22.80/6.68 YES 22.80/6.70 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 22.80/6.70 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 22.80/6.70 22.80/6.70 22.80/6.70 Termination w.r.t. Q of the given QTRS could be proven: 22.80/6.70 22.80/6.70 (0) QTRS 22.80/6.70 (1) DependencyPairsProof [EQUIVALENT, 108 ms] 22.80/6.70 (2) QDP 22.80/6.70 (3) DependencyGraphProof [EQUIVALENT, 7 ms] 22.80/6.70 (4) QDP 22.80/6.70 (5) QDPOrderProof [EQUIVALENT, 189 ms] 22.80/6.70 (6) QDP 22.80/6.70 (7) DependencyGraphProof [EQUIVALENT, 0 ms] 22.80/6.70 (8) TRUE 22.80/6.70 22.80/6.70 22.80/6.70 ---------------------------------------- 22.80/6.70 22.80/6.70 (0) 22.80/6.70 Obligation: 22.80/6.70 Q restricted rewrite system: 22.80/6.70 The TRS R consists of the following rules: 22.80/6.70 22.80/6.70 0(0(1(x1))) -> 0(1(2(0(x1)))) 22.80/6.70 0(0(1(x1))) -> 0(3(1(0(x1)))) 22.80/6.70 0(0(1(x1))) -> 1(0(4(0(x1)))) 22.80/6.70 0(0(1(x1))) -> 0(1(3(0(2(x1))))) 22.80/6.70 0(0(1(x1))) -> 0(1(3(0(4(x1))))) 22.80/6.70 0(0(1(x1))) -> 0(2(0(1(2(x1))))) 22.80/6.70 0(0(1(x1))) -> 0(3(0(1(2(x1))))) 22.80/6.70 0(0(1(x1))) -> 0(3(0(3(1(x1))))) 22.80/6.70 0(0(1(x1))) -> 0(4(0(4(1(x1))))) 22.80/6.70 0(0(1(x1))) -> 1(2(0(2(0(x1))))) 22.80/6.70 0(0(1(x1))) -> 1(2(2(0(0(x1))))) 22.80/6.70 0(0(1(x1))) -> 0(0(2(2(1(2(x1)))))) 22.80/6.70 0(0(1(x1))) -> 0(1(2(4(2(0(x1)))))) 22.80/6.70 0(0(1(x1))) -> 1(2(0(3(0(4(x1)))))) 22.80/6.70 0(1(1(x1))) -> 0(2(1(1(x1)))) 22.80/6.70 0(1(1(x1))) -> 0(3(1(1(x1)))) 22.80/6.70 0(1(1(x1))) -> 1(1(3(0(4(x1))))) 22.80/6.70 0(1(1(x1))) -> 1(2(0(2(1(x1))))) 22.80/6.70 0(1(1(x1))) -> 1(0(3(1(2(4(x1)))))) 22.80/6.70 0(1(1(x1))) -> 1(0(4(2(1(2(x1)))))) 22.80/6.70 0(1(1(x1))) -> 1(1(2(4(3(0(x1)))))) 22.80/6.70 0(1(1(x1))) -> 1(2(1(0(4(4(x1)))))) 22.80/6.70 0(1(1(x1))) -> 1(2(2(1(3(0(x1)))))) 22.80/6.70 0(5(1(x1))) -> 0(3(1(5(x1)))) 22.80/6.70 0(5(1(x1))) -> 0(4(5(1(x1)))) 22.80/6.70 0(5(1(x1))) -> 0(2(3(1(5(x1))))) 22.80/6.70 0(5(1(x1))) -> 0(3(1(5(2(x1))))) 22.80/6.70 0(5(1(x1))) -> 0(3(1(2(5(2(x1)))))) 22.80/6.70 5(0(1(x1))) -> 5(1(2(4(0(x1))))) 22.80/6.70 5(0(1(x1))) -> 5(0(2(1(2(4(x1)))))) 22.80/6.70 5(0(1(x1))) -> 5(1(2(3(0(4(x1)))))) 22.80/6.70 0(0(1(5(x1)))) -> 0(4(1(0(5(x1))))) 22.80/6.70 0(0(2(1(x1)))) -> 2(0(3(0(2(1(x1)))))) 22.80/6.70 0(0(2(1(x1)))) -> 2(3(0(2(0(1(x1)))))) 22.80/6.70 0(1(0(1(x1)))) -> 1(0(2(0(1(x1))))) 22.80/6.70 0(1(1(1(x1)))) -> 1(1(3(1(0(x1))))) 22.80/6.70 5(0(1(1(x1)))) -> 1(5(1(2(0(x1))))) 22.80/6.70 5(3(0(1(x1)))) -> 5(1(2(3(0(x1))))) 22.80/6.70 5(3(1(5(x1)))) -> 5(3(1(2(5(x1))))) 22.80/6.70 5(3(2(1(x1)))) -> 1(2(3(5(2(x1))))) 22.80/6.70 5(4(0(1(x1)))) -> 1(2(5(0(4(x1))))) 22.80/6.70 0(0(5(1(5(x1))))) -> 1(2(5(5(0(0(x1)))))) 22.80/6.70 0(5(3(0(1(x1))))) -> 1(0(5(3(0(4(x1)))))) 22.80/6.70 0(5(3(4(1(x1))))) -> 1(0(3(5(4(5(x1)))))) 22.80/6.70 0(5(4(0(1(x1))))) -> 0(1(3(0(4(5(x1)))))) 22.80/6.70 5(4(2(1(1(x1))))) -> 5(4(1(2(1(2(x1)))))) 22.80/6.70 22.80/6.70 Q is empty. 22.80/6.70 22.80/6.70 ---------------------------------------- 22.80/6.70 22.80/6.70 (1) DependencyPairsProof (EQUIVALENT) 22.80/6.70 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 22.80/6.70 ---------------------------------------- 22.80/6.70 22.80/6.70 (2) 22.80/6.70 Obligation: 22.80/6.70 Q DP problem: 22.80/6.70 The TRS P consists of the following rules: 22.80/6.70 22.80/6.70 0^1(0(1(x1))) -> 0^1(1(2(0(x1)))) 22.80/6.70 0^1(0(1(x1))) -> 0^1(x1) 22.80/6.70 0^1(0(1(x1))) -> 0^1(3(1(0(x1)))) 22.80/6.70 0^1(0(1(x1))) -> 0^1(4(0(x1))) 22.80/6.70 0^1(0(1(x1))) -> 0^1(1(3(0(2(x1))))) 22.80/6.70 0^1(0(1(x1))) -> 0^1(2(x1)) 22.80/6.70 0^1(0(1(x1))) -> 0^1(1(3(0(4(x1))))) 22.80/6.70 0^1(0(1(x1))) -> 0^1(4(x1)) 22.80/6.70 0^1(0(1(x1))) -> 0^1(2(0(1(2(x1))))) 22.80/6.70 0^1(0(1(x1))) -> 0^1(1(2(x1))) 22.80/6.70 0^1(0(1(x1))) -> 0^1(3(0(1(2(x1))))) 22.80/6.70 0^1(0(1(x1))) -> 0^1(3(0(3(1(x1))))) 22.80/6.70 0^1(0(1(x1))) -> 0^1(3(1(x1))) 22.80/6.70 0^1(0(1(x1))) -> 0^1(4(0(4(1(x1))))) 22.80/6.70 0^1(0(1(x1))) -> 0^1(4(1(x1))) 22.80/6.70 0^1(0(1(x1))) -> 0^1(2(0(x1))) 22.80/6.70 0^1(0(1(x1))) -> 0^1(0(x1)) 22.80/6.70 0^1(0(1(x1))) -> 0^1(0(2(2(1(2(x1)))))) 22.80/6.70 0^1(0(1(x1))) -> 0^1(2(2(1(2(x1))))) 22.80/6.70 0^1(0(1(x1))) -> 0^1(1(2(4(2(0(x1)))))) 22.80/6.70 0^1(0(1(x1))) -> 0^1(3(0(4(x1)))) 22.80/6.70 0^1(1(1(x1))) -> 0^1(2(1(1(x1)))) 22.80/6.70 0^1(1(1(x1))) -> 0^1(3(1(1(x1)))) 22.80/6.70 0^1(1(1(x1))) -> 0^1(4(x1)) 22.80/6.70 0^1(1(1(x1))) -> 0^1(2(1(x1))) 22.80/6.70 0^1(1(1(x1))) -> 0^1(3(1(2(4(x1))))) 22.80/6.70 0^1(1(1(x1))) -> 0^1(4(2(1(2(x1))))) 22.80/6.70 0^1(1(1(x1))) -> 0^1(x1) 22.80/6.70 0^1(1(1(x1))) -> 0^1(4(4(x1))) 22.80/6.70 0^1(5(1(x1))) -> 0^1(3(1(5(x1)))) 22.80/6.70 0^1(5(1(x1))) -> 5^1(x1) 22.80/6.70 0^1(5(1(x1))) -> 0^1(4(5(1(x1)))) 22.80/6.70 0^1(5(1(x1))) -> 0^1(2(3(1(5(x1))))) 22.80/6.70 0^1(5(1(x1))) -> 0^1(3(1(5(2(x1))))) 22.80/6.70 0^1(5(1(x1))) -> 5^1(2(x1)) 22.80/6.70 0^1(5(1(x1))) -> 0^1(3(1(2(5(2(x1)))))) 22.80/6.70 5^1(0(1(x1))) -> 5^1(1(2(4(0(x1))))) 22.80/6.70 5^1(0(1(x1))) -> 0^1(x1) 22.80/6.70 5^1(0(1(x1))) -> 5^1(0(2(1(2(4(x1)))))) 22.80/6.70 5^1(0(1(x1))) -> 0^1(2(1(2(4(x1))))) 22.80/6.70 5^1(0(1(x1))) -> 5^1(1(2(3(0(4(x1)))))) 22.80/6.70 5^1(0(1(x1))) -> 0^1(4(x1)) 22.80/6.70 0^1(0(1(5(x1)))) -> 0^1(4(1(0(5(x1))))) 22.80/6.70 0^1(0(1(5(x1)))) -> 0^1(5(x1)) 22.80/6.70 0^1(0(2(1(x1)))) -> 0^1(3(0(2(1(x1))))) 22.80/6.70 0^1(0(2(1(x1)))) -> 0^1(2(0(1(x1)))) 22.80/6.70 0^1(0(2(1(x1)))) -> 0^1(1(x1)) 22.80/6.70 0^1(1(0(1(x1)))) -> 0^1(2(0(1(x1)))) 22.80/6.70 0^1(1(1(1(x1)))) -> 0^1(x1) 22.80/6.70 5^1(0(1(1(x1)))) -> 5^1(1(2(0(x1)))) 22.80/6.70 5^1(0(1(1(x1)))) -> 0^1(x1) 22.80/6.70 5^1(3(0(1(x1)))) -> 5^1(1(2(3(0(x1))))) 22.80/6.70 5^1(3(0(1(x1)))) -> 0^1(x1) 22.80/6.70 5^1(3(1(5(x1)))) -> 5^1(3(1(2(5(x1))))) 22.80/6.70 5^1(3(2(1(x1)))) -> 5^1(2(x1)) 22.80/6.70 5^1(4(0(1(x1)))) -> 5^1(0(4(x1))) 22.80/6.70 5^1(4(0(1(x1)))) -> 0^1(4(x1)) 22.80/6.70 0^1(0(5(1(5(x1))))) -> 5^1(5(0(0(x1)))) 22.80/6.70 0^1(0(5(1(5(x1))))) -> 5^1(0(0(x1))) 22.80/6.70 0^1(0(5(1(5(x1))))) -> 0^1(0(x1)) 22.80/6.70 0^1(0(5(1(5(x1))))) -> 0^1(x1) 22.80/6.70 0^1(5(3(0(1(x1))))) -> 0^1(5(3(0(4(x1))))) 22.80/6.70 0^1(5(3(0(1(x1))))) -> 5^1(3(0(4(x1)))) 22.80/6.70 0^1(5(3(0(1(x1))))) -> 0^1(4(x1)) 22.80/6.70 0^1(5(3(4(1(x1))))) -> 0^1(3(5(4(5(x1))))) 22.80/6.70 0^1(5(3(4(1(x1))))) -> 5^1(4(5(x1))) 22.80/6.70 0^1(5(3(4(1(x1))))) -> 5^1(x1) 22.80/6.70 0^1(5(4(0(1(x1))))) -> 0^1(1(3(0(4(5(x1)))))) 22.80/6.70 0^1(5(4(0(1(x1))))) -> 0^1(4(5(x1))) 22.80/6.70 0^1(5(4(0(1(x1))))) -> 5^1(x1) 22.80/6.70 5^1(4(2(1(1(x1))))) -> 5^1(4(1(2(1(2(x1)))))) 22.80/6.70 22.80/6.70 The TRS R consists of the following rules: 22.80/6.70 22.80/6.70 0(0(1(x1))) -> 0(1(2(0(x1)))) 22.80/6.70 0(0(1(x1))) -> 0(3(1(0(x1)))) 22.80/6.70 0(0(1(x1))) -> 1(0(4(0(x1)))) 22.80/6.70 0(0(1(x1))) -> 0(1(3(0(2(x1))))) 22.80/6.70 0(0(1(x1))) -> 0(1(3(0(4(x1))))) 22.80/6.70 0(0(1(x1))) -> 0(2(0(1(2(x1))))) 22.80/6.70 0(0(1(x1))) -> 0(3(0(1(2(x1))))) 22.80/6.70 0(0(1(x1))) -> 0(3(0(3(1(x1))))) 22.80/6.70 0(0(1(x1))) -> 0(4(0(4(1(x1))))) 22.80/6.70 0(0(1(x1))) -> 1(2(0(2(0(x1))))) 22.80/6.70 0(0(1(x1))) -> 1(2(2(0(0(x1))))) 22.80/6.70 0(0(1(x1))) -> 0(0(2(2(1(2(x1)))))) 22.80/6.70 0(0(1(x1))) -> 0(1(2(4(2(0(x1)))))) 22.80/6.70 0(0(1(x1))) -> 1(2(0(3(0(4(x1)))))) 22.80/6.70 0(1(1(x1))) -> 0(2(1(1(x1)))) 22.80/6.70 0(1(1(x1))) -> 0(3(1(1(x1)))) 22.80/6.70 0(1(1(x1))) -> 1(1(3(0(4(x1))))) 22.80/6.70 0(1(1(x1))) -> 1(2(0(2(1(x1))))) 22.80/6.70 0(1(1(x1))) -> 1(0(3(1(2(4(x1)))))) 22.80/6.70 0(1(1(x1))) -> 1(0(4(2(1(2(x1)))))) 22.80/6.70 0(1(1(x1))) -> 1(1(2(4(3(0(x1)))))) 22.80/6.70 0(1(1(x1))) -> 1(2(1(0(4(4(x1)))))) 22.80/6.70 0(1(1(x1))) -> 1(2(2(1(3(0(x1)))))) 22.80/6.70 0(5(1(x1))) -> 0(3(1(5(x1)))) 22.80/6.70 0(5(1(x1))) -> 0(4(5(1(x1)))) 22.80/6.70 0(5(1(x1))) -> 0(2(3(1(5(x1))))) 22.80/6.70 0(5(1(x1))) -> 0(3(1(5(2(x1))))) 22.80/6.70 0(5(1(x1))) -> 0(3(1(2(5(2(x1)))))) 22.80/6.70 5(0(1(x1))) -> 5(1(2(4(0(x1))))) 22.80/6.70 5(0(1(x1))) -> 5(0(2(1(2(4(x1)))))) 22.80/6.70 5(0(1(x1))) -> 5(1(2(3(0(4(x1)))))) 22.80/6.70 0(0(1(5(x1)))) -> 0(4(1(0(5(x1))))) 22.80/6.70 0(0(2(1(x1)))) -> 2(0(3(0(2(1(x1)))))) 22.80/6.70 0(0(2(1(x1)))) -> 2(3(0(2(0(1(x1)))))) 22.80/6.70 0(1(0(1(x1)))) -> 1(0(2(0(1(x1))))) 22.80/6.70 0(1(1(1(x1)))) -> 1(1(3(1(0(x1))))) 22.80/6.70 5(0(1(1(x1)))) -> 1(5(1(2(0(x1))))) 22.80/6.70 5(3(0(1(x1)))) -> 5(1(2(3(0(x1))))) 22.80/6.70 5(3(1(5(x1)))) -> 5(3(1(2(5(x1))))) 22.80/6.70 5(3(2(1(x1)))) -> 1(2(3(5(2(x1))))) 22.80/6.70 5(4(0(1(x1)))) -> 1(2(5(0(4(x1))))) 22.80/6.70 0(0(5(1(5(x1))))) -> 1(2(5(5(0(0(x1)))))) 22.80/6.70 0(5(3(0(1(x1))))) -> 1(0(5(3(0(4(x1)))))) 22.80/6.70 0(5(3(4(1(x1))))) -> 1(0(3(5(4(5(x1)))))) 22.80/6.70 0(5(4(0(1(x1))))) -> 0(1(3(0(4(5(x1)))))) 22.80/6.70 5(4(2(1(1(x1))))) -> 5(4(1(2(1(2(x1)))))) 22.80/6.70 22.80/6.70 Q is empty. 22.80/6.70 We have to consider all minimal (P,Q,R)-chains. 22.80/6.70 ---------------------------------------- 22.80/6.70 22.80/6.70 (3) DependencyGraphProof (EQUIVALENT) 22.80/6.70 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 55 less nodes. 22.80/6.70 ---------------------------------------- 22.80/6.70 22.80/6.70 (4) 22.80/6.70 Obligation: 22.80/6.70 Q DP problem: 22.80/6.70 The TRS P consists of the following rules: 22.80/6.70 22.80/6.70 0^1(0(1(x1))) -> 0^1(0(x1)) 22.80/6.70 0^1(0(1(x1))) -> 0^1(x1) 22.80/6.70 0^1(1(1(x1))) -> 0^1(x1) 22.80/6.70 0^1(5(1(x1))) -> 5^1(x1) 22.80/6.70 5^1(0(1(x1))) -> 0^1(x1) 22.80/6.70 0^1(0(1(5(x1)))) -> 0^1(5(x1)) 22.80/6.70 0^1(0(2(1(x1)))) -> 0^1(1(x1)) 22.80/6.70 0^1(1(1(1(x1)))) -> 0^1(x1) 22.80/6.70 0^1(0(5(1(5(x1))))) -> 5^1(5(0(0(x1)))) 22.80/6.70 5^1(0(1(1(x1)))) -> 0^1(x1) 22.80/6.70 0^1(0(5(1(5(x1))))) -> 5^1(0(0(x1))) 22.80/6.70 0^1(0(5(1(5(x1))))) -> 0^1(0(x1)) 22.80/6.70 0^1(0(5(1(5(x1))))) -> 0^1(x1) 22.80/6.70 0^1(5(3(4(1(x1))))) -> 5^1(x1) 22.80/6.70 5^1(3(0(1(x1)))) -> 0^1(x1) 22.80/6.70 0^1(5(4(0(1(x1))))) -> 5^1(x1) 22.80/6.70 22.80/6.70 The TRS R consists of the following rules: 22.80/6.70 22.80/6.70 0(0(1(x1))) -> 0(1(2(0(x1)))) 22.80/6.70 0(0(1(x1))) -> 0(3(1(0(x1)))) 22.80/6.70 0(0(1(x1))) -> 1(0(4(0(x1)))) 22.80/6.70 0(0(1(x1))) -> 0(1(3(0(2(x1))))) 22.80/6.70 0(0(1(x1))) -> 0(1(3(0(4(x1))))) 22.80/6.70 0(0(1(x1))) -> 0(2(0(1(2(x1))))) 22.80/6.70 0(0(1(x1))) -> 0(3(0(1(2(x1))))) 22.80/6.70 0(0(1(x1))) -> 0(3(0(3(1(x1))))) 22.80/6.70 0(0(1(x1))) -> 0(4(0(4(1(x1))))) 22.80/6.70 0(0(1(x1))) -> 1(2(0(2(0(x1))))) 22.80/6.70 0(0(1(x1))) -> 1(2(2(0(0(x1))))) 22.80/6.70 0(0(1(x1))) -> 0(0(2(2(1(2(x1)))))) 22.80/6.70 0(0(1(x1))) -> 0(1(2(4(2(0(x1)))))) 22.80/6.70 0(0(1(x1))) -> 1(2(0(3(0(4(x1)))))) 22.80/6.70 0(1(1(x1))) -> 0(2(1(1(x1)))) 22.80/6.70 0(1(1(x1))) -> 0(3(1(1(x1)))) 22.80/6.70 0(1(1(x1))) -> 1(1(3(0(4(x1))))) 22.80/6.70 0(1(1(x1))) -> 1(2(0(2(1(x1))))) 22.80/6.70 0(1(1(x1))) -> 1(0(3(1(2(4(x1)))))) 22.80/6.70 0(1(1(x1))) -> 1(0(4(2(1(2(x1)))))) 22.80/6.70 0(1(1(x1))) -> 1(1(2(4(3(0(x1)))))) 22.80/6.70 0(1(1(x1))) -> 1(2(1(0(4(4(x1)))))) 22.80/6.70 0(1(1(x1))) -> 1(2(2(1(3(0(x1)))))) 22.80/6.70 0(5(1(x1))) -> 0(3(1(5(x1)))) 22.80/6.70 0(5(1(x1))) -> 0(4(5(1(x1)))) 22.80/6.70 0(5(1(x1))) -> 0(2(3(1(5(x1))))) 22.80/6.70 0(5(1(x1))) -> 0(3(1(5(2(x1))))) 22.80/6.70 0(5(1(x1))) -> 0(3(1(2(5(2(x1)))))) 22.80/6.70 5(0(1(x1))) -> 5(1(2(4(0(x1))))) 22.80/6.70 5(0(1(x1))) -> 5(0(2(1(2(4(x1)))))) 22.80/6.70 5(0(1(x1))) -> 5(1(2(3(0(4(x1)))))) 22.80/6.70 0(0(1(5(x1)))) -> 0(4(1(0(5(x1))))) 22.80/6.70 0(0(2(1(x1)))) -> 2(0(3(0(2(1(x1)))))) 22.80/6.70 0(0(2(1(x1)))) -> 2(3(0(2(0(1(x1)))))) 22.80/6.70 0(1(0(1(x1)))) -> 1(0(2(0(1(x1))))) 22.80/6.70 0(1(1(1(x1)))) -> 1(1(3(1(0(x1))))) 22.80/6.70 5(0(1(1(x1)))) -> 1(5(1(2(0(x1))))) 22.80/6.70 5(3(0(1(x1)))) -> 5(1(2(3(0(x1))))) 22.80/6.70 5(3(1(5(x1)))) -> 5(3(1(2(5(x1))))) 22.80/6.70 5(3(2(1(x1)))) -> 1(2(3(5(2(x1))))) 22.80/6.70 5(4(0(1(x1)))) -> 1(2(5(0(4(x1))))) 22.80/6.70 0(0(5(1(5(x1))))) -> 1(2(5(5(0(0(x1)))))) 22.80/6.70 0(5(3(0(1(x1))))) -> 1(0(5(3(0(4(x1)))))) 22.80/6.70 0(5(3(4(1(x1))))) -> 1(0(3(5(4(5(x1)))))) 22.80/6.70 0(5(4(0(1(x1))))) -> 0(1(3(0(4(5(x1)))))) 22.80/6.70 5(4(2(1(1(x1))))) -> 5(4(1(2(1(2(x1)))))) 22.80/6.70 22.80/6.70 Q is empty. 22.80/6.70 We have to consider all minimal (P,Q,R)-chains. 22.80/6.70 ---------------------------------------- 22.80/6.70 22.80/6.70 (5) QDPOrderProof (EQUIVALENT) 22.80/6.70 We use the reduction pair processor [LPAR04,JAR06]. 22.80/6.70 22.80/6.70 22.80/6.70 The following pairs can be oriented strictly and are deleted. 22.80/6.70 22.80/6.70 0^1(0(1(x1))) -> 0^1(0(x1)) 22.80/6.70 0^1(0(1(x1))) -> 0^1(x1) 22.80/6.70 0^1(1(1(x1))) -> 0^1(x1) 22.80/6.70 0^1(5(1(x1))) -> 5^1(x1) 22.80/6.70 5^1(0(1(x1))) -> 0^1(x1) 22.80/6.70 0^1(0(1(5(x1)))) -> 0^1(5(x1)) 22.80/6.70 0^1(1(1(1(x1)))) -> 0^1(x1) 22.80/6.70 0^1(0(5(1(5(x1))))) -> 5^1(5(0(0(x1)))) 22.80/6.70 5^1(0(1(1(x1)))) -> 0^1(x1) 22.80/6.70 0^1(0(5(1(5(x1))))) -> 5^1(0(0(x1))) 22.80/6.70 0^1(0(5(1(5(x1))))) -> 0^1(0(x1)) 22.80/6.70 0^1(0(5(1(5(x1))))) -> 0^1(x1) 22.80/6.70 0^1(5(3(4(1(x1))))) -> 5^1(x1) 22.80/6.70 5^1(3(0(1(x1)))) -> 0^1(x1) 22.80/6.70 0^1(5(4(0(1(x1))))) -> 5^1(x1) 22.80/6.70 The remaining pairs can at least be oriented weakly. 22.80/6.70 Used ordering: Polynomial interpretation [POLO]: 22.80/6.70 22.80/6.70 POL(0(x_1)) = x_1 22.80/6.70 POL(0^1(x_1)) = x_1 22.80/6.70 POL(1(x_1)) = 1 + x_1 22.80/6.70 POL(2(x_1)) = x_1 22.80/6.70 POL(3(x_1)) = x_1 22.80/6.70 POL(4(x_1)) = x_1 22.80/6.70 POL(5(x_1)) = x_1 22.80/6.70 POL(5^1(x_1)) = x_1 22.80/6.70 22.80/6.70 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 22.80/6.70 22.80/6.70 0(0(1(x1))) -> 0(1(2(0(x1)))) 22.80/6.70 0(0(1(x1))) -> 0(3(1(0(x1)))) 22.80/6.70 0(0(1(x1))) -> 1(0(4(0(x1)))) 22.80/6.70 0(0(1(x1))) -> 0(1(3(0(2(x1))))) 22.80/6.70 0(0(1(x1))) -> 0(1(3(0(4(x1))))) 22.80/6.70 0(0(1(x1))) -> 0(2(0(1(2(x1))))) 22.80/6.70 0(0(1(x1))) -> 0(3(0(1(2(x1))))) 22.80/6.70 0(0(1(x1))) -> 0(3(0(3(1(x1))))) 22.80/6.70 0(0(1(x1))) -> 0(4(0(4(1(x1))))) 22.80/6.70 0(0(1(x1))) -> 1(2(0(2(0(x1))))) 22.80/6.70 0(0(1(x1))) -> 1(2(2(0(0(x1))))) 22.80/6.70 0(0(1(x1))) -> 0(0(2(2(1(2(x1)))))) 22.80/6.70 0(0(1(x1))) -> 0(1(2(4(2(0(x1)))))) 22.80/6.70 0(0(1(x1))) -> 1(2(0(3(0(4(x1)))))) 22.80/6.70 0(1(1(x1))) -> 0(2(1(1(x1)))) 22.80/6.70 0(1(1(x1))) -> 0(3(1(1(x1)))) 22.80/6.70 0(1(1(x1))) -> 1(1(3(0(4(x1))))) 22.80/6.70 0(1(1(x1))) -> 1(2(0(2(1(x1))))) 22.80/6.70 0(1(1(x1))) -> 1(0(3(1(2(4(x1)))))) 22.80/6.70 0(1(1(x1))) -> 1(0(4(2(1(2(x1)))))) 22.80/6.70 0(1(1(x1))) -> 1(1(2(4(3(0(x1)))))) 22.80/6.70 0(1(1(x1))) -> 1(2(1(0(4(4(x1)))))) 22.80/6.70 0(1(1(x1))) -> 1(2(2(1(3(0(x1)))))) 22.80/6.70 0(5(1(x1))) -> 0(3(1(5(x1)))) 22.80/6.70 0(5(1(x1))) -> 0(4(5(1(x1)))) 22.80/6.70 0(5(1(x1))) -> 0(2(3(1(5(x1))))) 22.80/6.70 0(5(1(x1))) -> 0(3(1(5(2(x1))))) 22.80/6.70 0(5(1(x1))) -> 0(3(1(2(5(2(x1)))))) 22.80/6.70 0(0(1(5(x1)))) -> 0(4(1(0(5(x1))))) 22.80/6.70 0(0(2(1(x1)))) -> 2(0(3(0(2(1(x1)))))) 22.80/6.70 0(0(2(1(x1)))) -> 2(3(0(2(0(1(x1)))))) 22.80/6.70 0(1(0(1(x1)))) -> 1(0(2(0(1(x1))))) 22.80/6.70 0(1(1(1(x1)))) -> 1(1(3(1(0(x1))))) 22.80/6.70 0(0(5(1(5(x1))))) -> 1(2(5(5(0(0(x1)))))) 22.80/6.70 0(5(3(0(1(x1))))) -> 1(0(5(3(0(4(x1)))))) 22.80/6.70 0(5(3(4(1(x1))))) -> 1(0(3(5(4(5(x1)))))) 22.80/6.70 0(5(4(0(1(x1))))) -> 0(1(3(0(4(5(x1)))))) 22.80/6.70 5(0(1(x1))) -> 5(1(2(4(0(x1))))) 22.80/6.70 5(0(1(x1))) -> 5(0(2(1(2(4(x1)))))) 22.80/6.70 5(0(1(x1))) -> 5(1(2(3(0(4(x1)))))) 22.80/6.70 5(0(1(1(x1)))) -> 1(5(1(2(0(x1))))) 22.80/6.70 5(3(0(1(x1)))) -> 5(1(2(3(0(x1))))) 22.80/6.70 5(3(1(5(x1)))) -> 5(3(1(2(5(x1))))) 22.80/6.70 5(3(2(1(x1)))) -> 1(2(3(5(2(x1))))) 22.80/6.70 5(4(0(1(x1)))) -> 1(2(5(0(4(x1))))) 22.80/6.70 5(4(2(1(1(x1))))) -> 5(4(1(2(1(2(x1)))))) 22.80/6.70 22.80/6.70 22.80/6.70 ---------------------------------------- 22.80/6.70 22.80/6.70 (6) 22.80/6.70 Obligation: 22.80/6.70 Q DP problem: 22.80/6.70 The TRS P consists of the following rules: 22.80/6.70 22.80/6.70 0^1(0(2(1(x1)))) -> 0^1(1(x1)) 22.80/6.70 22.80/6.70 The TRS R consists of the following rules: 22.80/6.70 22.80/6.70 0(0(1(x1))) -> 0(1(2(0(x1)))) 22.80/6.70 0(0(1(x1))) -> 0(3(1(0(x1)))) 22.80/6.70 0(0(1(x1))) -> 1(0(4(0(x1)))) 22.80/6.70 0(0(1(x1))) -> 0(1(3(0(2(x1))))) 22.80/6.70 0(0(1(x1))) -> 0(1(3(0(4(x1))))) 22.80/6.70 0(0(1(x1))) -> 0(2(0(1(2(x1))))) 22.80/6.70 0(0(1(x1))) -> 0(3(0(1(2(x1))))) 22.80/6.70 0(0(1(x1))) -> 0(3(0(3(1(x1))))) 22.80/6.70 0(0(1(x1))) -> 0(4(0(4(1(x1))))) 22.80/6.70 0(0(1(x1))) -> 1(2(0(2(0(x1))))) 22.80/6.70 0(0(1(x1))) -> 1(2(2(0(0(x1))))) 22.80/6.70 0(0(1(x1))) -> 0(0(2(2(1(2(x1)))))) 22.80/6.70 0(0(1(x1))) -> 0(1(2(4(2(0(x1)))))) 22.80/6.70 0(0(1(x1))) -> 1(2(0(3(0(4(x1)))))) 22.80/6.70 0(1(1(x1))) -> 0(2(1(1(x1)))) 22.80/6.70 0(1(1(x1))) -> 0(3(1(1(x1)))) 22.80/6.70 0(1(1(x1))) -> 1(1(3(0(4(x1))))) 22.80/6.70 0(1(1(x1))) -> 1(2(0(2(1(x1))))) 22.80/6.70 0(1(1(x1))) -> 1(0(3(1(2(4(x1)))))) 22.80/6.70 0(1(1(x1))) -> 1(0(4(2(1(2(x1)))))) 22.80/6.70 0(1(1(x1))) -> 1(1(2(4(3(0(x1)))))) 22.80/6.70 0(1(1(x1))) -> 1(2(1(0(4(4(x1)))))) 22.80/6.70 0(1(1(x1))) -> 1(2(2(1(3(0(x1)))))) 22.80/6.70 0(5(1(x1))) -> 0(3(1(5(x1)))) 22.80/6.70 0(5(1(x1))) -> 0(4(5(1(x1)))) 22.80/6.70 0(5(1(x1))) -> 0(2(3(1(5(x1))))) 22.80/6.70 0(5(1(x1))) -> 0(3(1(5(2(x1))))) 22.80/6.70 0(5(1(x1))) -> 0(3(1(2(5(2(x1)))))) 22.80/6.70 5(0(1(x1))) -> 5(1(2(4(0(x1))))) 22.80/6.70 5(0(1(x1))) -> 5(0(2(1(2(4(x1)))))) 22.80/6.70 5(0(1(x1))) -> 5(1(2(3(0(4(x1)))))) 22.80/6.70 0(0(1(5(x1)))) -> 0(4(1(0(5(x1))))) 22.80/6.70 0(0(2(1(x1)))) -> 2(0(3(0(2(1(x1)))))) 22.80/6.70 0(0(2(1(x1)))) -> 2(3(0(2(0(1(x1)))))) 22.80/6.70 0(1(0(1(x1)))) -> 1(0(2(0(1(x1))))) 22.80/6.70 0(1(1(1(x1)))) -> 1(1(3(1(0(x1))))) 22.80/6.70 5(0(1(1(x1)))) -> 1(5(1(2(0(x1))))) 22.80/6.70 5(3(0(1(x1)))) -> 5(1(2(3(0(x1))))) 22.80/6.70 5(3(1(5(x1)))) -> 5(3(1(2(5(x1))))) 22.80/6.70 5(3(2(1(x1)))) -> 1(2(3(5(2(x1))))) 22.80/6.70 5(4(0(1(x1)))) -> 1(2(5(0(4(x1))))) 22.80/6.70 0(0(5(1(5(x1))))) -> 1(2(5(5(0(0(x1)))))) 22.80/6.70 0(5(3(0(1(x1))))) -> 1(0(5(3(0(4(x1)))))) 22.80/6.70 0(5(3(4(1(x1))))) -> 1(0(3(5(4(5(x1)))))) 22.80/6.70 0(5(4(0(1(x1))))) -> 0(1(3(0(4(5(x1)))))) 22.80/6.70 5(4(2(1(1(x1))))) -> 5(4(1(2(1(2(x1)))))) 22.80/6.70 22.80/6.70 Q is empty. 22.80/6.70 We have to consider all minimal (P,Q,R)-chains. 22.80/6.70 ---------------------------------------- 22.80/6.70 22.80/6.70 (7) DependencyGraphProof (EQUIVALENT) 22.80/6.70 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. 22.80/6.70 ---------------------------------------- 22.80/6.70 22.80/6.70 (8) 22.80/6.70 TRUE 23.16/6.79 EOF