YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 178 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 4171 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 4541 ms] (12) QDP (13) QDPOrderProof [EQUIVALENT, 4755 ms] (14) QDP (15) QDPOrderProof [EQUIVALENT, 3907 ms] (16) QDP (17) QDPOrderProof [EQUIVALENT, 3037 ms] (18) QDP (19) QDPOrderProof [EQUIVALENT, 1953 ms] (20) QDP (21) QDPOrderProof [EQUIVALENT, 3713 ms] (22) QDP (23) QDPOrderProof [EQUIVALENT, 2897 ms] (24) QDP (25) DependencyGraphProof [EQUIVALENT, 0 ms] (26) QDP (27) QDPOrderProof [EQUIVALENT, 24 ms] (28) QDP (29) QDPSizeChangeProof [EQUIVALENT, 0 ms] (30) 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) QTRS Reverse (EQUIVALENT) We applied the QTRS Reverse Processor [REVERSE]. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 2(1(0(x1))) -> 2(1(3(0(x1)))) 2(1(0(x1))) -> 1(2(0(1(x1)))) 2(1(0(x1))) -> 0(4(1(2(x1)))) 2(1(0(x1))) -> 3(3(1(2(0(x1))))) 2(1(0(x1))) -> 3(1(4(2(0(x1))))) 2(1(0(x1))) -> 4(5(2(0(1(x1))))) 2(1(0(x1))) -> 3(0(1(1(2(x1))))) 2(1(0(x1))) -> 0(4(3(1(2(x1))))) 2(1(0(x1))) -> 0(3(4(1(2(x1))))) 2(1(0(x1))) -> 2(1(3(0(4(x1))))) 2(1(0(x1))) -> 2(3(0(1(4(x1))))) 2(1(0(x1))) -> 0(1(2(4(4(x1))))) 2(1(0(x1))) -> 5(2(1(4(0(0(x1)))))) 2(1(0(x1))) -> 5(2(1(5(4(0(x1)))))) 2(1(0(x1))) -> 4(1(3(2(5(0(x1)))))) 2(1(0(x1))) -> 4(3(2(2(0(1(x1)))))) 2(1(0(x1))) -> 2(3(0(4(1(1(x1)))))) 2(1(0(x1))) -> 1(3(3(0(0(2(x1)))))) 2(1(0(x1))) -> 5(1(4(3(0(2(x1)))))) 2(1(0(x1))) -> 1(4(0(2(3(2(x1)))))) 2(1(0(x1))) -> 1(3(0(3(5(2(x1)))))) 2(1(0(x1))) -> 0(1(3(5(2(3(x1)))))) 2(1(0(x1))) -> 2(1(5(0(4(3(x1)))))) 2(1(0(x1))) -> 0(1(2(4(5(5(x1)))))) 2(1(0(0(x1)))) -> 1(4(2(0(0(x1))))) 2(1(0(0(x1)))) -> 4(0(5(1(2(0(x1)))))) 2(1(0(0(x1)))) -> 0(5(3(2(0(1(x1)))))) 2(1(0(0(x1)))) -> 0(4(0(2(1(5(x1)))))) 2(5(0(0(x1)))) -> 2(0(3(0(3(5(x1)))))) 0(2(1(0(x1)))) -> 0(0(4(2(1(x1))))) 0(2(1(0(x1)))) -> 2(0(0(1(4(0(x1)))))) 2(2(1(0(x1)))) -> 2(1(2(0(3(3(x1)))))) 2(5(1(0(x1)))) -> 3(0(1(5(2(x1))))) 2(5(1(0(x1)))) -> 2(1(3(5(5(0(x1)))))) 2(5(1(0(x1)))) -> 2(2(0(5(3(1(x1)))))) 2(2(5(0(x1)))) -> 2(5(1(2(0(3(x1)))))) 2(1(0(1(x1)))) -> 1(2(1(4(0(x1))))) 2(1(0(1(x1)))) -> 0(3(1(2(1(x1))))) 2(1(0(1(x1)))) -> 2(1(4(1(3(0(x1)))))) 2(1(0(1(x1)))) -> 2(0(1(4(1(4(x1)))))) 2(1(0(5(x1)))) -> 5(3(1(2(0(x1))))) 2(1(0(5(x1)))) -> 5(1(3(0(2(x1))))) 2(1(0(5(x1)))) -> 2(3(0(1(3(5(x1)))))) 2(1(0(5(x1)))) -> 4(2(0(1(5(5(x1)))))) 2(2(0(1(0(x1))))) -> 2(0(1(2(0(3(x1)))))) 2(5(2(1(0(x1))))) -> 0(2(2(1(5(3(x1)))))) 0(0(3(1(0(x1))))) -> 0(3(1(4(0(0(x1)))))) 2(2(4(1(0(x1))))) -> 2(5(1(4(2(0(x1)))))) 2(2(4(1(0(x1))))) -> 1(1(2(4(2(0(x1)))))) 2(1(0(3(1(x1))))) -> 2(0(1(4(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: 2^1(1(0(x1))) -> 2^1(1(3(0(x1)))) 2^1(1(0(x1))) -> 2^1(0(1(x1))) 2^1(1(0(x1))) -> 0^1(1(x1)) 2^1(1(0(x1))) -> 0^1(4(1(2(x1)))) 2^1(1(0(x1))) -> 2^1(x1) 2^1(1(0(x1))) -> 2^1(0(x1)) 2^1(1(0(x1))) -> 0^1(1(1(2(x1)))) 2^1(1(0(x1))) -> 0^1(4(3(1(2(x1))))) 2^1(1(0(x1))) -> 0^1(3(4(1(2(x1))))) 2^1(1(0(x1))) -> 2^1(1(3(0(4(x1))))) 2^1(1(0(x1))) -> 0^1(4(x1)) 2^1(1(0(x1))) -> 2^1(3(0(1(4(x1))))) 2^1(1(0(x1))) -> 0^1(1(4(x1))) 2^1(1(0(x1))) -> 0^1(1(2(4(4(x1))))) 2^1(1(0(x1))) -> 2^1(4(4(x1))) 2^1(1(0(x1))) -> 2^1(1(4(0(0(x1))))) 2^1(1(0(x1))) -> 0^1(0(x1)) 2^1(1(0(x1))) -> 2^1(1(5(4(0(x1))))) 2^1(1(0(x1))) -> 2^1(5(0(x1))) 2^1(1(0(x1))) -> 2^1(2(0(1(x1)))) 2^1(1(0(x1))) -> 2^1(3(0(4(1(1(x1)))))) 2^1(1(0(x1))) -> 0^1(4(1(1(x1)))) 2^1(1(0(x1))) -> 0^1(0(2(x1))) 2^1(1(0(x1))) -> 0^1(2(x1)) 2^1(1(0(x1))) -> 0^1(2(3(2(x1)))) 2^1(1(0(x1))) -> 2^1(3(2(x1))) 2^1(1(0(x1))) -> 0^1(3(5(2(x1)))) 2^1(1(0(x1))) -> 0^1(1(3(5(2(3(x1)))))) 2^1(1(0(x1))) -> 2^1(3(x1)) 2^1(1(0(x1))) -> 2^1(1(5(0(4(3(x1)))))) 2^1(1(0(x1))) -> 0^1(4(3(x1))) 2^1(1(0(x1))) -> 0^1(1(2(4(5(5(x1)))))) 2^1(1(0(x1))) -> 2^1(4(5(5(x1)))) 2^1(1(0(0(x1)))) -> 2^1(0(0(x1))) 2^1(1(0(0(x1)))) -> 0^1(5(1(2(0(x1))))) 2^1(1(0(0(x1)))) -> 2^1(0(x1)) 2^1(1(0(0(x1)))) -> 0^1(5(3(2(0(1(x1)))))) 2^1(1(0(0(x1)))) -> 2^1(0(1(x1))) 2^1(1(0(0(x1)))) -> 0^1(1(x1)) 2^1(1(0(0(x1)))) -> 0^1(4(0(2(1(5(x1)))))) 2^1(1(0(0(x1)))) -> 0^1(2(1(5(x1)))) 2^1(1(0(0(x1)))) -> 2^1(1(5(x1))) 2^1(5(0(0(x1)))) -> 2^1(0(3(0(3(5(x1)))))) 2^1(5(0(0(x1)))) -> 0^1(3(0(3(5(x1))))) 2^1(5(0(0(x1)))) -> 0^1(3(5(x1))) 0^1(2(1(0(x1)))) -> 0^1(0(4(2(1(x1))))) 0^1(2(1(0(x1)))) -> 0^1(4(2(1(x1)))) 0^1(2(1(0(x1)))) -> 2^1(1(x1)) 0^1(2(1(0(x1)))) -> 2^1(0(0(1(4(0(x1)))))) 0^1(2(1(0(x1)))) -> 0^1(0(1(4(0(x1))))) 0^1(2(1(0(x1)))) -> 0^1(1(4(0(x1)))) 2^1(2(1(0(x1)))) -> 2^1(1(2(0(3(3(x1)))))) 2^1(2(1(0(x1)))) -> 2^1(0(3(3(x1)))) 2^1(2(1(0(x1)))) -> 0^1(3(3(x1))) 2^1(5(1(0(x1)))) -> 0^1(1(5(2(x1)))) 2^1(5(1(0(x1)))) -> 2^1(x1) 2^1(5(1(0(x1)))) -> 2^1(1(3(5(5(0(x1)))))) 2^1(5(1(0(x1)))) -> 2^1(2(0(5(3(1(x1)))))) 2^1(5(1(0(x1)))) -> 2^1(0(5(3(1(x1))))) 2^1(5(1(0(x1)))) -> 0^1(5(3(1(x1)))) 2^1(2(5(0(x1)))) -> 2^1(5(1(2(0(3(x1)))))) 2^1(2(5(0(x1)))) -> 2^1(0(3(x1))) 2^1(2(5(0(x1)))) -> 0^1(3(x1)) 2^1(1(0(1(x1)))) -> 2^1(1(4(0(x1)))) 2^1(1(0(1(x1)))) -> 0^1(x1) 2^1(1(0(1(x1)))) -> 0^1(3(1(2(1(x1))))) 2^1(1(0(1(x1)))) -> 2^1(1(x1)) 2^1(1(0(1(x1)))) -> 2^1(1(4(1(3(0(x1)))))) 2^1(1(0(1(x1)))) -> 2^1(0(1(4(1(4(x1)))))) 2^1(1(0(1(x1)))) -> 0^1(1(4(1(4(x1))))) 2^1(1(0(5(x1)))) -> 2^1(0(x1)) 2^1(1(0(5(x1)))) -> 0^1(x1) 2^1(1(0(5(x1)))) -> 0^1(2(x1)) 2^1(1(0(5(x1)))) -> 2^1(x1) 2^1(1(0(5(x1)))) -> 2^1(3(0(1(3(5(x1)))))) 2^1(1(0(5(x1)))) -> 0^1(1(3(5(x1)))) 2^1(1(0(5(x1)))) -> 2^1(0(1(5(5(x1))))) 2^1(1(0(5(x1)))) -> 0^1(1(5(5(x1)))) 2^1(2(0(1(0(x1))))) -> 2^1(0(1(2(0(3(x1)))))) 2^1(2(0(1(0(x1))))) -> 0^1(1(2(0(3(x1))))) 2^1(2(0(1(0(x1))))) -> 2^1(0(3(x1))) 2^1(2(0(1(0(x1))))) -> 0^1(3(x1)) 2^1(5(2(1(0(x1))))) -> 0^1(2(2(1(5(3(x1)))))) 2^1(5(2(1(0(x1))))) -> 2^1(2(1(5(3(x1))))) 2^1(5(2(1(0(x1))))) -> 2^1(1(5(3(x1)))) 0^1(0(3(1(0(x1))))) -> 0^1(3(1(4(0(0(x1)))))) 0^1(0(3(1(0(x1))))) -> 0^1(0(x1)) 2^1(2(4(1(0(x1))))) -> 2^1(5(1(4(2(0(x1)))))) 2^1(2(4(1(0(x1))))) -> 2^1(0(x1)) 2^1(2(4(1(0(x1))))) -> 2^1(4(2(0(x1)))) 2^1(1(0(3(1(x1))))) -> 2^1(0(1(4(3(1(x1)))))) 2^1(1(0(3(1(x1))))) -> 0^1(1(4(3(1(x1))))) The TRS R consists of the following rules: 2(1(0(x1))) -> 2(1(3(0(x1)))) 2(1(0(x1))) -> 1(2(0(1(x1)))) 2(1(0(x1))) -> 0(4(1(2(x1)))) 2(1(0(x1))) -> 3(3(1(2(0(x1))))) 2(1(0(x1))) -> 3(1(4(2(0(x1))))) 2(1(0(x1))) -> 4(5(2(0(1(x1))))) 2(1(0(x1))) -> 3(0(1(1(2(x1))))) 2(1(0(x1))) -> 0(4(3(1(2(x1))))) 2(1(0(x1))) -> 0(3(4(1(2(x1))))) 2(1(0(x1))) -> 2(1(3(0(4(x1))))) 2(1(0(x1))) -> 2(3(0(1(4(x1))))) 2(1(0(x1))) -> 0(1(2(4(4(x1))))) 2(1(0(x1))) -> 5(2(1(4(0(0(x1)))))) 2(1(0(x1))) -> 5(2(1(5(4(0(x1)))))) 2(1(0(x1))) -> 4(1(3(2(5(0(x1)))))) 2(1(0(x1))) -> 4(3(2(2(0(1(x1)))))) 2(1(0(x1))) -> 2(3(0(4(1(1(x1)))))) 2(1(0(x1))) -> 1(3(3(0(0(2(x1)))))) 2(1(0(x1))) -> 5(1(4(3(0(2(x1)))))) 2(1(0(x1))) -> 1(4(0(2(3(2(x1)))))) 2(1(0(x1))) -> 1(3(0(3(5(2(x1)))))) 2(1(0(x1))) -> 0(1(3(5(2(3(x1)))))) 2(1(0(x1))) -> 2(1(5(0(4(3(x1)))))) 2(1(0(x1))) -> 0(1(2(4(5(5(x1)))))) 2(1(0(0(x1)))) -> 1(4(2(0(0(x1))))) 2(1(0(0(x1)))) -> 4(0(5(1(2(0(x1)))))) 2(1(0(0(x1)))) -> 0(5(3(2(0(1(x1)))))) 2(1(0(0(x1)))) -> 0(4(0(2(1(5(x1)))))) 2(5(0(0(x1)))) -> 2(0(3(0(3(5(x1)))))) 0(2(1(0(x1)))) -> 0(0(4(2(1(x1))))) 0(2(1(0(x1)))) -> 2(0(0(1(4(0(x1)))))) 2(2(1(0(x1)))) -> 2(1(2(0(3(3(x1)))))) 2(5(1(0(x1)))) -> 3(0(1(5(2(x1))))) 2(5(1(0(x1)))) -> 2(1(3(5(5(0(x1)))))) 2(5(1(0(x1)))) -> 2(2(0(5(3(1(x1)))))) 2(2(5(0(x1)))) -> 2(5(1(2(0(3(x1)))))) 2(1(0(1(x1)))) -> 1(2(1(4(0(x1))))) 2(1(0(1(x1)))) -> 0(3(1(2(1(x1))))) 2(1(0(1(x1)))) -> 2(1(4(1(3(0(x1)))))) 2(1(0(1(x1)))) -> 2(0(1(4(1(4(x1)))))) 2(1(0(5(x1)))) -> 5(3(1(2(0(x1))))) 2(1(0(5(x1)))) -> 5(1(3(0(2(x1))))) 2(1(0(5(x1)))) -> 2(3(0(1(3(5(x1)))))) 2(1(0(5(x1)))) -> 4(2(0(1(5(5(x1)))))) 2(2(0(1(0(x1))))) -> 2(0(1(2(0(3(x1)))))) 2(5(2(1(0(x1))))) -> 0(2(2(1(5(3(x1)))))) 0(0(3(1(0(x1))))) -> 0(3(1(4(0(0(x1)))))) 2(2(4(1(0(x1))))) -> 2(5(1(4(2(0(x1)))))) 2(2(4(1(0(x1))))) -> 1(1(2(4(2(0(x1)))))) 2(1(0(3(1(x1))))) -> 2(0(1(4(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 1 SCC with 74 less nodes. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: 2^1(1(0(x1))) -> 2^1(0(x1)) 2^1(1(0(x1))) -> 2^1(x1) 2^1(1(0(x1))) -> 0^1(0(x1)) 0^1(2(1(0(x1)))) -> 2^1(1(x1)) 2^1(1(0(x1))) -> 2^1(5(0(x1))) 2^1(5(1(0(x1)))) -> 2^1(x1) 2^1(1(0(x1))) -> 0^1(0(2(x1))) 0^1(0(3(1(0(x1))))) -> 0^1(0(x1)) 2^1(1(0(x1))) -> 0^1(2(x1)) 2^1(1(0(0(x1)))) -> 2^1(0(0(x1))) 2^1(1(0(0(x1)))) -> 2^1(0(x1)) 2^1(1(0(1(x1)))) -> 0^1(x1) 2^1(1(0(1(x1)))) -> 2^1(1(x1)) 2^1(1(0(5(x1)))) -> 2^1(0(x1)) 2^1(1(0(5(x1)))) -> 0^1(x1) 2^1(1(0(5(x1)))) -> 0^1(2(x1)) 2^1(1(0(5(x1)))) -> 2^1(x1) 2^1(2(4(1(0(x1))))) -> 2^1(0(x1)) The TRS R consists of the following rules: 2(1(0(x1))) -> 2(1(3(0(x1)))) 2(1(0(x1))) -> 1(2(0(1(x1)))) 2(1(0(x1))) -> 0(4(1(2(x1)))) 2(1(0(x1))) -> 3(3(1(2(0(x1))))) 2(1(0(x1))) -> 3(1(4(2(0(x1))))) 2(1(0(x1))) -> 4(5(2(0(1(x1))))) 2(1(0(x1))) -> 3(0(1(1(2(x1))))) 2(1(0(x1))) -> 0(4(3(1(2(x1))))) 2(1(0(x1))) -> 0(3(4(1(2(x1))))) 2(1(0(x1))) -> 2(1(3(0(4(x1))))) 2(1(0(x1))) -> 2(3(0(1(4(x1))))) 2(1(0(x1))) -> 0(1(2(4(4(x1))))) 2(1(0(x1))) -> 5(2(1(4(0(0(x1)))))) 2(1(0(x1))) -> 5(2(1(5(4(0(x1)))))) 2(1(0(x1))) -> 4(1(3(2(5(0(x1)))))) 2(1(0(x1))) -> 4(3(2(2(0(1(x1)))))) 2(1(0(x1))) -> 2(3(0(4(1(1(x1)))))) 2(1(0(x1))) -> 1(3(3(0(0(2(x1)))))) 2(1(0(x1))) -> 5(1(4(3(0(2(x1)))))) 2(1(0(x1))) -> 1(4(0(2(3(2(x1)))))) 2(1(0(x1))) -> 1(3(0(3(5(2(x1)))))) 2(1(0(x1))) -> 0(1(3(5(2(3(x1)))))) 2(1(0(x1))) -> 2(1(5(0(4(3(x1)))))) 2(1(0(x1))) -> 0(1(2(4(5(5(x1)))))) 2(1(0(0(x1)))) -> 1(4(2(0(0(x1))))) 2(1(0(0(x1)))) -> 4(0(5(1(2(0(x1)))))) 2(1(0(0(x1)))) -> 0(5(3(2(0(1(x1)))))) 2(1(0(0(x1)))) -> 0(4(0(2(1(5(x1)))))) 2(5(0(0(x1)))) -> 2(0(3(0(3(5(x1)))))) 0(2(1(0(x1)))) -> 0(0(4(2(1(x1))))) 0(2(1(0(x1)))) -> 2(0(0(1(4(0(x1)))))) 2(2(1(0(x1)))) -> 2(1(2(0(3(3(x1)))))) 2(5(1(0(x1)))) -> 3(0(1(5(2(x1))))) 2(5(1(0(x1)))) -> 2(1(3(5(5(0(x1)))))) 2(5(1(0(x1)))) -> 2(2(0(5(3(1(x1)))))) 2(2(5(0(x1)))) -> 2(5(1(2(0(3(x1)))))) 2(1(0(1(x1)))) -> 1(2(1(4(0(x1))))) 2(1(0(1(x1)))) -> 0(3(1(2(1(x1))))) 2(1(0(1(x1)))) -> 2(1(4(1(3(0(x1)))))) 2(1(0(1(x1)))) -> 2(0(1(4(1(4(x1)))))) 2(1(0(5(x1)))) -> 5(3(1(2(0(x1))))) 2(1(0(5(x1)))) -> 5(1(3(0(2(x1))))) 2(1(0(5(x1)))) -> 2(3(0(1(3(5(x1)))))) 2(1(0(5(x1)))) -> 4(2(0(1(5(5(x1)))))) 2(2(0(1(0(x1))))) -> 2(0(1(2(0(3(x1)))))) 2(5(2(1(0(x1))))) -> 0(2(2(1(5(3(x1)))))) 0(0(3(1(0(x1))))) -> 0(3(1(4(0(0(x1)))))) 2(2(4(1(0(x1))))) -> 2(5(1(4(2(0(x1)))))) 2(2(4(1(0(x1))))) -> 1(1(2(4(2(0(x1)))))) 2(1(0(3(1(x1))))) -> 2(0(1(4(3(1(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. 2^1(5(1(0(x1)))) -> 2^1(x1) The remaining pairs can at least be oriented weakly. Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: <<< POL(2^1(x_1)) = [[0A]] + [[0A, 0A, 0A]] * x_1 >>> <<< POL(1(x_1)) = [[0A], [0A], [1A]] + [[0A, 0A, -I], [0A, -I, -I], [0A, 0A, 0A]] * x_1 >>> <<< POL(0(x_1)) = [[1A], [0A], [0A]] + [[0A, 0A, 0A], [-I, -I, -I], [0A, 0A, 0A]] * x_1 >>> <<< POL(0^1(x_1)) = [[1A]] + [[-I, 0A, -I]] * x_1 >>> <<< POL(2(x_1)) = [[0A], [0A], [0A]] + [[-I, 0A, -I], [-I, 0A, -I], [0A, 1A, -I]] * x_1 >>> <<< POL(5(x_1)) = [[0A], [0A], [0A]] + [[-I, -I, -I], [-I, 0A, -I], [0A, 1A, 0A]] * x_1 >>> <<< POL(3(x_1)) = [[0A], [-I], [-I]] + [[-I, -I, -I], [-I, -I, -I], [-I, -I, -I]] * x_1 >>> <<< POL(4(x_1)) = [[0A], [0A], [0A]] + [[0A, 0A, -I], [-I, -I, -I], [0A, 0A, -I]] * x_1 >>> The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 0(2(1(0(x1)))) -> 0(0(4(2(1(x1))))) 0(2(1(0(x1)))) -> 2(0(0(1(4(0(x1)))))) 0(0(3(1(0(x1))))) -> 0(3(1(4(0(0(x1)))))) 2(1(0(x1))) -> 2(1(3(0(x1)))) 2(1(0(x1))) -> 1(2(0(1(x1)))) 2(1(0(x1))) -> 0(4(1(2(x1)))) 2(1(0(x1))) -> 3(3(1(2(0(x1))))) 2(1(0(x1))) -> 3(1(4(2(0(x1))))) 2(1(0(x1))) -> 4(5(2(0(1(x1))))) 2(1(0(x1))) -> 3(0(1(1(2(x1))))) 2(1(0(x1))) -> 0(4(3(1(2(x1))))) 2(1(0(x1))) -> 0(3(4(1(2(x1))))) 2(1(0(x1))) -> 2(1(3(0(4(x1))))) 2(1(0(x1))) -> 2(3(0(1(4(x1))))) 2(1(0(x1))) -> 0(1(2(4(4(x1))))) 2(1(0(x1))) -> 5(2(1(4(0(0(x1)))))) 2(1(0(x1))) -> 5(2(1(5(4(0(x1)))))) 2(1(0(x1))) -> 4(1(3(2(5(0(x1)))))) 2(1(0(x1))) -> 4(3(2(2(0(1(x1)))))) 2(1(0(x1))) -> 2(3(0(4(1(1(x1)))))) 2(1(0(x1))) -> 1(3(3(0(0(2(x1)))))) 2(1(0(x1))) -> 5(1(4(3(0(2(x1)))))) 2(1(0(x1))) -> 1(4(0(2(3(2(x1)))))) 2(1(0(x1))) -> 1(3(0(3(5(2(x1)))))) 2(1(0(x1))) -> 0(1(3(5(2(3(x1)))))) 2(1(0(x1))) -> 2(1(5(0(4(3(x1)))))) 2(1(0(x1))) -> 0(1(2(4(5(5(x1)))))) 2(1(0(0(x1)))) -> 1(4(2(0(0(x1))))) 2(1(0(0(x1)))) -> 4(0(5(1(2(0(x1)))))) 2(1(0(0(x1)))) -> 0(5(3(2(0(1(x1)))))) 2(1(0(0(x1)))) -> 0(4(0(2(1(5(x1)))))) 2(5(0(0(x1)))) -> 2(0(3(0(3(5(x1)))))) 2(2(1(0(x1)))) -> 2(1(2(0(3(3(x1)))))) 2(5(1(0(x1)))) -> 3(0(1(5(2(x1))))) 2(5(1(0(x1)))) -> 2(1(3(5(5(0(x1)))))) 2(5(1(0(x1)))) -> 2(2(0(5(3(1(x1)))))) 2(2(5(0(x1)))) -> 2(5(1(2(0(3(x1)))))) 2(1(0(1(x1)))) -> 1(2(1(4(0(x1))))) 2(1(0(1(x1)))) -> 0(3(1(2(1(x1))))) 2(1(0(1(x1)))) -> 2(1(4(1(3(0(x1)))))) 2(1(0(1(x1)))) -> 2(0(1(4(1(4(x1)))))) 2(1(0(5(x1)))) -> 5(3(1(2(0(x1))))) 2(1(0(5(x1)))) -> 5(1(3(0(2(x1))))) 2(1(0(5(x1)))) -> 2(3(0(1(3(5(x1)))))) 2(1(0(5(x1)))) -> 4(2(0(1(5(5(x1)))))) 2(2(0(1(0(x1))))) -> 2(0(1(2(0(3(x1)))))) 2(5(2(1(0(x1))))) -> 0(2(2(1(5(3(x1)))))) 2(2(4(1(0(x1))))) -> 2(5(1(4(2(0(x1)))))) 2(2(4(1(0(x1))))) -> 1(1(2(4(2(0(x1)))))) 2(1(0(3(1(x1))))) -> 2(0(1(4(3(1(x1)))))) ---------------------------------------- (8) Obligation: Q DP problem: The TRS P consists of the following rules: 2^1(1(0(x1))) -> 2^1(0(x1)) 2^1(1(0(x1))) -> 2^1(x1) 2^1(1(0(x1))) -> 0^1(0(x1)) 0^1(2(1(0(x1)))) -> 2^1(1(x1)) 2^1(1(0(x1))) -> 2^1(5(0(x1))) 2^1(1(0(x1))) -> 0^1(0(2(x1))) 0^1(0(3(1(0(x1))))) -> 0^1(0(x1)) 2^1(1(0(x1))) -> 0^1(2(x1)) 2^1(1(0(0(x1)))) -> 2^1(0(0(x1))) 2^1(1(0(0(x1)))) -> 2^1(0(x1)) 2^1(1(0(1(x1)))) -> 0^1(x1) 2^1(1(0(1(x1)))) -> 2^1(1(x1)) 2^1(1(0(5(x1)))) -> 2^1(0(x1)) 2^1(1(0(5(x1)))) -> 0^1(x1) 2^1(1(0(5(x1)))) -> 0^1(2(x1)) 2^1(1(0(5(x1)))) -> 2^1(x1) 2^1(2(4(1(0(x1))))) -> 2^1(0(x1)) The TRS R consists of the following rules: 2(1(0(x1))) -> 2(1(3(0(x1)))) 2(1(0(x1))) -> 1(2(0(1(x1)))) 2(1(0(x1))) -> 0(4(1(2(x1)))) 2(1(0(x1))) -> 3(3(1(2(0(x1))))) 2(1(0(x1))) -> 3(1(4(2(0(x1))))) 2(1(0(x1))) -> 4(5(2(0(1(x1))))) 2(1(0(x1))) -> 3(0(1(1(2(x1))))) 2(1(0(x1))) -> 0(4(3(1(2(x1))))) 2(1(0(x1))) -> 0(3(4(1(2(x1))))) 2(1(0(x1))) -> 2(1(3(0(4(x1))))) 2(1(0(x1))) -> 2(3(0(1(4(x1))))) 2(1(0(x1))) -> 0(1(2(4(4(x1))))) 2(1(0(x1))) -> 5(2(1(4(0(0(x1)))))) 2(1(0(x1))) -> 5(2(1(5(4(0(x1)))))) 2(1(0(x1))) -> 4(1(3(2(5(0(x1)))))) 2(1(0(x1))) -> 4(3(2(2(0(1(x1)))))) 2(1(0(x1))) -> 2(3(0(4(1(1(x1)))))) 2(1(0(x1))) -> 1(3(3(0(0(2(x1)))))) 2(1(0(x1))) -> 5(1(4(3(0(2(x1)))))) 2(1(0(x1))) -> 1(4(0(2(3(2(x1)))))) 2(1(0(x1))) -> 1(3(0(3(5(2(x1)))))) 2(1(0(x1))) -> 0(1(3(5(2(3(x1)))))) 2(1(0(x1))) -> 2(1(5(0(4(3(x1)))))) 2(1(0(x1))) -> 0(1(2(4(5(5(x1)))))) 2(1(0(0(x1)))) -> 1(4(2(0(0(x1))))) 2(1(0(0(x1)))) -> 4(0(5(1(2(0(x1)))))) 2(1(0(0(x1)))) -> 0(5(3(2(0(1(x1)))))) 2(1(0(0(x1)))) -> 0(4(0(2(1(5(x1)))))) 2(5(0(0(x1)))) -> 2(0(3(0(3(5(x1)))))) 0(2(1(0(x1)))) -> 0(0(4(2(1(x1))))) 0(2(1(0(x1)))) -> 2(0(0(1(4(0(x1)))))) 2(2(1(0(x1)))) -> 2(1(2(0(3(3(x1)))))) 2(5(1(0(x1)))) -> 3(0(1(5(2(x1))))) 2(5(1(0(x1)))) -> 2(1(3(5(5(0(x1)))))) 2(5(1(0(x1)))) -> 2(2(0(5(3(1(x1)))))) 2(2(5(0(x1)))) -> 2(5(1(2(0(3(x1)))))) 2(1(0(1(x1)))) -> 1(2(1(4(0(x1))))) 2(1(0(1(x1)))) -> 0(3(1(2(1(x1))))) 2(1(0(1(x1)))) -> 2(1(4(1(3(0(x1)))))) 2(1(0(1(x1)))) -> 2(0(1(4(1(4(x1)))))) 2(1(0(5(x1)))) -> 5(3(1(2(0(x1))))) 2(1(0(5(x1)))) -> 5(1(3(0(2(x1))))) 2(1(0(5(x1)))) -> 2(3(0(1(3(5(x1)))))) 2(1(0(5(x1)))) -> 4(2(0(1(5(5(x1)))))) 2(2(0(1(0(x1))))) -> 2(0(1(2(0(3(x1)))))) 2(5(2(1(0(x1))))) -> 0(2(2(1(5(3(x1)))))) 0(0(3(1(0(x1))))) -> 0(3(1(4(0(0(x1)))))) 2(2(4(1(0(x1))))) -> 2(5(1(4(2(0(x1)))))) 2(2(4(1(0(x1))))) -> 1(1(2(4(2(0(x1)))))) 2(1(0(3(1(x1))))) -> 2(0(1(4(3(1(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (9) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (10) Obligation: Q DP problem: The TRS P consists of the following rules: 2^1(1(0(x1))) -> 2^1(x1) 2^1(1(0(x1))) -> 2^1(0(x1)) 2^1(1(0(x1))) -> 0^1(0(x1)) 0^1(2(1(0(x1)))) -> 2^1(1(x1)) 2^1(1(0(x1))) -> 0^1(0(2(x1))) 0^1(0(3(1(0(x1))))) -> 0^1(0(x1)) 2^1(1(0(x1))) -> 0^1(2(x1)) 2^1(1(0(0(x1)))) -> 2^1(0(0(x1))) 2^1(1(0(0(x1)))) -> 2^1(0(x1)) 2^1(1(0(1(x1)))) -> 0^1(x1) 2^1(1(0(1(x1)))) -> 2^1(1(x1)) 2^1(1(0(5(x1)))) -> 2^1(0(x1)) 2^1(1(0(5(x1)))) -> 0^1(x1) 2^1(1(0(5(x1)))) -> 0^1(2(x1)) 2^1(1(0(5(x1)))) -> 2^1(x1) 2^1(2(4(1(0(x1))))) -> 2^1(0(x1)) The TRS R consists of the following rules: 2(1(0(x1))) -> 2(1(3(0(x1)))) 2(1(0(x1))) -> 1(2(0(1(x1)))) 2(1(0(x1))) -> 0(4(1(2(x1)))) 2(1(0(x1))) -> 3(3(1(2(0(x1))))) 2(1(0(x1))) -> 3(1(4(2(0(x1))))) 2(1(0(x1))) -> 4(5(2(0(1(x1))))) 2(1(0(x1))) -> 3(0(1(1(2(x1))))) 2(1(0(x1))) -> 0(4(3(1(2(x1))))) 2(1(0(x1))) -> 0(3(4(1(2(x1))))) 2(1(0(x1))) -> 2(1(3(0(4(x1))))) 2(1(0(x1))) -> 2(3(0(1(4(x1))))) 2(1(0(x1))) -> 0(1(2(4(4(x1))))) 2(1(0(x1))) -> 5(2(1(4(0(0(x1)))))) 2(1(0(x1))) -> 5(2(1(5(4(0(x1)))))) 2(1(0(x1))) -> 4(1(3(2(5(0(x1)))))) 2(1(0(x1))) -> 4(3(2(2(0(1(x1)))))) 2(1(0(x1))) -> 2(3(0(4(1(1(x1)))))) 2(1(0(x1))) -> 1(3(3(0(0(2(x1)))))) 2(1(0(x1))) -> 5(1(4(3(0(2(x1)))))) 2(1(0(x1))) -> 1(4(0(2(3(2(x1)))))) 2(1(0(x1))) -> 1(3(0(3(5(2(x1)))))) 2(1(0(x1))) -> 0(1(3(5(2(3(x1)))))) 2(1(0(x1))) -> 2(1(5(0(4(3(x1)))))) 2(1(0(x1))) -> 0(1(2(4(5(5(x1)))))) 2(1(0(0(x1)))) -> 1(4(2(0(0(x1))))) 2(1(0(0(x1)))) -> 4(0(5(1(2(0(x1)))))) 2(1(0(0(x1)))) -> 0(5(3(2(0(1(x1)))))) 2(1(0(0(x1)))) -> 0(4(0(2(1(5(x1)))))) 2(5(0(0(x1)))) -> 2(0(3(0(3(5(x1)))))) 0(2(1(0(x1)))) -> 0(0(4(2(1(x1))))) 0(2(1(0(x1)))) -> 2(0(0(1(4(0(x1)))))) 2(2(1(0(x1)))) -> 2(1(2(0(3(3(x1)))))) 2(5(1(0(x1)))) -> 3(0(1(5(2(x1))))) 2(5(1(0(x1)))) -> 2(1(3(5(5(0(x1)))))) 2(5(1(0(x1)))) -> 2(2(0(5(3(1(x1)))))) 2(2(5(0(x1)))) -> 2(5(1(2(0(3(x1)))))) 2(1(0(1(x1)))) -> 1(2(1(4(0(x1))))) 2(1(0(1(x1)))) -> 0(3(1(2(1(x1))))) 2(1(0(1(x1)))) -> 2(1(4(1(3(0(x1)))))) 2(1(0(1(x1)))) -> 2(0(1(4(1(4(x1)))))) 2(1(0(5(x1)))) -> 5(3(1(2(0(x1))))) 2(1(0(5(x1)))) -> 5(1(3(0(2(x1))))) 2(1(0(5(x1)))) -> 2(3(0(1(3(5(x1)))))) 2(1(0(5(x1)))) -> 4(2(0(1(5(5(x1)))))) 2(2(0(1(0(x1))))) -> 2(0(1(2(0(3(x1)))))) 2(5(2(1(0(x1))))) -> 0(2(2(1(5(3(x1)))))) 0(0(3(1(0(x1))))) -> 0(3(1(4(0(0(x1)))))) 2(2(4(1(0(x1))))) -> 2(5(1(4(2(0(x1)))))) 2(2(4(1(0(x1))))) -> 1(1(2(4(2(0(x1)))))) 2(1(0(3(1(x1))))) -> 2(0(1(4(3(1(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (11) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. 2^1(2(4(1(0(x1))))) -> 2^1(0(x1)) The remaining pairs can at least be oriented weakly. Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: <<< POL(2^1(x_1)) = [[0A]] + [[0A, 0A, 0A]] * x_1 >>> <<< POL(1(x_1)) = [[0A], [1A], [0A]] + [[0A, 0A, -I], [0A, 0A, -I], [0A, 0A, 0A]] * x_1 >>> <<< POL(0(x_1)) = [[0A], [0A], [0A]] + [[0A, 0A, -I], [0A, -I, 0A], [0A, 0A, 0A]] * x_1 >>> <<< POL(0^1(x_1)) = [[1A]] + [[0A, -I, -I]] * x_1 >>> <<< POL(2(x_1)) = [[0A], [1A], [0A]] + [[-I, 0A, 0A], [0A, -I, 0A], [0A, 1A, 0A]] * x_1 >>> <<< POL(3(x_1)) = [[0A], [0A], [0A]] + [[-I, -I, -I], [0A, -I, -I], [-I, -I, -I]] * x_1 >>> <<< POL(5(x_1)) = [[0A], [0A], [0A]] + [[-I, -I, -I], [0A, 0A, -I], [0A, -I, 0A]] * x_1 >>> <<< POL(4(x_1)) = [[0A], [0A], [0A]] + [[-I, -I, -I], [0A, -I, -I], [0A, -I, -I]] * x_1 >>> The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 0(2(1(0(x1)))) -> 0(0(4(2(1(x1))))) 0(2(1(0(x1)))) -> 2(0(0(1(4(0(x1)))))) 0(0(3(1(0(x1))))) -> 0(3(1(4(0(0(x1)))))) 2(1(0(x1))) -> 2(1(3(0(x1)))) 2(1(0(x1))) -> 1(2(0(1(x1)))) 2(1(0(x1))) -> 0(4(1(2(x1)))) 2(1(0(x1))) -> 3(3(1(2(0(x1))))) 2(1(0(x1))) -> 3(1(4(2(0(x1))))) 2(1(0(x1))) -> 4(5(2(0(1(x1))))) 2(1(0(x1))) -> 3(0(1(1(2(x1))))) 2(1(0(x1))) -> 0(4(3(1(2(x1))))) 2(1(0(x1))) -> 0(3(4(1(2(x1))))) 2(1(0(x1))) -> 2(1(3(0(4(x1))))) 2(1(0(x1))) -> 2(3(0(1(4(x1))))) 2(1(0(x1))) -> 0(1(2(4(4(x1))))) 2(1(0(x1))) -> 5(2(1(4(0(0(x1)))))) 2(1(0(x1))) -> 5(2(1(5(4(0(x1)))))) 2(1(0(x1))) -> 4(1(3(2(5(0(x1)))))) 2(1(0(x1))) -> 4(3(2(2(0(1(x1)))))) 2(1(0(x1))) -> 2(3(0(4(1(1(x1)))))) 2(1(0(x1))) -> 1(3(3(0(0(2(x1)))))) 2(1(0(x1))) -> 5(1(4(3(0(2(x1)))))) 2(1(0(x1))) -> 1(4(0(2(3(2(x1)))))) 2(1(0(x1))) -> 1(3(0(3(5(2(x1)))))) 2(1(0(x1))) -> 0(1(3(5(2(3(x1)))))) 2(1(0(x1))) -> 2(1(5(0(4(3(x1)))))) 2(1(0(x1))) -> 0(1(2(4(5(5(x1)))))) 2(1(0(0(x1)))) -> 1(4(2(0(0(x1))))) 2(1(0(0(x1)))) -> 4(0(5(1(2(0(x1)))))) 2(1(0(0(x1)))) -> 0(5(3(2(0(1(x1)))))) 2(1(0(0(x1)))) -> 0(4(0(2(1(5(x1)))))) 2(5(0(0(x1)))) -> 2(0(3(0(3(5(x1)))))) 2(2(1(0(x1)))) -> 2(1(2(0(3(3(x1)))))) 2(5(1(0(x1)))) -> 3(0(1(5(2(x1))))) 2(5(1(0(x1)))) -> 2(1(3(5(5(0(x1)))))) 2(5(1(0(x1)))) -> 2(2(0(5(3(1(x1)))))) 2(2(5(0(x1)))) -> 2(5(1(2(0(3(x1)))))) 2(1(0(1(x1)))) -> 1(2(1(4(0(x1))))) 2(1(0(1(x1)))) -> 0(3(1(2(1(x1))))) 2(1(0(1(x1)))) -> 2(1(4(1(3(0(x1)))))) 2(1(0(1(x1)))) -> 2(0(1(4(1(4(x1)))))) 2(1(0(5(x1)))) -> 5(3(1(2(0(x1))))) 2(1(0(5(x1)))) -> 5(1(3(0(2(x1))))) 2(1(0(5(x1)))) -> 2(3(0(1(3(5(x1)))))) 2(1(0(5(x1)))) -> 4(2(0(1(5(5(x1)))))) 2(2(0(1(0(x1))))) -> 2(0(1(2(0(3(x1)))))) 2(5(2(1(0(x1))))) -> 0(2(2(1(5(3(x1)))))) 2(2(4(1(0(x1))))) -> 2(5(1(4(2(0(x1)))))) 2(2(4(1(0(x1))))) -> 1(1(2(4(2(0(x1)))))) 2(1(0(3(1(x1))))) -> 2(0(1(4(3(1(x1)))))) ---------------------------------------- (12) Obligation: Q DP problem: The TRS P consists of the following rules: 2^1(1(0(x1))) -> 2^1(x1) 2^1(1(0(x1))) -> 2^1(0(x1)) 2^1(1(0(x1))) -> 0^1(0(x1)) 0^1(2(1(0(x1)))) -> 2^1(1(x1)) 2^1(1(0(x1))) -> 0^1(0(2(x1))) 0^1(0(3(1(0(x1))))) -> 0^1(0(x1)) 2^1(1(0(x1))) -> 0^1(2(x1)) 2^1(1(0(0(x1)))) -> 2^1(0(0(x1))) 2^1(1(0(0(x1)))) -> 2^1(0(x1)) 2^1(1(0(1(x1)))) -> 0^1(x1) 2^1(1(0(1(x1)))) -> 2^1(1(x1)) 2^1(1(0(5(x1)))) -> 2^1(0(x1)) 2^1(1(0(5(x1)))) -> 0^1(x1) 2^1(1(0(5(x1)))) -> 0^1(2(x1)) 2^1(1(0(5(x1)))) -> 2^1(x1) The TRS R consists of the following rules: 2(1(0(x1))) -> 2(1(3(0(x1)))) 2(1(0(x1))) -> 1(2(0(1(x1)))) 2(1(0(x1))) -> 0(4(1(2(x1)))) 2(1(0(x1))) -> 3(3(1(2(0(x1))))) 2(1(0(x1))) -> 3(1(4(2(0(x1))))) 2(1(0(x1))) -> 4(5(2(0(1(x1))))) 2(1(0(x1))) -> 3(0(1(1(2(x1))))) 2(1(0(x1))) -> 0(4(3(1(2(x1))))) 2(1(0(x1))) -> 0(3(4(1(2(x1))))) 2(1(0(x1))) -> 2(1(3(0(4(x1))))) 2(1(0(x1))) -> 2(3(0(1(4(x1))))) 2(1(0(x1))) -> 0(1(2(4(4(x1))))) 2(1(0(x1))) -> 5(2(1(4(0(0(x1)))))) 2(1(0(x1))) -> 5(2(1(5(4(0(x1)))))) 2(1(0(x1))) -> 4(1(3(2(5(0(x1)))))) 2(1(0(x1))) -> 4(3(2(2(0(1(x1)))))) 2(1(0(x1))) -> 2(3(0(4(1(1(x1)))))) 2(1(0(x1))) -> 1(3(3(0(0(2(x1)))))) 2(1(0(x1))) -> 5(1(4(3(0(2(x1)))))) 2(1(0(x1))) -> 1(4(0(2(3(2(x1)))))) 2(1(0(x1))) -> 1(3(0(3(5(2(x1)))))) 2(1(0(x1))) -> 0(1(3(5(2(3(x1)))))) 2(1(0(x1))) -> 2(1(5(0(4(3(x1)))))) 2(1(0(x1))) -> 0(1(2(4(5(5(x1)))))) 2(1(0(0(x1)))) -> 1(4(2(0(0(x1))))) 2(1(0(0(x1)))) -> 4(0(5(1(2(0(x1)))))) 2(1(0(0(x1)))) -> 0(5(3(2(0(1(x1)))))) 2(1(0(0(x1)))) -> 0(4(0(2(1(5(x1)))))) 2(5(0(0(x1)))) -> 2(0(3(0(3(5(x1)))))) 0(2(1(0(x1)))) -> 0(0(4(2(1(x1))))) 0(2(1(0(x1)))) -> 2(0(0(1(4(0(x1)))))) 2(2(1(0(x1)))) -> 2(1(2(0(3(3(x1)))))) 2(5(1(0(x1)))) -> 3(0(1(5(2(x1))))) 2(5(1(0(x1)))) -> 2(1(3(5(5(0(x1)))))) 2(5(1(0(x1)))) -> 2(2(0(5(3(1(x1)))))) 2(2(5(0(x1)))) -> 2(5(1(2(0(3(x1)))))) 2(1(0(1(x1)))) -> 1(2(1(4(0(x1))))) 2(1(0(1(x1)))) -> 0(3(1(2(1(x1))))) 2(1(0(1(x1)))) -> 2(1(4(1(3(0(x1)))))) 2(1(0(1(x1)))) -> 2(0(1(4(1(4(x1)))))) 2(1(0(5(x1)))) -> 5(3(1(2(0(x1))))) 2(1(0(5(x1)))) -> 5(1(3(0(2(x1))))) 2(1(0(5(x1)))) -> 2(3(0(1(3(5(x1)))))) 2(1(0(5(x1)))) -> 4(2(0(1(5(5(x1)))))) 2(2(0(1(0(x1))))) -> 2(0(1(2(0(3(x1)))))) 2(5(2(1(0(x1))))) -> 0(2(2(1(5(3(x1)))))) 0(0(3(1(0(x1))))) -> 0(3(1(4(0(0(x1)))))) 2(2(4(1(0(x1))))) -> 2(5(1(4(2(0(x1)))))) 2(2(4(1(0(x1))))) -> 1(1(2(4(2(0(x1)))))) 2(1(0(3(1(x1))))) -> 2(0(1(4(3(1(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (13) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. 2^1(1(0(5(x1)))) -> 0^1(x1) The remaining pairs can at least be oriented weakly. Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: <<< POL(2^1(x_1)) = [[0A]] + [[0A, -I, -I]] * x_1 >>> <<< POL(1(x_1)) = [[-I], [-I], [-I]] + [[0A, -I, 0A], [-I, -I, 0A], [-I, 0A, 0A]] * x_1 >>> <<< POL(0(x_1)) = [[-I], [1A], [-I]] + [[0A, 0A, -I], [0A, -I, 0A], [0A, -I, 0A]] * x_1 >>> <<< POL(0^1(x_1)) = [[0A]] + [[-I, -I, 0A]] * x_1 >>> <<< POL(2(x_1)) = [[0A], [-I], [0A]] + [[0A, -I, 0A], [0A, -I, 0A], [0A, -I, -I]] * x_1 >>> <<< POL(3(x_1)) = [[0A], [-I], [0A]] + [[0A, -I, -I], [-I, -I, -I], [-I, -I, -I]] * x_1 >>> <<< POL(5(x_1)) = [[0A], [1A], [0A]] + [[0A, -I, 0A], [0A, 0A, 1A], [-I, -I, -I]] * x_1 >>> <<< POL(4(x_1)) = [[0A], [0A], [0A]] + [[-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(2(1(0(x1)))) -> 0(0(4(2(1(x1))))) 0(2(1(0(x1)))) -> 2(0(0(1(4(0(x1)))))) 0(0(3(1(0(x1))))) -> 0(3(1(4(0(0(x1)))))) 2(1(0(x1))) -> 2(1(3(0(x1)))) 2(1(0(x1))) -> 1(2(0(1(x1)))) 2(1(0(x1))) -> 0(4(1(2(x1)))) 2(1(0(x1))) -> 3(3(1(2(0(x1))))) 2(1(0(x1))) -> 3(1(4(2(0(x1))))) 2(1(0(x1))) -> 4(5(2(0(1(x1))))) 2(1(0(x1))) -> 3(0(1(1(2(x1))))) 2(1(0(x1))) -> 0(4(3(1(2(x1))))) 2(1(0(x1))) -> 0(3(4(1(2(x1))))) 2(1(0(x1))) -> 2(1(3(0(4(x1))))) 2(1(0(x1))) -> 2(3(0(1(4(x1))))) 2(1(0(x1))) -> 0(1(2(4(4(x1))))) 2(1(0(x1))) -> 5(2(1(4(0(0(x1)))))) 2(1(0(x1))) -> 5(2(1(5(4(0(x1)))))) 2(1(0(x1))) -> 4(1(3(2(5(0(x1)))))) 2(1(0(x1))) -> 4(3(2(2(0(1(x1)))))) 2(1(0(x1))) -> 2(3(0(4(1(1(x1)))))) 2(1(0(x1))) -> 1(3(3(0(0(2(x1)))))) 2(1(0(x1))) -> 5(1(4(3(0(2(x1)))))) 2(1(0(x1))) -> 1(4(0(2(3(2(x1)))))) 2(1(0(x1))) -> 1(3(0(3(5(2(x1)))))) 2(1(0(x1))) -> 0(1(3(5(2(3(x1)))))) 2(1(0(x1))) -> 2(1(5(0(4(3(x1)))))) 2(1(0(x1))) -> 0(1(2(4(5(5(x1)))))) 2(1(0(0(x1)))) -> 1(4(2(0(0(x1))))) 2(1(0(0(x1)))) -> 4(0(5(1(2(0(x1)))))) 2(1(0(0(x1)))) -> 0(5(3(2(0(1(x1)))))) 2(1(0(0(x1)))) -> 0(4(0(2(1(5(x1)))))) 2(5(0(0(x1)))) -> 2(0(3(0(3(5(x1)))))) 2(2(1(0(x1)))) -> 2(1(2(0(3(3(x1)))))) 2(5(1(0(x1)))) -> 3(0(1(5(2(x1))))) 2(5(1(0(x1)))) -> 2(1(3(5(5(0(x1)))))) 2(5(1(0(x1)))) -> 2(2(0(5(3(1(x1)))))) 2(2(5(0(x1)))) -> 2(5(1(2(0(3(x1)))))) 2(1(0(1(x1)))) -> 1(2(1(4(0(x1))))) 2(1(0(1(x1)))) -> 0(3(1(2(1(x1))))) 2(1(0(1(x1)))) -> 2(1(4(1(3(0(x1)))))) 2(1(0(1(x1)))) -> 2(0(1(4(1(4(x1)))))) 2(1(0(5(x1)))) -> 5(3(1(2(0(x1))))) 2(1(0(5(x1)))) -> 5(1(3(0(2(x1))))) 2(1(0(5(x1)))) -> 2(3(0(1(3(5(x1)))))) 2(1(0(5(x1)))) -> 4(2(0(1(5(5(x1)))))) 2(2(0(1(0(x1))))) -> 2(0(1(2(0(3(x1)))))) 2(5(2(1(0(x1))))) -> 0(2(2(1(5(3(x1)))))) 2(2(4(1(0(x1))))) -> 2(5(1(4(2(0(x1)))))) 2(2(4(1(0(x1))))) -> 1(1(2(4(2(0(x1)))))) 2(1(0(3(1(x1))))) -> 2(0(1(4(3(1(x1)))))) ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: 2^1(1(0(x1))) -> 2^1(x1) 2^1(1(0(x1))) -> 2^1(0(x1)) 2^1(1(0(x1))) -> 0^1(0(x1)) 0^1(2(1(0(x1)))) -> 2^1(1(x1)) 2^1(1(0(x1))) -> 0^1(0(2(x1))) 0^1(0(3(1(0(x1))))) -> 0^1(0(x1)) 2^1(1(0(x1))) -> 0^1(2(x1)) 2^1(1(0(0(x1)))) -> 2^1(0(0(x1))) 2^1(1(0(0(x1)))) -> 2^1(0(x1)) 2^1(1(0(1(x1)))) -> 0^1(x1) 2^1(1(0(1(x1)))) -> 2^1(1(x1)) 2^1(1(0(5(x1)))) -> 2^1(0(x1)) 2^1(1(0(5(x1)))) -> 0^1(2(x1)) 2^1(1(0(5(x1)))) -> 2^1(x1) The TRS R consists of the following rules: 2(1(0(x1))) -> 2(1(3(0(x1)))) 2(1(0(x1))) -> 1(2(0(1(x1)))) 2(1(0(x1))) -> 0(4(1(2(x1)))) 2(1(0(x1))) -> 3(3(1(2(0(x1))))) 2(1(0(x1))) -> 3(1(4(2(0(x1))))) 2(1(0(x1))) -> 4(5(2(0(1(x1))))) 2(1(0(x1))) -> 3(0(1(1(2(x1))))) 2(1(0(x1))) -> 0(4(3(1(2(x1))))) 2(1(0(x1))) -> 0(3(4(1(2(x1))))) 2(1(0(x1))) -> 2(1(3(0(4(x1))))) 2(1(0(x1))) -> 2(3(0(1(4(x1))))) 2(1(0(x1))) -> 0(1(2(4(4(x1))))) 2(1(0(x1))) -> 5(2(1(4(0(0(x1)))))) 2(1(0(x1))) -> 5(2(1(5(4(0(x1)))))) 2(1(0(x1))) -> 4(1(3(2(5(0(x1)))))) 2(1(0(x1))) -> 4(3(2(2(0(1(x1)))))) 2(1(0(x1))) -> 2(3(0(4(1(1(x1)))))) 2(1(0(x1))) -> 1(3(3(0(0(2(x1)))))) 2(1(0(x1))) -> 5(1(4(3(0(2(x1)))))) 2(1(0(x1))) -> 1(4(0(2(3(2(x1)))))) 2(1(0(x1))) -> 1(3(0(3(5(2(x1)))))) 2(1(0(x1))) -> 0(1(3(5(2(3(x1)))))) 2(1(0(x1))) -> 2(1(5(0(4(3(x1)))))) 2(1(0(x1))) -> 0(1(2(4(5(5(x1)))))) 2(1(0(0(x1)))) -> 1(4(2(0(0(x1))))) 2(1(0(0(x1)))) -> 4(0(5(1(2(0(x1)))))) 2(1(0(0(x1)))) -> 0(5(3(2(0(1(x1)))))) 2(1(0(0(x1)))) -> 0(4(0(2(1(5(x1)))))) 2(5(0(0(x1)))) -> 2(0(3(0(3(5(x1)))))) 0(2(1(0(x1)))) -> 0(0(4(2(1(x1))))) 0(2(1(0(x1)))) -> 2(0(0(1(4(0(x1)))))) 2(2(1(0(x1)))) -> 2(1(2(0(3(3(x1)))))) 2(5(1(0(x1)))) -> 3(0(1(5(2(x1))))) 2(5(1(0(x1)))) -> 2(1(3(5(5(0(x1)))))) 2(5(1(0(x1)))) -> 2(2(0(5(3(1(x1)))))) 2(2(5(0(x1)))) -> 2(5(1(2(0(3(x1)))))) 2(1(0(1(x1)))) -> 1(2(1(4(0(x1))))) 2(1(0(1(x1)))) -> 0(3(1(2(1(x1))))) 2(1(0(1(x1)))) -> 2(1(4(1(3(0(x1)))))) 2(1(0(1(x1)))) -> 2(0(1(4(1(4(x1)))))) 2(1(0(5(x1)))) -> 5(3(1(2(0(x1))))) 2(1(0(5(x1)))) -> 5(1(3(0(2(x1))))) 2(1(0(5(x1)))) -> 2(3(0(1(3(5(x1)))))) 2(1(0(5(x1)))) -> 4(2(0(1(5(5(x1)))))) 2(2(0(1(0(x1))))) -> 2(0(1(2(0(3(x1)))))) 2(5(2(1(0(x1))))) -> 0(2(2(1(5(3(x1)))))) 0(0(3(1(0(x1))))) -> 0(3(1(4(0(0(x1)))))) 2(2(4(1(0(x1))))) -> 2(5(1(4(2(0(x1)))))) 2(2(4(1(0(x1))))) -> 1(1(2(4(2(0(x1)))))) 2(1(0(3(1(x1))))) -> 2(0(1(4(3(1(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. 2^1(1(0(5(x1)))) -> 0^1(2(x1)) The remaining pairs can at least be oriented weakly. Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: <<< POL(2^1(x_1)) = [[1A]] + [[0A, -I, 0A]] * x_1 >>> <<< POL(1(x_1)) = [[0A], [0A], [0A]] + [[0A, 0A, -I], [0A, 0A, -I], [-I, 0A, -I]] * x_1 >>> <<< POL(0(x_1)) = [[1A], [0A], [0A]] + [[0A, 0A, 0A], [-I, 0A, -I], [0A, -I, -I]] * x_1 >>> <<< POL(0^1(x_1)) = [[0A]] + [[0A, -I, -I]] * x_1 >>> <<< POL(2(x_1)) = [[0A], [0A], [1A]] + [[-I, 0A, -I], [-I, 0A, 0A], [0A, -I, 0A]] * x_1 >>> <<< POL(3(x_1)) = [[0A], [-I], [0A]] + [[-I, -I, -I], [-I, -I, -I], [0A, -I, 0A]] * x_1 >>> <<< POL(5(x_1)) = [[0A], [0A], [0A]] + [[0A, -I, -I], [-I, 0A, 0A], [0A, 1A, -I]] * x_1 >>> <<< POL(4(x_1)) = [[0A], [0A], [0A]] + [[-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(2(1(0(x1)))) -> 0(0(4(2(1(x1))))) 0(2(1(0(x1)))) -> 2(0(0(1(4(0(x1)))))) 0(0(3(1(0(x1))))) -> 0(3(1(4(0(0(x1)))))) 2(1(0(x1))) -> 2(1(3(0(x1)))) 2(1(0(x1))) -> 1(2(0(1(x1)))) 2(1(0(x1))) -> 0(4(1(2(x1)))) 2(1(0(x1))) -> 3(3(1(2(0(x1))))) 2(1(0(x1))) -> 3(1(4(2(0(x1))))) 2(1(0(x1))) -> 4(5(2(0(1(x1))))) 2(1(0(x1))) -> 3(0(1(1(2(x1))))) 2(1(0(x1))) -> 0(4(3(1(2(x1))))) 2(1(0(x1))) -> 0(3(4(1(2(x1))))) 2(1(0(x1))) -> 2(1(3(0(4(x1))))) 2(1(0(x1))) -> 2(3(0(1(4(x1))))) 2(1(0(x1))) -> 0(1(2(4(4(x1))))) 2(1(0(x1))) -> 5(2(1(4(0(0(x1)))))) 2(1(0(x1))) -> 5(2(1(5(4(0(x1)))))) 2(1(0(x1))) -> 4(1(3(2(5(0(x1)))))) 2(1(0(x1))) -> 4(3(2(2(0(1(x1)))))) 2(1(0(x1))) -> 2(3(0(4(1(1(x1)))))) 2(1(0(x1))) -> 1(3(3(0(0(2(x1)))))) 2(1(0(x1))) -> 5(1(4(3(0(2(x1)))))) 2(1(0(x1))) -> 1(4(0(2(3(2(x1)))))) 2(1(0(x1))) -> 1(3(0(3(5(2(x1)))))) 2(1(0(x1))) -> 0(1(3(5(2(3(x1)))))) 2(1(0(x1))) -> 2(1(5(0(4(3(x1)))))) 2(1(0(x1))) -> 0(1(2(4(5(5(x1)))))) 2(1(0(0(x1)))) -> 1(4(2(0(0(x1))))) 2(1(0(0(x1)))) -> 4(0(5(1(2(0(x1)))))) 2(1(0(0(x1)))) -> 0(5(3(2(0(1(x1)))))) 2(1(0(0(x1)))) -> 0(4(0(2(1(5(x1)))))) 2(5(0(0(x1)))) -> 2(0(3(0(3(5(x1)))))) 2(2(1(0(x1)))) -> 2(1(2(0(3(3(x1)))))) 2(5(1(0(x1)))) -> 3(0(1(5(2(x1))))) 2(5(1(0(x1)))) -> 2(1(3(5(5(0(x1)))))) 2(5(1(0(x1)))) -> 2(2(0(5(3(1(x1)))))) 2(2(5(0(x1)))) -> 2(5(1(2(0(3(x1)))))) 2(1(0(1(x1)))) -> 1(2(1(4(0(x1))))) 2(1(0(1(x1)))) -> 0(3(1(2(1(x1))))) 2(1(0(1(x1)))) -> 2(1(4(1(3(0(x1)))))) 2(1(0(1(x1)))) -> 2(0(1(4(1(4(x1)))))) 2(1(0(5(x1)))) -> 5(3(1(2(0(x1))))) 2(1(0(5(x1)))) -> 5(1(3(0(2(x1))))) 2(1(0(5(x1)))) -> 2(3(0(1(3(5(x1)))))) 2(1(0(5(x1)))) -> 4(2(0(1(5(5(x1)))))) 2(2(0(1(0(x1))))) -> 2(0(1(2(0(3(x1)))))) 2(5(2(1(0(x1))))) -> 0(2(2(1(5(3(x1)))))) 2(2(4(1(0(x1))))) -> 2(5(1(4(2(0(x1)))))) 2(2(4(1(0(x1))))) -> 1(1(2(4(2(0(x1)))))) 2(1(0(3(1(x1))))) -> 2(0(1(4(3(1(x1)))))) ---------------------------------------- (16) Obligation: Q DP problem: The TRS P consists of the following rules: 2^1(1(0(x1))) -> 2^1(x1) 2^1(1(0(x1))) -> 2^1(0(x1)) 2^1(1(0(x1))) -> 0^1(0(x1)) 0^1(2(1(0(x1)))) -> 2^1(1(x1)) 2^1(1(0(x1))) -> 0^1(0(2(x1))) 0^1(0(3(1(0(x1))))) -> 0^1(0(x1)) 2^1(1(0(x1))) -> 0^1(2(x1)) 2^1(1(0(0(x1)))) -> 2^1(0(0(x1))) 2^1(1(0(0(x1)))) -> 2^1(0(x1)) 2^1(1(0(1(x1)))) -> 0^1(x1) 2^1(1(0(1(x1)))) -> 2^1(1(x1)) 2^1(1(0(5(x1)))) -> 2^1(0(x1)) 2^1(1(0(5(x1)))) -> 2^1(x1) The TRS R consists of the following rules: 2(1(0(x1))) -> 2(1(3(0(x1)))) 2(1(0(x1))) -> 1(2(0(1(x1)))) 2(1(0(x1))) -> 0(4(1(2(x1)))) 2(1(0(x1))) -> 3(3(1(2(0(x1))))) 2(1(0(x1))) -> 3(1(4(2(0(x1))))) 2(1(0(x1))) -> 4(5(2(0(1(x1))))) 2(1(0(x1))) -> 3(0(1(1(2(x1))))) 2(1(0(x1))) -> 0(4(3(1(2(x1))))) 2(1(0(x1))) -> 0(3(4(1(2(x1))))) 2(1(0(x1))) -> 2(1(3(0(4(x1))))) 2(1(0(x1))) -> 2(3(0(1(4(x1))))) 2(1(0(x1))) -> 0(1(2(4(4(x1))))) 2(1(0(x1))) -> 5(2(1(4(0(0(x1)))))) 2(1(0(x1))) -> 5(2(1(5(4(0(x1)))))) 2(1(0(x1))) -> 4(1(3(2(5(0(x1)))))) 2(1(0(x1))) -> 4(3(2(2(0(1(x1)))))) 2(1(0(x1))) -> 2(3(0(4(1(1(x1)))))) 2(1(0(x1))) -> 1(3(3(0(0(2(x1)))))) 2(1(0(x1))) -> 5(1(4(3(0(2(x1)))))) 2(1(0(x1))) -> 1(4(0(2(3(2(x1)))))) 2(1(0(x1))) -> 1(3(0(3(5(2(x1)))))) 2(1(0(x1))) -> 0(1(3(5(2(3(x1)))))) 2(1(0(x1))) -> 2(1(5(0(4(3(x1)))))) 2(1(0(x1))) -> 0(1(2(4(5(5(x1)))))) 2(1(0(0(x1)))) -> 1(4(2(0(0(x1))))) 2(1(0(0(x1)))) -> 4(0(5(1(2(0(x1)))))) 2(1(0(0(x1)))) -> 0(5(3(2(0(1(x1)))))) 2(1(0(0(x1)))) -> 0(4(0(2(1(5(x1)))))) 2(5(0(0(x1)))) -> 2(0(3(0(3(5(x1)))))) 0(2(1(0(x1)))) -> 0(0(4(2(1(x1))))) 0(2(1(0(x1)))) -> 2(0(0(1(4(0(x1)))))) 2(2(1(0(x1)))) -> 2(1(2(0(3(3(x1)))))) 2(5(1(0(x1)))) -> 3(0(1(5(2(x1))))) 2(5(1(0(x1)))) -> 2(1(3(5(5(0(x1)))))) 2(5(1(0(x1)))) -> 2(2(0(5(3(1(x1)))))) 2(2(5(0(x1)))) -> 2(5(1(2(0(3(x1)))))) 2(1(0(1(x1)))) -> 1(2(1(4(0(x1))))) 2(1(0(1(x1)))) -> 0(3(1(2(1(x1))))) 2(1(0(1(x1)))) -> 2(1(4(1(3(0(x1)))))) 2(1(0(1(x1)))) -> 2(0(1(4(1(4(x1)))))) 2(1(0(5(x1)))) -> 5(3(1(2(0(x1))))) 2(1(0(5(x1)))) -> 5(1(3(0(2(x1))))) 2(1(0(5(x1)))) -> 2(3(0(1(3(5(x1)))))) 2(1(0(5(x1)))) -> 4(2(0(1(5(5(x1)))))) 2(2(0(1(0(x1))))) -> 2(0(1(2(0(3(x1)))))) 2(5(2(1(0(x1))))) -> 0(2(2(1(5(3(x1)))))) 0(0(3(1(0(x1))))) -> 0(3(1(4(0(0(x1)))))) 2(2(4(1(0(x1))))) -> 2(5(1(4(2(0(x1)))))) 2(2(4(1(0(x1))))) -> 1(1(2(4(2(0(x1)))))) 2(1(0(3(1(x1))))) -> 2(0(1(4(3(1(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. 2^1(1(0(5(x1)))) -> 2^1(x1) The remaining pairs can at least be oriented weakly. Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: <<< POL(2^1(x_1)) = [[0A]] + [[-I, -I, 0A]] * x_1 >>> <<< POL(1(x_1)) = [[0A], [0A], [-I]] + [[0A, 0A, 0A], [0A, 0A, -I], [-I, 0A, 0A]] * x_1 >>> <<< POL(0(x_1)) = [[1A], [0A], [0A]] + [[-I, 0A, 0A], [-I, -I, 0A], [0A, 0A, 0A]] * x_1 >>> <<< POL(0^1(x_1)) = [[0A]] + [[-I, 0A, 0A]] * x_1 >>> <<< POL(2(x_1)) = [[0A], [0A], [0A]] + [[-I, 0A, -I], [-I, 0A, -I], [-I, 0A, -I]] * x_1 >>> <<< POL(3(x_1)) = [[0A], [-I], [-I]] + [[0A, -I, -I], [-I, -I, -I], [-I, -I, -I]] * x_1 >>> <<< POL(5(x_1)) = [[0A], [0A], [1A]] + [[-I, 0A, 0A], [-I, 0A, 1A], [0A, 0A, 0A]] * x_1 >>> <<< POL(4(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(2(1(0(x1)))) -> 0(0(4(2(1(x1))))) 0(2(1(0(x1)))) -> 2(0(0(1(4(0(x1)))))) 0(0(3(1(0(x1))))) -> 0(3(1(4(0(0(x1)))))) 2(1(0(x1))) -> 2(1(3(0(x1)))) 2(1(0(x1))) -> 1(2(0(1(x1)))) 2(1(0(x1))) -> 0(4(1(2(x1)))) 2(1(0(x1))) -> 3(3(1(2(0(x1))))) 2(1(0(x1))) -> 3(1(4(2(0(x1))))) 2(1(0(x1))) -> 4(5(2(0(1(x1))))) 2(1(0(x1))) -> 3(0(1(1(2(x1))))) 2(1(0(x1))) -> 0(4(3(1(2(x1))))) 2(1(0(x1))) -> 0(3(4(1(2(x1))))) 2(1(0(x1))) -> 2(1(3(0(4(x1))))) 2(1(0(x1))) -> 2(3(0(1(4(x1))))) 2(1(0(x1))) -> 0(1(2(4(4(x1))))) 2(1(0(x1))) -> 5(2(1(4(0(0(x1)))))) 2(1(0(x1))) -> 5(2(1(5(4(0(x1)))))) 2(1(0(x1))) -> 4(1(3(2(5(0(x1)))))) 2(1(0(x1))) -> 4(3(2(2(0(1(x1)))))) 2(1(0(x1))) -> 2(3(0(4(1(1(x1)))))) 2(1(0(x1))) -> 1(3(3(0(0(2(x1)))))) 2(1(0(x1))) -> 5(1(4(3(0(2(x1)))))) 2(1(0(x1))) -> 1(4(0(2(3(2(x1)))))) 2(1(0(x1))) -> 1(3(0(3(5(2(x1)))))) 2(1(0(x1))) -> 0(1(3(5(2(3(x1)))))) 2(1(0(x1))) -> 2(1(5(0(4(3(x1)))))) 2(1(0(x1))) -> 0(1(2(4(5(5(x1)))))) 2(1(0(0(x1)))) -> 1(4(2(0(0(x1))))) 2(1(0(0(x1)))) -> 4(0(5(1(2(0(x1)))))) 2(1(0(0(x1)))) -> 0(5(3(2(0(1(x1)))))) 2(1(0(0(x1)))) -> 0(4(0(2(1(5(x1)))))) 2(5(0(0(x1)))) -> 2(0(3(0(3(5(x1)))))) 2(2(1(0(x1)))) -> 2(1(2(0(3(3(x1)))))) 2(5(1(0(x1)))) -> 3(0(1(5(2(x1))))) 2(5(1(0(x1)))) -> 2(1(3(5(5(0(x1)))))) 2(5(1(0(x1)))) -> 2(2(0(5(3(1(x1)))))) 2(2(5(0(x1)))) -> 2(5(1(2(0(3(x1)))))) 2(1(0(1(x1)))) -> 1(2(1(4(0(x1))))) 2(1(0(1(x1)))) -> 0(3(1(2(1(x1))))) 2(1(0(1(x1)))) -> 2(1(4(1(3(0(x1)))))) 2(1(0(1(x1)))) -> 2(0(1(4(1(4(x1)))))) 2(1(0(5(x1)))) -> 5(3(1(2(0(x1))))) 2(1(0(5(x1)))) -> 5(1(3(0(2(x1))))) 2(1(0(5(x1)))) -> 2(3(0(1(3(5(x1)))))) 2(1(0(5(x1)))) -> 4(2(0(1(5(5(x1)))))) 2(2(0(1(0(x1))))) -> 2(0(1(2(0(3(x1)))))) 2(5(2(1(0(x1))))) -> 0(2(2(1(5(3(x1)))))) 2(2(4(1(0(x1))))) -> 2(5(1(4(2(0(x1)))))) 2(2(4(1(0(x1))))) -> 1(1(2(4(2(0(x1)))))) 2(1(0(3(1(x1))))) -> 2(0(1(4(3(1(x1)))))) ---------------------------------------- (18) Obligation: Q DP problem: The TRS P consists of the following rules: 2^1(1(0(x1))) -> 2^1(x1) 2^1(1(0(x1))) -> 2^1(0(x1)) 2^1(1(0(x1))) -> 0^1(0(x1)) 0^1(2(1(0(x1)))) -> 2^1(1(x1)) 2^1(1(0(x1))) -> 0^1(0(2(x1))) 0^1(0(3(1(0(x1))))) -> 0^1(0(x1)) 2^1(1(0(x1))) -> 0^1(2(x1)) 2^1(1(0(0(x1)))) -> 2^1(0(0(x1))) 2^1(1(0(0(x1)))) -> 2^1(0(x1)) 2^1(1(0(1(x1)))) -> 0^1(x1) 2^1(1(0(1(x1)))) -> 2^1(1(x1)) 2^1(1(0(5(x1)))) -> 2^1(0(x1)) The TRS R consists of the following rules: 2(1(0(x1))) -> 2(1(3(0(x1)))) 2(1(0(x1))) -> 1(2(0(1(x1)))) 2(1(0(x1))) -> 0(4(1(2(x1)))) 2(1(0(x1))) -> 3(3(1(2(0(x1))))) 2(1(0(x1))) -> 3(1(4(2(0(x1))))) 2(1(0(x1))) -> 4(5(2(0(1(x1))))) 2(1(0(x1))) -> 3(0(1(1(2(x1))))) 2(1(0(x1))) -> 0(4(3(1(2(x1))))) 2(1(0(x1))) -> 0(3(4(1(2(x1))))) 2(1(0(x1))) -> 2(1(3(0(4(x1))))) 2(1(0(x1))) -> 2(3(0(1(4(x1))))) 2(1(0(x1))) -> 0(1(2(4(4(x1))))) 2(1(0(x1))) -> 5(2(1(4(0(0(x1)))))) 2(1(0(x1))) -> 5(2(1(5(4(0(x1)))))) 2(1(0(x1))) -> 4(1(3(2(5(0(x1)))))) 2(1(0(x1))) -> 4(3(2(2(0(1(x1)))))) 2(1(0(x1))) -> 2(3(0(4(1(1(x1)))))) 2(1(0(x1))) -> 1(3(3(0(0(2(x1)))))) 2(1(0(x1))) -> 5(1(4(3(0(2(x1)))))) 2(1(0(x1))) -> 1(4(0(2(3(2(x1)))))) 2(1(0(x1))) -> 1(3(0(3(5(2(x1)))))) 2(1(0(x1))) -> 0(1(3(5(2(3(x1)))))) 2(1(0(x1))) -> 2(1(5(0(4(3(x1)))))) 2(1(0(x1))) -> 0(1(2(4(5(5(x1)))))) 2(1(0(0(x1)))) -> 1(4(2(0(0(x1))))) 2(1(0(0(x1)))) -> 4(0(5(1(2(0(x1)))))) 2(1(0(0(x1)))) -> 0(5(3(2(0(1(x1)))))) 2(1(0(0(x1)))) -> 0(4(0(2(1(5(x1)))))) 2(5(0(0(x1)))) -> 2(0(3(0(3(5(x1)))))) 0(2(1(0(x1)))) -> 0(0(4(2(1(x1))))) 0(2(1(0(x1)))) -> 2(0(0(1(4(0(x1)))))) 2(2(1(0(x1)))) -> 2(1(2(0(3(3(x1)))))) 2(5(1(0(x1)))) -> 3(0(1(5(2(x1))))) 2(5(1(0(x1)))) -> 2(1(3(5(5(0(x1)))))) 2(5(1(0(x1)))) -> 2(2(0(5(3(1(x1)))))) 2(2(5(0(x1)))) -> 2(5(1(2(0(3(x1)))))) 2(1(0(1(x1)))) -> 1(2(1(4(0(x1))))) 2(1(0(1(x1)))) -> 0(3(1(2(1(x1))))) 2(1(0(1(x1)))) -> 2(1(4(1(3(0(x1)))))) 2(1(0(1(x1)))) -> 2(0(1(4(1(4(x1)))))) 2(1(0(5(x1)))) -> 5(3(1(2(0(x1))))) 2(1(0(5(x1)))) -> 5(1(3(0(2(x1))))) 2(1(0(5(x1)))) -> 2(3(0(1(3(5(x1)))))) 2(1(0(5(x1)))) -> 4(2(0(1(5(5(x1)))))) 2(2(0(1(0(x1))))) -> 2(0(1(2(0(3(x1)))))) 2(5(2(1(0(x1))))) -> 0(2(2(1(5(3(x1)))))) 0(0(3(1(0(x1))))) -> 0(3(1(4(0(0(x1)))))) 2(2(4(1(0(x1))))) -> 2(5(1(4(2(0(x1)))))) 2(2(4(1(0(x1))))) -> 1(1(2(4(2(0(x1)))))) 2(1(0(3(1(x1))))) -> 2(0(1(4(3(1(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. 2^1(1(0(0(x1)))) -> 2^1(0(x1)) The remaining pairs can at least be oriented weakly. Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: <<< POL(2^1(x_1)) = [[0A]] + [[-I, 0A, -I]] * x_1 >>> <<< POL(1(x_1)) = [[0A], [-I], [-I]] + [[-I, -I, 0A], [-I, 0A, -I], [-I, -I, 0A]] * x_1 >>> <<< POL(0(x_1)) = [[0A], [-I], [1A]] + [[0A, 0A, 0A], [-I, 1A, 0A], [0A, -I, -I]] * x_1 >>> <<< POL(0^1(x_1)) = [[0A]] + [[-I, 0A, -I]] * x_1 >>> <<< POL(2(x_1)) = [[1A], [-I], [-I]] + [[0A, 0A, 0A], [-I, 0A, -I], [-I, 0A, 0A]] * x_1 >>> <<< POL(3(x_1)) = [[0A], [-I], [-I]] + [[0A, -I, -I], [-I, -I, -I], [-I, 0A, -I]] * x_1 >>> <<< POL(5(x_1)) = [[0A], [-I], [-I]] + [[0A, 0A, -I], [-I, 0A, -I], [-I, 0A, 0A]] * x_1 >>> <<< POL(4(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(2(1(0(x1)))) -> 0(0(4(2(1(x1))))) 0(2(1(0(x1)))) -> 2(0(0(1(4(0(x1)))))) 0(0(3(1(0(x1))))) -> 0(3(1(4(0(0(x1)))))) 2(1(0(x1))) -> 2(1(3(0(x1)))) 2(1(0(x1))) -> 1(2(0(1(x1)))) 2(1(0(x1))) -> 0(4(1(2(x1)))) 2(1(0(x1))) -> 3(3(1(2(0(x1))))) 2(1(0(x1))) -> 3(1(4(2(0(x1))))) 2(1(0(x1))) -> 4(5(2(0(1(x1))))) 2(1(0(x1))) -> 3(0(1(1(2(x1))))) 2(1(0(x1))) -> 0(4(3(1(2(x1))))) 2(1(0(x1))) -> 0(3(4(1(2(x1))))) 2(1(0(x1))) -> 2(1(3(0(4(x1))))) 2(1(0(x1))) -> 2(3(0(1(4(x1))))) 2(1(0(x1))) -> 0(1(2(4(4(x1))))) 2(1(0(x1))) -> 5(2(1(4(0(0(x1)))))) 2(1(0(x1))) -> 5(2(1(5(4(0(x1)))))) 2(1(0(x1))) -> 4(1(3(2(5(0(x1)))))) 2(1(0(x1))) -> 4(3(2(2(0(1(x1)))))) 2(1(0(x1))) -> 2(3(0(4(1(1(x1)))))) 2(1(0(x1))) -> 1(3(3(0(0(2(x1)))))) 2(1(0(x1))) -> 5(1(4(3(0(2(x1)))))) 2(1(0(x1))) -> 1(4(0(2(3(2(x1)))))) 2(1(0(x1))) -> 1(3(0(3(5(2(x1)))))) 2(1(0(x1))) -> 0(1(3(5(2(3(x1)))))) 2(1(0(x1))) -> 2(1(5(0(4(3(x1)))))) 2(1(0(x1))) -> 0(1(2(4(5(5(x1)))))) 2(1(0(0(x1)))) -> 1(4(2(0(0(x1))))) 2(1(0(0(x1)))) -> 4(0(5(1(2(0(x1)))))) 2(1(0(0(x1)))) -> 0(5(3(2(0(1(x1)))))) 2(1(0(0(x1)))) -> 0(4(0(2(1(5(x1)))))) 2(5(0(0(x1)))) -> 2(0(3(0(3(5(x1)))))) 2(2(1(0(x1)))) -> 2(1(2(0(3(3(x1)))))) 2(5(1(0(x1)))) -> 3(0(1(5(2(x1))))) 2(5(1(0(x1)))) -> 2(1(3(5(5(0(x1)))))) 2(5(1(0(x1)))) -> 2(2(0(5(3(1(x1)))))) 2(2(5(0(x1)))) -> 2(5(1(2(0(3(x1)))))) 2(1(0(1(x1)))) -> 1(2(1(4(0(x1))))) 2(1(0(1(x1)))) -> 0(3(1(2(1(x1))))) 2(1(0(1(x1)))) -> 2(1(4(1(3(0(x1)))))) 2(1(0(1(x1)))) -> 2(0(1(4(1(4(x1)))))) 2(1(0(5(x1)))) -> 5(3(1(2(0(x1))))) 2(1(0(5(x1)))) -> 5(1(3(0(2(x1))))) 2(1(0(5(x1)))) -> 2(3(0(1(3(5(x1)))))) 2(1(0(5(x1)))) -> 4(2(0(1(5(5(x1)))))) 2(2(0(1(0(x1))))) -> 2(0(1(2(0(3(x1)))))) 2(5(2(1(0(x1))))) -> 0(2(2(1(5(3(x1)))))) 2(2(4(1(0(x1))))) -> 2(5(1(4(2(0(x1)))))) 2(2(4(1(0(x1))))) -> 1(1(2(4(2(0(x1)))))) 2(1(0(3(1(x1))))) -> 2(0(1(4(3(1(x1)))))) ---------------------------------------- (20) Obligation: Q DP problem: The TRS P consists of the following rules: 2^1(1(0(x1))) -> 2^1(x1) 2^1(1(0(x1))) -> 2^1(0(x1)) 2^1(1(0(x1))) -> 0^1(0(x1)) 0^1(2(1(0(x1)))) -> 2^1(1(x1)) 2^1(1(0(x1))) -> 0^1(0(2(x1))) 0^1(0(3(1(0(x1))))) -> 0^1(0(x1)) 2^1(1(0(x1))) -> 0^1(2(x1)) 2^1(1(0(0(x1)))) -> 2^1(0(0(x1))) 2^1(1(0(1(x1)))) -> 0^1(x1) 2^1(1(0(1(x1)))) -> 2^1(1(x1)) 2^1(1(0(5(x1)))) -> 2^1(0(x1)) The TRS R consists of the following rules: 2(1(0(x1))) -> 2(1(3(0(x1)))) 2(1(0(x1))) -> 1(2(0(1(x1)))) 2(1(0(x1))) -> 0(4(1(2(x1)))) 2(1(0(x1))) -> 3(3(1(2(0(x1))))) 2(1(0(x1))) -> 3(1(4(2(0(x1))))) 2(1(0(x1))) -> 4(5(2(0(1(x1))))) 2(1(0(x1))) -> 3(0(1(1(2(x1))))) 2(1(0(x1))) -> 0(4(3(1(2(x1))))) 2(1(0(x1))) -> 0(3(4(1(2(x1))))) 2(1(0(x1))) -> 2(1(3(0(4(x1))))) 2(1(0(x1))) -> 2(3(0(1(4(x1))))) 2(1(0(x1))) -> 0(1(2(4(4(x1))))) 2(1(0(x1))) -> 5(2(1(4(0(0(x1)))))) 2(1(0(x1))) -> 5(2(1(5(4(0(x1)))))) 2(1(0(x1))) -> 4(1(3(2(5(0(x1)))))) 2(1(0(x1))) -> 4(3(2(2(0(1(x1)))))) 2(1(0(x1))) -> 2(3(0(4(1(1(x1)))))) 2(1(0(x1))) -> 1(3(3(0(0(2(x1)))))) 2(1(0(x1))) -> 5(1(4(3(0(2(x1)))))) 2(1(0(x1))) -> 1(4(0(2(3(2(x1)))))) 2(1(0(x1))) -> 1(3(0(3(5(2(x1)))))) 2(1(0(x1))) -> 0(1(3(5(2(3(x1)))))) 2(1(0(x1))) -> 2(1(5(0(4(3(x1)))))) 2(1(0(x1))) -> 0(1(2(4(5(5(x1)))))) 2(1(0(0(x1)))) -> 1(4(2(0(0(x1))))) 2(1(0(0(x1)))) -> 4(0(5(1(2(0(x1)))))) 2(1(0(0(x1)))) -> 0(5(3(2(0(1(x1)))))) 2(1(0(0(x1)))) -> 0(4(0(2(1(5(x1)))))) 2(5(0(0(x1)))) -> 2(0(3(0(3(5(x1)))))) 0(2(1(0(x1)))) -> 0(0(4(2(1(x1))))) 0(2(1(0(x1)))) -> 2(0(0(1(4(0(x1)))))) 2(2(1(0(x1)))) -> 2(1(2(0(3(3(x1)))))) 2(5(1(0(x1)))) -> 3(0(1(5(2(x1))))) 2(5(1(0(x1)))) -> 2(1(3(5(5(0(x1)))))) 2(5(1(0(x1)))) -> 2(2(0(5(3(1(x1)))))) 2(2(5(0(x1)))) -> 2(5(1(2(0(3(x1)))))) 2(1(0(1(x1)))) -> 1(2(1(4(0(x1))))) 2(1(0(1(x1)))) -> 0(3(1(2(1(x1))))) 2(1(0(1(x1)))) -> 2(1(4(1(3(0(x1)))))) 2(1(0(1(x1)))) -> 2(0(1(4(1(4(x1)))))) 2(1(0(5(x1)))) -> 5(3(1(2(0(x1))))) 2(1(0(5(x1)))) -> 5(1(3(0(2(x1))))) 2(1(0(5(x1)))) -> 2(3(0(1(3(5(x1)))))) 2(1(0(5(x1)))) -> 4(2(0(1(5(5(x1)))))) 2(2(0(1(0(x1))))) -> 2(0(1(2(0(3(x1)))))) 2(5(2(1(0(x1))))) -> 0(2(2(1(5(3(x1)))))) 0(0(3(1(0(x1))))) -> 0(3(1(4(0(0(x1)))))) 2(2(4(1(0(x1))))) -> 2(5(1(4(2(0(x1)))))) 2(2(4(1(0(x1))))) -> 1(1(2(4(2(0(x1)))))) 2(1(0(3(1(x1))))) -> 2(0(1(4(3(1(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. 2^1(1(0(0(x1)))) -> 2^1(0(0(x1))) The remaining pairs can at least be oriented weakly. Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: <<< POL(2^1(x_1)) = [[0A]] + [[-I, 0A, 0A]] * x_1 >>> <<< POL(1(x_1)) = [[0A], [0A], [-I]] + [[0A, 0A, 0A], [0A, 0A, 0A], [-I, -I, 0A]] * x_1 >>> <<< POL(0(x_1)) = [[0A], [0A], [0A]] + [[0A, 0A, 1A], [-I, -I, -I], [-I, -I, -I]] * x_1 >>> <<< POL(0^1(x_1)) = [[0A]] + [[-I, 0A, 0A]] * x_1 >>> <<< POL(2(x_1)) = [[0A], [0A], [-I]] + [[0A, 0A, 0A], [-I, 0A, 0A], [-I, -I, 0A]] * x_1 >>> <<< POL(3(x_1)) = [[0A], [0A], [-I]] + [[-I, -I, -I], [0A, -I, 0A], [-I, -I, 0A]] * x_1 >>> <<< POL(5(x_1)) = [[0A], [0A], [-I]] + [[0A, 0A, -I], [0A, 0A, 0A], [-I, -I, 0A]] * x_1 >>> <<< POL(4(x_1)) = [[0A], [0A], [-I]] + [[-I, 0A, 0A], [-I, 0A, 0A], [-I, -I, -I]] * x_1 >>> The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 0(2(1(0(x1)))) -> 0(0(4(2(1(x1))))) 0(2(1(0(x1)))) -> 2(0(0(1(4(0(x1)))))) 0(0(3(1(0(x1))))) -> 0(3(1(4(0(0(x1)))))) 2(1(0(x1))) -> 2(1(3(0(x1)))) 2(1(0(x1))) -> 1(2(0(1(x1)))) 2(1(0(x1))) -> 0(4(1(2(x1)))) 2(1(0(x1))) -> 3(3(1(2(0(x1))))) 2(1(0(x1))) -> 3(1(4(2(0(x1))))) 2(1(0(x1))) -> 4(5(2(0(1(x1))))) 2(1(0(x1))) -> 3(0(1(1(2(x1))))) 2(1(0(x1))) -> 0(4(3(1(2(x1))))) 2(1(0(x1))) -> 0(3(4(1(2(x1))))) 2(1(0(x1))) -> 2(1(3(0(4(x1))))) 2(1(0(x1))) -> 2(3(0(1(4(x1))))) 2(1(0(x1))) -> 0(1(2(4(4(x1))))) 2(1(0(x1))) -> 5(2(1(4(0(0(x1)))))) 2(1(0(x1))) -> 5(2(1(5(4(0(x1)))))) 2(1(0(x1))) -> 4(1(3(2(5(0(x1)))))) 2(1(0(x1))) -> 4(3(2(2(0(1(x1)))))) 2(1(0(x1))) -> 2(3(0(4(1(1(x1)))))) 2(1(0(x1))) -> 1(3(3(0(0(2(x1)))))) 2(1(0(x1))) -> 5(1(4(3(0(2(x1)))))) 2(1(0(x1))) -> 1(4(0(2(3(2(x1)))))) 2(1(0(x1))) -> 1(3(0(3(5(2(x1)))))) 2(1(0(x1))) -> 0(1(3(5(2(3(x1)))))) 2(1(0(x1))) -> 2(1(5(0(4(3(x1)))))) 2(1(0(x1))) -> 0(1(2(4(5(5(x1)))))) 2(1(0(0(x1)))) -> 1(4(2(0(0(x1))))) 2(1(0(0(x1)))) -> 4(0(5(1(2(0(x1)))))) 2(1(0(0(x1)))) -> 0(5(3(2(0(1(x1)))))) 2(1(0(0(x1)))) -> 0(4(0(2(1(5(x1)))))) 2(5(0(0(x1)))) -> 2(0(3(0(3(5(x1)))))) 2(2(1(0(x1)))) -> 2(1(2(0(3(3(x1)))))) 2(5(1(0(x1)))) -> 3(0(1(5(2(x1))))) 2(5(1(0(x1)))) -> 2(1(3(5(5(0(x1)))))) 2(5(1(0(x1)))) -> 2(2(0(5(3(1(x1)))))) 2(2(5(0(x1)))) -> 2(5(1(2(0(3(x1)))))) 2(1(0(1(x1)))) -> 1(2(1(4(0(x1))))) 2(1(0(1(x1)))) -> 0(3(1(2(1(x1))))) 2(1(0(1(x1)))) -> 2(1(4(1(3(0(x1)))))) 2(1(0(1(x1)))) -> 2(0(1(4(1(4(x1)))))) 2(1(0(5(x1)))) -> 5(3(1(2(0(x1))))) 2(1(0(5(x1)))) -> 5(1(3(0(2(x1))))) 2(1(0(5(x1)))) -> 2(3(0(1(3(5(x1)))))) 2(1(0(5(x1)))) -> 4(2(0(1(5(5(x1)))))) 2(2(0(1(0(x1))))) -> 2(0(1(2(0(3(x1)))))) 2(5(2(1(0(x1))))) -> 0(2(2(1(5(3(x1)))))) 2(2(4(1(0(x1))))) -> 2(5(1(4(2(0(x1)))))) 2(2(4(1(0(x1))))) -> 1(1(2(4(2(0(x1)))))) 2(1(0(3(1(x1))))) -> 2(0(1(4(3(1(x1)))))) ---------------------------------------- (22) Obligation: Q DP problem: The TRS P consists of the following rules: 2^1(1(0(x1))) -> 2^1(x1) 2^1(1(0(x1))) -> 2^1(0(x1)) 2^1(1(0(x1))) -> 0^1(0(x1)) 0^1(2(1(0(x1)))) -> 2^1(1(x1)) 2^1(1(0(x1))) -> 0^1(0(2(x1))) 0^1(0(3(1(0(x1))))) -> 0^1(0(x1)) 2^1(1(0(x1))) -> 0^1(2(x1)) 2^1(1(0(1(x1)))) -> 0^1(x1) 2^1(1(0(1(x1)))) -> 2^1(1(x1)) 2^1(1(0(5(x1)))) -> 2^1(0(x1)) The TRS R consists of the following rules: 2(1(0(x1))) -> 2(1(3(0(x1)))) 2(1(0(x1))) -> 1(2(0(1(x1)))) 2(1(0(x1))) -> 0(4(1(2(x1)))) 2(1(0(x1))) -> 3(3(1(2(0(x1))))) 2(1(0(x1))) -> 3(1(4(2(0(x1))))) 2(1(0(x1))) -> 4(5(2(0(1(x1))))) 2(1(0(x1))) -> 3(0(1(1(2(x1))))) 2(1(0(x1))) -> 0(4(3(1(2(x1))))) 2(1(0(x1))) -> 0(3(4(1(2(x1))))) 2(1(0(x1))) -> 2(1(3(0(4(x1))))) 2(1(0(x1))) -> 2(3(0(1(4(x1))))) 2(1(0(x1))) -> 0(1(2(4(4(x1))))) 2(1(0(x1))) -> 5(2(1(4(0(0(x1)))))) 2(1(0(x1))) -> 5(2(1(5(4(0(x1)))))) 2(1(0(x1))) -> 4(1(3(2(5(0(x1)))))) 2(1(0(x1))) -> 4(3(2(2(0(1(x1)))))) 2(1(0(x1))) -> 2(3(0(4(1(1(x1)))))) 2(1(0(x1))) -> 1(3(3(0(0(2(x1)))))) 2(1(0(x1))) -> 5(1(4(3(0(2(x1)))))) 2(1(0(x1))) -> 1(4(0(2(3(2(x1)))))) 2(1(0(x1))) -> 1(3(0(3(5(2(x1)))))) 2(1(0(x1))) -> 0(1(3(5(2(3(x1)))))) 2(1(0(x1))) -> 2(1(5(0(4(3(x1)))))) 2(1(0(x1))) -> 0(1(2(4(5(5(x1)))))) 2(1(0(0(x1)))) -> 1(4(2(0(0(x1))))) 2(1(0(0(x1)))) -> 4(0(5(1(2(0(x1)))))) 2(1(0(0(x1)))) -> 0(5(3(2(0(1(x1)))))) 2(1(0(0(x1)))) -> 0(4(0(2(1(5(x1)))))) 2(5(0(0(x1)))) -> 2(0(3(0(3(5(x1)))))) 0(2(1(0(x1)))) -> 0(0(4(2(1(x1))))) 0(2(1(0(x1)))) -> 2(0(0(1(4(0(x1)))))) 2(2(1(0(x1)))) -> 2(1(2(0(3(3(x1)))))) 2(5(1(0(x1)))) -> 3(0(1(5(2(x1))))) 2(5(1(0(x1)))) -> 2(1(3(5(5(0(x1)))))) 2(5(1(0(x1)))) -> 2(2(0(5(3(1(x1)))))) 2(2(5(0(x1)))) -> 2(5(1(2(0(3(x1)))))) 2(1(0(1(x1)))) -> 1(2(1(4(0(x1))))) 2(1(0(1(x1)))) -> 0(3(1(2(1(x1))))) 2(1(0(1(x1)))) -> 2(1(4(1(3(0(x1)))))) 2(1(0(1(x1)))) -> 2(0(1(4(1(4(x1)))))) 2(1(0(5(x1)))) -> 5(3(1(2(0(x1))))) 2(1(0(5(x1)))) -> 5(1(3(0(2(x1))))) 2(1(0(5(x1)))) -> 2(3(0(1(3(5(x1)))))) 2(1(0(5(x1)))) -> 4(2(0(1(5(5(x1)))))) 2(2(0(1(0(x1))))) -> 2(0(1(2(0(3(x1)))))) 2(5(2(1(0(x1))))) -> 0(2(2(1(5(3(x1)))))) 0(0(3(1(0(x1))))) -> 0(3(1(4(0(0(x1)))))) 2(2(4(1(0(x1))))) -> 2(5(1(4(2(0(x1)))))) 2(2(4(1(0(x1))))) -> 1(1(2(4(2(0(x1)))))) 2(1(0(3(1(x1))))) -> 2(0(1(4(3(1(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (23) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. 2^1(1(0(x1))) -> 2^1(x1) 2^1(1(0(x1))) -> 0^1(0(x1)) 2^1(1(0(x1))) -> 0^1(0(2(x1))) 0^1(0(3(1(0(x1))))) -> 0^1(0(x1)) 2^1(1(0(x1))) -> 0^1(2(x1)) 2^1(1(0(1(x1)))) -> 0^1(x1) 2^1(1(0(1(x1)))) -> 2^1(1(x1)) The remaining pairs can at least be oriented weakly. Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: <<< POL(2^1(x_1)) = [[0A]] + [[1A, -I, 0A]] * x_1 >>> <<< POL(1(x_1)) = [[1A], [-I], [0A]] + [[-I, 0A, 0A], [0A, 0A, 1A], [0A, 0A, 0A]] * x_1 >>> <<< POL(0(x_1)) = [[0A], [0A], [1A]] + [[-I, 1A, 0A], [-I, 0A, 0A], [1A, 1A, 1A]] * x_1 >>> <<< POL(0^1(x_1)) = [[0A]] + [[0A, 0A, 0A]] * x_1 >>> <<< POL(2(x_1)) = [[0A], [0A], [0A]] + [[0A, 0A, 0A], [0A, 0A, -I], [-I, 0A, -I]] * x_1 >>> <<< POL(3(x_1)) = [[0A], [-I], [0A]] + [[0A, -I, -I], [-I, -I, -I], [0A, 0A, -I]] * x_1 >>> <<< POL(5(x_1)) = [[0A], [0A], [0A]] + [[0A, -I, 0A], [0A, 0A, -I], [0A, 0A, 0A]] * x_1 >>> <<< POL(4(x_1)) = [[0A], [-I], [-I]] + [[0A, 0A, 0A], [0A, 0A, 0A], [-I, -I, -I]] * x_1 >>> The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 0(2(1(0(x1)))) -> 0(0(4(2(1(x1))))) 0(2(1(0(x1)))) -> 2(0(0(1(4(0(x1)))))) 0(0(3(1(0(x1))))) -> 0(3(1(4(0(0(x1)))))) 2(1(0(x1))) -> 2(1(3(0(x1)))) 2(1(0(x1))) -> 1(2(0(1(x1)))) 2(1(0(x1))) -> 0(4(1(2(x1)))) 2(1(0(x1))) -> 3(3(1(2(0(x1))))) 2(1(0(x1))) -> 3(1(4(2(0(x1))))) 2(1(0(x1))) -> 4(5(2(0(1(x1))))) 2(1(0(x1))) -> 3(0(1(1(2(x1))))) 2(1(0(x1))) -> 0(4(3(1(2(x1))))) 2(1(0(x1))) -> 0(3(4(1(2(x1))))) 2(1(0(x1))) -> 2(1(3(0(4(x1))))) 2(1(0(x1))) -> 2(3(0(1(4(x1))))) 2(1(0(x1))) -> 0(1(2(4(4(x1))))) 2(1(0(x1))) -> 5(2(1(4(0(0(x1)))))) 2(1(0(x1))) -> 5(2(1(5(4(0(x1)))))) 2(1(0(x1))) -> 4(1(3(2(5(0(x1)))))) 2(1(0(x1))) -> 4(3(2(2(0(1(x1)))))) 2(1(0(x1))) -> 2(3(0(4(1(1(x1)))))) 2(1(0(x1))) -> 1(3(3(0(0(2(x1)))))) 2(1(0(x1))) -> 5(1(4(3(0(2(x1)))))) 2(1(0(x1))) -> 1(4(0(2(3(2(x1)))))) 2(1(0(x1))) -> 1(3(0(3(5(2(x1)))))) 2(1(0(x1))) -> 0(1(3(5(2(3(x1)))))) 2(1(0(x1))) -> 2(1(5(0(4(3(x1)))))) 2(1(0(x1))) -> 0(1(2(4(5(5(x1)))))) 2(1(0(0(x1)))) -> 1(4(2(0(0(x1))))) 2(1(0(0(x1)))) -> 4(0(5(1(2(0(x1)))))) 2(1(0(0(x1)))) -> 0(5(3(2(0(1(x1)))))) 2(1(0(0(x1)))) -> 0(4(0(2(1(5(x1)))))) 2(5(0(0(x1)))) -> 2(0(3(0(3(5(x1)))))) 2(2(1(0(x1)))) -> 2(1(2(0(3(3(x1)))))) 2(5(1(0(x1)))) -> 3(0(1(5(2(x1))))) 2(5(1(0(x1)))) -> 2(1(3(5(5(0(x1)))))) 2(5(1(0(x1)))) -> 2(2(0(5(3(1(x1)))))) 2(2(5(0(x1)))) -> 2(5(1(2(0(3(x1)))))) 2(1(0(1(x1)))) -> 1(2(1(4(0(x1))))) 2(1(0(1(x1)))) -> 0(3(1(2(1(x1))))) 2(1(0(1(x1)))) -> 2(1(4(1(3(0(x1)))))) 2(1(0(1(x1)))) -> 2(0(1(4(1(4(x1)))))) 2(1(0(5(x1)))) -> 5(3(1(2(0(x1))))) 2(1(0(5(x1)))) -> 5(1(3(0(2(x1))))) 2(1(0(5(x1)))) -> 2(3(0(1(3(5(x1)))))) 2(1(0(5(x1)))) -> 4(2(0(1(5(5(x1)))))) 2(2(0(1(0(x1))))) -> 2(0(1(2(0(3(x1)))))) 2(5(2(1(0(x1))))) -> 0(2(2(1(5(3(x1)))))) 2(2(4(1(0(x1))))) -> 2(5(1(4(2(0(x1)))))) 2(2(4(1(0(x1))))) -> 1(1(2(4(2(0(x1)))))) 2(1(0(3(1(x1))))) -> 2(0(1(4(3(1(x1)))))) ---------------------------------------- (24) Obligation: Q DP problem: The TRS P consists of the following rules: 2^1(1(0(x1))) -> 2^1(0(x1)) 0^1(2(1(0(x1)))) -> 2^1(1(x1)) 2^1(1(0(5(x1)))) -> 2^1(0(x1)) The TRS R consists of the following rules: 2(1(0(x1))) -> 2(1(3(0(x1)))) 2(1(0(x1))) -> 1(2(0(1(x1)))) 2(1(0(x1))) -> 0(4(1(2(x1)))) 2(1(0(x1))) -> 3(3(1(2(0(x1))))) 2(1(0(x1))) -> 3(1(4(2(0(x1))))) 2(1(0(x1))) -> 4(5(2(0(1(x1))))) 2(1(0(x1))) -> 3(0(1(1(2(x1))))) 2(1(0(x1))) -> 0(4(3(1(2(x1))))) 2(1(0(x1))) -> 0(3(4(1(2(x1))))) 2(1(0(x1))) -> 2(1(3(0(4(x1))))) 2(1(0(x1))) -> 2(3(0(1(4(x1))))) 2(1(0(x1))) -> 0(1(2(4(4(x1))))) 2(1(0(x1))) -> 5(2(1(4(0(0(x1)))))) 2(1(0(x1))) -> 5(2(1(5(4(0(x1)))))) 2(1(0(x1))) -> 4(1(3(2(5(0(x1)))))) 2(1(0(x1))) -> 4(3(2(2(0(1(x1)))))) 2(1(0(x1))) -> 2(3(0(4(1(1(x1)))))) 2(1(0(x1))) -> 1(3(3(0(0(2(x1)))))) 2(1(0(x1))) -> 5(1(4(3(0(2(x1)))))) 2(1(0(x1))) -> 1(4(0(2(3(2(x1)))))) 2(1(0(x1))) -> 1(3(0(3(5(2(x1)))))) 2(1(0(x1))) -> 0(1(3(5(2(3(x1)))))) 2(1(0(x1))) -> 2(1(5(0(4(3(x1)))))) 2(1(0(x1))) -> 0(1(2(4(5(5(x1)))))) 2(1(0(0(x1)))) -> 1(4(2(0(0(x1))))) 2(1(0(0(x1)))) -> 4(0(5(1(2(0(x1)))))) 2(1(0(0(x1)))) -> 0(5(3(2(0(1(x1)))))) 2(1(0(0(x1)))) -> 0(4(0(2(1(5(x1)))))) 2(5(0(0(x1)))) -> 2(0(3(0(3(5(x1)))))) 0(2(1(0(x1)))) -> 0(0(4(2(1(x1))))) 0(2(1(0(x1)))) -> 2(0(0(1(4(0(x1)))))) 2(2(1(0(x1)))) -> 2(1(2(0(3(3(x1)))))) 2(5(1(0(x1)))) -> 3(0(1(5(2(x1))))) 2(5(1(0(x1)))) -> 2(1(3(5(5(0(x1)))))) 2(5(1(0(x1)))) -> 2(2(0(5(3(1(x1)))))) 2(2(5(0(x1)))) -> 2(5(1(2(0(3(x1)))))) 2(1(0(1(x1)))) -> 1(2(1(4(0(x1))))) 2(1(0(1(x1)))) -> 0(3(1(2(1(x1))))) 2(1(0(1(x1)))) -> 2(1(4(1(3(0(x1)))))) 2(1(0(1(x1)))) -> 2(0(1(4(1(4(x1)))))) 2(1(0(5(x1)))) -> 5(3(1(2(0(x1))))) 2(1(0(5(x1)))) -> 5(1(3(0(2(x1))))) 2(1(0(5(x1)))) -> 2(3(0(1(3(5(x1)))))) 2(1(0(5(x1)))) -> 4(2(0(1(5(5(x1)))))) 2(2(0(1(0(x1))))) -> 2(0(1(2(0(3(x1)))))) 2(5(2(1(0(x1))))) -> 0(2(2(1(5(3(x1)))))) 0(0(3(1(0(x1))))) -> 0(3(1(4(0(0(x1)))))) 2(2(4(1(0(x1))))) -> 2(5(1(4(2(0(x1)))))) 2(2(4(1(0(x1))))) -> 1(1(2(4(2(0(x1)))))) 2(1(0(3(1(x1))))) -> 2(0(1(4(3(1(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (25) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (26) Obligation: Q DP problem: The TRS P consists of the following rules: 2^1(1(0(5(x1)))) -> 2^1(0(x1)) 2^1(1(0(x1))) -> 2^1(0(x1)) The TRS R consists of the following rules: 2(1(0(x1))) -> 2(1(3(0(x1)))) 2(1(0(x1))) -> 1(2(0(1(x1)))) 2(1(0(x1))) -> 0(4(1(2(x1)))) 2(1(0(x1))) -> 3(3(1(2(0(x1))))) 2(1(0(x1))) -> 3(1(4(2(0(x1))))) 2(1(0(x1))) -> 4(5(2(0(1(x1))))) 2(1(0(x1))) -> 3(0(1(1(2(x1))))) 2(1(0(x1))) -> 0(4(3(1(2(x1))))) 2(1(0(x1))) -> 0(3(4(1(2(x1))))) 2(1(0(x1))) -> 2(1(3(0(4(x1))))) 2(1(0(x1))) -> 2(3(0(1(4(x1))))) 2(1(0(x1))) -> 0(1(2(4(4(x1))))) 2(1(0(x1))) -> 5(2(1(4(0(0(x1)))))) 2(1(0(x1))) -> 5(2(1(5(4(0(x1)))))) 2(1(0(x1))) -> 4(1(3(2(5(0(x1)))))) 2(1(0(x1))) -> 4(3(2(2(0(1(x1)))))) 2(1(0(x1))) -> 2(3(0(4(1(1(x1)))))) 2(1(0(x1))) -> 1(3(3(0(0(2(x1)))))) 2(1(0(x1))) -> 5(1(4(3(0(2(x1)))))) 2(1(0(x1))) -> 1(4(0(2(3(2(x1)))))) 2(1(0(x1))) -> 1(3(0(3(5(2(x1)))))) 2(1(0(x1))) -> 0(1(3(5(2(3(x1)))))) 2(1(0(x1))) -> 2(1(5(0(4(3(x1)))))) 2(1(0(x1))) -> 0(1(2(4(5(5(x1)))))) 2(1(0(0(x1)))) -> 1(4(2(0(0(x1))))) 2(1(0(0(x1)))) -> 4(0(5(1(2(0(x1)))))) 2(1(0(0(x1)))) -> 0(5(3(2(0(1(x1)))))) 2(1(0(0(x1)))) -> 0(4(0(2(1(5(x1)))))) 2(5(0(0(x1)))) -> 2(0(3(0(3(5(x1)))))) 0(2(1(0(x1)))) -> 0(0(4(2(1(x1))))) 0(2(1(0(x1)))) -> 2(0(0(1(4(0(x1)))))) 2(2(1(0(x1)))) -> 2(1(2(0(3(3(x1)))))) 2(5(1(0(x1)))) -> 3(0(1(5(2(x1))))) 2(5(1(0(x1)))) -> 2(1(3(5(5(0(x1)))))) 2(5(1(0(x1)))) -> 2(2(0(5(3(1(x1)))))) 2(2(5(0(x1)))) -> 2(5(1(2(0(3(x1)))))) 2(1(0(1(x1)))) -> 1(2(1(4(0(x1))))) 2(1(0(1(x1)))) -> 0(3(1(2(1(x1))))) 2(1(0(1(x1)))) -> 2(1(4(1(3(0(x1)))))) 2(1(0(1(x1)))) -> 2(0(1(4(1(4(x1)))))) 2(1(0(5(x1)))) -> 5(3(1(2(0(x1))))) 2(1(0(5(x1)))) -> 5(1(3(0(2(x1))))) 2(1(0(5(x1)))) -> 2(3(0(1(3(5(x1)))))) 2(1(0(5(x1)))) -> 4(2(0(1(5(5(x1)))))) 2(2(0(1(0(x1))))) -> 2(0(1(2(0(3(x1)))))) 2(5(2(1(0(x1))))) -> 0(2(2(1(5(3(x1)))))) 0(0(3(1(0(x1))))) -> 0(3(1(4(0(0(x1)))))) 2(2(4(1(0(x1))))) -> 2(5(1(4(2(0(x1)))))) 2(2(4(1(0(x1))))) -> 1(1(2(4(2(0(x1)))))) 2(1(0(3(1(x1))))) -> 2(0(1(4(3(1(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (27) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. 2^1(1(0(5(x1)))) -> 2^1(0(x1)) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial interpretation [POLO]: POL(0(x_1)) = x_1 POL(1(x_1)) = x_1 POL(2(x_1)) = 0 POL(2^1(x_1)) = x_1 POL(3(x_1)) = 0 POL(4(x_1)) = 0 POL(5(x_1)) = 1 + x_1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 0(2(1(0(x1)))) -> 0(0(4(2(1(x1))))) 0(2(1(0(x1)))) -> 2(0(0(1(4(0(x1)))))) 0(0(3(1(0(x1))))) -> 0(3(1(4(0(0(x1)))))) ---------------------------------------- (28) Obligation: Q DP problem: The TRS P consists of the following rules: 2^1(1(0(x1))) -> 2^1(0(x1)) The TRS R consists of the following rules: 2(1(0(x1))) -> 2(1(3(0(x1)))) 2(1(0(x1))) -> 1(2(0(1(x1)))) 2(1(0(x1))) -> 0(4(1(2(x1)))) 2(1(0(x1))) -> 3(3(1(2(0(x1))))) 2(1(0(x1))) -> 3(1(4(2(0(x1))))) 2(1(0(x1))) -> 4(5(2(0(1(x1))))) 2(1(0(x1))) -> 3(0(1(1(2(x1))))) 2(1(0(x1))) -> 0(4(3(1(2(x1))))) 2(1(0(x1))) -> 0(3(4(1(2(x1))))) 2(1(0(x1))) -> 2(1(3(0(4(x1))))) 2(1(0(x1))) -> 2(3(0(1(4(x1))))) 2(1(0(x1))) -> 0(1(2(4(4(x1))))) 2(1(0(x1))) -> 5(2(1(4(0(0(x1)))))) 2(1(0(x1))) -> 5(2(1(5(4(0(x1)))))) 2(1(0(x1))) -> 4(1(3(2(5(0(x1)))))) 2(1(0(x1))) -> 4(3(2(2(0(1(x1)))))) 2(1(0(x1))) -> 2(3(0(4(1(1(x1)))))) 2(1(0(x1))) -> 1(3(3(0(0(2(x1)))))) 2(1(0(x1))) -> 5(1(4(3(0(2(x1)))))) 2(1(0(x1))) -> 1(4(0(2(3(2(x1)))))) 2(1(0(x1))) -> 1(3(0(3(5(2(x1)))))) 2(1(0(x1))) -> 0(1(3(5(2(3(x1)))))) 2(1(0(x1))) -> 2(1(5(0(4(3(x1)))))) 2(1(0(x1))) -> 0(1(2(4(5(5(x1)))))) 2(1(0(0(x1)))) -> 1(4(2(0(0(x1))))) 2(1(0(0(x1)))) -> 4(0(5(1(2(0(x1)))))) 2(1(0(0(x1)))) -> 0(5(3(2(0(1(x1)))))) 2(1(0(0(x1)))) -> 0(4(0(2(1(5(x1)))))) 2(5(0(0(x1)))) -> 2(0(3(0(3(5(x1)))))) 0(2(1(0(x1)))) -> 0(0(4(2(1(x1))))) 0(2(1(0(x1)))) -> 2(0(0(1(4(0(x1)))))) 2(2(1(0(x1)))) -> 2(1(2(0(3(3(x1)))))) 2(5(1(0(x1)))) -> 3(0(1(5(2(x1))))) 2(5(1(0(x1)))) -> 2(1(3(5(5(0(x1)))))) 2(5(1(0(x1)))) -> 2(2(0(5(3(1(x1)))))) 2(2(5(0(x1)))) -> 2(5(1(2(0(3(x1)))))) 2(1(0(1(x1)))) -> 1(2(1(4(0(x1))))) 2(1(0(1(x1)))) -> 0(3(1(2(1(x1))))) 2(1(0(1(x1)))) -> 2(1(4(1(3(0(x1)))))) 2(1(0(1(x1)))) -> 2(0(1(4(1(4(x1)))))) 2(1(0(5(x1)))) -> 5(3(1(2(0(x1))))) 2(1(0(5(x1)))) -> 5(1(3(0(2(x1))))) 2(1(0(5(x1)))) -> 2(3(0(1(3(5(x1)))))) 2(1(0(5(x1)))) -> 4(2(0(1(5(5(x1)))))) 2(2(0(1(0(x1))))) -> 2(0(1(2(0(3(x1)))))) 2(5(2(1(0(x1))))) -> 0(2(2(1(5(3(x1)))))) 0(0(3(1(0(x1))))) -> 0(3(1(4(0(0(x1)))))) 2(2(4(1(0(x1))))) -> 2(5(1(4(2(0(x1)))))) 2(2(4(1(0(x1))))) -> 1(1(2(4(2(0(x1)))))) 2(1(0(3(1(x1))))) -> 2(0(1(4(3(1(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (29) 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: *2^1(1(0(x1))) -> 2^1(0(x1)) The graph contains the following edges 1 > 1 ---------------------------------------- (30) YES