(3(0(x1))) 1^1(0(1(2(x1)))) -> 0^1(x1) 1^1(0(1(2(x1)))) -> 0^1(3(1(4(1(2(x1)))))) 1^1(0(1(2(x1)))) -> 1^1(4(1(2(x1)))) 1^1(0(1(2(x1)))) -> 1^1(4(1(0(2(x1))))) 1^1(0(1(2(x1)))) -> 1^1(0(2(x1))) 1^1(0(1(2(x1)))) -> 0^1(2(x1)) 5^1(0(1(2(x1)))) -> 0^1(2(1(3(5(x1))))) 5^1(0(1(2(x1)))) -> 1^1(3(5(x1))) 5^1(0(1(2(x1)))) -> 5^1(x1) 5^1(0(1(2(x1)))) -> 0^1(3(1(5(x1)))) 5^1(0(1(2(x1)))) -> 1^1(5(x1)) 5^1(0(1(2(x1)))) -> 5^1(3(1(0(3(2(x1)))))) 5^1(0(1(2(x1)))) -> 1^1(0(3(2(x1)))) 5^1(0(1(2(x1)))) -> 0^1(3(2(x1))) 5^1(0(1(2(x1)))) -> 5^1(5(1(0(2(4(x1)))))) 5^1(0(1(2(x1)))) -> 5^1(1(0(2(4(x1))))) 5^1(0(1(2(x1)))) -> 1^1(0(2(4(x1)))) 5^1(0(1(2(x1)))) -> 0^1(2(4(x1))) 0^1(1(0(2(2(x1))))) -> 0^1(2(1(0(2(x1))))) 0^1(1(0(2(2(x1))))) -> 1^1(0(2(x1))) 0^1(1(0(2(2(x1))))) -> 0^1(2(x1)) 0^1(1(2(5(2(x1))))) -> 5^1(1(2(2(0(x1))))) 0^1(1(2(5(2(x1))))) -> 1^1(2(2(0(x1)))) 0^1(1(2(5(2(x1))))) -> 0^1(x1) 0^1(1(3(0(0(x1))))) -> 0^1(0(4(1(3(0(x1)))))) 0^1(1(3(0(0(x1))))) -> 0^1(4(1(3(0(x1))))) 0^1(1(3(0(0(x1))))) -> 1^1(3(0(x1))) 0^1(1(4(2(2(x1))))) -> 0^1(2(4(1(5(2(x1)))))) 0^1(1(4(2(2(x1))))) -> 1^1(5(2(x1))) 0^1(1(4(2(2(x1))))) -> 5^1(2(x1)) 0^1(1(4(2(2(x1))))) -> 0^1(2(4(2(1(1(x1)))))) 0^1(1(4(2(2(x1))))) -> 1^1(1(x1)) 0^1(1(4(2(2(x1))))) -> 1^1(x1) 1^1(3(0(1(2(x1))))) -> 1^1(3(4(1(0(2(x1)))))) 1^1(3(0(1(2(x1))))) -> 1^1(0(2(x1))) 1^1(3(0(1(2(x1))))) -> 0^1(2(x1)) The TRS R consists of the following rules: 0(1(2(x1))) -> 0(3(1(2(x1)))) 0(1(2(x1))) -> 1(0(2(1(x1)))) 0(1(2(x1))) -> 2(1(4(0(x1)))) 0(1(2(x1))) -> 0(2(1(3(3(x1))))) 0(1(2(x1))) -> 0(2(4(1(3(x1))))) 0(1(2(x1))) -> 1(0(2(5(4(x1))))) 0(1(2(x1))) -> 2(1(1(0(3(x1))))) 0(1(2(x1))) -> 2(1(3(4(0(x1))))) 0(1(2(x1))) -> 2(1(4(3(0(x1))))) 0(1(2(x1))) -> 4(0(3(1(2(x1))))) 0(1(2(x1))) -> 4(1(0(3(2(x1))))) 0(1(2(x1))) -> 4(4(2(1(0(x1))))) 0(1(2(x1))) -> 0(0(4(1(2(5(x1)))))) 0(1(2(x1))) -> 0(4(5(1(2(5(x1)))))) 0(1(2(x1))) -> 0(5(2(3(1(4(x1)))))) 0(1(2(x1))) -> 1(0(2(2(3(4(x1)))))) 0(1(2(x1))) -> 1(1(4(0(3(2(x1)))))) 0(1(2(x1))) -> 2(0(0(3(3(1(x1)))))) 0(1(2(x1))) -> 2(0(3(4(1(5(x1)))))) 0(1(2(x1))) -> 2(3(2(0(4(1(x1)))))) 0(1(2(x1))) -> 2(5(3(0(3(1(x1)))))) 0(1(2(x1))) -> 3(2(5(3(1(0(x1)))))) 0(1(2(x1))) -> 3(4(0(5(1(2(x1)))))) 0(1(2(x1))) -> 5(5(4(2(1(0(x1)))))) 0(0(1(2(x1)))) -> 0(0(2(4(1(x1))))) 0(0(1(2(x1)))) -> 0(2(1(5(0(4(x1)))))) 0(0(1(2(x1)))) -> 1(0(2(3(5(0(x1)))))) 0(0(1(2(x1)))) -> 5(1(2(0(4(0(x1)))))) 0(0(5(2(x1)))) -> 5(3(0(3(0(2(x1)))))) 0(1(2(0(x1)))) -> 1(2(4(0(0(x1))))) 0(1(2(0(x1)))) -> 0(4(1(0(0(2(x1)))))) 0(1(2(2(x1)))) -> 3(3(0(2(1(2(x1)))))) 0(1(5(2(x1)))) -> 2(5(1(0(3(x1))))) 0(1(5(2(x1)))) -> 0(5(5(3(1(2(x1)))))) 0(1(5(2(x1)))) -> 1(3(5(0(2(2(x1)))))) 0(5(2(2(x1)))) -> 3(0(2(1(5(2(x1)))))) 1(0(1(2(x1)))) -> 0(4(1(2(1(x1))))) 1(0(1(2(x1)))) -> 1(2(1(3(0(x1))))) 1(0(1(2(x1)))) -> 0(3(1(4(1(2(x1)))))) 1(0(1(2(x1)))) -> 4(1(4(1(0(2(x1)))))) 5(0(1(2(x1)))) -> 0(2(1(3(5(x1))))) 5(0(1(2(x1)))) -> 2(0(3(1(5(x1))))) 5(0(1(2(x1)))) -> 5(3(1(0(3(2(x1)))))) 5(0(1(2(x1)))) -> 5(5(1(0(2(4(x1)))))) 0(1(0(2(2(x1))))) -> 3(0(2(1(0(2(x1)))))) 0(1(2(5(2(x1))))) -> 3(5(1(2(2(0(x1)))))) 0(1(3(0(0(x1))))) -> 0(0(4(1(3(0(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(1(5(2(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(2(1(1(x1)))))) 1(3(0(1(2(x1))))) -> 1(3(4(1(0(2(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 109 less nodes. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: 0^1(1(2(x1))) -> 1^1(x1) 1^1(0(1(2(x1)))) -> 1^1(x1) 1^1(0(1(2(x1)))) -> 0^1(x1) 0^1(1(2(x1))) -> 0^1(x1) 0^1(1(2(x1))) -> 1^1(0(x1)) 0^1(1(2(x1))) -> 5^1(x1) 5^1(0(1(2(x1)))) -> 5^1(x1) 5^1(0(1(2(x1)))) -> 1^1(5(x1)) 0^1(1(2(x1))) -> 1^1(5(x1)) 0^1(0(1(2(x1)))) -> 1^1(x1) 0^1(0(1(2(x1)))) -> 5^1(0(x1)) 0^1(0(1(2(x1)))) -> 0^1(x1) 0^1(1(2(0(x1)))) -> 0^1(0(x1)) 0^1(1(2(5(2(x1))))) -> 0^1(x1) 0^1(1(4(2(2(x1))))) -> 1^1(1(x1)) 0^1(1(4(2(2(x1))))) -> 1^1(x1) The TRS R consists of the following rules: 0(1(2(x1))) -> 0(3(1(2(x1)))) 0(1(2(x1))) -> 1(0(2(1(x1)))) 0(1(2(x1))) -> 2(1(4(0(x1)))) 0(1(2(x1))) -> 0(2(1(3(3(x1))))) 0(1(2(x1))) -> 0(2(4(1(3(x1))))) 0(1(2(x1))) -> 1(0(2(5(4(x1))))) 0(1(2(x1))) -> 2(1(1(0(3(x1))))) 0(1(2(x1))) -> 2(1(3(4(0(x1))))) 0(1(2(x1))) -> 2(1(4(3(0(x1))))) 0(1(2(x1))) -> 4(0(3(1(2(x1))))) 0(1(2(x1))) -> 4(1(0(3(2(x1))))) 0(1(2(x1))) -> 4(4(2(1(0(x1))))) 0(1(2(x1))) -> 0(0(4(1(2(5(x1)))))) 0(1(2(x1))) -> 0(4(5(1(2(5(x1)))))) 0(1(2(x1))) -> 0(5(2(3(1(4(x1)))))) 0(1(2(x1))) -> 1(0(2(2(3(4(x1)))))) 0(1(2(x1))) -> 1(1(4(0(3(2(x1)))))) 0(1(2(x1))) -> 2(0(0(3(3(1(x1)))))) 0(1(2(x1))) -> 2(0(3(4(1(5(x1)))))) 0(1(2(x1))) -> 2(3(2(0(4(1(x1)))))) 0(1(2(x1))) -> 2(5(3(0(3(1(x1)))))) 0(1(2(x1))) -> 3(2(5(3(1(0(x1)))))) 0(1(2(x1))) -> 3(4(0(5(1(2(x1)))))) 0(1(2(x1))) -> 5(5(4(2(1(0(x1)))))) 0(0(1(2(x1)))) -> 0(0(2(4(1(x1))))) 0(0(1(2(x1)))) -> 0(2(1(5(0(4(x1)))))) 0(0(1(2(x1)))) -> 1(0(2(3(5(0(x1)))))) 0(0(1(2(x1)))) -> 5(1(2(0(4(0(x1)))))) 0(0(5(2(x1)))) -> 5(3(0(3(0(2(x1)))))) 0(1(2(0(x1)))) -> 1(2(4(0(0(x1))))) 0(1(2(0(x1)))) -> 0(4(1(0(0(2(x1)))))) 0(1(2(2(x1)))) -> 3(3(0(2(1(2(x1)))))) 0(1(5(2(x1)))) -> 2(5(1(0(3(x1))))) 0(1(5(2(x1)))) -> 0(5(5(3(1(2(x1)))))) 0(1(5(2(x1)))) -> 1(3(5(0(2(2(x1)))))) 0(5(2(2(x1)))) -> 3(0(2(1(5(2(x1)))))) 1(0(1(2(x1)))) -> 0(4(1(2(1(x1))))) 1(0(1(2(x1)))) -> 1(2(1(3(0(x1))))) 1(0(1(2(x1)))) -> 0(3(1(4(1(2(x1)))))) 1(0(1(2(x1)))) -> 4(1(4(1(0(2(x1)))))) 5(0(1(2(x1)))) -> 0(2(1(3(5(x1))))) 5(0(1(2(x1)))) -> 2(0(3(1(5(x1))))) 5(0(1(2(x1)))) -> 5(3(1(0(3(2(x1)))))) 5(0(1(2(x1)))) -> 5(5(1(0(2(4(x1)))))) 0(1(0(2(2(x1))))) -> 3(0(2(1(0(2(x1)))))) 0(1(2(5(2(x1))))) -> 3(5(1(2(2(0(x1)))))) 0(1(3(0(0(x1))))) -> 0(0(4(1(3(0(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(1(5(2(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(2(1(1(x1)))))) 1(3(0(1(2(x1))))) -> 1(3(4(1(0(2(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. 0^1(1(2(5(2(x1))))) -> 0^1(x1) The remaining pairs can at least be oriented weakly. Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: <<< POL(0^1(x_1)) = [[0A]] + [[0A, 0A, -I]] * x_1 >>> <<< POL(1(x_1)) = [[-I], [-I], [-I]] + [[0A, -I, 0A], [-I, 0A, 1A], [0A, -I, -I]] * x_1 >>> <<< POL(2(x_1)) = [[-I], [0A], [-I]] + [[0A, -I, -I], [0A, 0A, -I], [1A, 0A, 1A]] * x_1 >>> <<< POL(1^1(x_1)) = [[0A]] + [[0A, -I, 0A]] * x_1 >>> <<< POL(0(x_1)) = [[-I], [-I], [-I]] + [[0A, -I, -I], [-I, 0A, -I], [0A, -I, -I]] * x_1 >>> <<< POL(5^1(x_1)) = [[0A]] + [[0A, -I, 0A]] * x_1 >>> <<< POL(5(x_1)) = [[-I], [-I], [-I]] + [[0A, -I, 0A], [-I, 0A, 0A], [0A, -I, -I]] * x_1 >>> <<< POL(4(x_1)) = [[-I], [-I], [-I]] + [[0A, -I, 0A], [0A, -I, 0A], [0A, -I, -I]] * x_1 >>> <<< POL(3(x_1)) = [[-I], [-I], [-I]] + [[0A, -I, -I], [0A, -I, 0A], [-I, -I, -I]] * x_1 >>> The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 0(1(2(x1))) -> 0(3(1(2(x1)))) 0(1(2(x1))) -> 1(0(2(1(x1)))) 0(1(2(x1))) -> 2(1(4(0(x1)))) 0(1(2(x1))) -> 0(2(1(3(3(x1))))) 0(1(2(x1))) -> 0(2(4(1(3(x1))))) 0(1(2(x1))) -> 1(0(2(5(4(x1))))) 0(1(2(x1))) -> 2(1(1(0(3(x1))))) 0(1(2(x1))) -> 2(1(3(4(0(x1))))) 0(1(2(x1))) -> 2(1(4(3(0(x1))))) 0(1(2(x1))) -> 4(0(3(1(2(x1))))) 0(1(2(x1))) -> 4(1(0(3(2(x1))))) 0(1(2(x1))) -> 4(4(2(1(0(x1))))) 0(1(2(x1))) -> 0(0(4(1(2(5(x1)))))) 0(1(2(x1))) -> 0(4(5(1(2(5(x1)))))) 0(1(2(x1))) -> 0(5(2(3(1(4(x1)))))) 0(1(2(x1))) -> 1(0(2(2(3(4(x1)))))) 0(1(2(x1))) -> 1(1(4(0(3(2(x1)))))) 0(1(2(x1))) -> 2(0(0(3(3(1(x1)))))) 0(1(2(x1))) -> 2(0(3(4(1(5(x1)))))) 0(1(2(x1))) -> 2(3(2(0(4(1(x1)))))) 0(1(2(x1))) -> 2(5(3(0(3(1(x1)))))) 0(1(2(x1))) -> 3(2(5(3(1(0(x1)))))) 0(1(2(x1))) -> 3(4(0(5(1(2(x1)))))) 0(1(2(x1))) -> 5(5(4(2(1(0(x1)))))) 0(0(1(2(x1)))) -> 0(0(2(4(1(x1))))) 0(0(1(2(x1)))) -> 0(2(1(5(0(4(x1)))))) 0(0(1(2(x1)))) -> 1(0(2(3(5(0(x1)))))) 0(0(1(2(x1)))) -> 5(1(2(0(4(0(x1)))))) 0(0(5(2(x1)))) -> 5(3(0(3(0(2(x1)))))) 0(1(2(0(x1)))) -> 1(2(4(0(0(x1))))) 0(1(2(0(x1)))) -> 0(4(1(0(0(2(x1)))))) 0(1(2(2(x1)))) -> 3(3(0(2(1(2(x1)))))) 0(1(5(2(x1)))) -> 2(5(1(0(3(x1))))) 0(1(5(2(x1)))) -> 0(5(5(3(1(2(x1)))))) 0(1(5(2(x1)))) -> 1(3(5(0(2(2(x1)))))) 0(5(2(2(x1)))) -> 3(0(2(1(5(2(x1)))))) 0(1(0(2(2(x1))))) -> 3(0(2(1(0(2(x1)))))) 0(1(2(5(2(x1))))) -> 3(5(1(2(2(0(x1)))))) 0(1(3(0(0(x1))))) -> 0(0(4(1(3(0(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(1(5(2(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(2(1(1(x1)))))) 5(0(1(2(x1)))) -> 0(2(1(3(5(x1))))) 5(0(1(2(x1)))) -> 2(0(3(1(5(x1))))) 5(0(1(2(x1)))) -> 5(3(1(0(3(2(x1)))))) 5(0(1(2(x1)))) -> 5(5(1(0(2(4(x1)))))) 1(0(1(2(x1)))) -> 0(4(1(2(1(x1))))) 1(0(1(2(x1)))) -> 1(2(1(3(0(x1))))) 1(0(1(2(x1)))) -> 0(3(1(4(1(2(x1)))))) 1(0(1(2(x1)))) -> 4(1(4(1(0(2(x1)))))) 1(3(0(1(2(x1))))) -> 1(3(4(1(0(2(x1)))))) ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: 0^1(1(2(x1))) -> 1^1(x1) 1^1(0(1(2(x1)))) -> 1^1(x1) 1^1(0(1(2(x1)))) -> 0^1(x1) 0^1(1(2(x1))) -> 0^1(x1) 0^1(1(2(x1))) -> 1^1(0(x1)) 0^1(1(2(x1))) -> 5^1(x1) 5^1(0(1(2(x1)))) -> 5^1(x1) 5^1(0(1(2(x1)))) -> 1^1(5(x1)) 0^1(1(2(x1))) -> 1^1(5(x1)) 0^1(0(1(2(x1)))) -> 1^1(x1) 0^1(0(1(2(x1)))) -> 5^1(0(x1)) 0^1(0(1(2(x1)))) -> 0^1(x1) 0^1(1(2(0(x1)))) -> 0^1(0(x1)) 0^1(1(4(2(2(x1))))) -> 1^1(1(x1)) 0^1(1(4(2(2(x1))))) -> 1^1(x1) The TRS R consists of the following rules: 0(1(2(x1))) -> 0(3(1(2(x1)))) 0(1(2(x1))) -> 1(0(2(1(x1)))) 0(1(2(x1))) -> 2(1(4(0(x1)))) 0(1(2(x1))) -> 0(2(1(3(3(x1))))) 0(1(2(x1))) -> 0(2(4(1(3(x1))))) 0(1(2(x1))) -> 1(0(2(5(4(x1))))) 0(1(2(x1))) -> 2(1(1(0(3(x1))))) 0(1(2(x1))) -> 2(1(3(4(0(x1))))) 0(1(2(x1))) -> 2(1(4(3(0(x1))))) 0(1(2(x1))) -> 4(0(3(1(2(x1))))) 0(1(2(x1))) -> 4(1(0(3(2(x1))))) 0(1(2(x1))) -> 4(4(2(1(0(x1))))) 0(1(2(x1))) -> 0(0(4(1(2(5(x1)))))) 0(1(2(x1))) -> 0(4(5(1(2(5(x1)))))) 0(1(2(x1))) -> 0(5(2(3(1(4(x1)))))) 0(1(2(x1))) -> 1(0(2(2(3(4(x1)))))) 0(1(2(x1))) -> 1(1(4(0(3(2(x1)))))) 0(1(2(x1))) -> 2(0(0(3(3(1(x1)))))) 0(1(2(x1))) -> 2(0(3(4(1(5(x1)))))) 0(1(2(x1))) -> 2(3(2(0(4(1(x1)))))) 0(1(2(x1))) -> 2(5(3(0(3(1(x1)))))) 0(1(2(x1))) -> 3(2(5(3(1(0(x1)))))) 0(1(2(x1))) -> 3(4(0(5(1(2(x1)))))) 0(1(2(x1))) -> 5(5(4(2(1(0(x1)))))) 0(0(1(2(x1)))) -> 0(0(2(4(1(x1))))) 0(0(1(2(x1)))) -> 0(2(1(5(0(4(x1)))))) 0(0(1(2(x1)))) -> 1(0(2(3(5(0(x1)))))) 0(0(1(2(x1)))) -> 5(1(2(0(4(0(x1)))))) 0(0(5(2(x1)))) -> 5(3(0(3(0(2(x1)))))) 0(1(2(0(x1)))) -> 1(2(4(0(0(x1))))) 0(1(2(0(x1)))) -> 0(4(1(0(0(2(x1)))))) 0(1(2(2(x1)))) -> 3(3(0(2(1(2(x1)))))) 0(1(5(2(x1)))) -> 2(5(1(0(3(x1))))) 0(1(5(2(x1)))) -> 0(5(5(3(1(2(x1)))))) 0(1(5(2(x1)))) -> 1(3(5(0(2(2(x1)))))) 0(5(2(2(x1)))) -> 3(0(2(1(5(2(x1)))))) 1(0(1(2(x1)))) -> 0(4(1(2(1(x1))))) 1(0(1(2(x1)))) -> 1(2(1(3(0(x1))))) 1(0(1(2(x1)))) -> 0(3(1(4(1(2(x1)))))) 1(0(1(2(x1)))) -> 4(1(4(1(0(2(x1)))))) 5(0(1(2(x1)))) -> 0(2(1(3(5(x1))))) 5(0(1(2(x1)))) -> 2(0(3(1(5(x1))))) 5(0(1(2(x1)))) -> 5(3(1(0(3(2(x1)))))) 5(0(1(2(x1)))) -> 5(5(1(0(2(4(x1)))))) 0(1(0(2(2(x1))))) -> 3(0(2(1(0(2(x1)))))) 0(1(2(5(2(x1))))) -> 3(5(1(2(2(0(x1)))))) 0(1(3(0(0(x1))))) -> 0(0(4(1(3(0(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(1(5(2(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(2(1(1(x1)))))) 1(3(0(1(2(x1))))) -> 1(3(4(1(0(2(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (7) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. 0^1(0(1(2(x1)))) -> 5^1(0(x1)) The remaining pairs can at least be oriented weakly. Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: <<< POL(0^1(x_1)) = [[0A]] + [[-I, 0A, -I]] * x_1 >>> <<< POL(1(x_1)) = [[1A], [0A], [-I]] + [[0A, 0A, 0A], [0A, 0A, 0A], [-I, 0A, -I]] * x_1 >>> <<< POL(2(x_1)) = [[0A], [-I], [-I]] + [[-I, 0A, 0A], [0A, 0A, 0A], [0A, 0A, 0A]] * x_1 >>> <<< POL(1^1(x_1)) = [[-I]] + [[0A, -I, 0A]] * x_1 >>> <<< POL(0(x_1)) = [[-I], [0A], [-I]] + [[0A, -I, -I], [0A, 0A, 0A], [0A, -I, -I]] * x_1 >>> <<< POL(5^1(x_1)) = [[0A]] + [[-I, -I, -I]] * x_1 >>> <<< POL(5(x_1)) = [[0A], [-I], [0A]] + [[-I, -I, -I], [-I, -I, 0A], [-I, -I, -I]] * x_1 >>> <<< POL(4(x_1)) = [[1A], [0A], [0A]] + [[0A, 0A, 0A], [0A, 0A, 0A], [-I, 0A, 0A]] * x_1 >>> <<< POL(3(x_1)) = [[0A], [-I], [-I]] + [[-I, -I, -I], [-I, -I, -I], [-I, -I, -I]] * x_1 >>> The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 0(1(2(x1))) -> 0(3(1(2(x1)))) 0(1(2(x1))) -> 1(0(2(1(x1)))) 0(1(2(x1))) -> 2(1(4(0(x1)))) 0(1(2(x1))) -> 0(2(1(3(3(x1))))) 0(1(2(x1))) -> 0(2(4(1(3(x1))))) 0(1(2(x1))) -> 1(0(2(5(4(x1))))) 0(1(2(x1))) -> 2(1(1(0(3(x1))))) 0(1(2(x1))) -> 2(1(3(4(0(x1))))) 0(1(2(x1))) -> 2(1(4(3(0(x1))))) 0(1(2(x1))) -> 4(0(3(1(2(x1))))) 0(1(2(x1))) -> 4(1(0(3(2(x1))))) 0(1(2(x1))) -> 4(4(2(1(0(x1))))) 0(1(2(x1))) -> 0(0(4(1(2(5(x1)))))) 0(1(2(x1))) -> 0(4(5(1(2(5(x1)))))) 0(1(2(x1))) -> 0(5(2(3(1(4(x1)))))) 0(1(2(x1))) -> 1(0(2(2(3(4(x1)))))) 0(1(2(x1))) -> 1(1(4(0(3(2(x1)))))) 0(1(2(x1))) -> 2(0(0(3(3(1(x1)))))) 0(1(2(x1))) -> 2(0(3(4(1(5(x1)))))) 0(1(2(x1))) -> 2(3(2(0(4(1(x1)))))) 0(1(2(x1))) -> 2(5(3(0(3(1(x1)))))) 0(1(2(x1))) -> 3(2(5(3(1(0(x1)))))) 0(1(2(x1))) -> 3(4(0(5(1(2(x1)))))) 0(1(2(x1))) -> 5(5(4(2(1(0(x1)))))) 0(0(1(2(x1)))) -> 0(0(2(4(1(x1))))) 0(0(1(2(x1)))) -> 0(2(1(5(0(4(x1)))))) 0(0(1(2(x1)))) -> 1(0(2(3(5(0(x1)))))) 0(0(1(2(x1)))) -> 5(1(2(0(4(0(x1)))))) 0(0(5(2(x1)))) -> 5(3(0(3(0(2(x1)))))) 0(1(2(0(x1)))) -> 1(2(4(0(0(x1))))) 0(1(2(0(x1)))) -> 0(4(1(0(0(2(x1)))))) 0(1(2(2(x1)))) -> 3(3(0(2(1(2(x1)))))) 0(1(5(2(x1)))) -> 2(5(1(0(3(x1))))) 0(1(5(2(x1)))) -> 0(5(5(3(1(2(x1)))))) 0(1(5(2(x1)))) -> 1(3(5(0(2(2(x1)))))) 0(5(2(2(x1)))) -> 3(0(2(1(5(2(x1)))))) 0(1(0(2(2(x1))))) -> 3(0(2(1(0(2(x1)))))) 0(1(2(5(2(x1))))) -> 3(5(1(2(2(0(x1)))))) 0(1(3(0(0(x1))))) -> 0(0(4(1(3(0(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(1(5(2(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(2(1(1(x1)))))) 5(0(1(2(x1)))) -> 0(2(1(3(5(x1))))) 5(0(1(2(x1)))) -> 2(0(3(1(5(x1))))) 5(0(1(2(x1)))) -> 5(3(1(0(3(2(x1)))))) 5(0(1(2(x1)))) -> 5(5(1(0(2(4(x1)))))) 1(0(1(2(x1)))) -> 0(4(1(2(1(x1))))) 1(0(1(2(x1)))) -> 1(2(1(3(0(x1))))) 1(0(1(2(x1)))) -> 0(3(1(4(1(2(x1)))))) 1(0(1(2(x1)))) -> 4(1(4(1(0(2(x1)))))) 1(3(0(1(2(x1))))) -> 1(3(4(1(0(2(x1)))))) ---------------------------------------- (8) Obligation: Q DP problem: The TRS P consists of the following rules: 0^1(1(2(x1))) -> 1^1(x1) 1^1(0(1(2(x1)))) -> 1^1(x1) 1^1(0(1(2(x1)))) -> 0^1(x1) 0^1(1(2(x1))) -> 0^1(x1) 0^1(1(2(x1))) -> 1^1(0(x1)) 0^1(1(2(x1))) -> 5^1(x1) 5^1(0(1(2(x1)))) -> 5^1(x1) 5^1(0(1(2(x1)))) -> 1^1(5(x1)) 0^1(1(2(x1))) -> 1^1(5(x1)) 0^1(0(1(2(x1)))) -> 1^1(x1) 0^1(0(1(2(x1)))) -> 0^1(x1) 0^1(1(2(0(x1)))) -> 0^1(0(x1)) 0^1(1(4(2(2(x1))))) -> 1^1(1(x1)) 0^1(1(4(2(2(x1))))) -> 1^1(x1) The TRS R consists of the following rules: 0(1(2(x1))) -> 0(3(1(2(x1)))) 0(1(2(x1))) -> 1(0(2(1(x1)))) 0(1(2(x1))) -> 2(1(4(0(x1)))) 0(1(2(x1))) -> 0(2(1(3(3(x1))))) 0(1(2(x1))) -> 0(2(4(1(3(x1))))) 0(1(2(x1))) -> 1(0(2(5(4(x1))))) 0(1(2(x1))) -> 2(1(1(0(3(x1))))) 0(1(2(x1))) -> 2(1(3(4(0(x1))))) 0(1(2(x1))) -> 2(1(4(3(0(x1))))) 0(1(2(x1))) -> 4(0(3(1(2(x1))))) 0(1(2(x1))) -> 4(1(0(3(2(x1))))) 0(1(2(x1))) -> 4(4(2(1(0(x1))))) 0(1(2(x1))) -> 0(0(4(1(2(5(x1)))))) 0(1(2(x1))) -> 0(4(5(1(2(5(x1)))))) 0(1(2(x1))) -> 0(5(2(3(1(4(x1)))))) 0(1(2(x1))) -> 1(0(2(2(3(4(x1)))))) 0(1(2(x1))) -> 1(1(4(0(3(2(x1)))))) 0(1(2(x1))) -> 2(0(0(3(3(1(x1)))))) 0(1(2(x1))) -> 2(0(3(4(1(5(x1)))))) 0(1(2(x1))) -> 2(3(2(0(4(1(x1)))))) 0(1(2(x1))) -> 2(5(3(0(3(1(x1)))))) 0(1(2(x1))) -> 3(2(5(3(1(0(x1)))))) 0(1(2(x1))) -> 3(4(0(5(1(2(x1)))))) 0(1(2(x1))) -> 5(5(4(2(1(0(x1)))))) 0(0(1(2(x1)))) -> 0(0(2(4(1(x1))))) 0(0(1(2(x1)))) -> 0(2(1(5(0(4(x1)))))) 0(0(1(2(x1)))) -> 1(0(2(3(5(0(x1)))))) 0(0(1(2(x1)))) -> 5(1(2(0(4(0(x1)))))) 0(0(5(2(x1)))) -> 5(3(0(3(0(2(x1)))))) 0(1(2(0(x1)))) -> 1(2(4(0(0(x1))))) 0(1(2(0(x1)))) -> 0(4(1(0(0(2(x1)))))) 0(1(2(2(x1)))) -> 3(3(0(2(1(2(x1)))))) 0(1(5(2(x1)))) -> 2(5(1(0(3(x1))))) 0(1(5(2(x1)))) -> 0(5(5(3(1(2(x1)))))) 0(1(5(2(x1)))) -> 1(3(5(0(2(2(x1)))))) 0(5(2(2(x1)))) -> 3(0(2(1(5(2(x1)))))) 1(0(1(2(x1)))) -> 0(4(1(2(1(x1))))) 1(0(1(2(x1)))) -> 1(2(1(3(0(x1))))) 1(0(1(2(x1)))) -> 0(3(1(4(1(2(x1)))))) 1(0(1(2(x1)))) -> 4(1(4(1(0(2(x1)))))) 5(0(1(2(x1)))) -> 0(2(1(3(5(x1))))) 5(0(1(2(x1)))) -> 2(0(3(1(5(x1))))) 5(0(1(2(x1)))) -> 5(3(1(0(3(2(x1)))))) 5(0(1(2(x1)))) -> 5(5(1(0(2(4(x1)))))) 0(1(0(2(2(x1))))) -> 3(0(2(1(0(2(x1)))))) 0(1(2(5(2(x1))))) -> 3(5(1(2(2(0(x1)))))) 0(1(3(0(0(x1))))) -> 0(0(4(1(3(0(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(1(5(2(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(2(1(1(x1)))))) 1(3(0(1(2(x1))))) -> 1(3(4(1(0(2(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (9) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. 5^1(0(1(2(x1)))) -> 5^1(x1) 5^1(0(1(2(x1)))) -> 1^1(5(x1)) The remaining pairs can at least be oriented weakly. Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: <<< POL(0^1(x_1)) = [[0A]] + [[-I, 0A, 0A]] * x_1 >>> <<< POL(1(x_1)) = [[0A], [0A], [1A]] + [[-I, 0A, -I], [-I, -I, -I], [0A, 0A, 0A]] * x_1 >>> <<< POL(2(x_1)) = [[0A], [-I], [-I]] + [[0A, 0A, 0A], [-I, 0A, -I], [-I, -I, -I]] * x_1 >>> <<< POL(1^1(x_1)) = [[1A]] + [[-I, 0A, -I]] * x_1 >>> <<< POL(0(x_1)) = [[0A], [-I], [-I]] + [[0A, 0A, 1A], [-I, 0A, 0A], [-I, -I, 0A]] * x_1 >>> <<< POL(5^1(x_1)) = [[1A]] + [[0A, 0A, 0A]] * x_1 >>> <<< POL(5(x_1)) = [[0A], [0A], [0A]] + [[-I, -I, 0A], [0A, 0A, 0A], [0A, 0A, 0A]] * x_1 >>> <<< POL(4(x_1)) = [[0A], [-I], [-I]] + [[-I, -I, -I], [-I, 0A, -I], [-I, -I, -I]] * x_1 >>> <<< POL(3(x_1)) = [[0A], [-I], [-I]] + [[0A, -I, -I], [-I, -I, -I], [-I, -I, -I]] * x_1 >>> The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 0(1(2(x1))) -> 0(3(1(2(x1)))) 0(1(2(x1))) -> 1(0(2(1(x1)))) 0(1(2(x1))) -> 2(1(4(0(x1)))) 0(1(2(x1))) -> 0(2(1(3(3(x1))))) 0(1(2(x1))) -> 0(2(4(1(3(x1))))) 0(1(2(x1))) -> 1(0(2(5(4(x1))))) 0(1(2(x1))) -> 2(1(1(0(3(x1))))) 0(1(2(x1))) -> 2(1(3(4(0(x1))))) 0(1(2(x1))) -> 2(1(4(3(0(x1))))) 0(1(2(x1))) -> 4(0(3(1(2(x1))))) 0(1(2(x1))) -> 4(1(0(3(2(x1))))) 0(1(2(x1))) -> 4(4(2(1(0(x1))))) 0(1(2(x1))) -> 0(0(4(1(2(5(x1)))))) 0(1(2(x1))) -> 0(4(5(1(2(5(x1)))))) 0(1(2(x1))) -> 0(5(2(3(1(4(x1)))))) 0(1(2(x1))) -> 1(0(2(2(3(4(x1)))))) 0(1(2(x1))) -> 1(1(4(0(3(2(x1)))))) 0(1(2(x1))) -> 2(0(0(3(3(1(x1)))))) 0(1(2(x1))) -> 2(0(3(4(1(5(x1)))))) 0(1(2(x1))) -> 2(3(2(0(4(1(x1)))))) 0(1(2(x1))) -> 2(5(3(0(3(1(x1)))))) 0(1(2(x1))) -> 3(2(5(3(1(0(x1)))))) 0(1(2(x1))) -> 3(4(0(5(1(2(x1)))))) 0(1(2(x1))) -> 5(5(4(2(1(0(x1)))))) 0(0(1(2(x1)))) -> 0(0(2(4(1(x1))))) 0(0(1(2(x1)))) -> 0(2(1(5(0(4(x1)))))) 0(0(1(2(x1)))) -> 1(0(2(3(5(0(x1)))))) 0(0(1(2(x1)))) -> 5(1(2(0(4(0(x1)))))) 0(0(5(2(x1)))) -> 5(3(0(3(0(2(x1)))))) 0(1(2(0(x1)))) -> 1(2(4(0(0(x1))))) 0(1(2(0(x1)))) -> 0(4(1(0(0(2(x1)))))) 0(1(2(2(x1)))) -> 3(3(0(2(1(2(x1)))))) 0(1(5(2(x1)))) -> 2(5(1(0(3(x1))))) 0(1(5(2(x1)))) -> 0(5(5(3(1(2(x1)))))) 0(1(5(2(x1)))) -> 1(3(5(0(2(2(x1)))))) 0(5(2(2(x1)))) -> 3(0(2(1(5(2(x1)))))) 0(1(0(2(2(x1))))) -> 3(0(2(1(0(2(x1)))))) 0(1(2(5(2(x1))))) -> 3(5(1(2(2(0(x1)))))) 0(1(3(0(0(x1))))) -> 0(0(4(1(3(0(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(1(5(2(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(2(1(1(x1)))))) 5(0(1(2(x1)))) -> 0(2(1(3(5(x1))))) 5(0(1(2(x1)))) -> 2(0(3(1(5(x1))))) 5(0(1(2(x1)))) -> 5(3(1(0(3(2(x1)))))) 5(0(1(2(x1)))) -> 5(5(1(0(2(4(x1)))))) 1(0(1(2(x1)))) -> 0(4(1(2(1(x1))))) 1(0(1(2(x1)))) -> 1(2(1(3(0(x1))))) 1(0(1(2(x1)))) -> 0(3(1(4(1(2(x1)))))) 1(0(1(2(x1)))) -> 4(1(4(1(0(2(x1)))))) 1(3(0(1(2(x1))))) -> 1(3(4(1(0(2(x1)))))) ---------------------------------------- (10) Obligation: Q DP problem: The TRS P consists of the following rules: 0^1(1(2(x1))) -> 1^1(x1) 1^1(0(1(2(x1)))) -> 1^1(x1) 1^1(0(1(2(x1)))) -> 0^1(x1) 0^1(1(2(x1))) -> 0^1(x1) 0^1(1(2(x1))) -> 1^1(0(x1)) 0^1(1(2(x1))) -> 5^1(x1) 0^1(1(2(x1))) -> 1^1(5(x1)) 0^1(0(1(2(x1)))) -> 1^1(x1) 0^1(0(1(2(x1)))) -> 0^1(x1) 0^1(1(2(0(x1)))) -> 0^1(0(x1)) 0^1(1(4(2(2(x1))))) -> 1^1(1(x1)) 0^1(1(4(2(2(x1))))) -> 1^1(x1) The TRS R consists of the following rules: 0(1(2(x1))) -> 0(3(1(2(x1)))) 0(1(2(x1))) -> 1(0(2(1(x1)))) 0(1(2(x1))) -> 2(1(4(0(x1)))) 0(1(2(x1))) -> 0(2(1(3(3(x1))))) 0(1(2(x1))) -> 0(2(4(1(3(x1))))) 0(1(2(x1))) -> 1(0(2(5(4(x1))))) 0(1(2(x1))) -> 2(1(1(0(3(x1))))) 0(1(2(x1))) -> 2(1(3(4(0(x1))))) 0(1(2(x1))) -> 2(1(4(3(0(x1))))) 0(1(2(x1))) -> 4(0(3(1(2(x1))))) 0(1(2(x1))) -> 4(1(0(3(2(x1))))) 0(1(2(x1))) -> 4(4(2(1(0(x1))))) 0(1(2(x1))) -> 0(0(4(1(2(5(x1)))))) 0(1(2(x1))) -> 0(4(5(1(2(5(x1)))))) 0(1(2(x1))) -> 0(5(2(3(1(4(x1)))))) 0(1(2(x1))) -> 1(0(2(2(3(4(x1)))))) 0(1(2(x1))) -> 1(1(4(0(3(2(x1)))))) 0(1(2(x1))) -> 2(0(0(3(3(1(x1)))))) 0(1(2(x1))) -> 2(0(3(4(1(5(x1)))))) 0(1(2(x1))) -> 2(3(2(0(4(1(x1)))))) 0(1(2(x1))) -> 2(5(3(0(3(1(x1)))))) 0(1(2(x1))) -> 3(2(5(3(1(0(x1)))))) 0(1(2(x1))) -> 3(4(0(5(1(2(x1)))))) 0(1(2(x1))) -> 5(5(4(2(1(0(x1)))))) 0(0(1(2(x1)))) -> 0(0(2(4(1(x1))))) 0(0(1(2(x1)))) -> 0(2(1(5(0(4(x1)))))) 0(0(1(2(x1)))) -> 1(0(2(3(5(0(x1)))))) 0(0(1(2(x1)))) -> 5(1(2(0(4(0(x1)))))) 0(0(5(2(x1)))) -> 5(3(0(3(0(2(x1)))))) 0(1(2(0(x1)))) -> 1(2(4(0(0(x1))))) 0(1(2(0(x1)))) -> 0(4(1(0(0(2(x1)))))) 0(1(2(2(x1)))) -> 3(3(0(2(1(2(x1)))))) 0(1(5(2(x1)))) -> 2(5(1(0(3(x1))))) 0(1(5(2(x1)))) -> 0(5(5(3(1(2(x1)))))) 0(1(5(2(x1)))) -> 1(3(5(0(2(2(x1)))))) 0(5(2(2(x1)))) -> 3(0(2(1(5(2(x1)))))) 1(0(1(2(x1)))) -> 0(4(1(2(1(x1))))) 1(0(1(2(x1)))) -> 1(2(1(3(0(x1))))) 1(0(1(2(x1)))) -> 0(3(1(4(1(2(x1)))))) 1(0(1(2(x1)))) -> 4(1(4(1(0(2(x1)))))) 5(0(1(2(x1)))) -> 0(2(1(3(5(x1))))) 5(0(1(2(x1)))) -> 2(0(3(1(5(x1))))) 5(0(1(2(x1)))) -> 5(3(1(0(3(2(x1)))))) 5(0(1(2(x1)))) -> 5(5(1(0(2(4(x1)))))) 0(1(0(2(2(x1))))) -> 3(0(2(1(0(2(x1)))))) 0(1(2(5(2(x1))))) -> 3(5(1(2(2(0(x1)))))) 0(1(3(0(0(x1))))) -> 0(0(4(1(3(0(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(1(5(2(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(2(1(1(x1)))))) 1(3(0(1(2(x1))))) -> 1(3(4(1(0(2(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (11) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (12) Obligation: Q DP problem: The TRS P consists of the following rules: 1^1(0(1(2(x1)))) -> 1^1(x1) 1^1(0(1(2(x1)))) -> 0^1(x1) 0^1(1(2(x1))) -> 1^1(x1) 0^1(1(2(x1))) -> 0^1(x1) 0^1(1(2(x1))) -> 1^1(0(x1)) 0^1(1(2(x1))) -> 1^1(5(x1)) 0^1(0(1(2(x1)))) -> 1^1(x1) 0^1(0(1(2(x1)))) -> 0^1(x1) 0^1(1(2(0(x1)))) -> 0^1(0(x1)) 0^1(1(4(2(2(x1))))) -> 1^1(1(x1)) 0^1(1(4(2(2(x1))))) -> 1^1(x1) The TRS R consists of the following rules: 0(1(2(x1))) -> 0(3(1(2(x1)))) 0(1(2(x1))) -> 1(0(2(1(x1)))) 0(1(2(x1))) -> 2(1(4(0(x1)))) 0(1(2(x1))) -> 0(2(1(3(3(x1))))) 0(1(2(x1))) -> 0(2(4(1(3(x1))))) 0(1(2(x1))) -> 1(0(2(5(4(x1))))) 0(1(2(x1))) -> 2(1(1(0(3(x1))))) 0(1(2(x1))) -> 2(1(3(4(0(x1))))) 0(1(2(x1))) -> 2(1(4(3(0(x1))))) 0(1(2(x1))) -> 4(0(3(1(2(x1))))) 0(1(2(x1))) -> 4(1(0(3(2(x1))))) 0(1(2(x1))) -> 4(4(2(1(0(x1))))) 0(1(2(x1))) -> 0(0(4(1(2(5(x1)))))) 0(1(2(x1))) -> 0(4(5(1(2(5(x1)))))) 0(1(2(x1))) -> 0(5(2(3(1(4(x1)))))) 0(1(2(x1))) -> 1(0(2(2(3(4(x1)))))) 0(1(2(x1))) -> 1(1(4(0(3(2(x1)))))) 0(1(2(x1))) -> 2(0(0(3(3(1(x1)))))) 0(1(2(x1))) -> 2(0(3(4(1(5(x1)))))) 0(1(2(x1))) -> 2(3(2(0(4(1(x1)))))) 0(1(2(x1))) -> 2(5(3(0(3(1(x1)))))) 0(1(2(x1))) -> 3(2(5(3(1(0(x1)))))) 0(1(2(x1))) -> 3(4(0(5(1(2(x1)))))) 0(1(2(x1))) -> 5(5(4(2(1(0(x1)))))) 0(0(1(2(x1)))) -> 0(0(2(4(1(x1))))) 0(0(1(2(x1)))) -> 0(2(1(5(0(4(x1)))))) 0(0(1(2(x1)))) -> 1(0(2(3(5(0(x1)))))) 0(0(1(2(x1)))) -> 5(1(2(0(4(0(x1)))))) 0(0(5(2(x1)))) -> 5(3(0(3(0(2(x1)))))) 0(1(2(0(x1)))) -> 1(2(4(0(0(x1))))) 0(1(2(0(x1)))) -> 0(4(1(0(0(2(x1)))))) 0(1(2(2(x1)))) -> 3(3(0(2(1(2(x1)))))) 0(1(5(2(x1)))) -> 2(5(1(0(3(x1))))) 0(1(5(2(x1)))) -> 0(5(5(3(1(2(x1)))))) 0(1(5(2(x1)))) -> 1(3(5(0(2(2(x1)))))) 0(5(2(2(x1)))) -> 3(0(2(1(5(2(x1)))))) 1(0(1(2(x1)))) -> 0(4(1(2(1(x1))))) 1(0(1(2(x1)))) -> 1(2(1(3(0(x1))))) 1(0(1(2(x1)))) -> 0(3(1(4(1(2(x1)))))) 1(0(1(2(x1)))) -> 4(1(4(1(0(2(x1)))))) 5(0(1(2(x1)))) -> 0(2(1(3(5(x1))))) 5(0(1(2(x1)))) -> 2(0(3(1(5(x1))))) 5(0(1(2(x1)))) -> 5(3(1(0(3(2(x1)))))) 5(0(1(2(x1)))) -> 5(5(1(0(2(4(x1)))))) 0(1(0(2(2(x1))))) -> 3(0(2(1(0(2(x1)))))) 0(1(2(5(2(x1))))) -> 3(5(1(2(2(0(x1)))))) 0(1(3(0(0(x1))))) -> 0(0(4(1(3(0(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(1(5(2(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(2(1(1(x1)))))) 1(3(0(1(2(x1))))) -> 1(3(4(1(0(2(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. 0^1(0(1(2(x1)))) -> 1^1(x1) 0^1(0(1(2(x1)))) -> 0^1(x1) The remaining pairs can at least be oriented weakly. Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: <<< POL(1^1(x_1)) = [[1A]] + [[0A, -I, 0A]] * x_1 >>> <<< POL(0(x_1)) = [[0A], [0A], [-I]] + [[0A, 0A, -I], [1A, 0A, 0A], [-I, 0A, -I]] * x_1 >>> <<< POL(1(x_1)) = [[1A], [1A], [0A]] + [[0A, -I, 0A], [0A, -I, 0A], [0A, -I, 0A]] * x_1 >>> <<< POL(2(x_1)) = [[0A], [1A], [0A]] + [[0A, 0A, 0A], [0A, 0A, 0A], [0A, 0A, 0A]] * x_1 >>> <<< POL(0^1(x_1)) = [[0A]] + [[0A, 0A, 0A]] * x_1 >>> <<< POL(5(x_1)) = [[1A], [0A], [1A]] + [[-I, -I, -I], [0A, 0A, -I], [-I, -I, -I]] * x_1 >>> <<< POL(4(x_1)) = [[0A], [0A], [0A]] + [[-I, -I, -I], [-I, -I, -I], [-I, -I, 0A]] * x_1 >>> <<< POL(3(x_1)) = [[0A], [-I], [0A]] + [[-I, -I, -I], [-I, -I, -I], [-I, -I, 0A]] * x_1 >>> The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 0(1(2(x1))) -> 0(3(1(2(x1)))) 0(1(2(x1))) -> 1(0(2(1(x1)))) 0(1(2(x1))) -> 2(1(4(0(x1)))) 0(1(2(x1))) -> 0(2(1(3(3(x1))))) 0(1(2(x1))) -> 0(2(4(1(3(x1))))) 0(1(2(x1))) -> 1(0(2(5(4(x1))))) 0(1(2(x1))) -> 2(1(1(0(3(x1))))) 0(1(2(x1))) -> 2(1(3(4(0(x1))))) 0(1(2(x1))) -> 2(1(4(3(0(x1))))) 0(1(2(x1))) -> 4(0(3(1(2(x1))))) 0(1(2(x1))) -> 4(1(0(3(2(x1))))) 0(1(2(x1))) -> 4(4(2(1(0(x1))))) 0(1(2(x1))) -> 0(0(4(1(2(5(x1)))))) 0(1(2(x1))) -> 0(4(5(1(2(5(x1)))))) 0(1(2(x1))) -> 0(5(2(3(1(4(x1)))))) 0(1(2(x1))) -> 1(0(2(2(3(4(x1)))))) 0(1(2(x1))) -> 1(1(4(0(3(2(x1)))))) 0(1(2(x1))) -> 2(0(0(3(3(1(x1)))))) 0(1(2(x1))) -> 2(0(3(4(1(5(x1)))))) 0(1(2(x1))) -> 2(3(2(0(4(1(x1)))))) 0(1(2(x1))) -> 2(5(3(0(3(1(x1)))))) 0(1(2(x1))) -> 3(2(5(3(1(0(x1)))))) 0(1(2(x1))) -> 3(4(0(5(1(2(x1)))))) 0(1(2(x1))) -> 5(5(4(2(1(0(x1)))))) 0(0(1(2(x1)))) -> 0(0(2(4(1(x1))))) 0(0(1(2(x1)))) -> 0(2(1(5(0(4(x1)))))) 0(0(1(2(x1)))) -> 1(0(2(3(5(0(x1)))))) 0(0(1(2(x1)))) -> 5(1(2(0(4(0(x1)))))) 0(0(5(2(x1)))) -> 5(3(0(3(0(2(x1)))))) 0(1(2(0(x1)))) -> 1(2(4(0(0(x1))))) 0(1(2(0(x1)))) -> 0(4(1(0(0(2(x1)))))) 0(1(2(2(x1)))) -> 3(3(0(2(1(2(x1)))))) 0(1(5(2(x1)))) -> 2(5(1(0(3(x1))))) 0(1(5(2(x1)))) -> 0(5(5(3(1(2(x1)))))) 0(1(5(2(x1)))) -> 1(3(5(0(2(2(x1)))))) 0(5(2(2(x1)))) -> 3(0(2(1(5(2(x1)))))) 0(1(0(2(2(x1))))) -> 3(0(2(1(0(2(x1)))))) 0(1(2(5(2(x1))))) -> 3(5(1(2(2(0(x1)))))) 0(1(3(0(0(x1))))) -> 0(0(4(1(3(0(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(1(5(2(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(2(1(1(x1)))))) 5(0(1(2(x1)))) -> 0(2(1(3(5(x1))))) 5(0(1(2(x1)))) -> 2(0(3(1(5(x1))))) 5(0(1(2(x1)))) -> 5(3(1(0(3(2(x1)))))) 5(0(1(2(x1)))) -> 5(5(1(0(2(4(x1)))))) 1(0(1(2(x1)))) -> 0(4(1(2(1(x1))))) 1(0(1(2(x1)))) -> 1(2(1(3(0(x1))))) 1(0(1(2(x1)))) -> 0(3(1(4(1(2(x1)))))) 1(0(1(2(x1)))) -> 4(1(4(1(0(2(x1)))))) 1(3(0(1(2(x1))))) -> 1(3(4(1(0(2(x1)))))) ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: 1^1(0(1(2(x1)))) -> 1^1(x1) 1^1(0(1(2(x1)))) -> 0^1(x1) 0^1(1(2(x1))) -> 1^1(x1) 0^1(1(2(x1))) -> 0^1(x1) 0^1(1(2(x1))) -> 1^1(0(x1)) 0^1(1(2(x1))) -> 1^1(5(x1)) 0^1(1(2(0(x1)))) -> 0^1(0(x1)) 0^1(1(4(2(2(x1))))) -> 1^1(1(x1)) 0^1(1(4(2(2(x1))))) -> 1^1(x1) The TRS R consists of the following rules: 0(1(2(x1))) -> 0(3(1(2(x1)))) 0(1(2(x1))) -> 1(0(2(1(x1)))) 0(1(2(x1))) -> 2(1(4(0(x1)))) 0(1(2(x1))) -> 0(2(1(3(3(x1))))) 0(1(2(x1))) -> 0(2(4(1(3(x1))))) 0(1(2(x1))) -> 1(0(2(5(4(x1))))) 0(1(2(x1))) -> 2(1(1(0(3(x1))))) 0(1(2(x1))) -> 2(1(3(4(0(x1))))) 0(1(2(x1))) -> 2(1(4(3(0(x1))))) 0(1(2(x1))) -> 4(0(3(1(2(x1))))) 0(1(2(x1))) -> 4(1(0(3(2(x1))))) 0(1(2(x1))) -> 4(4(2(1(0(x1))))) 0(1(2(x1))) -> 0(0(4(1(2(5(x1)))))) 0(1(2(x1))) -> 0(4(5(1(2(5(x1)))))) 0(1(2(x1))) -> 0(5(2(3(1(4(x1)))))) 0(1(2(x1))) -> 1(0(2(2(3(4(x1)))))) 0(1(2(x1))) -> 1(1(4(0(3(2(x1)))))) 0(1(2(x1))) -> 2(0(0(3(3(1(x1)))))) 0(1(2(x1))) -> 2(0(3(4(1(5(x1)))))) 0(1(2(x1))) -> 2(3(2(0(4(1(x1)))))) 0(1(2(x1))) -> 2(5(3(0(3(1(x1)))))) 0(1(2(x1))) -> 3(2(5(3(1(0(x1)))))) 0(1(2(x1))) -> 3(4(0(5(1(2(x1)))))) 0(1(2(x1))) -> 5(5(4(2(1(0(x1)))))) 0(0(1(2(x1)))) -> 0(0(2(4(1(x1))))) 0(0(1(2(x1)))) -> 0(2(1(5(0(4(x1)))))) 0(0(1(2(x1)))) -> 1(0(2(3(5(0(x1)))))) 0(0(1(2(x1)))) -> 5(1(2(0(4(0(x1)))))) 0(0(5(2(x1)))) -> 5(3(0(3(0(2(x1)))))) 0(1(2(0(x1)))) -> 1(2(4(0(0(x1))))) 0(1(2(0(x1)))) -> 0(4(1(0(0(2(x1)))))) 0(1(2(2(x1)))) -> 3(3(0(2(1(2(x1)))))) 0(1(5(2(x1)))) -> 2(5(1(0(3(x1))))) 0(1(5(2(x1)))) -> 0(5(5(3(1(2(x1)))))) 0(1(5(2(x1)))) -> 1(3(5(0(2(2(x1)))))) 0(5(2(2(x1)))) -> 3(0(2(1(5(2(x1)))))) 1(0(1(2(x1)))) -> 0(4(1(2(1(x1))))) 1(0(1(2(x1)))) -> 1(2(1(3(0(x1))))) 1(0(1(2(x1)))) -> 0(3(1(4(1(2(x1)))))) 1(0(1(2(x1)))) -> 4(1(4(1(0(2(x1)))))) 5(0(1(2(x1)))) -> 0(2(1(3(5(x1))))) 5(0(1(2(x1)))) -> 2(0(3(1(5(x1))))) 5(0(1(2(x1)))) -> 5(3(1(0(3(2(x1)))))) 5(0(1(2(x1)))) -> 5(5(1(0(2(4(x1)))))) 0(1(0(2(2(x1))))) -> 3(0(2(1(0(2(x1)))))) 0(1(2(5(2(x1))))) -> 3(5(1(2(2(0(x1)))))) 0(1(3(0(0(x1))))) -> 0(0(4(1(3(0(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(1(5(2(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(2(1(1(x1)))))) 1(3(0(1(2(x1))))) -> 1(3(4(1(0(2(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (15) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. 1^1(0(1(2(x1)))) -> 1^1(x1) The remaining pairs can at least be oriented weakly. Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: <<< POL(1^1(x_1)) = [[1A]] + [[0A, 0A, -I]] * x_1 >>> <<< POL(0(x_1)) = [[-I], [-I], [0A]] + [[0A, 1A, 0A], [-I, -I, 0A], [-I, -I, 0A]] * x_1 >>> <<< POL(1(x_1)) = [[0A], [1A], [1A]] + [[0A, 0A, 0A], [-I, 0A, 0A], [0A, 0A, 0A]] * x_1 >>> <<< POL(2(x_1)) = [[0A], [-I], [0A]] + [[0A, 0A, 0A], [-I, -I, -I], [0A, 0A, -I]] * x_1 >>> <<< POL(0^1(x_1)) = [[1A]] + [[1A, 0A, 0A]] * x_1 >>> <<< POL(5(x_1)) = [[0A], [0A], [0A]] + [[0A, 0A, -I], [-I, 0A, -I], [-I, 0A, -I]] * x_1 >>> <<< POL(4(x_1)) = [[0A], [0A], [0A]] + [[-I, 0A, 0A], [-I, -I, -I], [-I, 0A, -I]] * x_1 >>> <<< POL(3(x_1)) = [[0A], [0A], [0A]] + [[-I, 0A, 0A], [-I, -I, -I], [-I, -I, -I]] * x_1 >>> The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 0(1(2(x1))) -> 0(3(1(2(x1)))) 0(1(2(x1))) -> 1(0(2(1(x1)))) 0(1(2(x1))) -> 2(1(4(0(x1)))) 0(1(2(x1))) -> 0(2(1(3(3(x1))))) 0(1(2(x1))) -> 0(2(4(1(3(x1))))) 0(1(2(x1))) -> 1(0(2(5(4(x1))))) 0(1(2(x1))) -> 2(1(1(0(3(x1))))) 0(1(2(x1))) -> 2(1(3(4(0(x1))))) 0(1(2(x1))) -> 2(1(4(3(0(x1))))) 0(1(2(x1))) -> 4(0(3(1(2(x1))))) 0(1(2(x1))) -> 4(1(0(3(2(x1))))) 0(1(2(x1))) -> 4(4(2(1(0(x1))))) 0(1(2(x1))) -> 0(0(4(1(2(5(x1)))))) 0(1(2(x1))) -> 0(4(5(1(2(5(x1)))))) 0(1(2(x1))) -> 0(5(2(3(1(4(x1)))))) 0(1(2(x1))) -> 1(0(2(2(3(4(x1)))))) 0(1(2(x1))) -> 1(1(4(0(3(2(x1)))))) 0(1(2(x1))) -> 2(0(0(3(3(1(x1)))))) 0(1(2(x1))) -> 2(0(3(4(1(5(x1)))))) 0(1(2(x1))) -> 2(3(2(0(4(1(x1)))))) 0(1(2(x1))) -> 2(5(3(0(3(1(x1)))))) 0(1(2(x1))) -> 3(2(5(3(1(0(x1)))))) 0(1(2(x1))) -> 3(4(0(5(1(2(x1)))))) 0(1(2(x1))) -> 5(5(4(2(1(0(x1)))))) 0(0(1(2(x1)))) -> 0(0(2(4(1(x1))))) 0(0(1(2(x1)))) -> 0(2(1(5(0(4(x1)))))) 0(0(1(2(x1)))) -> 1(0(2(3(5(0(x1)))))) 0(0(1(2(x1)))) -> 5(1(2(0(4(0(x1)))))) 0(0(5(2(x1)))) -> 5(3(0(3(0(2(x1)))))) 0(1(2(0(x1)))) -> 1(2(4(0(0(x1))))) 0(1(2(0(x1)))) -> 0(4(1(0(0(2(x1)))))) 0(1(2(2(x1)))) -> 3(3(0(2(1(2(x1)))))) 0(1(5(2(x1)))) -> 2(5(1(0(3(x1))))) 0(1(5(2(x1)))) -> 0(5(5(3(1(2(x1)))))) 0(1(5(2(x1)))) -> 1(3(5(0(2(2(x1)))))) 0(5(2(2(x1)))) -> 3(0(2(1(5(2(x1)))))) 0(1(0(2(2(x1))))) -> 3(0(2(1(0(2(x1)))))) 0(1(2(5(2(x1))))) -> 3(5(1(2(2(0(x1)))))) 0(1(3(0(0(x1))))) -> 0(0(4(1(3(0(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(1(5(2(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(2(1(1(x1)))))) 5(0(1(2(x1)))) -> 0(2(1(3(5(x1))))) 5(0(1(2(x1)))) -> 2(0(3(1(5(x1))))) 5(0(1(2(x1)))) -> 5(3(1(0(3(2(x1)))))) 5(0(1(2(x1)))) -> 5(5(1(0(2(4(x1)))))) 1(0(1(2(x1)))) -> 0(4(1(2(1(x1))))) 1(0(1(2(x1)))) -> 1(2(1(3(0(x1))))) 1(0(1(2(x1)))) -> 0(3(1(4(1(2(x1)))))) 1(0(1(2(x1)))) -> 4(1(4(1(0(2(x1)))))) 1(3(0(1(2(x1))))) -> 1(3(4(1(0(2(x1)))))) ---------------------------------------- (16) Obligation: Q DP problem: The TRS P consists of the following rules: 1^1(0(1(2(x1)))) -> 0^1(x1) 0^1(1(2(x1))) -> 1^1(x1) 0^1(1(2(x1))) -> 0^1(x1) 0^1(1(2(x1))) -> 1^1(0(x1)) 0^1(1(2(x1))) -> 1^1(5(x1)) 0^1(1(2(0(x1)))) -> 0^1(0(x1)) 0^1(1(4(2(2(x1))))) -> 1^1(1(x1)) 0^1(1(4(2(2(x1))))) -> 1^1(x1) The TRS R consists of the following rules: 0(1(2(x1))) -> 0(3(1(2(x1)))) 0(1(2(x1))) -> 1(0(2(1(x1)))) 0(1(2(x1))) -> 2(1(4(0(x1)))) 0(1(2(x1))) -> 0(2(1(3(3(x1))))) 0(1(2(x1))) -> 0(2(4(1(3(x1))))) 0(1(2(x1))) -> 1(0(2(5(4(x1))))) 0(1(2(x1))) -> 2(1(1(0(3(x1))))) 0(1(2(x1))) -> 2(1(3(4(0(x1))))) 0(1(2(x1))) -> 2(1(4(3(0(x1))))) 0(1(2(x1))) -> 4(0(3(1(2(x1))))) 0(1(2(x1))) -> 4(1(0(3(2(x1))))) 0(1(2(x1))) -> 4(4(2(1(0(x1))))) 0(1(2(x1))) -> 0(0(4(1(2(5(x1)))))) 0(1(2(x1))) -> 0(4(5(1(2(5(x1)))))) 0(1(2(x1))) -> 0(5(2(3(1(4(x1)))))) 0(1(2(x1))) -> 1(0(2(2(3(4(x1)))))) 0(1(2(x1))) -> 1(1(4(0(3(2(x1)))))) 0(1(2(x1))) -> 2(0(0(3(3(1(x1)))))) 0(1(2(x1))) -> 2(0(3(4(1(5(x1)))))) 0(1(2(x1))) -> 2(3(2(0(4(1(x1)))))) 0(1(2(x1))) -> 2(5(3(0(3(1(x1)))))) 0(1(2(x1))) -> 3(2(5(3(1(0(x1)))))) 0(1(2(x1))) -> 3(4(0(5(1(2(x1)))))) 0(1(2(x1))) -> 5(5(4(2(1(0(x1)))))) 0(0(1(2(x1)))) -> 0(0(2(4(1(x1))))) 0(0(1(2(x1)))) -> 0(2(1(5(0(4(x1)))))) 0(0(1(2(x1)))) -> 1(0(2(3(5(0(x1)))))) 0(0(1(2(x1)))) -> 5(1(2(0(4(0(x1)))))) 0(0(5(2(x1)))) -> 5(3(0(3(0(2(x1)))))) 0(1(2(0(x1)))) -> 1(2(4(0(0(x1))))) 0(1(2(0(x1)))) -> 0(4(1(0(0(2(x1)))))) 0(1(2(2(x1)))) -> 3(3(0(2(1(2(x1)))))) 0(1(5(2(x1)))) -> 2(5(1(0(3(x1))))) 0(1(5(2(x1)))) -> 0(5(5(3(1(2(x1)))))) 0(1(5(2(x1)))) -> 1(3(5(0(2(2(x1)))))) 0(5(2(2(x1)))) -> 3(0(2(1(5(2(x1)))))) 1(0(1(2(x1)))) -> 0(4(1(2(1(x1))))) 1(0(1(2(x1)))) -> 1(2(1(3(0(x1))))) 1(0(1(2(x1)))) -> 0(3(1(4(1(2(x1)))))) 1(0(1(2(x1)))) -> 4(1(4(1(0(2(x1)))))) 5(0(1(2(x1)))) -> 0(2(1(3(5(x1))))) 5(0(1(2(x1)))) -> 2(0(3(1(5(x1))))) 5(0(1(2(x1)))) -> 5(3(1(0(3(2(x1)))))) 5(0(1(2(x1)))) -> 5(5(1(0(2(4(x1)))))) 0(1(0(2(2(x1))))) -> 3(0(2(1(0(2(x1)))))) 0(1(2(5(2(x1))))) -> 3(5(1(2(2(0(x1)))))) 0(1(3(0(0(x1))))) -> 0(0(4(1(3(0(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(1(5(2(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(2(1(1(x1)))))) 1(3(0(1(2(x1))))) -> 1(3(4(1(0(2(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. 0^1(1(2(0(x1)))) -> 0^1(0(x1)) The remaining pairs can at least be oriented weakly. Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: <<< POL(1^1(x_1)) = [[0A]] + [[0A, -I, -I]] * x_1 >>> <<< POL(0(x_1)) = [[-I], [0A], [0A]] + [[0A, -I, -I], [0A, -I, -I], [0A, -I, -I]] * x_1 >>> <<< POL(1(x_1)) = [[-I], [0A], [0A]] + [[0A, -I, 0A], [0A, -I, 0A], [0A, 0A, 0A]] * x_1 >>> <<< POL(2(x_1)) = [[0A], [-I], [1A]] + [[0A, -I, 0A], [-I, -I, -I], [0A, 1A, -I]] * x_1 >>> <<< POL(0^1(x_1)) = [[0A]] + [[0A, 0A, 0A]] * x_1 >>> <<< POL(5(x_1)) = [[0A], [0A], [0A]] + [[0A, -I, 0A], [0A, -I, 0A], [0A, -I, 0A]] * x_1 >>> <<< POL(4(x_1)) = [[0A], [0A], [0A]] + [[-I, -I, -I], [-I, -I, 0A], [-I, -I, -I]] * x_1 >>> <<< POL(3(x_1)) = [[0A], [-I], [-I]] + [[-I, -I, -I], [-I, -I, -I], [-I, -I, -I]] * x_1 >>> The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 0(1(2(x1))) -> 0(3(1(2(x1)))) 0(1(2(x1))) -> 1(0(2(1(x1)))) 0(1(2(x1))) -> 2(1(4(0(x1)))) 0(1(2(x1))) -> 0(2(1(3(3(x1))))) 0(1(2(x1))) -> 0(2(4(1(3(x1))))) 0(1(2(x1))) -> 1(0(2(5(4(x1))))) 0(1(2(x1))) -> 2(1(1(0(3(x1))))) 0(1(2(x1))) -> 2(1(3(4(0(x1))))) 0(1(2(x1))) -> 2(1(4(3(0(x1))))) 0(1(2(x1))) -> 4(0(3(1(2(x1))))) 0(1(2(x1))) -> 4(1(0(3(2(x1))))) 0(1(2(x1))) -> 4(4(2(1(0(x1))))) 0(1(2(x1))) -> 0(0(4(1(2(5(x1)))))) 0(1(2(x1))) -> 0(4(5(1(2(5(x1)))))) 0(1(2(x1))) -> 0(5(2(3(1(4(x1)))))) 0(1(2(x1))) -> 1(0(2(2(3(4(x1)))))) 0(1(2(x1))) -> 1(1(4(0(3(2(x1)))))) 0(1(2(x1))) -> 2(0(0(3(3(1(x1)))))) 0(1(2(x1))) -> 2(0(3(4(1(5(x1)))))) 0(1(2(x1))) -> 2(3(2(0(4(1(x1)))))) 0(1(2(x1))) -> 2(5(3(0(3(1(x1)))))) 0(1(2(x1))) -> 3(2(5(3(1(0(x1)))))) 0(1(2(x1))) -> 3(4(0(5(1(2(x1)))))) 0(1(2(x1))) -> 5(5(4(2(1(0(x1)))))) 0(0(1(2(x1)))) -> 0(0(2(4(1(x1))))) 0(0(1(2(x1)))) -> 0(2(1(5(0(4(x1)))))) 0(0(1(2(x1)))) -> 1(0(2(3(5(0(x1)))))) 0(0(1(2(x1)))) -> 5(1(2(0(4(0(x1)))))) 0(0(5(2(x1)))) -> 5(3(0(3(0(2(x1)))))) 0(1(2(0(x1)))) -> 1(2(4(0(0(x1))))) 0(1(2(0(x1)))) -> 0(4(1(0(0(2(x1)))))) 0(1(2(2(x1)))) -> 3(3(0(2(1(2(x1)))))) 0(1(5(2(x1)))) -> 2(5(1(0(3(x1))))) 0(1(5(2(x1)))) -> 0(5(5(3(1(2(x1)))))) 0(1(5(2(x1)))) -> 1(3(5(0(2(2(x1)))))) 0(5(2(2(x1)))) -> 3(0(2(1(5(2(x1)))))) 0(1(0(2(2(x1))))) -> 3(0(2(1(0(2(x1)))))) 0(1(2(5(2(x1))))) -> 3(5(1(2(2(0(x1)))))) 0(1(3(0(0(x1))))) -> 0(0(4(1(3(0(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(1(5(2(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(2(1(1(x1)))))) 5(0(1(2(x1)))) -> 0(2(1(3(5(x1))))) 5(0(1(2(x1)))) -> 2(0(3(1(5(x1))))) 5(0(1(2(x1)))) -> 5(3(1(0(3(2(x1)))))) 5(0(1(2(x1)))) -> 5(5(1(0(2(4(x1)))))) 1(0(1(2(x1)))) -> 0(4(1(2(1(x1))))) 1(0(1(2(x1)))) -> 1(2(1(3(0(x1))))) 1(0(1(2(x1)))) -> 0(3(1(4(1(2(x1)))))) 1(0(1(2(x1)))) -> 4(1(4(1(0(2(x1)))))) 1(3(0(1(2(x1))))) -> 1(3(4(1(0(2(x1)))))) ---------------------------------------- (18) Obligation: Q DP problem: The TRS P consists of the following rules: 1^1(0(1(2(x1)))) -> 0^1(x1) 0^1(1(2(x1))) -> 1^1(x1) 0^1(1(2(x1))) -> 0^1(x1) 0^1(1(2(x1))) -> 1^1(0(x1)) 0^1(1(2(x1))) -> 1^1(5(x1)) 0^1(1(4(2(2(x1))))) -> 1^1(1(x1)) 0^1(1(4(2(2(x1))))) -> 1^1(x1) The TRS R consists of the following rules: 0(1(2(x1))) -> 0(3(1(2(x1)))) 0(1(2(x1))) -> 1(0(2(1(x1)))) 0(1(2(x1))) -> 2(1(4(0(x1)))) 0(1(2(x1))) -> 0(2(1(3(3(x1))))) 0(1(2(x1))) -> 0(2(4(1(3(x1))))) 0(1(2(x1))) -> 1(0(2(5(4(x1))))) 0(1(2(x1))) -> 2(1(1(0(3(x1))))) 0(1(2(x1))) -> 2(1(3(4(0(x1))))) 0(1(2(x1))) -> 2(1(4(3(0(x1))))) 0(1(2(x1))) -> 4(0(3(1(2(x1))))) 0(1(2(x1))) -> 4(1(0(3(2(x1))))) 0(1(2(x1))) -> 4(4(2(1(0(x1))))) 0(1(2(x1))) -> 0(0(4(1(2(5(x1)))))) 0(1(2(x1))) -> 0(4(5(1(2(5(x1)))))) 0(1(2(x1))) -> 0(5(2(3(1(4(x1)))))) 0(1(2(x1))) -> 1(0(2(2(3(4(x1)))))) 0(1(2(x1))) -> 1(1(4(0(3(2(x1)))))) 0(1(2(x1))) -> 2(0(0(3(3(1(x1)))))) 0(1(2(x1))) -> 2(0(3(4(1(5(x1)))))) 0(1(2(x1))) -> 2(3(2(0(4(1(x1)))))) 0(1(2(x1))) -> 2(5(3(0(3(1(x1)))))) 0(1(2(x1))) -> 3(2(5(3(1(0(x1)))))) 0(1(2(x1))) -> 3(4(0(5(1(2(x1)))))) 0(1(2(x1))) -> 5(5(4(2(1(0(x1)))))) 0(0(1(2(x1)))) -> 0(0(2(4(1(x1))))) 0(0(1(2(x1)))) -> 0(2(1(5(0(4(x1)))))) 0(0(1(2(x1)))) -> 1(0(2(3(5(0(x1)))))) 0(0(1(2(x1)))) -> 5(1(2(0(4(0(x1)))))) 0(0(5(2(x1)))) -> 5(3(0(3(0(2(x1)))))) 0(1(2(0(x1)))) -> 1(2(4(0(0(x1))))) 0(1(2(0(x1)))) -> 0(4(1(0(0(2(x1)))))) 0(1(2(2(x1)))) -> 3(3(0(2(1(2(x1)))))) 0(1(5(2(x1)))) -> 2(5(1(0(3(x1))))) 0(1(5(2(x1)))) -> 0(5(5(3(1(2(x1)))))) 0(1(5(2(x1)))) -> 1(3(5(0(2(2(x1)))))) 0(5(2(2(x1)))) -> 3(0(2(1(5(2(x1)))))) 1(0(1(2(x1)))) -> 0(4(1(2(1(x1))))) 1(0(1(2(x1)))) -> 1(2(1(3(0(x1))))) 1(0(1(2(x1)))) -> 0(3(1(4(1(2(x1)))))) 1(0(1(2(x1)))) -> 4(1(4(1(0(2(x1)))))) 5(0(1(2(x1)))) -> 0(2(1(3(5(x1))))) 5(0(1(2(x1)))) -> 2(0(3(1(5(x1))))) 5(0(1(2(x1)))) -> 5(3(1(0(3(2(x1)))))) 5(0(1(2(x1)))) -> 5(5(1(0(2(4(x1)))))) 0(1(0(2(2(x1))))) -> 3(0(2(1(0(2(x1)))))) 0(1(2(5(2(x1))))) -> 3(5(1(2(2(0(x1)))))) 0(1(3(0(0(x1))))) -> 0(0(4(1(3(0(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(1(5(2(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(2(1(1(x1)))))) 1(3(0(1(2(x1))))) -> 1(3(4(1(0(2(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (19) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. 0^1(1(4(2(2(x1))))) -> 1^1(1(x1)) The remaining pairs can at least be oriented weakly. Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: <<< POL(1^1(x_1)) = [[0A]] + [[0A, -I, -I]] * x_1 >>> <<< POL(0(x_1)) = [[0A], [0A], [0A]] + [[-I, 0A, -I], [0A, 0A, 0A], [0A, 0A, 0A]] * x_1 >>> <<< POL(1(x_1)) = [[0A], [0A], [0A]] + [[-I, -I, -I], [0A, 0A, 0A], [0A, 0A, 0A]] * x_1 >>> <<< POL(2(x_1)) = [[-I], [1A], [0A]] + [[0A, 0A, 0A], [-I, 0A, 0A], [0A, 0A, 0A]] * x_1 >>> <<< POL(0^1(x_1)) = [[0A]] + [[0A, 0A, 0A]] * x_1 >>> <<< POL(5(x_1)) = [[0A], [0A], [0A]] + [[0A, 0A, 0A], [0A, 0A, 0A], [-I, -I, 0A]] * x_1 >>> <<< POL(4(x_1)) = [[0A], [0A], [0A]] + [[0A, -I, -I], [0A, -I, -I], [-I, -I, -I]] * x_1 >>> <<< POL(3(x_1)) = [[0A], [0A], [0A]] + [[0A, 0A, 0A], [0A, 0A, 0A], [0A, 0A, 0A]] * x_1 >>> The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 0(1(2(x1))) -> 0(3(1(2(x1)))) 0(1(2(x1))) -> 1(0(2(1(x1)))) 0(1(2(x1))) -> 2(1(4(0(x1)))) 0(1(2(x1))) -> 0(2(1(3(3(x1))))) 0(1(2(x1))) -> 0(2(4(1(3(x1))))) 0(1(2(x1))) -> 1(0(2(5(4(x1))))) 0(1(2(x1))) -> 2(1(1(0(3(x1))))) 0(1(2(x1))) -> 2(1(3(4(0(x1))))) 0(1(2(x1))) -> 2(1(4(3(0(x1))))) 0(1(2(x1))) -> 4(0(3(1(2(x1))))) 0(1(2(x1))) -> 4(1(0(3(2(x1))))) 0(1(2(x1))) -> 4(4(2(1(0(x1))))) 0(1(2(x1))) -> 0(0(4(1(2(5(x1)))))) 0(1(2(x1))) -> 0(4(5(1(2(5(x1)))))) 0(1(2(x1))) -> 0(5(2(3(1(4(x1)))))) 0(1(2(x1))) -> 1(0(2(2(3(4(x1)))))) 0(1(2(x1))) -> 1(1(4(0(3(2(x1)))))) 0(1(2(x1))) -> 2(0(0(3(3(1(x1)))))) 0(1(2(x1))) -> 2(0(3(4(1(5(x1)))))) 0(1(2(x1))) -> 2(3(2(0(4(1(x1)))))) 0(1(2(x1))) -> 2(5(3(0(3(1(x1)))))) 0(1(2(x1))) -> 3(2(5(3(1(0(x1)))))) 0(1(2(x1))) -> 3(4(0(5(1(2(x1)))))) 0(1(2(x1))) -> 5(5(4(2(1(0(x1)))))) 0(0(1(2(x1)))) -> 0(0(2(4(1(x1))))) 0(0(1(2(x1)))) -> 0(2(1(5(0(4(x1)))))) 0(0(1(2(x1)))) -> 1(0(2(3(5(0(x1)))))) 0(0(1(2(x1)))) -> 5(1(2(0(4(0(x1)))))) 0(0(5(2(x1)))) -> 5(3(0(3(0(2(x1)))))) 0(1(2(0(x1)))) -> 1(2(4(0(0(x1))))) 0(1(2(0(x1)))) -> 0(4(1(0(0(2(x1)))))) 0(1(2(2(x1)))) -> 3(3(0(2(1(2(x1)))))) 0(1(5(2(x1)))) -> 2(5(1(0(3(x1))))) 0(1(5(2(x1)))) -> 0(5(5(3(1(2(x1)))))) 0(1(5(2(x1)))) -> 1(3(5(0(2(2(x1)))))) 0(5(2(2(x1)))) -> 3(0(2(1(5(2(x1)))))) 0(1(0(2(2(x1))))) -> 3(0(2(1(0(2(x1)))))) 0(1(2(5(2(x1))))) -> 3(5(1(2(2(0(x1)))))) 0(1(3(0(0(x1))))) -> 0(0(4(1(3(0(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(1(5(2(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(2(1(1(x1)))))) 5(0(1(2(x1)))) -> 0(2(1(3(5(x1))))) 5(0(1(2(x1)))) -> 2(0(3(1(5(x1))))) 5(0(1(2(x1)))) -> 5(3(1(0(3(2(x1)))))) 5(0(1(2(x1)))) -> 5(5(1(0(2(4(x1)))))) 1(0(1(2(x1)))) -> 0(4(1(2(1(x1))))) 1(0(1(2(x1)))) -> 1(2(1(3(0(x1))))) 1(0(1(2(x1)))) -> 0(3(1(4(1(2(x1)))))) 1(0(1(2(x1)))) -> 4(1(4(1(0(2(x1)))))) 1(3(0(1(2(x1))))) -> 1(3(4(1(0(2(x1)))))) ---------------------------------------- (20) Obligation: Q DP problem: The TRS P consists of the following rules: 1^1(0(1(2(x1)))) -> 0^1(x1) 0^1(1(2(x1))) -> 1^1(x1) 0^1(1(2(x1))) -> 0^1(x1) 0^1(1(2(x1))) -> 1^1(0(x1)) 0^1(1(2(x1))) -> 1^1(5(x1)) 0^1(1(4(2(2(x1))))) -> 1^1(x1) The TRS R consists of the following rules: 0(1(2(x1))) -> 0(3(1(2(x1)))) 0(1(2(x1))) -> 1(0(2(1(x1)))) 0(1(2(x1))) -> 2(1(4(0(x1)))) 0(1(2(x1))) -> 0(2(1(3(3(x1))))) 0(1(2(x1))) -> 0(2(4(1(3(x1))))) 0(1(2(x1))) -> 1(0(2(5(4(x1))))) 0(1(2(x1))) -> 2(1(1(0(3(x1))))) 0(1(2(x1))) -> 2(1(3(4(0(x1))))) 0(1(2(x1))) -> 2(1(4(3(0(x1))))) 0(1(2(x1))) -> 4(0(3(1(2(x1))))) 0(1(2(x1))) -> 4(1(0(3(2(x1))))) 0(1(2(x1))) -> 4(4(2(1(0(x1))))) 0(1(2(x1))) -> 0(0(4(1(2(5(x1)))))) 0(1(2(x1))) -> 0(4(5(1(2(5(x1)))))) 0(1(2(x1))) -> 0(5(2(3(1(4(x1)))))) 0(1(2(x1))) -> 1(0(2(2(3(4(x1)))))) 0(1(2(x1))) -> 1(1(4(0(3(2(x1)))))) 0(1(2(x1))) -> 2(0(0(3(3(1(x1)))))) 0(1(2(x1))) -> 2(0(3(4(1(5(x1)))))) 0(1(2(x1))) -> 2(3(2(0(4(1(x1)))))) 0(1(2(x1))) -> 2(5(3(0(3(1(x1)))))) 0(1(2(x1))) -> 3(2(5(3(1(0(x1)))))) 0(1(2(x1))) -> 3(4(0(5(1(2(x1)))))) 0(1(2(x1))) -> 5(5(4(2(1(0(x1)))))) 0(0(1(2(x1)))) -> 0(0(2(4(1(x1))))) 0(0(1(2(x1)))) -> 0(2(1(5(0(4(x1)))))) 0(0(1(2(x1)))) -> 1(0(2(3(5(0(x1)))))) 0(0(1(2(x1)))) -> 5(1(2(0(4(0(x1)))))) 0(0(5(2(x1)))) -> 5(3(0(3(0(2(x1)))))) 0(1(2(0(x1)))) -> 1(2(4(0(0(x1))))) 0(1(2(0(x1)))) -> 0(4(1(0(0(2(x1)))))) 0(1(2(2(x1)))) -> 3(3(0(2(1(2(x1)))))) 0(1(5(2(x1)))) -> 2(5(1(0(3(x1))))) 0(1(5(2(x1)))) -> 0(5(5(3(1(2(x1)))))) 0(1(5(2(x1)))) -> 1(3(5(0(2(2(x1)))))) 0(5(2(2(x1)))) -> 3(0(2(1(5(2(x1)))))) 1(0(1(2(x1)))) -> 0(4(1(2(1(x1))))) 1(0(1(2(x1)))) -> 1(2(1(3(0(x1))))) 1(0(1(2(x1)))) -> 0(3(1(4(1(2(x1)))))) 1(0(1(2(x1)))) -> 4(1(4(1(0(2(x1)))))) 5(0(1(2(x1)))) -> 0(2(1(3(5(x1))))) 5(0(1(2(x1)))) -> 2(0(3(1(5(x1))))) 5(0(1(2(x1)))) -> 5(3(1(0(3(2(x1)))))) 5(0(1(2(x1)))) -> 5(5(1(0(2(4(x1)))))) 0(1(0(2(2(x1))))) -> 3(0(2(1(0(2(x1)))))) 0(1(2(5(2(x1))))) -> 3(5(1(2(2(0(x1)))))) 0(1(3(0(0(x1))))) -> 0(0(4(1(3(0(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(1(5(2(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(2(1(1(x1)))))) 1(3(0(1(2(x1))))) -> 1(3(4(1(0(2(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (21) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. 1^1(0(1(2(x1)))) -> 0^1(x1) The remaining pairs can at least be oriented weakly. Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: <<< POL(1^1(x_1)) = [[-I]] + [[0A, -I, 0A]] * x_1 >>> <<< POL(0(x_1)) = [[-I], [-I], [-I]] + [[0A, 0A, -I], [-I, 1A, 0A], [0A, 1A, -I]] * x_1 >>> <<< POL(1(x_1)) = [[-I], [-I], [-I]] + [[0A, -I, -I], [-I, -I, 0A], [0A, -I, -I]] * x_1 >>> <<< POL(2(x_1)) = [[-I], [-I], [-I]] + [[0A, -I, 0A], [-I, -I, 0A], [0A, 1A, -I]] * x_1 >>> <<< POL(0^1(x_1)) = [[-I]] + [[0A, 0A, -I]] * x_1 >>> <<< POL(5(x_1)) = [[-I], [-I], [-I]] + [[0A, -I, -I], [-I, 0A, 0A], [-I, 0A, 0A]] * x_1 >>> <<< POL(4(x_1)) = [[-I], [-I], [-I]] + [[0A, -I, -I], [-I, -I, -I], [-I, -I, -I]] * x_1 >>> <<< POL(3(x_1)) = [[-I], [-I], [-I]] + [[0A, -I, -I], [-I, -I, -I], [-I, -I, -I]] * x_1 >>> The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 0(1(2(x1))) -> 0(3(1(2(x1)))) 0(1(2(x1))) -> 1(0(2(1(x1)))) 0(1(2(x1))) -> 2(1(4(0(x1)))) 0(1(2(x1))) -> 0(2(1(3(3(x1))))) 0(1(2(x1))) -> 0(2(4(1(3(x1))))) 0(1(2(x1))) -> 1(0(2(5(4(x1))))) 0(1(2(x1))) -> 2(1(1(0(3(x1))))) 0(1(2(x1))) -> 2(1(3(4(0(x1))))) 0(1(2(x1))) -> 2(1(4(3(0(x1))))) 0(1(2(x1))) -> 4(0(3(1(2(x1))))) 0(1(2(x1))) -> 4(1(0(3(2(x1))))) 0(1(2(x1))) -> 4(4(2(1(0(x1))))) 0(1(2(x1))) -> 0(0(4(1(2(5(x1)))))) 0(1(2(x1))) -> 0(4(5(1(2(5(x1)))))) 0(1(2(x1))) -> 0(5(2(3(1(4(x1)))))) 0(1(2(x1))) -> 1(0(2(2(3(4(x1)))))) 0(1(2(x1))) -> 1(1(4(0(3(2(x1)))))) 0(1(2(x1))) -> 2(0(0(3(3(1(x1)))))) 0(1(2(x1))) -> 2(0(3(4(1(5(x1)))))) 0(1(2(x1))) -> 2(3(2(0(4(1(x1)))))) 0(1(2(x1))) -> 2(5(3(0(3(1(x1)))))) 0(1(2(x1))) -> 3(2(5(3(1(0(x1)))))) 0(1(2(x1))) -> 3(4(0(5(1(2(x1)))))) 0(1(2(x1))) -> 5(5(4(2(1(0(x1)))))) 0(0(1(2(x1)))) -> 0(0(2(4(1(x1))))) 0(0(1(2(x1)))) -> 0(2(1(5(0(4(x1)))))) 0(0(1(2(x1)))) -> 1(0(2(3(5(0(x1)))))) 0(0(1(2(x1)))) -> 5(1(2(0(4(0(x1)))))) 0(0(5(2(x1)))) -> 5(3(0(3(0(2(x1)))))) 0(1(2(0(x1)))) -> 1(2(4(0(0(x1))))) 0(1(2(0(x1)))) -> 0(4(1(0(0(2(x1)))))) 0(1(2(2(x1)))) -> 3(3(0(2(1(2(x1)))))) 0(1(5(2(x1)))) -> 2(5(1(0(3(x1))))) 0(1(5(2(x1)))) -> 0(5(5(3(1(2(x1)))))) 0(1(5(2(x1)))) -> 1(3(5(0(2(2(x1)))))) 0(5(2(2(x1)))) -> 3(0(2(1(5(2(x1)))))) 0(1(0(2(2(x1))))) -> 3(0(2(1(0(2(x1)))))) 0(1(2(5(2(x1))))) -> 3(5(1(2(2(0(x1)))))) 0(1(3(0(0(x1))))) -> 0(0(4(1(3(0(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(1(5(2(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(2(1(1(x1)))))) 5(0(1(2(x1)))) -> 0(2(1(3(5(x1))))) 5(0(1(2(x1)))) -> 2(0(3(1(5(x1))))) 5(0(1(2(x1)))) -> 5(3(1(0(3(2(x1)))))) 5(0(1(2(x1)))) -> 5(5(1(0(2(4(x1)))))) 1(0(1(2(x1)))) -> 0(4(1(2(1(x1))))) 1(0(1(2(x1)))) -> 1(2(1(3(0(x1))))) 1(0(1(2(x1)))) -> 0(3(1(4(1(2(x1)))))) 1(0(1(2(x1)))) -> 4(1(4(1(0(2(x1)))))) 1(3(0(1(2(x1))))) -> 1(3(4(1(0(2(x1)))))) ---------------------------------------- (22) Obligation: Q DP problem: The TRS P consists of the following rules: 0^1(1(2(x1))) -> 1^1(x1) 0^1(1(2(x1))) -> 0^1(x1) 0^1(1(2(x1))) -> 1^1(0(x1)) 0^1(1(2(x1))) -> 1^1(5(x1)) 0^1(1(4(2(2(x1))))) -> 1^1(x1) The TRS R consists of the following rules: 0(1(2(x1))) -> 0(3(1(2(x1)))) 0(1(2(x1))) -> 1(0(2(1(x1)))) 0(1(2(x1))) -> 2(1(4(0(x1)))) 0(1(2(x1))) -> 0(2(1(3(3(x1))))) 0(1(2(x1))) -> 0(2(4(1(3(x1))))) 0(1(2(x1))) -> 1(0(2(5(4(x1))))) 0(1(2(x1))) -> 2(1(1(0(3(x1))))) 0(1(2(x1))) -> 2(1(3(4(0(x1))))) 0(1(2(x1))) -> 2(1(4(3(0(x1))))) 0(1(2(x1))) -> 4(0(3(1(2(x1))))) 0(1(2(x1))) -> 4(1(0(3(2(x1))))) 0(1(2(x1))) -> 4(4(2(1(0(x1))))) 0(1(2(x1))) -> 0(0(4(1(2(5(x1)))))) 0(1(2(x1))) -> 0(4(5(1(2(5(x1)))))) 0(1(2(x1))) -> 0(5(2(3(1(4(x1)))))) 0(1(2(x1))) -> 1(0(2(2(3(4(x1)))))) 0(1(2(x1))) -> 1(1(4(0(3(2(x1)))))) 0(1(2(x1))) -> 2(0(0(3(3(1(x1)))))) 0(1(2(x1))) -> 2(0(3(4(1(5(x1)))))) 0(1(2(x1))) -> 2(3(2(0(4(1(x1)))))) 0(1(2(x1))) -> 2(5(3(0(3(1(x1)))))) 0(1(2(x1))) -> 3(2(5(3(1(0(x1)))))) 0(1(2(x1))) -> 3(4(0(5(1(2(x1)))))) 0(1(2(x1))) -> 5(5(4(2(1(0(x1)))))) 0(0(1(2(x1)))) -> 0(0(2(4(1(x1))))) 0(0(1(2(x1)))) -> 0(2(1(5(0(4(x1)))))) 0(0(1(2(x1)))) -> 1(0(2(3(5(0(x1)))))) 0(0(1(2(x1)))) -> 5(1(2(0(4(0(x1)))))) 0(0(5(2(x1)))) -> 5(3(0(3(0(2(x1)))))) 0(1(2(0(x1)))) -> 1(2(4(0(0(x1))))) 0(1(2(0(x1)))) -> 0(4(1(0(0(2(x1)))))) 0(1(2(2(x1)))) -> 3(3(0(2(1(2(x1)))))) 0(1(5(2(x1)))) -> 2(5(1(0(3(x1))))) 0(1(5(2(x1)))) -> 0(5(5(3(1(2(x1)))))) 0(1(5(2(x1)))) -> 1(3(5(0(2(2(x1)))))) 0(5(2(2(x1)))) -> 3(0(2(1(5(2(x1)))))) 1(0(1(2(x1)))) -> 0(4(1(2(1(x1))))) 1(0(1(2(x1)))) -> 1(2(1(3(0(x1))))) 1(0(1(2(x1)))) -> 0(3(1(4(1(2(x1)))))) 1(0(1(2(x1)))) -> 4(1(4(1(0(2(x1)))))) 5(0(1(2(x1)))) -> 0(2(1(3(5(x1))))) 5(0(1(2(x1)))) -> 2(0(3(1(5(x1))))) 5(0(1(2(x1)))) -> 5(3(1(0(3(2(x1)))))) 5(0(1(2(x1)))) -> 5(5(1(0(2(4(x1)))))) 0(1(0(2(2(x1))))) -> 3(0(2(1(0(2(x1)))))) 0(1(2(5(2(x1))))) -> 3(5(1(2(2(0(x1)))))) 0(1(3(0(0(x1))))) -> 0(0(4(1(3(0(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(1(5(2(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(2(1(1(x1)))))) 1(3(0(1(2(x1))))) -> 1(3(4(1(0(2(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 4 less nodes. ---------------------------------------- (24) Obligation: Q DP problem: The TRS P consists of the following rules: 0^1(1(2(x1))) -> 0^1(x1) The TRS R consists of the following rules: 0(1(2(x1))) -> 0(3(1(2(x1)))) 0(1(2(x1))) -> 1(0(2(1(x1)))) 0(1(2(x1))) -> 2(1(4(0(x1)))) 0(1(2(x1))) -> 0(2(1(3(3(x1))))) 0(1(2(x1))) -> 0(2(4(1(3(x1))))) 0(1(2(x1))) -> 1(0(2(5(4(x1))))) 0(1(2(x1))) -> 2(1(1(0(3(x1))))) 0(1(2(x1))) -> 2(1(3(4(0(x1))))) 0(1(2(x1))) -> 2(1(4(3(0(x1))))) 0(1(2(x1))) -> 4(0(3(1(2(x1))))) 0(1(2(x1))) -> 4(1(0(3(2(x1))))) 0(1(2(x1))) -> 4(4(2(1(0(x1))))) 0(1(2(x1))) -> 0(0(4(1(2(5(x1)))))) 0(1(2(x1))) -> 0(4(5(1(2(5(x1)))))) 0(1(2(x1))) -> 0(5(2(3(1(4(x1)))))) 0(1(2(x1))) -> 1(0(2(2(3(4(x1)))))) 0(1(2(x1))) -> 1(1(4(0(3(2(x1)))))) 0(1(2(x1))) -> 2(0(0(3(3(1(x1)))))) 0(1(2(x1))) -> 2(0(3(4(1(5(x1)))))) 0(1(2(x1))) -> 2(3(2(0(4(1(x1)))))) 0(1(2(x1))) -> 2(5(3(0(3(1(x1)))))) 0(1(2(x1))) -> 3(2(5(3(1(0(x1)))))) 0(1(2(x1))) -> 3(4(0(5(1(2(x1)))))) 0(1(2(x1))) -> 5(5(4(2(1(0(x1)))))) 0(0(1(2(x1)))) -> 0(0(2(4(1(x1))))) 0(0(1(2(x1)))) -> 0(2(1(5(0(4(x1)))))) 0(0(1(2(x1)))) -> 1(0(2(3(5(0(x1)))))) 0(0(1(2(x1)))) -> 5(1(2(0(4(0(x1)))))) 0(0(5(2(x1)))) -> 5(3(0(3(0(2(x1)))))) 0(1(2(0(x1)))) -> 1(2(4(0(0(x1))))) 0(1(2(0(x1)))) -> 0(4(1(0(0(2(x1)))))) 0(1(2(2(x1)))) -> 3(3(0(2(1(2(x1)))))) 0(1(5(2(x1)))) -> 2(5(1(0(3(x1))))) 0(1(5(2(x1)))) -> 0(5(5(3(1(2(x1)))))) 0(1(5(2(x1)))) -> 1(3(5(0(2(2(x1)))))) 0(5(2(2(x1)))) -> 3(0(2(1(5(2(x1)))))) 1(0(1(2(x1)))) -> 0(4(1(2(1(x1))))) 1(0(1(2(x1)))) -> 1(2(1(3(0(x1))))) 1(0(1(2(x1)))) -> 0(3(1(4(1(2(x1)))))) 1(0(1(2(x1)))) -> 4(1(4(1(0(2(x1)))))) 5(0(1(2(x1)))) -> 0(2(1(3(5(x1))))) 5(0(1(2(x1)))) -> 2(0(3(1(5(x1))))) 5(0(1(2(x1)))) -> 5(3(1(0(3(2(x1)))))) 5(0(1(2(x1)))) -> 5(5(1(0(2(4(x1)))))) 0(1(0(2(2(x1))))) -> 3(0(2(1(0(2(x1)))))) 0(1(2(5(2(x1))))) -> 3(5(1(2(2(0(x1)))))) 0(1(3(0(0(x1))))) -> 0(0(4(1(3(0(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(1(5(2(x1)))))) 0(1(4(2(2(x1))))) -> 0(2(4(2(1(1(x1)))))) 1(3(0(1(2(x1))))) -> 1(3(4(1(0(2(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (25) 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. ---------------------------------------- (26) Obligation: Q DP problem: The TRS P consists of the following rules: 0^1(1(2(x1))) -> 0^1(x1) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (27) 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: *0^1(1(2(x1))) -> 0^1(x1) The graph contains the following edges 1 > 1 ---------------------------------------- (28) YES