13.66/4.41 YES 13.66/4.47 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 13.66/4.47 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 13.66/4.47 13.66/4.47 13.66/4.47 Termination w.r.t. Q of the given QTRS could be proven: 13.66/4.47 13.66/4.47 (0) QTRS 13.66/4.47 (1) QTRS Reverse [EQUIVALENT, 0 ms] 13.66/4.47 (2) QTRS 13.66/4.47 (3) DependencyPairsProof [EQUIVALENT, 139 ms] 13.66/4.47 (4) QDP 13.66/4.47 (5) DependencyGraphProof [EQUIVALENT, 0 ms] 13.66/4.47 (6) QDP 13.66/4.47 (7) UsableRulesProof [EQUIVALENT, 4 ms] 13.66/4.47 (8) QDP 13.66/4.47 (9) QDPSizeChangeProof [EQUIVALENT, 0 ms] 13.66/4.47 (10) YES 13.66/4.47 13.66/4.47 13.66/4.47 ---------------------------------------- 13.66/4.47 13.66/4.47 (0) 13.66/4.47 Obligation: 13.66/4.47 Q restricted rewrite system: 13.66/4.47 The TRS R consists of the following rules: 13.66/4.47 13.66/4.47 0(0(x1)) -> 0(1(0(2(x1)))) 13.66/4.47 0(0(x1)) -> 1(0(2(0(x1)))) 13.66/4.47 0(0(x1)) -> 1(0(1(0(1(x1))))) 13.66/4.47 0(0(x1)) -> 1(0(1(2(0(x1))))) 13.66/4.47 0(0(x1)) -> 1(0(2(0(3(x1))))) 13.66/4.47 0(0(x1)) -> 1(0(2(2(0(x1))))) 13.66/4.47 0(0(x1)) -> 2(1(0(2(0(x1))))) 13.66/4.47 0(0(x1)) -> 0(1(0(2(1(2(x1)))))) 13.66/4.47 0(0(x1)) -> 1(0(1(0(2(2(x1)))))) 13.66/4.47 0(0(x1)) -> 1(0(1(3(0(1(x1)))))) 13.66/4.47 0(0(x1)) -> 1(0(4(1(0(2(x1)))))) 13.66/4.47 0(0(x1)) -> 1(1(1(0(2(0(x1)))))) 13.66/4.47 0(0(x1)) -> 3(0(4(0(2(2(x1)))))) 13.66/4.47 0(0(x1)) -> 3(1(0(1(0(4(x1)))))) 13.66/4.47 0(0(0(x1))) -> 0(1(0(4(0(4(x1)))))) 13.66/4.47 0(0(0(x1))) -> 3(0(0(1(0(2(x1)))))) 13.66/4.47 3(0(0(x1))) -> 3(0(2(0(3(x1))))) 13.66/4.47 3(0(0(x1))) -> 3(0(2(4(0(2(x1)))))) 13.66/4.47 5(2(0(x1))) -> 0(2(3(5(x1)))) 13.66/4.47 5(2(0(x1))) -> 3(5(0(2(x1)))) 13.66/4.47 5(2(0(x1))) -> 0(2(3(3(5(x1))))) 13.66/4.47 5(2(0(x1))) -> 1(0(2(3(5(x1))))) 13.66/4.47 5(2(0(x1))) -> 5(1(0(2(4(x1))))) 13.66/4.47 5(2(0(x1))) -> 5(0(1(2(2(2(x1)))))) 13.66/4.47 5(2(0(x1))) -> 5(3(5(1(0(2(x1)))))) 13.66/4.47 0(5(2(0(x1)))) -> 0(5(0(2(2(x1))))) 13.66/4.47 3(4(0(0(x1)))) -> 0(3(3(0(4(5(x1)))))) 13.66/4.47 3(4(0(0(x1)))) -> 3(0(4(5(3(0(x1)))))) 13.66/4.47 5(1(0(0(x1)))) -> 0(3(1(0(1(5(x1)))))) 13.66/4.47 5(1(4(0(x1)))) -> 0(1(5(2(4(x1))))) 13.66/4.47 5(1(5(0(x1)))) -> 5(1(0(3(5(x1))))) 13.66/4.47 5(2(2(0(x1)))) -> 0(2(1(2(4(5(x1)))))) 13.66/4.47 5(3(2(0(x1)))) -> 5(3(0(1(2(x1))))) 13.66/4.47 5(3(2(0(x1)))) -> 3(3(5(3(0(2(x1)))))) 13.66/4.47 5(4(0(0(x1)))) -> 0(4(5(5(0(2(x1)))))) 13.66/4.47 5(4(2(0(x1)))) -> 2(4(3(5(0(x1))))) 13.66/4.47 5(4(2(0(x1)))) -> 5(0(2(2(4(x1))))) 13.66/4.47 5(4(2(0(x1)))) -> 5(4(5(0(2(x1))))) 13.66/4.47 0(3(5(2(0(x1))))) -> 3(0(2(5(3(0(x1)))))) 13.66/4.47 3(3(5(2(0(x1))))) -> 3(5(2(3(0(2(x1)))))) 13.66/4.47 3(3(5(2(0(x1))))) -> 4(3(3(5(0(2(x1)))))) 13.66/4.47 5(0(5(2(0(x1))))) -> 0(3(5(5(0(2(x1)))))) 13.66/4.47 5(1(4(0(0(x1))))) -> 0(2(5(0(1(4(x1)))))) 13.66/4.47 5(3(3(2(0(x1))))) -> 5(2(3(3(0(2(x1)))))) 13.66/4.47 5(4(3(0(0(x1))))) -> 1(0(4(0(3(5(x1)))))) 13.66/4.47 13.66/4.47 Q is empty. 13.66/4.47 13.66/4.47 ---------------------------------------- 13.66/4.47 13.66/4.47 (1) QTRS Reverse (EQUIVALENT) 13.66/4.47 We applied the QTRS Reverse Processor [REVERSE]. 13.66/4.47 ---------------------------------------- 13.66/4.47 13.66/4.47 (2) 13.66/4.47 Obligation: 13.66/4.47 Q restricted rewrite system: 13.66/4.47 The TRS R consists of the following rules: 13.66/4.47 13.66/4.47 0(0(x1)) -> 2(0(1(0(x1)))) 13.66/4.47 0(0(x1)) -> 0(2(0(1(x1)))) 13.66/4.47 0(0(x1)) -> 1(0(1(0(1(x1))))) 13.66/4.47 0(0(x1)) -> 0(2(1(0(1(x1))))) 13.66/4.47 0(0(x1)) -> 3(0(2(0(1(x1))))) 13.66/4.47 0(0(x1)) -> 0(2(2(0(1(x1))))) 13.66/4.47 0(0(x1)) -> 0(2(0(1(2(x1))))) 13.66/4.47 0(0(x1)) -> 2(1(2(0(1(0(x1)))))) 13.66/4.47 0(0(x1)) -> 2(2(0(1(0(1(x1)))))) 13.66/4.47 0(0(x1)) -> 1(0(3(1(0(1(x1)))))) 13.66/4.47 0(0(x1)) -> 2(0(1(4(0(1(x1)))))) 13.66/4.47 0(0(x1)) -> 0(2(0(1(1(1(x1)))))) 13.66/4.47 0(0(x1)) -> 2(2(0(4(0(3(x1)))))) 13.66/4.47 0(0(x1)) -> 4(0(1(0(1(3(x1)))))) 13.66/4.47 0(0(0(x1))) -> 4(0(4(0(1(0(x1)))))) 13.66/4.47 0(0(0(x1))) -> 2(0(1(0(0(3(x1)))))) 13.66/4.47 0(0(3(x1))) -> 3(0(2(0(3(x1))))) 13.66/4.47 0(0(3(x1))) -> 2(0(4(2(0(3(x1)))))) 13.66/4.47 0(2(5(x1))) -> 5(3(2(0(x1)))) 13.66/4.47 0(2(5(x1))) -> 2(0(5(3(x1)))) 13.66/4.47 0(2(5(x1))) -> 5(3(3(2(0(x1))))) 13.66/4.47 0(2(5(x1))) -> 5(3(2(0(1(x1))))) 13.66/4.47 0(2(5(x1))) -> 4(2(0(1(5(x1))))) 13.66/4.47 0(2(5(x1))) -> 2(2(2(1(0(5(x1)))))) 13.66/4.47 0(2(5(x1))) -> 2(0(1(5(3(5(x1)))))) 13.66/4.47 0(2(5(0(x1)))) -> 2(2(0(5(0(x1))))) 13.66/4.47 0(0(4(3(x1)))) -> 5(4(0(3(3(0(x1)))))) 13.66/4.47 0(0(4(3(x1)))) -> 0(3(5(4(0(3(x1)))))) 13.66/4.47 0(0(1(5(x1)))) -> 5(1(0(1(3(0(x1)))))) 13.66/4.47 0(4(1(5(x1)))) -> 4(2(5(1(0(x1))))) 13.66/4.47 0(5(1(5(x1)))) -> 5(3(0(1(5(x1))))) 13.66/4.47 0(2(2(5(x1)))) -> 5(4(2(1(2(0(x1)))))) 13.66/4.47 0(2(3(5(x1)))) -> 2(1(0(3(5(x1))))) 13.66/4.47 0(2(3(5(x1)))) -> 2(0(3(5(3(3(x1)))))) 13.66/4.47 0(0(4(5(x1)))) -> 2(0(5(5(4(0(x1)))))) 13.66/4.47 0(2(4(5(x1)))) -> 0(5(3(4(2(x1))))) 13.66/4.47 0(2(4(5(x1)))) -> 4(2(2(0(5(x1))))) 13.66/4.47 0(2(4(5(x1)))) -> 2(0(5(4(5(x1))))) 13.66/4.47 0(2(5(3(0(x1))))) -> 0(3(5(2(0(3(x1)))))) 13.66/4.47 0(2(5(3(3(x1))))) -> 2(0(3(2(5(3(x1)))))) 13.66/4.47 0(2(5(3(3(x1))))) -> 2(0(5(3(3(4(x1)))))) 13.66/4.47 0(2(5(0(5(x1))))) -> 2(0(5(5(3(0(x1)))))) 13.66/4.47 0(0(4(1(5(x1))))) -> 4(1(0(5(2(0(x1)))))) 13.66/4.47 0(2(3(3(5(x1))))) -> 2(0(3(3(2(5(x1)))))) 13.66/4.47 0(0(3(4(5(x1))))) -> 5(3(0(4(0(1(x1)))))) 13.66/4.47 13.66/4.47 Q is empty. 13.66/4.47 13.66/4.47 ---------------------------------------- 13.66/4.47 13.66/4.47 (3) DependencyPairsProof (EQUIVALENT) 13.66/4.47 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 13.66/4.47 ---------------------------------------- 13.66/4.47 13.66/4.47 (4) 13.66/4.47 Obligation: 13.66/4.47 Q DP problem: 13.66/4.47 The TRS P consists of the following rules: 13.66/4.47 13.66/4.47 0^1(0(x1)) -> 0^1(1(0(x1))) 13.66/4.47 0^1(0(x1)) -> 0^1(2(0(1(x1)))) 13.66/4.47 0^1(0(x1)) -> 0^1(1(x1)) 13.66/4.47 0^1(0(x1)) -> 0^1(1(0(1(x1)))) 13.66/4.47 0^1(0(x1)) -> 0^1(2(1(0(1(x1))))) 13.66/4.47 0^1(0(x1)) -> 0^1(2(2(0(1(x1))))) 13.66/4.47 0^1(0(x1)) -> 0^1(2(0(1(2(x1))))) 13.66/4.47 0^1(0(x1)) -> 0^1(1(2(x1))) 13.66/4.47 0^1(0(x1)) -> 0^1(3(1(0(1(x1))))) 13.66/4.47 0^1(0(x1)) -> 0^1(1(4(0(1(x1))))) 13.66/4.47 0^1(0(x1)) -> 0^1(2(0(1(1(1(x1)))))) 13.66/4.47 0^1(0(x1)) -> 0^1(1(1(1(x1)))) 13.66/4.47 0^1(0(x1)) -> 0^1(4(0(3(x1)))) 13.66/4.47 0^1(0(x1)) -> 0^1(3(x1)) 13.66/4.47 0^1(0(x1)) -> 0^1(1(0(1(3(x1))))) 13.66/4.47 0^1(0(x1)) -> 0^1(1(3(x1))) 13.66/4.47 0^1(0(0(x1))) -> 0^1(4(0(1(0(x1))))) 13.66/4.47 0^1(0(0(x1))) -> 0^1(1(0(x1))) 13.66/4.47 0^1(0(0(x1))) -> 0^1(1(0(0(3(x1))))) 13.66/4.47 0^1(0(0(x1))) -> 0^1(0(3(x1))) 13.66/4.47 0^1(0(0(x1))) -> 0^1(3(x1)) 13.66/4.47 0^1(0(3(x1))) -> 0^1(2(0(3(x1)))) 13.66/4.47 0^1(0(3(x1))) -> 0^1(4(2(0(3(x1))))) 13.66/4.47 0^1(2(5(x1))) -> 0^1(x1) 13.66/4.47 0^1(2(5(x1))) -> 0^1(5(3(x1))) 13.66/4.47 0^1(2(5(x1))) -> 0^1(1(x1)) 13.66/4.47 0^1(2(5(x1))) -> 0^1(1(5(x1))) 13.66/4.47 0^1(2(5(x1))) -> 0^1(5(x1)) 13.66/4.47 0^1(2(5(x1))) -> 0^1(1(5(3(5(x1))))) 13.66/4.47 0^1(2(5(0(x1)))) -> 0^1(5(0(x1))) 13.66/4.47 0^1(0(4(3(x1)))) -> 0^1(3(3(0(x1)))) 13.66/4.47 0^1(0(4(3(x1)))) -> 0^1(x1) 13.66/4.47 0^1(0(4(3(x1)))) -> 0^1(3(5(4(0(3(x1)))))) 13.66/4.47 0^1(0(4(3(x1)))) -> 0^1(3(x1)) 13.66/4.47 0^1(0(1(5(x1)))) -> 0^1(1(3(0(x1)))) 13.66/4.47 0^1(0(1(5(x1)))) -> 0^1(x1) 13.66/4.47 0^1(4(1(5(x1)))) -> 0^1(x1) 13.66/4.47 0^1(5(1(5(x1)))) -> 0^1(1(5(x1))) 13.66/4.47 0^1(2(2(5(x1)))) -> 0^1(x1) 13.66/4.47 0^1(2(3(5(x1)))) -> 0^1(3(5(x1))) 13.66/4.47 0^1(2(3(5(x1)))) -> 0^1(3(5(3(3(x1))))) 13.66/4.47 0^1(0(4(5(x1)))) -> 0^1(5(5(4(0(x1))))) 13.66/4.47 0^1(0(4(5(x1)))) -> 0^1(x1) 13.66/4.47 0^1(2(4(5(x1)))) -> 0^1(5(3(4(2(x1))))) 13.66/4.47 0^1(2(4(5(x1)))) -> 0^1(5(x1)) 13.66/4.47 0^1(2(4(5(x1)))) -> 0^1(5(4(5(x1)))) 13.66/4.47 0^1(2(5(3(0(x1))))) -> 0^1(3(5(2(0(3(x1)))))) 13.66/4.47 0^1(2(5(3(0(x1))))) -> 0^1(3(x1)) 13.66/4.47 0^1(2(5(3(3(x1))))) -> 0^1(3(2(5(3(x1))))) 13.66/4.47 0^1(2(5(3(3(x1))))) -> 0^1(5(3(3(4(x1))))) 13.66/4.47 0^1(2(5(0(5(x1))))) -> 0^1(5(5(3(0(x1))))) 13.66/4.47 0^1(2(5(0(5(x1))))) -> 0^1(x1) 13.66/4.47 0^1(0(4(1(5(x1))))) -> 0^1(5(2(0(x1)))) 13.66/4.47 0^1(0(4(1(5(x1))))) -> 0^1(x1) 13.66/4.47 0^1(2(3(3(5(x1))))) -> 0^1(3(3(2(5(x1))))) 13.66/4.47 0^1(0(3(4(5(x1))))) -> 0^1(4(0(1(x1)))) 13.66/4.47 0^1(0(3(4(5(x1))))) -> 0^1(1(x1)) 13.66/4.47 13.66/4.47 The TRS R consists of the following rules: 13.66/4.47 13.66/4.47 0(0(x1)) -> 2(0(1(0(x1)))) 13.66/4.47 0(0(x1)) -> 0(2(0(1(x1)))) 13.66/4.47 0(0(x1)) -> 1(0(1(0(1(x1))))) 13.66/4.47 0(0(x1)) -> 0(2(1(0(1(x1))))) 13.66/4.47 0(0(x1)) -> 3(0(2(0(1(x1))))) 13.66/4.47 0(0(x1)) -> 0(2(2(0(1(x1))))) 13.66/4.47 0(0(x1)) -> 0(2(0(1(2(x1))))) 13.66/4.47 0(0(x1)) -> 2(1(2(0(1(0(x1)))))) 13.66/4.47 0(0(x1)) -> 2(2(0(1(0(1(x1)))))) 13.66/4.47 0(0(x1)) -> 1(0(3(1(0(1(x1)))))) 13.66/4.47 0(0(x1)) -> 2(0(1(4(0(1(x1)))))) 13.66/4.47 0(0(x1)) -> 0(2(0(1(1(1(x1)))))) 13.66/4.47 0(0(x1)) -> 2(2(0(4(0(3(x1)))))) 13.66/4.47 0(0(x1)) -> 4(0(1(0(1(3(x1)))))) 13.66/4.47 0(0(0(x1))) -> 4(0(4(0(1(0(x1)))))) 13.66/4.47 0(0(0(x1))) -> 2(0(1(0(0(3(x1)))))) 13.66/4.47 0(0(3(x1))) -> 3(0(2(0(3(x1))))) 13.66/4.47 0(0(3(x1))) -> 2(0(4(2(0(3(x1)))))) 13.66/4.47 0(2(5(x1))) -> 5(3(2(0(x1)))) 13.66/4.47 0(2(5(x1))) -> 2(0(5(3(x1)))) 13.66/4.47 0(2(5(x1))) -> 5(3(3(2(0(x1))))) 13.66/4.47 0(2(5(x1))) -> 5(3(2(0(1(x1))))) 13.66/4.47 0(2(5(x1))) -> 4(2(0(1(5(x1))))) 13.66/4.47 0(2(5(x1))) -> 2(2(2(1(0(5(x1)))))) 13.66/4.47 0(2(5(x1))) -> 2(0(1(5(3(5(x1)))))) 13.66/4.47 0(2(5(0(x1)))) -> 2(2(0(5(0(x1))))) 13.66/4.47 0(0(4(3(x1)))) -> 5(4(0(3(3(0(x1)))))) 13.66/4.47 0(0(4(3(x1)))) -> 0(3(5(4(0(3(x1)))))) 13.66/4.47 0(0(1(5(x1)))) -> 5(1(0(1(3(0(x1)))))) 13.66/4.47 0(4(1(5(x1)))) -> 4(2(5(1(0(x1))))) 13.66/4.47 0(5(1(5(x1)))) -> 5(3(0(1(5(x1))))) 13.66/4.47 0(2(2(5(x1)))) -> 5(4(2(1(2(0(x1)))))) 13.66/4.47 0(2(3(5(x1)))) -> 2(1(0(3(5(x1))))) 13.66/4.47 0(2(3(5(x1)))) -> 2(0(3(5(3(3(x1)))))) 13.66/4.47 0(0(4(5(x1)))) -> 2(0(5(5(4(0(x1)))))) 13.66/4.47 0(2(4(5(x1)))) -> 0(5(3(4(2(x1))))) 13.66/4.47 0(2(4(5(x1)))) -> 4(2(2(0(5(x1))))) 13.66/4.47 0(2(4(5(x1)))) -> 2(0(5(4(5(x1))))) 13.66/4.47 0(2(5(3(0(x1))))) -> 0(3(5(2(0(3(x1)))))) 13.66/4.47 0(2(5(3(3(x1))))) -> 2(0(3(2(5(3(x1)))))) 13.66/4.47 0(2(5(3(3(x1))))) -> 2(0(5(3(3(4(x1)))))) 13.66/4.47 0(2(5(0(5(x1))))) -> 2(0(5(5(3(0(x1)))))) 13.66/4.47 0(0(4(1(5(x1))))) -> 4(1(0(5(2(0(x1)))))) 13.66/4.47 0(2(3(3(5(x1))))) -> 2(0(3(3(2(5(x1)))))) 13.66/4.47 0(0(3(4(5(x1))))) -> 5(3(0(4(0(1(x1)))))) 13.66/4.47 13.66/4.47 Q is empty. 13.66/4.47 We have to consider all minimal (P,Q,R)-chains. 13.66/4.47 ---------------------------------------- 13.66/4.47 13.66/4.47 (5) DependencyGraphProof (EQUIVALENT) 13.66/4.47 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 49 less nodes. 13.66/4.47 ---------------------------------------- 13.66/4.47 13.66/4.47 (6) 13.66/4.47 Obligation: 13.66/4.47 Q DP problem: 13.66/4.47 The TRS P consists of the following rules: 13.66/4.47 13.66/4.47 0^1(0(4(3(x1)))) -> 0^1(x1) 13.66/4.47 0^1(2(5(x1))) -> 0^1(x1) 13.66/4.47 0^1(0(1(5(x1)))) -> 0^1(x1) 13.66/4.47 0^1(4(1(5(x1)))) -> 0^1(x1) 13.66/4.47 0^1(2(2(5(x1)))) -> 0^1(x1) 13.66/4.47 0^1(0(4(5(x1)))) -> 0^1(x1) 13.66/4.47 0^1(2(5(0(5(x1))))) -> 0^1(x1) 13.66/4.47 0^1(0(4(1(5(x1))))) -> 0^1(x1) 13.66/4.47 13.66/4.47 The TRS R consists of the following rules: 13.66/4.47 13.66/4.47 0(0(x1)) -> 2(0(1(0(x1)))) 13.66/4.47 0(0(x1)) -> 0(2(0(1(x1)))) 13.66/4.47 0(0(x1)) -> 1(0(1(0(1(x1))))) 13.66/4.47 0(0(x1)) -> 0(2(1(0(1(x1))))) 13.66/4.47 0(0(x1)) -> 3(0(2(0(1(x1))))) 13.66/4.47 0(0(x1)) -> 0(2(2(0(1(x1))))) 13.66/4.47 0(0(x1)) -> 0(2(0(1(2(x1))))) 13.66/4.47 0(0(x1)) -> 2(1(2(0(1(0(x1)))))) 13.66/4.47 0(0(x1)) -> 2(2(0(1(0(1(x1)))))) 13.66/4.47 0(0(x1)) -> 1(0(3(1(0(1(x1)))))) 13.66/4.47 0(0(x1)) -> 2(0(1(4(0(1(x1)))))) 13.66/4.47 0(0(x1)) -> 0(2(0(1(1(1(x1)))))) 13.66/4.47 0(0(x1)) -> 2(2(0(4(0(3(x1)))))) 13.66/4.47 0(0(x1)) -> 4(0(1(0(1(3(x1)))))) 13.66/4.47 0(0(0(x1))) -> 4(0(4(0(1(0(x1)))))) 13.66/4.47 0(0(0(x1))) -> 2(0(1(0(0(3(x1)))))) 13.66/4.47 0(0(3(x1))) -> 3(0(2(0(3(x1))))) 13.66/4.47 0(0(3(x1))) -> 2(0(4(2(0(3(x1)))))) 13.66/4.47 0(2(5(x1))) -> 5(3(2(0(x1)))) 13.66/4.47 0(2(5(x1))) -> 2(0(5(3(x1)))) 13.66/4.47 0(2(5(x1))) -> 5(3(3(2(0(x1))))) 13.66/4.47 0(2(5(x1))) -> 5(3(2(0(1(x1))))) 13.66/4.47 0(2(5(x1))) -> 4(2(0(1(5(x1))))) 13.66/4.47 0(2(5(x1))) -> 2(2(2(1(0(5(x1)))))) 13.66/4.47 0(2(5(x1))) -> 2(0(1(5(3(5(x1)))))) 13.66/4.47 0(2(5(0(x1)))) -> 2(2(0(5(0(x1))))) 13.66/4.47 0(0(4(3(x1)))) -> 5(4(0(3(3(0(x1)))))) 13.66/4.47 0(0(4(3(x1)))) -> 0(3(5(4(0(3(x1)))))) 13.66/4.47 0(0(1(5(x1)))) -> 5(1(0(1(3(0(x1)))))) 13.66/4.47 0(4(1(5(x1)))) -> 4(2(5(1(0(x1))))) 13.66/4.47 0(5(1(5(x1)))) -> 5(3(0(1(5(x1))))) 13.66/4.47 0(2(2(5(x1)))) -> 5(4(2(1(2(0(x1)))))) 13.66/4.47 0(2(3(5(x1)))) -> 2(1(0(3(5(x1))))) 13.66/4.47 0(2(3(5(x1)))) -> 2(0(3(5(3(3(x1)))))) 13.66/4.47 0(0(4(5(x1)))) -> 2(0(5(5(4(0(x1)))))) 13.66/4.47 0(2(4(5(x1)))) -> 0(5(3(4(2(x1))))) 13.66/4.47 0(2(4(5(x1)))) -> 4(2(2(0(5(x1))))) 13.66/4.47 0(2(4(5(x1)))) -> 2(0(5(4(5(x1))))) 13.66/4.47 0(2(5(3(0(x1))))) -> 0(3(5(2(0(3(x1)))))) 13.66/4.47 0(2(5(3(3(x1))))) -> 2(0(3(2(5(3(x1)))))) 13.66/4.47 0(2(5(3(3(x1))))) -> 2(0(5(3(3(4(x1)))))) 13.66/4.47 0(2(5(0(5(x1))))) -> 2(0(5(5(3(0(x1)))))) 13.66/4.47 0(0(4(1(5(x1))))) -> 4(1(0(5(2(0(x1)))))) 13.66/4.47 0(2(3(3(5(x1))))) -> 2(0(3(3(2(5(x1)))))) 13.66/4.47 0(0(3(4(5(x1))))) -> 5(3(0(4(0(1(x1)))))) 13.66/4.47 13.66/4.47 Q is empty. 13.66/4.47 We have to consider all minimal (P,Q,R)-chains. 13.66/4.47 ---------------------------------------- 13.66/4.47 13.66/4.47 (7) UsableRulesProof (EQUIVALENT) 13.66/4.47 We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. 13.66/4.47 ---------------------------------------- 13.66/4.47 13.66/4.47 (8) 13.66/4.47 Obligation: 13.66/4.47 Q DP problem: 13.66/4.47 The TRS P consists of the following rules: 13.66/4.47 13.66/4.47 0^1(0(4(3(x1)))) -> 0^1(x1) 13.66/4.47 0^1(2(5(x1))) -> 0^1(x1) 13.66/4.47 0^1(0(1(5(x1)))) -> 0^1(x1) 13.66/4.47 0^1(4(1(5(x1)))) -> 0^1(x1) 13.66/4.47 0^1(2(2(5(x1)))) -> 0^1(x1) 13.66/4.47 0^1(0(4(5(x1)))) -> 0^1(x1) 13.66/4.47 0^1(2(5(0(5(x1))))) -> 0^1(x1) 13.66/4.47 0^1(0(4(1(5(x1))))) -> 0^1(x1) 13.66/4.47 13.66/4.47 R is empty. 13.66/4.47 Q is empty. 13.66/4.47 We have to consider all minimal (P,Q,R)-chains. 13.66/4.47 ---------------------------------------- 13.66/4.47 13.66/4.47 (9) QDPSizeChangeProof (EQUIVALENT) 13.66/4.47 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. 13.66/4.47 13.66/4.47 From the DPs we obtained the following set of size-change graphs: 13.66/4.47 *0^1(0(4(3(x1)))) -> 0^1(x1) 13.66/4.47 The graph contains the following edges 1 > 1 13.66/4.47 13.66/4.47 13.66/4.47 *0^1(2(5(x1))) -> 0^1(x1) 13.66/4.47 The graph contains the following edges 1 > 1 13.66/4.47 13.66/4.47 13.66/4.47 *0^1(0(1(5(x1)))) -> 0^1(x1) 13.66/4.47 The graph contains the following edges 1 > 1 13.66/4.47 13.66/4.47 13.66/4.47 *0^1(4(1(5(x1)))) -> 0^1(x1) 13.66/4.47 The graph contains the following edges 1 > 1 13.66/4.47 13.66/4.47 13.66/4.47 *0^1(2(2(5(x1)))) -> 0^1(x1) 13.66/4.47 The graph contains the following edges 1 > 1 13.66/4.47 13.66/4.47 13.66/4.47 *0^1(0(4(5(x1)))) -> 0^1(x1) 13.66/4.47 The graph contains the following edges 1 > 1 13.66/4.47 13.66/4.47 13.66/4.47 *0^1(2(5(0(5(x1))))) -> 0^1(x1) 13.66/4.47 The graph contains the following edges 1 > 1 13.66/4.47 13.66/4.47 13.66/4.47 *0^1(0(4(1(5(x1))))) -> 0^1(x1) 13.66/4.47 The graph contains the following edges 1 > 1 13.66/4.47 13.66/4.47 13.66/4.47 ---------------------------------------- 13.66/4.47 13.66/4.47 (10) 13.66/4.47 YES 14.00/4.56 EOF