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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 164 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 6 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 290 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 340 ms] (10) QDP (11) PisEmptyProof [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(1(2(2(x1)))) -> 0(1(0(2(2(x1))))) 0(1(2(2(x1)))) -> 0(1(2(3(2(x1))))) 0(1(2(2(x1)))) -> 0(2(2(1(3(x1))))) 0(1(2(2(x1)))) -> 1(0(3(2(2(x1))))) 0(1(2(2(x1)))) -> 1(2(0(3(2(x1))))) 0(1(2(2(x1)))) -> 1(3(0(2(2(x1))))) 0(1(2(2(x1)))) -> 1(3(2(0(2(x1))))) 0(1(2(2(x1)))) -> 0(1(0(4(2(2(x1)))))) 0(1(2(2(x1)))) -> 0(2(1(3(2(3(x1)))))) 0(1(2(2(x1)))) -> 1(2(1(0(4(2(x1)))))) 0(1(2(2(x1)))) -> 1(5(0(4(2(2(x1)))))) 0(1(2(2(x1)))) -> 2(0(3(1(3(2(x1)))))) 0(1(2(2(x1)))) -> 2(1(1(0(4(2(x1)))))) 0(1(2(2(x1)))) -> 2(1(3(0(2(0(x1)))))) 0(1(2(2(x1)))) -> 2(1(3(3(2(0(x1)))))) 0(1(2(2(x1)))) -> 2(1(5(3(0(2(x1)))))) 0(1(2(2(x1)))) -> 2(2(1(3(0(5(x1)))))) 0(1(2(2(x1)))) -> 2(4(1(3(2(0(x1)))))) 0(1(4(5(x1)))) -> 1(5(0(4(1(x1))))) 0(1(4(5(x1)))) -> 5(0(4(1(5(x1))))) 0(1(4(5(x1)))) -> 5(4(1(5(0(x1))))) 0(1(4(5(x1)))) -> 1(1(5(0(4(1(x1)))))) 0(1(4(5(x1)))) -> 5(4(1(5(5(0(x1)))))) 5(1(2(2(x1)))) -> 1(0(2(2(5(x1))))) 5(1(2(2(x1)))) -> 1(3(5(2(2(x1))))) 5(1(2(2(x1)))) -> 1(5(2(3(2(x1))))) 5(1(2(2(x1)))) -> 1(5(0(2(2(3(x1)))))) 5(1(2(2(x1)))) -> 2(1(0(3(2(5(x1)))))) 5(1(2(2(x1)))) -> 3(1(3(5(2(2(x1)))))) 5(1(2(2(x1)))) -> 4(1(3(2(2(5(x1)))))) 5(1(2(2(x1)))) -> 5(1(0(4(2(2(x1)))))) 5(1(2(2(x1)))) -> 5(1(2(0(4(2(x1)))))) 0(1(1(4(5(x1))))) -> 3(1(0(4(1(5(x1)))))) 0(1(2(2(2(x1))))) -> 1(0(2(2(5(2(x1)))))) 0(1(2(2(5(x1))))) -> 1(5(0(4(2(2(x1)))))) 0(1(2(4(5(x1))))) -> 2(5(1(0(4(5(x1)))))) 0(1(4(5(2(x1))))) -> 1(0(4(2(0(5(x1)))))) 0(1(4(5(5(x1))))) -> 5(0(4(0(1(5(x1)))))) 0(1(5(4(5(x1))))) -> 1(5(0(4(1(5(x1)))))) 0(5(1(2(2(x1))))) -> 0(1(3(2(5(2(x1)))))) 3(3(1(2(2(x1))))) -> 1(3(2(0(3(2(x1)))))) 3(4(4(0(5(x1))))) -> 3(5(4(5(0(4(x1)))))) 5(0(1(2(2(x1))))) -> 1(3(2(0(5(2(x1)))))) 5(1(2(2(5(x1))))) -> 1(5(2(3(2(5(x1)))))) 5(2(1(2(2(x1))))) -> 2(1(3(5(2(2(x1)))))) 5(2(4(0(5(x1))))) -> 0(4(2(5(5(5(x1)))))) 5(2(4(0(5(x1))))) -> 0(4(5(4(2(5(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(2(x1)))) -> 0^1(1(0(2(2(x1))))) 0^1(1(2(2(x1)))) -> 0^1(2(2(x1))) 0^1(1(2(2(x1)))) -> 0^1(1(2(3(2(x1))))) 0^1(1(2(2(x1)))) -> 3^1(2(x1)) 0^1(1(2(2(x1)))) -> 0^1(2(2(1(3(x1))))) 0^1(1(2(2(x1)))) -> 3^1(x1) 0^1(1(2(2(x1)))) -> 0^1(3(2(2(x1)))) 0^1(1(2(2(x1)))) -> 3^1(2(2(x1))) 0^1(1(2(2(x1)))) -> 0^1(3(2(x1))) 0^1(1(2(2(x1)))) -> 3^1(0(2(2(x1)))) 0^1(1(2(2(x1)))) -> 3^1(2(0(2(x1)))) 0^1(1(2(2(x1)))) -> 0^1(2(x1)) 0^1(1(2(2(x1)))) -> 0^1(1(0(4(2(2(x1)))))) 0^1(1(2(2(x1)))) -> 0^1(4(2(2(x1)))) 0^1(1(2(2(x1)))) -> 0^1(2(1(3(2(3(x1)))))) 0^1(1(2(2(x1)))) -> 3^1(2(3(x1))) 0^1(1(2(2(x1)))) -> 0^1(4(2(x1))) 0^1(1(2(2(x1)))) -> 5^1(0(4(2(2(x1))))) 0^1(1(2(2(x1)))) -> 0^1(3(1(3(2(x1))))) 0^1(1(2(2(x1)))) -> 3^1(1(3(2(x1)))) 0^1(1(2(2(x1)))) -> 3^1(0(2(0(x1)))) 0^1(1(2(2(x1)))) -> 0^1(2(0(x1))) 0^1(1(2(2(x1)))) -> 0^1(x1) 0^1(1(2(2(x1)))) -> 3^1(3(2(0(x1)))) 0^1(1(2(2(x1)))) -> 3^1(2(0(x1))) 0^1(1(2(2(x1)))) -> 5^1(3(0(2(x1)))) 0^1(1(2(2(x1)))) -> 3^1(0(2(x1))) 0^1(1(2(2(x1)))) -> 3^1(0(5(x1))) 0^1(1(2(2(x1)))) -> 0^1(5(x1)) 0^1(1(2(2(x1)))) -> 5^1(x1) 0^1(1(4(5(x1)))) -> 5^1(0(4(1(x1)))) 0^1(1(4(5(x1)))) -> 0^1(4(1(x1))) 0^1(1(4(5(x1)))) -> 5^1(0(4(1(5(x1))))) 0^1(1(4(5(x1)))) -> 0^1(4(1(5(x1)))) 0^1(1(4(5(x1)))) -> 5^1(4(1(5(0(x1))))) 0^1(1(4(5(x1)))) -> 5^1(0(x1)) 0^1(1(4(5(x1)))) -> 0^1(x1) 0^1(1(4(5(x1)))) -> 5^1(4(1(5(5(0(x1)))))) 0^1(1(4(5(x1)))) -> 5^1(5(0(x1))) 5^1(1(2(2(x1)))) -> 0^1(2(2(5(x1)))) 5^1(1(2(2(x1)))) -> 5^1(x1) 5^1(1(2(2(x1)))) -> 3^1(5(2(2(x1)))) 5^1(1(2(2(x1)))) -> 5^1(2(2(x1))) 5^1(1(2(2(x1)))) -> 5^1(2(3(2(x1)))) 5^1(1(2(2(x1)))) -> 3^1(2(x1)) 5^1(1(2(2(x1)))) -> 5^1(0(2(2(3(x1))))) 5^1(1(2(2(x1)))) -> 0^1(2(2(3(x1)))) 5^1(1(2(2(x1)))) -> 3^1(x1) 5^1(1(2(2(x1)))) -> 0^1(3(2(5(x1)))) 5^1(1(2(2(x1)))) -> 3^1(2(5(x1))) 5^1(1(2(2(x1)))) -> 3^1(1(3(5(2(2(x1)))))) 5^1(1(2(2(x1)))) -> 3^1(2(2(5(x1)))) 5^1(1(2(2(x1)))) -> 5^1(1(0(4(2(2(x1)))))) 5^1(1(2(2(x1)))) -> 0^1(4(2(2(x1)))) 5^1(1(2(2(x1)))) -> 5^1(1(2(0(4(2(x1)))))) 5^1(1(2(2(x1)))) -> 0^1(4(2(x1))) 0^1(1(1(4(5(x1))))) -> 3^1(1(0(4(1(5(x1)))))) 0^1(1(1(4(5(x1))))) -> 0^1(4(1(5(x1)))) 0^1(1(2(2(2(x1))))) -> 0^1(2(2(5(2(x1))))) 0^1(1(2(2(2(x1))))) -> 5^1(2(x1)) 0^1(1(2(2(5(x1))))) -> 5^1(0(4(2(2(x1))))) 0^1(1(2(2(5(x1))))) -> 0^1(4(2(2(x1)))) 0^1(1(2(4(5(x1))))) -> 5^1(1(0(4(5(x1))))) 0^1(1(2(4(5(x1))))) -> 0^1(4(5(x1))) 0^1(1(4(5(2(x1))))) -> 0^1(4(2(0(5(x1))))) 0^1(1(4(5(2(x1))))) -> 0^1(5(x1)) 0^1(1(4(5(2(x1))))) -> 5^1(x1) 0^1(1(4(5(5(x1))))) -> 5^1(0(4(0(1(5(x1)))))) 0^1(1(4(5(5(x1))))) -> 0^1(4(0(1(5(x1))))) 0^1(1(4(5(5(x1))))) -> 0^1(1(5(x1))) 0^1(1(5(4(5(x1))))) -> 5^1(0(4(1(5(x1))))) 0^1(1(5(4(5(x1))))) -> 0^1(4(1(5(x1)))) 0^1(5(1(2(2(x1))))) -> 0^1(1(3(2(5(2(x1)))))) 0^1(5(1(2(2(x1))))) -> 3^1(2(5(2(x1)))) 0^1(5(1(2(2(x1))))) -> 5^1(2(x1)) 3^1(3(1(2(2(x1))))) -> 3^1(2(0(3(2(x1))))) 3^1(3(1(2(2(x1))))) -> 0^1(3(2(x1))) 3^1(3(1(2(2(x1))))) -> 3^1(2(x1)) 3^1(4(4(0(5(x1))))) -> 3^1(5(4(5(0(4(x1)))))) 3^1(4(4(0(5(x1))))) -> 5^1(4(5(0(4(x1))))) 3^1(4(4(0(5(x1))))) -> 5^1(0(4(x1))) 3^1(4(4(0(5(x1))))) -> 0^1(4(x1)) 5^1(0(1(2(2(x1))))) -> 3^1(2(0(5(2(x1))))) 5^1(0(1(2(2(x1))))) -> 0^1(5(2(x1))) 5^1(0(1(2(2(x1))))) -> 5^1(2(x1)) 5^1(1(2(2(5(x1))))) -> 5^1(2(3(2(5(x1))))) 5^1(1(2(2(5(x1))))) -> 3^1(2(5(x1))) 5^1(2(1(2(2(x1))))) -> 3^1(5(2(2(x1)))) 5^1(2(1(2(2(x1))))) -> 5^1(2(2(x1))) 5^1(2(4(0(5(x1))))) -> 0^1(4(2(5(5(5(x1)))))) 5^1(2(4(0(5(x1))))) -> 5^1(5(5(x1))) 5^1(2(4(0(5(x1))))) -> 5^1(5(x1)) 5^1(2(4(0(5(x1))))) -> 0^1(4(5(4(2(5(x1)))))) 5^1(2(4(0(5(x1))))) -> 5^1(4(2(5(x1)))) The TRS R consists of the following rules: 0(1(2(2(x1)))) -> 0(1(0(2(2(x1))))) 0(1(2(2(x1)))) -> 0(1(2(3(2(x1))))) 0(1(2(2(x1)))) -> 0(2(2(1(3(x1))))) 0(1(2(2(x1)))) -> 1(0(3(2(2(x1))))) 0(1(2(2(x1)))) -> 1(2(0(3(2(x1))))) 0(1(2(2(x1)))) -> 1(3(0(2(2(x1))))) 0(1(2(2(x1)))) -> 1(3(2(0(2(x1))))) 0(1(2(2(x1)))) -> 0(1(0(4(2(2(x1)))))) 0(1(2(2(x1)))) -> 0(2(1(3(2(3(x1)))))) 0(1(2(2(x1)))) -> 1(2(1(0(4(2(x1)))))) 0(1(2(2(x1)))) -> 1(5(0(4(2(2(x1)))))) 0(1(2(2(x1)))) -> 2(0(3(1(3(2(x1)))))) 0(1(2(2(x1)))) -> 2(1(1(0(4(2(x1)))))) 0(1(2(2(x1)))) -> 2(1(3(0(2(0(x1)))))) 0(1(2(2(x1)))) -> 2(1(3(3(2(0(x1)))))) 0(1(2(2(x1)))) -> 2(1(5(3(0(2(x1)))))) 0(1(2(2(x1)))) -> 2(2(1(3(0(5(x1)))))) 0(1(2(2(x1)))) -> 2(4(1(3(2(0(x1)))))) 0(1(4(5(x1)))) -> 1(5(0(4(1(x1))))) 0(1(4(5(x1)))) -> 5(0(4(1(5(x1))))) 0(1(4(5(x1)))) -> 5(4(1(5(0(x1))))) 0(1(4(5(x1)))) -> 1(1(5(0(4(1(x1)))))) 0(1(4(5(x1)))) -> 5(4(1(5(5(0(x1)))))) 5(1(2(2(x1)))) -> 1(0(2(2(5(x1))))) 5(1(2(2(x1)))) -> 1(3(5(2(2(x1))))) 5(1(2(2(x1)))) -> 1(5(2(3(2(x1))))) 5(1(2(2(x1)))) -> 1(5(0(2(2(3(x1)))))) 5(1(2(2(x1)))) -> 2(1(0(3(2(5(x1)))))) 5(1(2(2(x1)))) -> 3(1(3(5(2(2(x1)))))) 5(1(2(2(x1)))) -> 4(1(3(2(2(5(x1)))))) 5(1(2(2(x1)))) -> 5(1(0(4(2(2(x1)))))) 5(1(2(2(x1)))) -> 5(1(2(0(4(2(x1)))))) 0(1(1(4(5(x1))))) -> 3(1(0(4(1(5(x1)))))) 0(1(2(2(2(x1))))) -> 1(0(2(2(5(2(x1)))))) 0(1(2(2(5(x1))))) -> 1(5(0(4(2(2(x1)))))) 0(1(2(4(5(x1))))) -> 2(5(1(0(4(5(x1)))))) 0(1(4(5(2(x1))))) -> 1(0(4(2(0(5(x1)))))) 0(1(4(5(5(x1))))) -> 5(0(4(0(1(5(x1)))))) 0(1(5(4(5(x1))))) -> 1(5(0(4(1(5(x1)))))) 0(5(1(2(2(x1))))) -> 0(1(3(2(5(2(x1)))))) 3(3(1(2(2(x1))))) -> 1(3(2(0(3(2(x1)))))) 3(4(4(0(5(x1))))) -> 3(5(4(5(0(4(x1)))))) 5(0(1(2(2(x1))))) -> 1(3(2(0(5(2(x1)))))) 5(1(2(2(5(x1))))) -> 1(5(2(3(2(5(x1)))))) 5(2(1(2(2(x1))))) -> 2(1(3(5(2(2(x1)))))) 5(2(4(0(5(x1))))) -> 0(4(2(5(5(5(x1)))))) 5(2(4(0(5(x1))))) -> 0(4(5(4(2(5(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 78 less nodes. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: 0^1(1(2(2(x1)))) -> 0^1(5(x1)) 0^1(1(2(2(x1)))) -> 0^1(x1) 0^1(1(2(2(x1)))) -> 5^1(x1) 5^1(1(2(2(x1)))) -> 5^1(x1) 5^1(0(1(2(2(x1))))) -> 0^1(5(2(x1))) 0^1(1(4(5(x1)))) -> 5^1(0(x1)) 5^1(0(1(2(2(x1))))) -> 5^1(2(x1)) 5^1(2(4(0(5(x1))))) -> 5^1(5(5(x1))) 5^1(2(4(0(5(x1))))) -> 5^1(5(x1)) 0^1(1(4(5(x1)))) -> 0^1(x1) 0^1(1(4(5(x1)))) -> 5^1(5(0(x1))) 0^1(1(2(2(2(x1))))) -> 5^1(2(x1)) 0^1(1(4(5(2(x1))))) -> 0^1(5(x1)) 0^1(1(4(5(2(x1))))) -> 5^1(x1) 0^1(1(4(5(5(x1))))) -> 0^1(1(5(x1))) 0^1(5(1(2(2(x1))))) -> 5^1(2(x1)) The TRS R consists of the following rules: 0(1(2(2(x1)))) -> 0(1(0(2(2(x1))))) 0(1(2(2(x1)))) -> 0(1(2(3(2(x1))))) 0(1(2(2(x1)))) -> 0(2(2(1(3(x1))))) 0(1(2(2(x1)))) -> 1(0(3(2(2(x1))))) 0(1(2(2(x1)))) -> 1(2(0(3(2(x1))))) 0(1(2(2(x1)))) -> 1(3(0(2(2(x1))))) 0(1(2(2(x1)))) -> 1(3(2(0(2(x1))))) 0(1(2(2(x1)))) -> 0(1(0(4(2(2(x1)))))) 0(1(2(2(x1)))) -> 0(2(1(3(2(3(x1)))))) 0(1(2(2(x1)))) -> 1(2(1(0(4(2(x1)))))) 0(1(2(2(x1)))) -> 1(5(0(4(2(2(x1)))))) 0(1(2(2(x1)))) -> 2(0(3(1(3(2(x1)))))) 0(1(2(2(x1)))) -> 2(1(1(0(4(2(x1)))))) 0(1(2(2(x1)))) -> 2(1(3(0(2(0(x1)))))) 0(1(2(2(x1)))) -> 2(1(3(3(2(0(x1)))))) 0(1(2(2(x1)))) -> 2(1(5(3(0(2(x1)))))) 0(1(2(2(x1)))) -> 2(2(1(3(0(5(x1)))))) 0(1(2(2(x1)))) -> 2(4(1(3(2(0(x1)))))) 0(1(4(5(x1)))) -> 1(5(0(4(1(x1))))) 0(1(4(5(x1)))) -> 5(0(4(1(5(x1))))) 0(1(4(5(x1)))) -> 5(4(1(5(0(x1))))) 0(1(4(5(x1)))) -> 1(1(5(0(4(1(x1)))))) 0(1(4(5(x1)))) -> 5(4(1(5(5(0(x1)))))) 5(1(2(2(x1)))) -> 1(0(2(2(5(x1))))) 5(1(2(2(x1)))) -> 1(3(5(2(2(x1))))) 5(1(2(2(x1)))) -> 1(5(2(3(2(x1))))) 5(1(2(2(x1)))) -> 1(5(0(2(2(3(x1)))))) 5(1(2(2(x1)))) -> 2(1(0(3(2(5(x1)))))) 5(1(2(2(x1)))) -> 3(1(3(5(2(2(x1)))))) 5(1(2(2(x1)))) -> 4(1(3(2(2(5(x1)))))) 5(1(2(2(x1)))) -> 5(1(0(4(2(2(x1)))))) 5(1(2(2(x1)))) -> 5(1(2(0(4(2(x1)))))) 0(1(1(4(5(x1))))) -> 3(1(0(4(1(5(x1)))))) 0(1(2(2(2(x1))))) -> 1(0(2(2(5(2(x1)))))) 0(1(2(2(5(x1))))) -> 1(5(0(4(2(2(x1)))))) 0(1(2(4(5(x1))))) -> 2(5(1(0(4(5(x1)))))) 0(1(4(5(2(x1))))) -> 1(0(4(2(0(5(x1)))))) 0(1(4(5(5(x1))))) -> 5(0(4(0(1(5(x1)))))) 0(1(5(4(5(x1))))) -> 1(5(0(4(1(5(x1)))))) 0(5(1(2(2(x1))))) -> 0(1(3(2(5(2(x1)))))) 3(3(1(2(2(x1))))) -> 1(3(2(0(3(2(x1)))))) 3(4(4(0(5(x1))))) -> 3(5(4(5(0(4(x1)))))) 5(0(1(2(2(x1))))) -> 1(3(2(0(5(2(x1)))))) 5(1(2(2(5(x1))))) -> 1(5(2(3(2(5(x1)))))) 5(2(1(2(2(x1))))) -> 2(1(3(5(2(2(x1)))))) 5(2(4(0(5(x1))))) -> 0(4(2(5(5(5(x1)))))) 5(2(4(0(5(x1))))) -> 0(4(5(4(2(5(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(2(x1)))) -> 0^1(5(x1)) 0^1(1(2(2(x1)))) -> 0^1(x1) 0^1(1(2(2(x1)))) -> 5^1(x1) 5^1(1(2(2(x1)))) -> 5^1(x1) 0^1(1(4(5(x1)))) -> 5^1(0(x1)) 5^1(0(1(2(2(x1))))) -> 5^1(2(x1)) 5^1(2(4(0(5(x1))))) -> 5^1(5(5(x1))) 5^1(2(4(0(5(x1))))) -> 5^1(5(x1)) 0^1(1(4(5(x1)))) -> 5^1(5(0(x1))) 0^1(1(2(2(2(x1))))) -> 5^1(2(x1)) 0^1(1(4(5(2(x1))))) -> 0^1(5(x1)) 0^1(1(4(5(2(x1))))) -> 5^1(x1) 0^1(5(1(2(2(x1))))) -> 5^1(2(x1)) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial interpretation [POLO]: POL(0(x_1)) = x_1 POL(0^1(x_1)) = 1 + x_1 POL(1(x_1)) = x_1 POL(2(x_1)) = 1 + x_1 POL(3(x_1)) = x_1 POL(4(x_1)) = x_1 POL(5(x_1)) = x_1 POL(5^1(x_1)) = x_1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 5(1(2(2(x1)))) -> 1(0(2(2(5(x1))))) 5(1(2(2(x1)))) -> 1(3(5(2(2(x1))))) 5(1(2(2(x1)))) -> 1(5(2(3(2(x1))))) 5(1(2(2(x1)))) -> 1(5(0(2(2(3(x1)))))) 5(1(2(2(x1)))) -> 2(1(0(3(2(5(x1)))))) 5(1(2(2(x1)))) -> 3(1(3(5(2(2(x1)))))) 5(1(2(2(x1)))) -> 4(1(3(2(2(5(x1)))))) 5(1(2(2(x1)))) -> 5(1(0(4(2(2(x1)))))) 5(1(2(2(x1)))) -> 5(1(2(0(4(2(x1)))))) 5(0(1(2(2(x1))))) -> 1(3(2(0(5(2(x1)))))) 5(1(2(2(5(x1))))) -> 1(5(2(3(2(5(x1)))))) 5(2(1(2(2(x1))))) -> 2(1(3(5(2(2(x1)))))) 5(2(4(0(5(x1))))) -> 0(4(2(5(5(5(x1)))))) 5(2(4(0(5(x1))))) -> 0(4(5(4(2(5(x1)))))) 0(1(2(2(x1)))) -> 0(1(0(2(2(x1))))) 0(1(2(2(x1)))) -> 0(1(2(3(2(x1))))) 0(1(2(2(x1)))) -> 0(2(2(1(3(x1))))) 0(1(2(2(x1)))) -> 1(0(3(2(2(x1))))) 0(1(2(2(x1)))) -> 1(2(0(3(2(x1))))) 0(1(2(2(x1)))) -> 1(3(0(2(2(x1))))) 0(1(2(2(x1)))) -> 1(3(2(0(2(x1))))) 0(1(2(2(x1)))) -> 0(1(0(4(2(2(x1)))))) 0(1(2(2(x1)))) -> 0(2(1(3(2(3(x1)))))) 0(1(2(2(x1)))) -> 1(2(1(0(4(2(x1)))))) 0(1(2(2(x1)))) -> 1(5(0(4(2(2(x1)))))) 0(1(2(2(x1)))) -> 2(0(3(1(3(2(x1)))))) 0(1(2(2(x1)))) -> 2(1(1(0(4(2(x1)))))) 0(1(2(2(x1)))) -> 2(1(3(0(2(0(x1)))))) 0(1(2(2(x1)))) -> 2(1(3(3(2(0(x1)))))) 0(1(2(2(x1)))) -> 2(1(5(3(0(2(x1)))))) 0(1(2(2(x1)))) -> 2(2(1(3(0(5(x1)))))) 0(1(2(2(x1)))) -> 2(4(1(3(2(0(x1)))))) 0(1(4(5(x1)))) -> 1(5(0(4(1(x1))))) 0(1(4(5(x1)))) -> 5(0(4(1(5(x1))))) 0(1(4(5(x1)))) -> 5(4(1(5(0(x1))))) 0(1(4(5(x1)))) -> 1(1(5(0(4(1(x1)))))) 0(1(4(5(x1)))) -> 5(4(1(5(5(0(x1)))))) 0(1(1(4(5(x1))))) -> 3(1(0(4(1(5(x1)))))) 0(1(2(2(2(x1))))) -> 1(0(2(2(5(2(x1)))))) 0(1(2(2(5(x1))))) -> 1(5(0(4(2(2(x1)))))) 0(1(2(4(5(x1))))) -> 2(5(1(0(4(5(x1)))))) 0(1(4(5(2(x1))))) -> 1(0(4(2(0(5(x1)))))) 0(1(4(5(5(x1))))) -> 5(0(4(0(1(5(x1)))))) 0(1(5(4(5(x1))))) -> 1(5(0(4(1(5(x1)))))) 0(5(1(2(2(x1))))) -> 0(1(3(2(5(2(x1)))))) 3(3(1(2(2(x1))))) -> 1(3(2(0(3(2(x1)))))) 3(4(4(0(5(x1))))) -> 3(5(4(5(0(4(x1)))))) ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: 5^1(0(1(2(2(x1))))) -> 0^1(5(2(x1))) 0^1(1(4(5(x1)))) -> 0^1(x1) 0^1(1(4(5(5(x1))))) -> 0^1(1(5(x1))) The TRS R consists of the following rules: 0(1(2(2(x1)))) -> 0(1(0(2(2(x1))))) 0(1(2(2(x1)))) -> 0(1(2(3(2(x1))))) 0(1(2(2(x1)))) -> 0(2(2(1(3(x1))))) 0(1(2(2(x1)))) -> 1(0(3(2(2(x1))))) 0(1(2(2(x1)))) -> 1(2(0(3(2(x1))))) 0(1(2(2(x1)))) -> 1(3(0(2(2(x1))))) 0(1(2(2(x1)))) -> 1(3(2(0(2(x1))))) 0(1(2(2(x1)))) -> 0(1(0(4(2(2(x1)))))) 0(1(2(2(x1)))) -> 0(2(1(3(2(3(x1)))))) 0(1(2(2(x1)))) -> 1(2(1(0(4(2(x1)))))) 0(1(2(2(x1)))) -> 1(5(0(4(2(2(x1)))))) 0(1(2(2(x1)))) -> 2(0(3(1(3(2(x1)))))) 0(1(2(2(x1)))) -> 2(1(1(0(4(2(x1)))))) 0(1(2(2(x1)))) -> 2(1(3(0(2(0(x1)))))) 0(1(2(2(x1)))) -> 2(1(3(3(2(0(x1)))))) 0(1(2(2(x1)))) -> 2(1(5(3(0(2(x1)))))) 0(1(2(2(x1)))) -> 2(2(1(3(0(5(x1)))))) 0(1(2(2(x1)))) -> 2(4(1(3(2(0(x1)))))) 0(1(4(5(x1)))) -> 1(5(0(4(1(x1))))) 0(1(4(5(x1)))) -> 5(0(4(1(5(x1))))) 0(1(4(5(x1)))) -> 5(4(1(5(0(x1))))) 0(1(4(5(x1)))) -> 1(1(5(0(4(1(x1)))))) 0(1(4(5(x1)))) -> 5(4(1(5(5(0(x1)))))) 5(1(2(2(x1)))) -> 1(0(2(2(5(x1))))) 5(1(2(2(x1)))) -> 1(3(5(2(2(x1))))) 5(1(2(2(x1)))) -> 1(5(2(3(2(x1))))) 5(1(2(2(x1)))) -> 1(5(0(2(2(3(x1)))))) 5(1(2(2(x1)))) -> 2(1(0(3(2(5(x1)))))) 5(1(2(2(x1)))) -> 3(1(3(5(2(2(x1)))))) 5(1(2(2(x1)))) -> 4(1(3(2(2(5(x1)))))) 5(1(2(2(x1)))) -> 5(1(0(4(2(2(x1)))))) 5(1(2(2(x1)))) -> 5(1(2(0(4(2(x1)))))) 0(1(1(4(5(x1))))) -> 3(1(0(4(1(5(x1)))))) 0(1(2(2(2(x1))))) -> 1(0(2(2(5(2(x1)))))) 0(1(2(2(5(x1))))) -> 1(5(0(4(2(2(x1)))))) 0(1(2(4(5(x1))))) -> 2(5(1(0(4(5(x1)))))) 0(1(4(5(2(x1))))) -> 1(0(4(2(0(5(x1)))))) 0(1(4(5(5(x1))))) -> 5(0(4(0(1(5(x1)))))) 0(1(5(4(5(x1))))) -> 1(5(0(4(1(5(x1)))))) 0(5(1(2(2(x1))))) -> 0(1(3(2(5(2(x1)))))) 3(3(1(2(2(x1))))) -> 1(3(2(0(3(2(x1)))))) 3(4(4(0(5(x1))))) -> 3(5(4(5(0(4(x1)))))) 5(0(1(2(2(x1))))) -> 1(3(2(0(5(2(x1)))))) 5(1(2(2(5(x1))))) -> 1(5(2(3(2(5(x1)))))) 5(2(1(2(2(x1))))) -> 2(1(3(5(2(2(x1)))))) 5(2(4(0(5(x1))))) -> 0(4(2(5(5(5(x1)))))) 5(2(4(0(5(x1))))) -> 0(4(5(4(2(5(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (7) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (8) Obligation: Q DP problem: The TRS P consists of the following rules: 0^1(1(4(5(5(x1))))) -> 0^1(1(5(x1))) 0^1(1(4(5(x1)))) -> 0^1(x1) The TRS R consists of the following rules: 0(1(2(2(x1)))) -> 0(1(0(2(2(x1))))) 0(1(2(2(x1)))) -> 0(1(2(3(2(x1))))) 0(1(2(2(x1)))) -> 0(2(2(1(3(x1))))) 0(1(2(2(x1)))) -> 1(0(3(2(2(x1))))) 0(1(2(2(x1)))) -> 1(2(0(3(2(x1))))) 0(1(2(2(x1)))) -> 1(3(0(2(2(x1))))) 0(1(2(2(x1)))) -> 1(3(2(0(2(x1))))) 0(1(2(2(x1)))) -> 0(1(0(4(2(2(x1)))))) 0(1(2(2(x1)))) -> 0(2(1(3(2(3(x1)))))) 0(1(2(2(x1)))) -> 1(2(1(0(4(2(x1)))))) 0(1(2(2(x1)))) -> 1(5(0(4(2(2(x1)))))) 0(1(2(2(x1)))) -> 2(0(3(1(3(2(x1)))))) 0(1(2(2(x1)))) -> 2(1(1(0(4(2(x1)))))) 0(1(2(2(x1)))) -> 2(1(3(0(2(0(x1)))))) 0(1(2(2(x1)))) -> 2(1(3(3(2(0(x1)))))) 0(1(2(2(x1)))) -> 2(1(5(3(0(2(x1)))))) 0(1(2(2(x1)))) -> 2(2(1(3(0(5(x1)))))) 0(1(2(2(x1)))) -> 2(4(1(3(2(0(x1)))))) 0(1(4(5(x1)))) -> 1(5(0(4(1(x1))))) 0(1(4(5(x1)))) -> 5(0(4(1(5(x1))))) 0(1(4(5(x1)))) -> 5(4(1(5(0(x1))))) 0(1(4(5(x1)))) -> 1(1(5(0(4(1(x1)))))) 0(1(4(5(x1)))) -> 5(4(1(5(5(0(x1)))))) 5(1(2(2(x1)))) -> 1(0(2(2(5(x1))))) 5(1(2(2(x1)))) -> 1(3(5(2(2(x1))))) 5(1(2(2(x1)))) -> 1(5(2(3(2(x1))))) 5(1(2(2(x1)))) -> 1(5(0(2(2(3(x1)))))) 5(1(2(2(x1)))) -> 2(1(0(3(2(5(x1)))))) 5(1(2(2(x1)))) -> 3(1(3(5(2(2(x1)))))) 5(1(2(2(x1)))) -> 4(1(3(2(2(5(x1)))))) 5(1(2(2(x1)))) -> 5(1(0(4(2(2(x1)))))) 5(1(2(2(x1)))) -> 5(1(2(0(4(2(x1)))))) 0(1(1(4(5(x1))))) -> 3(1(0(4(1(5(x1)))))) 0(1(2(2(2(x1))))) -> 1(0(2(2(5(2(x1)))))) 0(1(2(2(5(x1))))) -> 1(5(0(4(2(2(x1)))))) 0(1(2(4(5(x1))))) -> 2(5(1(0(4(5(x1)))))) 0(1(4(5(2(x1))))) -> 1(0(4(2(0(5(x1)))))) 0(1(4(5(5(x1))))) -> 5(0(4(0(1(5(x1)))))) 0(1(5(4(5(x1))))) -> 1(5(0(4(1(5(x1)))))) 0(5(1(2(2(x1))))) -> 0(1(3(2(5(2(x1)))))) 3(3(1(2(2(x1))))) -> 1(3(2(0(3(2(x1)))))) 3(4(4(0(5(x1))))) -> 3(5(4(5(0(4(x1)))))) 5(0(1(2(2(x1))))) -> 1(3(2(0(5(2(x1)))))) 5(1(2(2(5(x1))))) -> 1(5(2(3(2(5(x1)))))) 5(2(1(2(2(x1))))) -> 2(1(3(5(2(2(x1)))))) 5(2(4(0(5(x1))))) -> 0(4(2(5(5(5(x1)))))) 5(2(4(0(5(x1))))) -> 0(4(5(4(2(5(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. 0^1(1(4(5(5(x1))))) -> 0^1(1(5(x1))) 0^1(1(4(5(x1)))) -> 0^1(x1) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation: POL( 0^1_1(x_1) ) = max{0, x_1 - 2} POL( 1_1(x_1) ) = x_1 + 2 POL( 5_1(x_1) ) = 2x_1 POL( 2_1(x_1) ) = x_1 + 2 POL( 0_1(x_1) ) = 2 POL( 3_1(x_1) ) = 2 POL( 4_1(x_1) ) = x_1 + 2 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 5(1(2(2(x1)))) -> 1(0(2(2(5(x1))))) 5(1(2(2(x1)))) -> 1(3(5(2(2(x1))))) 5(1(2(2(x1)))) -> 1(5(2(3(2(x1))))) 5(1(2(2(x1)))) -> 1(5(0(2(2(3(x1)))))) 5(1(2(2(x1)))) -> 2(1(0(3(2(5(x1)))))) 5(1(2(2(x1)))) -> 3(1(3(5(2(2(x1)))))) 5(1(2(2(x1)))) -> 4(1(3(2(2(5(x1)))))) 5(1(2(2(x1)))) -> 5(1(0(4(2(2(x1)))))) 5(1(2(2(x1)))) -> 5(1(2(0(4(2(x1)))))) 5(0(1(2(2(x1))))) -> 1(3(2(0(5(2(x1)))))) 5(1(2(2(5(x1))))) -> 1(5(2(3(2(5(x1)))))) 5(2(1(2(2(x1))))) -> 2(1(3(5(2(2(x1)))))) 5(2(4(0(5(x1))))) -> 0(4(2(5(5(5(x1)))))) 5(2(4(0(5(x1))))) -> 0(4(5(4(2(5(x1)))))) ---------------------------------------- (10) Obligation: Q DP problem: P is empty. The TRS R consists of the following rules: 0(1(2(2(x1)))) -> 0(1(0(2(2(x1))))) 0(1(2(2(x1)))) -> 0(1(2(3(2(x1))))) 0(1(2(2(x1)))) -> 0(2(2(1(3(x1))))) 0(1(2(2(x1)))) -> 1(0(3(2(2(x1))))) 0(1(2(2(x1)))) -> 1(2(0(3(2(x1))))) 0(1(2(2(x1)))) -> 1(3(0(2(2(x1))))) 0(1(2(2(x1)))) -> 1(3(2(0(2(x1))))) 0(1(2(2(x1)))) -> 0(1(0(4(2(2(x1)))))) 0(1(2(2(x1)))) -> 0(2(1(3(2(3(x1)))))) 0(1(2(2(x1)))) -> 1(2(1(0(4(2(x1)))))) 0(1(2(2(x1)))) -> 1(5(0(4(2(2(x1)))))) 0(1(2(2(x1)))) -> 2(0(3(1(3(2(x1)))))) 0(1(2(2(x1)))) -> 2(1(1(0(4(2(x1)))))) 0(1(2(2(x1)))) -> 2(1(3(0(2(0(x1)))))) 0(1(2(2(x1)))) -> 2(1(3(3(2(0(x1)))))) 0(1(2(2(x1)))) -> 2(1(5(3(0(2(x1)))))) 0(1(2(2(x1)))) -> 2(2(1(3(0(5(x1)))))) 0(1(2(2(x1)))) -> 2(4(1(3(2(0(x1)))))) 0(1(4(5(x1)))) -> 1(5(0(4(1(x1))))) 0(1(4(5(x1)))) -> 5(0(4(1(5(x1))))) 0(1(4(5(x1)))) -> 5(4(1(5(0(x1))))) 0(1(4(5(x1)))) -> 1(1(5(0(4(1(x1)))))) 0(1(4(5(x1)))) -> 5(4(1(5(5(0(x1)))))) 5(1(2(2(x1)))) -> 1(0(2(2(5(x1))))) 5(1(2(2(x1)))) -> 1(3(5(2(2(x1))))) 5(1(2(2(x1)))) -> 1(5(2(3(2(x1))))) 5(1(2(2(x1)))) -> 1(5(0(2(2(3(x1)))))) 5(1(2(2(x1)))) -> 2(1(0(3(2(5(x1)))))) 5(1(2(2(x1)))) -> 3(1(3(5(2(2(x1)))))) 5(1(2(2(x1)))) -> 4(1(3(2(2(5(x1)))))) 5(1(2(2(x1)))) -> 5(1(0(4(2(2(x1)))))) 5(1(2(2(x1)))) -> 5(1(2(0(4(2(x1)))))) 0(1(1(4(5(x1))))) -> 3(1(0(4(1(5(x1)))))) 0(1(2(2(2(x1))))) -> 1(0(2(2(5(2(x1)))))) 0(1(2(2(5(x1))))) -> 1(5(0(4(2(2(x1)))))) 0(1(2(4(5(x1))))) -> 2(5(1(0(4(5(x1)))))) 0(1(4(5(2(x1))))) -> 1(0(4(2(0(5(x1)))))) 0(1(4(5(5(x1))))) -> 5(0(4(0(1(5(x1)))))) 0(1(5(4(5(x1))))) -> 1(5(0(4(1(5(x1)))))) 0(5(1(2(2(x1))))) -> 0(1(3(2(5(2(x1)))))) 3(3(1(2(2(x1))))) -> 1(3(2(0(3(2(x1)))))) 3(4(4(0(5(x1))))) -> 3(5(4(5(0(4(x1)))))) 5(0(1(2(2(x1))))) -> 1(3(2(0(5(2(x1)))))) 5(1(2(2(5(x1))))) -> 1(5(2(3(2(5(x1)))))) 5(2(1(2(2(x1))))) -> 2(1(3(5(2(2(x1)))))) 5(2(4(0(5(x1))))) -> 0(4(2(5(5(5(x1)))))) 5(2(4(0(5(x1))))) -> 0(4(5(4(2(5(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (11) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (12) YES