32.27/9.54 YES 34.51/10.06 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 34.51/10.06 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 34.51/10.06 34.51/10.06 34.51/10.06 Termination w.r.t. Q of the given QTRS could be proven: 34.51/10.06 34.51/10.06 (0) QTRS 34.51/10.06 (1) QTRS Reverse [EQUIVALENT, 0 ms] 34.51/10.06 (2) QTRS 34.51/10.06 (3) DependencyPairsProof [EQUIVALENT, 318 ms] 34.51/10.06 (4) QDP 34.51/10.06 (5) DependencyGraphProof [EQUIVALENT, 10 ms] 34.51/10.06 (6) AND 34.51/10.06 (7) QDP 34.51/10.06 (8) UsableRulesProof [EQUIVALENT, 0 ms] 34.51/10.06 (9) QDP 34.51/10.06 (10) QDPSizeChangeProof [EQUIVALENT, 0 ms] 34.51/10.06 (11) YES 34.51/10.06 (12) QDP 34.51/10.06 (13) QDPOrderProof [EQUIVALENT, 227 ms] 34.51/10.06 (14) QDP 34.51/10.06 (15) PisEmptyProof [EQUIVALENT, 0 ms] 34.51/10.06 (16) YES 34.51/10.06 (17) QDP 34.51/10.06 (18) UsableRulesProof [EQUIVALENT, 0 ms] 34.51/10.06 (19) QDP 34.51/10.06 (20) QDPSizeChangeProof [EQUIVALENT, 0 ms] 34.51/10.06 (21) YES 34.51/10.06 34.51/10.06 34.51/10.06 ---------------------------------------- 34.51/10.06 34.51/10.06 (0) 34.51/10.06 Obligation: 34.51/10.06 Q restricted rewrite system: 34.51/10.06 The TRS R consists of the following rules: 34.51/10.06 34.51/10.06 0(1(2(x1))) -> 2(3(0(3(1(x1))))) 34.51/10.06 0(0(0(2(x1)))) -> 3(0(0(3(0(2(x1)))))) 34.51/10.06 0(0(2(2(x1)))) -> 2(3(0(0(3(2(x1)))))) 34.51/10.06 0(1(1(2(x1)))) -> 2(3(0(3(1(1(x1)))))) 34.51/10.06 0(1(2(1(x1)))) -> 0(1(2(4(1(3(x1)))))) 34.51/10.06 0(1(2(2(x1)))) -> 2(3(0(3(1(2(x1)))))) 34.51/10.06 0(1(2(5(x1)))) -> 2(0(5(3(1(x1))))) 34.51/10.06 0(1(5(1(x1)))) -> 1(0(3(5(3(1(x1)))))) 34.51/10.06 0(1(5(2(x1)))) -> 2(4(1(0(3(5(x1)))))) 34.51/10.06 0(1(5(2(x1)))) -> 2(4(3(1(0(5(x1)))))) 34.51/10.06 0(1(5(2(x1)))) -> 3(1(3(0(5(2(x1)))))) 34.51/10.06 0(1(5(5(x1)))) -> 1(3(0(5(5(x1))))) 34.51/10.06 0(2(1(2(x1)))) -> 2(2(3(0(3(1(x1)))))) 34.51/10.06 0(2(5(2(x1)))) -> 2(2(3(0(5(x1))))) 34.51/10.06 1(1(4(5(x1)))) -> 5(4(1(3(1(x1))))) 34.51/10.06 1(5(1(5(x1)))) -> 1(3(5(5(3(1(x1)))))) 34.51/10.06 1(5(5(1(x1)))) -> 1(5(3(1(5(3(x1)))))) 34.51/10.06 2(0(1(2(x1)))) -> 2(2(3(0(3(1(x1)))))) 34.51/10.06 2(0(1(5(x1)))) -> 2(1(3(0(5(x1))))) 34.51/10.06 5(0(1(2(x1)))) -> 3(0(5(3(1(2(x1)))))) 34.51/10.06 5(0(1(2(x1)))) -> 4(2(3(0(5(1(x1)))))) 34.51/10.06 0(0(0(0(1(x1))))) -> 0(0(1(0(0(3(x1)))))) 34.51/10.06 0(0(1(2(5(x1))))) -> 2(0(0(5(3(1(x1)))))) 34.51/10.06 0(0(1(5(2(x1))))) -> 0(1(0(3(5(2(x1)))))) 34.51/10.06 0(1(0(4(5(x1))))) -> 5(4(0(0(3(1(x1)))))) 34.51/10.06 0(1(1(1(2(x1))))) -> 1(0(3(1(1(2(x1)))))) 34.51/10.06 0(1(2(1(5(x1))))) -> 2(1(0(5(3(1(x1)))))) 34.51/10.06 0(1(3(5(2(x1))))) -> 3(0(4(1(5(2(x1)))))) 34.51/10.06 0(1(4(2(5(x1))))) -> 2(4(3(0(5(1(x1)))))) 34.51/10.06 0(1(4(4(2(x1))))) -> 1(0(4(4(4(2(x1)))))) 34.51/10.06 0(1(5(0(1(x1))))) -> 0(1(1(0(5(3(x1)))))) 34.51/10.06 0(1(5(0(5(x1))))) -> 3(5(1(0(5(0(x1)))))) 34.51/10.06 0(2(4(2(1(x1))))) -> 2(1(2(4(3(0(x1)))))) 34.51/10.06 0(4(0(2(1(x1))))) -> 3(0(4(1(2(0(x1)))))) 34.51/10.06 0(5(0(1(5(x1))))) -> 0(5(0(5(3(1(x1)))))) 34.51/10.06 1(0(0(1(5(x1))))) -> 5(1(0(0(3(1(x1)))))) 34.51/10.06 1(0(1(4(5(x1))))) -> 1(4(4(1(0(5(x1)))))) 34.51/10.06 1(4(0(1(5(x1))))) -> 1(4(1(3(0(5(x1)))))) 34.51/10.06 2(0(1(5(2(x1))))) -> 2(1(0(3(5(2(x1)))))) 34.51/10.06 2(0(4(2(1(x1))))) -> 2(1(4(2(3(0(x1)))))) 34.51/10.06 2(0(5(1(2(x1))))) -> 0(3(1(5(2(2(x1)))))) 34.51/10.06 2(2(1(1(2(x1))))) -> 2(2(1(3(1(2(x1)))))) 34.51/10.06 2(5(1(5(2(x1))))) -> 2(4(1(5(5(2(x1)))))) 34.51/10.06 5(0(1(4(5(x1))))) -> 4(1(0(3(5(5(x1)))))) 34.51/10.06 5(1(0(1(5(x1))))) -> 3(1(5(1(0(5(x1)))))) 34.51/10.06 5(4(0(2(1(x1))))) -> 4(1(3(5(2(0(x1)))))) 34.51/10.06 5(5(1(4(5(x1))))) -> 5(4(1(3(5(5(x1)))))) 34.51/10.06 34.51/10.06 Q is empty. 34.51/10.06 34.51/10.06 ---------------------------------------- 34.51/10.06 34.51/10.06 (1) QTRS Reverse (EQUIVALENT) 34.51/10.06 We applied the QTRS Reverse Processor [REVERSE]. 34.51/10.06 ---------------------------------------- 34.51/10.06 34.51/10.06 (2) 34.51/10.06 Obligation: 34.51/10.06 Q restricted rewrite system: 34.51/10.06 The TRS R consists of the following rules: 34.51/10.06 34.51/10.06 2(1(0(x1))) -> 1(3(0(3(2(x1))))) 34.51/10.06 2(0(0(0(x1)))) -> 2(0(3(0(0(3(x1)))))) 34.51/10.06 2(2(0(0(x1)))) -> 2(3(0(0(3(2(x1)))))) 34.51/10.06 2(1(1(0(x1)))) -> 1(1(3(0(3(2(x1)))))) 34.51/10.06 1(2(1(0(x1)))) -> 3(1(4(2(1(0(x1)))))) 34.51/10.06 2(2(1(0(x1)))) -> 2(1(3(0(3(2(x1)))))) 34.51/10.06 5(2(1(0(x1)))) -> 1(3(5(0(2(x1))))) 34.51/10.06 1(5(1(0(x1)))) -> 1(3(5(3(0(1(x1)))))) 34.51/10.06 2(5(1(0(x1)))) -> 5(3(0(1(4(2(x1)))))) 34.51/10.06 2(5(1(0(x1)))) -> 5(0(1(3(4(2(x1)))))) 34.51/10.06 2(5(1(0(x1)))) -> 2(5(0(3(1(3(x1)))))) 34.51/10.06 5(5(1(0(x1)))) -> 5(5(0(3(1(x1))))) 34.51/10.06 2(1(2(0(x1)))) -> 1(3(0(3(2(2(x1)))))) 34.51/10.06 2(5(2(0(x1)))) -> 5(0(3(2(2(x1))))) 34.51/10.06 5(4(1(1(x1)))) -> 1(3(1(4(5(x1))))) 34.51/10.06 5(1(5(1(x1)))) -> 1(3(5(5(3(1(x1)))))) 34.51/10.06 1(5(5(1(x1)))) -> 3(5(1(3(5(1(x1)))))) 34.51/10.06 2(1(0(2(x1)))) -> 1(3(0(3(2(2(x1)))))) 34.51/10.06 5(1(0(2(x1)))) -> 5(0(3(1(2(x1))))) 34.51/10.06 2(1(0(5(x1)))) -> 2(1(3(5(0(3(x1)))))) 34.51/10.06 2(1(0(5(x1)))) -> 1(5(0(3(2(4(x1)))))) 34.51/10.06 1(0(0(0(0(x1))))) -> 3(0(0(1(0(0(x1)))))) 34.51/10.06 5(2(1(0(0(x1))))) -> 1(3(5(0(0(2(x1)))))) 34.51/10.06 2(5(1(0(0(x1))))) -> 2(5(3(0(1(0(x1)))))) 34.51/10.06 5(4(0(1(0(x1))))) -> 1(3(0(0(4(5(x1)))))) 34.51/10.06 2(1(1(1(0(x1))))) -> 2(1(1(3(0(1(x1)))))) 34.51/10.06 5(1(2(1(0(x1))))) -> 1(3(5(0(1(2(x1)))))) 34.51/10.06 2(5(3(1(0(x1))))) -> 2(5(1(4(0(3(x1)))))) 34.51/10.06 5(2(4(1(0(x1))))) -> 1(5(0(3(4(2(x1)))))) 34.51/10.06 2(4(4(1(0(x1))))) -> 2(4(4(4(0(1(x1)))))) 34.51/10.06 1(0(5(1(0(x1))))) -> 3(5(0(1(1(0(x1)))))) 34.51/10.06 5(0(5(1(0(x1))))) -> 0(5(0(1(5(3(x1)))))) 34.51/10.06 1(2(4(2(0(x1))))) -> 0(3(4(2(1(2(x1)))))) 34.51/10.06 1(2(0(4(0(x1))))) -> 0(2(1(4(0(3(x1)))))) 34.51/10.06 5(1(0(5(0(x1))))) -> 1(3(5(0(5(0(x1)))))) 34.51/10.06 5(1(0(0(1(x1))))) -> 1(3(0(0(1(5(x1)))))) 34.51/10.06 5(4(1(0(1(x1))))) -> 5(0(1(4(4(1(x1)))))) 34.51/10.06 5(1(0(4(1(x1))))) -> 5(0(3(1(4(1(x1)))))) 34.51/10.06 2(5(1(0(2(x1))))) -> 2(5(3(0(1(2(x1)))))) 34.51/10.06 1(2(4(0(2(x1))))) -> 0(3(2(4(1(2(x1)))))) 34.51/10.06 2(1(5(0(2(x1))))) -> 2(2(5(1(3(0(x1)))))) 34.51/10.06 2(1(1(2(2(x1))))) -> 2(1(3(1(2(2(x1)))))) 34.51/10.06 2(5(1(5(2(x1))))) -> 2(5(5(1(4(2(x1)))))) 34.51/10.06 5(4(1(0(5(x1))))) -> 5(5(3(0(1(4(x1)))))) 34.51/10.06 5(1(0(1(5(x1))))) -> 5(0(1(5(1(3(x1)))))) 34.51/10.06 1(2(0(4(5(x1))))) -> 0(2(5(3(1(4(x1)))))) 34.51/10.06 5(4(1(5(5(x1))))) -> 5(5(3(1(4(5(x1)))))) 34.51/10.06 34.51/10.06 Q is empty. 34.51/10.06 34.51/10.06 ---------------------------------------- 34.51/10.06 34.51/10.06 (3) DependencyPairsProof (EQUIVALENT) 34.51/10.06 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 34.51/10.06 ---------------------------------------- 34.51/10.06 34.51/10.06 (4) 34.51/10.06 Obligation: 34.51/10.06 Q DP problem: 34.51/10.06 The TRS P consists of the following rules: 34.51/10.06 34.51/10.06 2^1(1(0(x1))) -> 1^1(3(0(3(2(x1))))) 34.51/10.06 2^1(1(0(x1))) -> 2^1(x1) 34.51/10.06 2^1(0(0(0(x1)))) -> 2^1(0(3(0(0(3(x1)))))) 34.51/10.06 2^1(2(0(0(x1)))) -> 2^1(3(0(0(3(2(x1)))))) 34.51/10.06 2^1(2(0(0(x1)))) -> 2^1(x1) 34.51/10.06 2^1(1(1(0(x1)))) -> 1^1(1(3(0(3(2(x1)))))) 34.51/10.06 2^1(1(1(0(x1)))) -> 1^1(3(0(3(2(x1))))) 34.51/10.06 2^1(1(1(0(x1)))) -> 2^1(x1) 34.51/10.06 1^1(2(1(0(x1)))) -> 1^1(4(2(1(0(x1))))) 34.51/10.06 2^1(2(1(0(x1)))) -> 2^1(1(3(0(3(2(x1)))))) 34.51/10.06 2^1(2(1(0(x1)))) -> 1^1(3(0(3(2(x1))))) 34.51/10.06 2^1(2(1(0(x1)))) -> 2^1(x1) 34.51/10.06 5^1(2(1(0(x1)))) -> 1^1(3(5(0(2(x1))))) 34.51/10.06 5^1(2(1(0(x1)))) -> 5^1(0(2(x1))) 34.51/10.06 5^1(2(1(0(x1)))) -> 2^1(x1) 34.51/10.06 1^1(5(1(0(x1)))) -> 1^1(3(5(3(0(1(x1)))))) 34.51/10.06 1^1(5(1(0(x1)))) -> 5^1(3(0(1(x1)))) 34.51/10.06 1^1(5(1(0(x1)))) -> 1^1(x1) 34.51/10.06 2^1(5(1(0(x1)))) -> 5^1(3(0(1(4(2(x1)))))) 34.51/10.06 2^1(5(1(0(x1)))) -> 1^1(4(2(x1))) 34.51/10.06 2^1(5(1(0(x1)))) -> 2^1(x1) 34.51/10.06 2^1(5(1(0(x1)))) -> 5^1(0(1(3(4(2(x1)))))) 34.51/10.06 2^1(5(1(0(x1)))) -> 1^1(3(4(2(x1)))) 34.51/10.06 2^1(5(1(0(x1)))) -> 2^1(5(0(3(1(3(x1)))))) 34.51/10.06 2^1(5(1(0(x1)))) -> 5^1(0(3(1(3(x1))))) 34.51/10.06 2^1(5(1(0(x1)))) -> 1^1(3(x1)) 34.51/10.06 5^1(5(1(0(x1)))) -> 5^1(5(0(3(1(x1))))) 34.51/10.06 5^1(5(1(0(x1)))) -> 5^1(0(3(1(x1)))) 34.51/10.06 5^1(5(1(0(x1)))) -> 1^1(x1) 34.51/10.06 2^1(1(2(0(x1)))) -> 1^1(3(0(3(2(2(x1)))))) 34.51/10.06 2^1(1(2(0(x1)))) -> 2^1(2(x1)) 34.51/10.06 2^1(1(2(0(x1)))) -> 2^1(x1) 34.51/10.06 2^1(5(2(0(x1)))) -> 5^1(0(3(2(2(x1))))) 34.51/10.06 2^1(5(2(0(x1)))) -> 2^1(2(x1)) 34.51/10.06 2^1(5(2(0(x1)))) -> 2^1(x1) 34.51/10.06 5^1(4(1(1(x1)))) -> 1^1(3(1(4(5(x1))))) 34.51/10.06 5^1(4(1(1(x1)))) -> 1^1(4(5(x1))) 34.51/10.06 5^1(4(1(1(x1)))) -> 5^1(x1) 34.51/10.06 5^1(1(5(1(x1)))) -> 1^1(3(5(5(3(1(x1)))))) 34.51/10.06 5^1(1(5(1(x1)))) -> 5^1(5(3(1(x1)))) 34.51/10.06 5^1(1(5(1(x1)))) -> 5^1(3(1(x1))) 34.51/10.06 1^1(5(5(1(x1)))) -> 5^1(1(3(5(1(x1))))) 34.51/10.06 1^1(5(5(1(x1)))) -> 1^1(3(5(1(x1)))) 34.51/10.06 2^1(1(0(2(x1)))) -> 1^1(3(0(3(2(2(x1)))))) 34.51/10.06 2^1(1(0(2(x1)))) -> 2^1(2(x1)) 34.51/10.06 5^1(1(0(2(x1)))) -> 5^1(0(3(1(2(x1))))) 34.51/10.06 5^1(1(0(2(x1)))) -> 1^1(2(x1)) 34.51/10.06 2^1(1(0(5(x1)))) -> 2^1(1(3(5(0(3(x1)))))) 34.51/10.06 2^1(1(0(5(x1)))) -> 1^1(3(5(0(3(x1))))) 34.51/10.06 2^1(1(0(5(x1)))) -> 5^1(0(3(x1))) 34.51/10.06 2^1(1(0(5(x1)))) -> 1^1(5(0(3(2(4(x1)))))) 34.51/10.06 2^1(1(0(5(x1)))) -> 5^1(0(3(2(4(x1))))) 34.51/10.06 2^1(1(0(5(x1)))) -> 2^1(4(x1)) 34.51/10.06 1^1(0(0(0(0(x1))))) -> 1^1(0(0(x1))) 34.51/10.06 5^1(2(1(0(0(x1))))) -> 1^1(3(5(0(0(2(x1)))))) 34.51/10.06 5^1(2(1(0(0(x1))))) -> 5^1(0(0(2(x1)))) 34.51/10.06 5^1(2(1(0(0(x1))))) -> 2^1(x1) 34.51/10.06 2^1(5(1(0(0(x1))))) -> 2^1(5(3(0(1(0(x1)))))) 34.51/10.06 2^1(5(1(0(0(x1))))) -> 5^1(3(0(1(0(x1))))) 34.51/10.06 2^1(5(1(0(0(x1))))) -> 1^1(0(x1)) 34.51/10.06 5^1(4(0(1(0(x1))))) -> 1^1(3(0(0(4(5(x1)))))) 34.51/10.06 5^1(4(0(1(0(x1))))) -> 5^1(x1) 34.51/10.06 2^1(1(1(1(0(x1))))) -> 2^1(1(1(3(0(1(x1)))))) 34.51/10.06 2^1(1(1(1(0(x1))))) -> 1^1(1(3(0(1(x1))))) 34.51/10.06 2^1(1(1(1(0(x1))))) -> 1^1(3(0(1(x1)))) 34.51/10.06 2^1(1(1(1(0(x1))))) -> 1^1(x1) 34.62/10.07 5^1(1(2(1(0(x1))))) -> 1^1(3(5(0(1(2(x1)))))) 34.62/10.07 5^1(1(2(1(0(x1))))) -> 5^1(0(1(2(x1)))) 34.62/10.07 5^1(1(2(1(0(x1))))) -> 1^1(2(x1)) 34.62/10.07 5^1(1(2(1(0(x1))))) -> 2^1(x1) 34.62/10.07 2^1(5(3(1(0(x1))))) -> 2^1(5(1(4(0(3(x1)))))) 34.62/10.07 2^1(5(3(1(0(x1))))) -> 5^1(1(4(0(3(x1))))) 34.62/10.07 2^1(5(3(1(0(x1))))) -> 1^1(4(0(3(x1)))) 34.62/10.07 5^1(2(4(1(0(x1))))) -> 1^1(5(0(3(4(2(x1)))))) 34.62/10.07 5^1(2(4(1(0(x1))))) -> 5^1(0(3(4(2(x1))))) 34.62/10.07 5^1(2(4(1(0(x1))))) -> 2^1(x1) 34.62/10.07 2^1(4(4(1(0(x1))))) -> 2^1(4(4(4(0(1(x1)))))) 34.62/10.07 2^1(4(4(1(0(x1))))) -> 1^1(x1) 34.62/10.07 1^1(0(5(1(0(x1))))) -> 5^1(0(1(1(0(x1))))) 34.62/10.07 1^1(0(5(1(0(x1))))) -> 1^1(1(0(x1))) 34.62/10.07 5^1(0(5(1(0(x1))))) -> 5^1(0(1(5(3(x1))))) 34.62/10.07 5^1(0(5(1(0(x1))))) -> 1^1(5(3(x1))) 34.62/10.07 5^1(0(5(1(0(x1))))) -> 5^1(3(x1)) 34.62/10.07 1^1(2(4(2(0(x1))))) -> 2^1(1(2(x1))) 34.62/10.07 1^1(2(4(2(0(x1))))) -> 1^1(2(x1)) 34.62/10.07 1^1(2(4(2(0(x1))))) -> 2^1(x1) 34.62/10.07 1^1(2(0(4(0(x1))))) -> 2^1(1(4(0(3(x1))))) 34.62/10.07 1^1(2(0(4(0(x1))))) -> 1^1(4(0(3(x1)))) 34.62/10.07 5^1(1(0(5(0(x1))))) -> 1^1(3(5(0(5(0(x1)))))) 34.62/10.07 5^1(1(0(5(0(x1))))) -> 5^1(0(5(0(x1)))) 34.62/10.07 5^1(1(0(0(1(x1))))) -> 1^1(3(0(0(1(5(x1)))))) 34.62/10.07 5^1(1(0(0(1(x1))))) -> 1^1(5(x1)) 34.62/10.07 5^1(1(0(0(1(x1))))) -> 5^1(x1) 34.62/10.07 5^1(4(1(0(1(x1))))) -> 5^1(0(1(4(4(1(x1)))))) 34.62/10.07 5^1(4(1(0(1(x1))))) -> 1^1(4(4(1(x1)))) 34.62/10.07 5^1(1(0(4(1(x1))))) -> 5^1(0(3(1(4(1(x1)))))) 34.62/10.07 5^1(1(0(4(1(x1))))) -> 1^1(4(1(x1))) 34.62/10.07 2^1(5(1(0(2(x1))))) -> 2^1(5(3(0(1(2(x1)))))) 34.62/10.07 2^1(5(1(0(2(x1))))) -> 5^1(3(0(1(2(x1))))) 34.62/10.07 2^1(5(1(0(2(x1))))) -> 1^1(2(x1)) 34.62/10.07 1^1(2(4(0(2(x1))))) -> 2^1(4(1(2(x1)))) 34.62/10.07 1^1(2(4(0(2(x1))))) -> 1^1(2(x1)) 34.62/10.07 2^1(1(5(0(2(x1))))) -> 2^1(2(5(1(3(0(x1)))))) 34.62/10.07 2^1(1(5(0(2(x1))))) -> 2^1(5(1(3(0(x1))))) 34.62/10.07 2^1(1(5(0(2(x1))))) -> 5^1(1(3(0(x1)))) 34.62/10.07 2^1(1(5(0(2(x1))))) -> 1^1(3(0(x1))) 34.62/10.07 2^1(1(1(2(2(x1))))) -> 2^1(1(3(1(2(2(x1)))))) 34.62/10.07 2^1(1(1(2(2(x1))))) -> 1^1(3(1(2(2(x1))))) 34.62/10.07 2^1(5(1(5(2(x1))))) -> 2^1(5(5(1(4(2(x1)))))) 34.62/10.07 2^1(5(1(5(2(x1))))) -> 5^1(5(1(4(2(x1))))) 34.62/10.07 2^1(5(1(5(2(x1))))) -> 5^1(1(4(2(x1)))) 34.62/10.07 2^1(5(1(5(2(x1))))) -> 1^1(4(2(x1))) 34.62/10.07 5^1(4(1(0(5(x1))))) -> 5^1(5(3(0(1(4(x1)))))) 34.62/10.07 5^1(4(1(0(5(x1))))) -> 5^1(3(0(1(4(x1))))) 34.62/10.07 5^1(4(1(0(5(x1))))) -> 1^1(4(x1)) 34.62/10.07 5^1(1(0(1(5(x1))))) -> 5^1(0(1(5(1(3(x1)))))) 34.62/10.07 5^1(1(0(1(5(x1))))) -> 1^1(5(1(3(x1)))) 34.62/10.07 5^1(1(0(1(5(x1))))) -> 5^1(1(3(x1))) 34.62/10.07 5^1(1(0(1(5(x1))))) -> 1^1(3(x1)) 34.62/10.07 1^1(2(0(4(5(x1))))) -> 2^1(5(3(1(4(x1))))) 34.62/10.07 1^1(2(0(4(5(x1))))) -> 5^1(3(1(4(x1)))) 34.62/10.07 1^1(2(0(4(5(x1))))) -> 1^1(4(x1)) 34.62/10.07 5^1(4(1(5(5(x1))))) -> 5^1(5(3(1(4(5(x1)))))) 34.62/10.07 5^1(4(1(5(5(x1))))) -> 5^1(3(1(4(5(x1))))) 34.62/10.07 5^1(4(1(5(5(x1))))) -> 1^1(4(5(x1))) 34.62/10.07 34.62/10.07 The TRS R consists of the following rules: 34.62/10.07 34.62/10.07 2(1(0(x1))) -> 1(3(0(3(2(x1))))) 34.62/10.07 2(0(0(0(x1)))) -> 2(0(3(0(0(3(x1)))))) 34.62/10.07 2(2(0(0(x1)))) -> 2(3(0(0(3(2(x1)))))) 34.62/10.07 2(1(1(0(x1)))) -> 1(1(3(0(3(2(x1)))))) 34.62/10.07 1(2(1(0(x1)))) -> 3(1(4(2(1(0(x1)))))) 34.62/10.07 2(2(1(0(x1)))) -> 2(1(3(0(3(2(x1)))))) 34.62/10.07 5(2(1(0(x1)))) -> 1(3(5(0(2(x1))))) 34.62/10.07 1(5(1(0(x1)))) -> 1(3(5(3(0(1(x1)))))) 34.62/10.07 2(5(1(0(x1)))) -> 5(3(0(1(4(2(x1)))))) 34.62/10.07 2(5(1(0(x1)))) -> 5(0(1(3(4(2(x1)))))) 34.62/10.07 2(5(1(0(x1)))) -> 2(5(0(3(1(3(x1)))))) 34.62/10.07 5(5(1(0(x1)))) -> 5(5(0(3(1(x1))))) 34.62/10.07 2(1(2(0(x1)))) -> 1(3(0(3(2(2(x1)))))) 34.62/10.07 2(5(2(0(x1)))) -> 5(0(3(2(2(x1))))) 34.62/10.07 5(4(1(1(x1)))) -> 1(3(1(4(5(x1))))) 34.62/10.07 5(1(5(1(x1)))) -> 1(3(5(5(3(1(x1)))))) 34.62/10.07 1(5(5(1(x1)))) -> 3(5(1(3(5(1(x1)))))) 34.62/10.07 2(1(0(2(x1)))) -> 1(3(0(3(2(2(x1)))))) 34.62/10.07 5(1(0(2(x1)))) -> 5(0(3(1(2(x1))))) 34.62/10.07 2(1(0(5(x1)))) -> 2(1(3(5(0(3(x1)))))) 34.62/10.07 2(1(0(5(x1)))) -> 1(5(0(3(2(4(x1)))))) 34.62/10.07 1(0(0(0(0(x1))))) -> 3(0(0(1(0(0(x1)))))) 34.62/10.07 5(2(1(0(0(x1))))) -> 1(3(5(0(0(2(x1)))))) 34.62/10.07 2(5(1(0(0(x1))))) -> 2(5(3(0(1(0(x1)))))) 34.62/10.07 5(4(0(1(0(x1))))) -> 1(3(0(0(4(5(x1)))))) 34.62/10.07 2(1(1(1(0(x1))))) -> 2(1(1(3(0(1(x1)))))) 34.62/10.07 5(1(2(1(0(x1))))) -> 1(3(5(0(1(2(x1)))))) 34.62/10.07 2(5(3(1(0(x1))))) -> 2(5(1(4(0(3(x1)))))) 34.62/10.07 5(2(4(1(0(x1))))) -> 1(5(0(3(4(2(x1)))))) 34.62/10.07 2(4(4(1(0(x1))))) -> 2(4(4(4(0(1(x1)))))) 34.62/10.07 1(0(5(1(0(x1))))) -> 3(5(0(1(1(0(x1)))))) 34.62/10.07 5(0(5(1(0(x1))))) -> 0(5(0(1(5(3(x1)))))) 34.62/10.07 1(2(4(2(0(x1))))) -> 0(3(4(2(1(2(x1)))))) 34.62/10.07 1(2(0(4(0(x1))))) -> 0(2(1(4(0(3(x1)))))) 34.62/10.07 5(1(0(5(0(x1))))) -> 1(3(5(0(5(0(x1)))))) 34.62/10.07 5(1(0(0(1(x1))))) -> 1(3(0(0(1(5(x1)))))) 34.62/10.07 5(4(1(0(1(x1))))) -> 5(0(1(4(4(1(x1)))))) 34.62/10.07 5(1(0(4(1(x1))))) -> 5(0(3(1(4(1(x1)))))) 34.62/10.07 2(5(1(0(2(x1))))) -> 2(5(3(0(1(2(x1)))))) 34.62/10.07 1(2(4(0(2(x1))))) -> 0(3(2(4(1(2(x1)))))) 34.62/10.07 2(1(5(0(2(x1))))) -> 2(2(5(1(3(0(x1)))))) 34.62/10.07 2(1(1(2(2(x1))))) -> 2(1(3(1(2(2(x1)))))) 34.62/10.07 2(5(1(5(2(x1))))) -> 2(5(5(1(4(2(x1)))))) 34.62/10.07 5(4(1(0(5(x1))))) -> 5(5(3(0(1(4(x1)))))) 34.62/10.07 5(1(0(1(5(x1))))) -> 5(0(1(5(1(3(x1)))))) 34.62/10.07 1(2(0(4(5(x1))))) -> 0(2(5(3(1(4(x1)))))) 34.62/10.07 5(4(1(5(5(x1))))) -> 5(5(3(1(4(5(x1)))))) 34.62/10.07 34.62/10.07 Q is empty. 34.62/10.07 We have to consider all minimal (P,Q,R)-chains. 34.62/10.07 ---------------------------------------- 34.62/10.07 34.62/10.07 (5) DependencyGraphProof (EQUIVALENT) 34.62/10.07 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 100 less nodes. 34.62/10.07 ---------------------------------------- 34.62/10.07 34.62/10.07 (6) 34.62/10.07 Complex Obligation (AND) 34.62/10.07 34.62/10.07 ---------------------------------------- 34.62/10.07 34.62/10.07 (7) 34.62/10.07 Obligation: 34.62/10.07 Q DP problem: 34.62/10.07 The TRS P consists of the following rules: 34.62/10.07 34.62/10.07 1^1(0(0(0(0(x1))))) -> 1^1(0(0(x1))) 34.62/10.07 34.62/10.07 The TRS R consists of the following rules: 34.62/10.07 34.62/10.07 2(1(0(x1))) -> 1(3(0(3(2(x1))))) 34.62/10.07 2(0(0(0(x1)))) -> 2(0(3(0(0(3(x1)))))) 34.62/10.07 2(2(0(0(x1)))) -> 2(3(0(0(3(2(x1)))))) 34.62/10.07 2(1(1(0(x1)))) -> 1(1(3(0(3(2(x1)))))) 34.62/10.07 1(2(1(0(x1)))) -> 3(1(4(2(1(0(x1)))))) 34.62/10.07 2(2(1(0(x1)))) -> 2(1(3(0(3(2(x1)))))) 34.62/10.07 5(2(1(0(x1)))) -> 1(3(5(0(2(x1))))) 34.62/10.07 1(5(1(0(x1)))) -> 1(3(5(3(0(1(x1)))))) 34.62/10.07 2(5(1(0(x1)))) -> 5(3(0(1(4(2(x1)))))) 34.62/10.07 2(5(1(0(x1)))) -> 5(0(1(3(4(2(x1)))))) 34.62/10.07 2(5(1(0(x1)))) -> 2(5(0(3(1(3(x1)))))) 34.62/10.07 5(5(1(0(x1)))) -> 5(5(0(3(1(x1))))) 34.62/10.07 2(1(2(0(x1)))) -> 1(3(0(3(2(2(x1)))))) 34.62/10.07 2(5(2(0(x1)))) -> 5(0(3(2(2(x1))))) 34.62/10.07 5(4(1(1(x1)))) -> 1(3(1(4(5(x1))))) 34.62/10.07 5(1(5(1(x1)))) -> 1(3(5(5(3(1(x1)))))) 34.62/10.07 1(5(5(1(x1)))) -> 3(5(1(3(5(1(x1)))))) 34.62/10.07 2(1(0(2(x1)))) -> 1(3(0(3(2(2(x1)))))) 34.62/10.07 5(1(0(2(x1)))) -> 5(0(3(1(2(x1))))) 34.62/10.07 2(1(0(5(x1)))) -> 2(1(3(5(0(3(x1)))))) 34.62/10.07 2(1(0(5(x1)))) -> 1(5(0(3(2(4(x1)))))) 34.62/10.07 1(0(0(0(0(x1))))) -> 3(0(0(1(0(0(x1)))))) 34.62/10.07 5(2(1(0(0(x1))))) -> 1(3(5(0(0(2(x1)))))) 34.62/10.07 2(5(1(0(0(x1))))) -> 2(5(3(0(1(0(x1)))))) 34.62/10.07 5(4(0(1(0(x1))))) -> 1(3(0(0(4(5(x1)))))) 34.62/10.07 2(1(1(1(0(x1))))) -> 2(1(1(3(0(1(x1)))))) 34.62/10.07 5(1(2(1(0(x1))))) -> 1(3(5(0(1(2(x1)))))) 34.62/10.07 2(5(3(1(0(x1))))) -> 2(5(1(4(0(3(x1)))))) 34.62/10.07 5(2(4(1(0(x1))))) -> 1(5(0(3(4(2(x1)))))) 34.62/10.07 2(4(4(1(0(x1))))) -> 2(4(4(4(0(1(x1)))))) 34.62/10.07 1(0(5(1(0(x1))))) -> 3(5(0(1(1(0(x1)))))) 34.62/10.07 5(0(5(1(0(x1))))) -> 0(5(0(1(5(3(x1)))))) 34.62/10.07 1(2(4(2(0(x1))))) -> 0(3(4(2(1(2(x1)))))) 34.62/10.07 1(2(0(4(0(x1))))) -> 0(2(1(4(0(3(x1)))))) 34.62/10.07 5(1(0(5(0(x1))))) -> 1(3(5(0(5(0(x1)))))) 34.62/10.07 5(1(0(0(1(x1))))) -> 1(3(0(0(1(5(x1)))))) 34.62/10.07 5(4(1(0(1(x1))))) -> 5(0(1(4(4(1(x1)))))) 34.62/10.07 5(1(0(4(1(x1))))) -> 5(0(3(1(4(1(x1)))))) 34.62/10.07 2(5(1(0(2(x1))))) -> 2(5(3(0(1(2(x1)))))) 34.62/10.07 1(2(4(0(2(x1))))) -> 0(3(2(4(1(2(x1)))))) 34.62/10.07 2(1(5(0(2(x1))))) -> 2(2(5(1(3(0(x1)))))) 34.62/10.07 2(1(1(2(2(x1))))) -> 2(1(3(1(2(2(x1)))))) 34.62/10.07 2(5(1(5(2(x1))))) -> 2(5(5(1(4(2(x1)))))) 34.62/10.07 5(4(1(0(5(x1))))) -> 5(5(3(0(1(4(x1)))))) 34.62/10.07 5(1(0(1(5(x1))))) -> 5(0(1(5(1(3(x1)))))) 34.62/10.07 1(2(0(4(5(x1))))) -> 0(2(5(3(1(4(x1)))))) 34.62/10.07 5(4(1(5(5(x1))))) -> 5(5(3(1(4(5(x1)))))) 34.62/10.07 34.62/10.07 Q is empty. 34.62/10.07 We have to consider all minimal (P,Q,R)-chains. 34.62/10.07 ---------------------------------------- 34.62/10.07 34.62/10.07 (8) UsableRulesProof (EQUIVALENT) 34.62/10.07 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. 34.62/10.07 ---------------------------------------- 34.62/10.07 34.62/10.07 (9) 34.62/10.07 Obligation: 34.62/10.07 Q DP problem: 34.62/10.07 The TRS P consists of the following rules: 34.62/10.07 34.62/10.07 1^1(0(0(0(0(x1))))) -> 1^1(0(0(x1))) 34.62/10.07 34.62/10.07 R is empty. 34.62/10.07 Q is empty. 34.62/10.07 We have to consider all minimal (P,Q,R)-chains. 34.62/10.07 ---------------------------------------- 34.62/10.07 34.62/10.07 (10) QDPSizeChangeProof (EQUIVALENT) 34.62/10.07 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. 34.62/10.07 34.62/10.07 From the DPs we obtained the following set of size-change graphs: 34.62/10.07 *1^1(0(0(0(0(x1))))) -> 1^1(0(0(x1))) 34.62/10.07 The graph contains the following edges 1 > 1 34.62/10.07 34.62/10.07 34.62/10.07 ---------------------------------------- 34.62/10.07 34.62/10.07 (11) 34.62/10.07 YES 34.62/10.07 34.62/10.07 ---------------------------------------- 34.62/10.07 34.62/10.07 (12) 34.62/10.07 Obligation: 34.62/10.07 Q DP problem: 34.62/10.07 The TRS P consists of the following rules: 34.62/10.07 34.62/10.07 2^1(2(0(0(x1)))) -> 2^1(x1) 34.62/10.07 2^1(1(0(x1))) -> 2^1(x1) 34.62/10.07 2^1(1(1(0(x1)))) -> 2^1(x1) 34.62/10.07 2^1(2(1(0(x1)))) -> 2^1(x1) 34.62/10.07 2^1(5(1(0(x1)))) -> 2^1(x1) 34.62/10.07 2^1(1(2(0(x1)))) -> 2^1(2(x1)) 34.62/10.07 2^1(1(2(0(x1)))) -> 2^1(x1) 34.62/10.07 2^1(5(2(0(x1)))) -> 2^1(2(x1)) 34.62/10.07 2^1(5(2(0(x1)))) -> 2^1(x1) 34.62/10.07 2^1(1(0(2(x1)))) -> 2^1(2(x1)) 34.62/10.07 2^1(1(0(5(x1)))) -> 2^1(4(x1)) 34.62/10.07 2^1(4(4(1(0(x1))))) -> 1^1(x1) 34.62/10.07 1^1(5(1(0(x1)))) -> 1^1(x1) 34.62/10.07 1^1(0(5(1(0(x1))))) -> 1^1(1(0(x1))) 34.62/10.07 1^1(2(4(2(0(x1))))) -> 2^1(1(2(x1))) 34.62/10.07 2^1(5(1(0(0(x1))))) -> 1^1(0(x1)) 34.62/10.07 2^1(1(1(1(0(x1))))) -> 1^1(x1) 34.62/10.07 1^1(2(4(2(0(x1))))) -> 1^1(2(x1)) 34.62/10.07 1^1(2(4(2(0(x1))))) -> 2^1(x1) 34.62/10.07 2^1(5(1(0(2(x1))))) -> 1^1(2(x1)) 34.62/10.07 1^1(2(4(0(2(x1))))) -> 1^1(2(x1)) 34.62/10.07 34.62/10.07 The TRS R consists of the following rules: 34.62/10.07 34.62/10.07 2(1(0(x1))) -> 1(3(0(3(2(x1))))) 34.62/10.07 2(0(0(0(x1)))) -> 2(0(3(0(0(3(x1)))))) 34.62/10.07 2(2(0(0(x1)))) -> 2(3(0(0(3(2(x1)))))) 34.62/10.07 2(1(1(0(x1)))) -> 1(1(3(0(3(2(x1)))))) 34.62/10.07 1(2(1(0(x1)))) -> 3(1(4(2(1(0(x1)))))) 34.62/10.07 2(2(1(0(x1)))) -> 2(1(3(0(3(2(x1)))))) 34.62/10.07 5(2(1(0(x1)))) -> 1(3(5(0(2(x1))))) 34.62/10.07 1(5(1(0(x1)))) -> 1(3(5(3(0(1(x1)))))) 34.62/10.07 2(5(1(0(x1)))) -> 5(3(0(1(4(2(x1)))))) 34.62/10.07 2(5(1(0(x1)))) -> 5(0(1(3(4(2(x1)))))) 34.62/10.07 2(5(1(0(x1)))) -> 2(5(0(3(1(3(x1)))))) 34.62/10.07 5(5(1(0(x1)))) -> 5(5(0(3(1(x1))))) 34.62/10.07 2(1(2(0(x1)))) -> 1(3(0(3(2(2(x1)))))) 34.62/10.07 2(5(2(0(x1)))) -> 5(0(3(2(2(x1))))) 34.62/10.07 5(4(1(1(x1)))) -> 1(3(1(4(5(x1))))) 34.62/10.07 5(1(5(1(x1)))) -> 1(3(5(5(3(1(x1)))))) 34.62/10.07 1(5(5(1(x1)))) -> 3(5(1(3(5(1(x1)))))) 34.62/10.07 2(1(0(2(x1)))) -> 1(3(0(3(2(2(x1)))))) 34.62/10.07 5(1(0(2(x1)))) -> 5(0(3(1(2(x1))))) 34.62/10.07 2(1(0(5(x1)))) -> 2(1(3(5(0(3(x1)))))) 34.62/10.07 2(1(0(5(x1)))) -> 1(5(0(3(2(4(x1)))))) 34.62/10.07 1(0(0(0(0(x1))))) -> 3(0(0(1(0(0(x1)))))) 34.62/10.07 5(2(1(0(0(x1))))) -> 1(3(5(0(0(2(x1)))))) 34.62/10.07 2(5(1(0(0(x1))))) -> 2(5(3(0(1(0(x1)))))) 34.62/10.07 5(4(0(1(0(x1))))) -> 1(3(0(0(4(5(x1)))))) 34.62/10.07 2(1(1(1(0(x1))))) -> 2(1(1(3(0(1(x1)))))) 34.62/10.07 5(1(2(1(0(x1))))) -> 1(3(5(0(1(2(x1)))))) 34.62/10.07 2(5(3(1(0(x1))))) -> 2(5(1(4(0(3(x1)))))) 34.62/10.07 5(2(4(1(0(x1))))) -> 1(5(0(3(4(2(x1)))))) 34.62/10.07 2(4(4(1(0(x1))))) -> 2(4(4(4(0(1(x1)))))) 34.62/10.07 1(0(5(1(0(x1))))) -> 3(5(0(1(1(0(x1)))))) 34.62/10.07 5(0(5(1(0(x1))))) -> 0(5(0(1(5(3(x1)))))) 34.62/10.07 1(2(4(2(0(x1))))) -> 0(3(4(2(1(2(x1)))))) 34.62/10.07 1(2(0(4(0(x1))))) -> 0(2(1(4(0(3(x1)))))) 34.62/10.07 5(1(0(5(0(x1))))) -> 1(3(5(0(5(0(x1)))))) 34.62/10.07 5(1(0(0(1(x1))))) -> 1(3(0(0(1(5(x1)))))) 34.62/10.07 5(4(1(0(1(x1))))) -> 5(0(1(4(4(1(x1)))))) 34.62/10.07 5(1(0(4(1(x1))))) -> 5(0(3(1(4(1(x1)))))) 34.62/10.07 2(5(1(0(2(x1))))) -> 2(5(3(0(1(2(x1)))))) 34.62/10.07 1(2(4(0(2(x1))))) -> 0(3(2(4(1(2(x1)))))) 34.62/10.07 2(1(5(0(2(x1))))) -> 2(2(5(1(3(0(x1)))))) 34.62/10.07 2(1(1(2(2(x1))))) -> 2(1(3(1(2(2(x1)))))) 34.62/10.07 2(5(1(5(2(x1))))) -> 2(5(5(1(4(2(x1)))))) 34.62/10.07 5(4(1(0(5(x1))))) -> 5(5(3(0(1(4(x1)))))) 34.62/10.07 5(1(0(1(5(x1))))) -> 5(0(1(5(1(3(x1)))))) 34.62/10.07 1(2(0(4(5(x1))))) -> 0(2(5(3(1(4(x1)))))) 34.62/10.07 5(4(1(5(5(x1))))) -> 5(5(3(1(4(5(x1)))))) 34.62/10.07 34.62/10.07 Q is empty. 34.62/10.07 We have to consider all minimal (P,Q,R)-chains. 34.62/10.07 ---------------------------------------- 34.62/10.07 34.62/10.07 (13) QDPOrderProof (EQUIVALENT) 34.62/10.07 We use the reduction pair processor [LPAR04,JAR06]. 34.62/10.07 34.62/10.07 34.62/10.07 The following pairs can be oriented strictly and are deleted. 34.62/10.07 34.62/10.07 2^1(2(0(0(x1)))) -> 2^1(x1) 34.62/10.07 2^1(1(0(x1))) -> 2^1(x1) 34.62/10.07 2^1(1(1(0(x1)))) -> 2^1(x1) 34.62/10.07 2^1(2(1(0(x1)))) -> 2^1(x1) 34.62/10.07 2^1(5(1(0(x1)))) -> 2^1(x1) 34.62/10.07 2^1(1(2(0(x1)))) -> 2^1(2(x1)) 34.62/10.07 2^1(1(2(0(x1)))) -> 2^1(x1) 34.62/10.07 2^1(5(2(0(x1)))) -> 2^1(2(x1)) 34.62/10.07 2^1(5(2(0(x1)))) -> 2^1(x1) 34.62/10.07 2^1(1(0(2(x1)))) -> 2^1(2(x1)) 34.62/10.07 2^1(1(0(5(x1)))) -> 2^1(4(x1)) 34.62/10.07 2^1(4(4(1(0(x1))))) -> 1^1(x1) 34.62/10.07 1^1(5(1(0(x1)))) -> 1^1(x1) 34.62/10.07 1^1(0(5(1(0(x1))))) -> 1^1(1(0(x1))) 34.62/10.07 1^1(2(4(2(0(x1))))) -> 2^1(1(2(x1))) 34.62/10.07 2^1(5(1(0(0(x1))))) -> 1^1(0(x1)) 34.62/10.07 2^1(1(1(1(0(x1))))) -> 1^1(x1) 34.62/10.07 1^1(2(4(2(0(x1))))) -> 1^1(2(x1)) 34.62/10.07 1^1(2(4(2(0(x1))))) -> 2^1(x1) 34.62/10.07 2^1(5(1(0(2(x1))))) -> 1^1(2(x1)) 34.62/10.07 1^1(2(4(0(2(x1))))) -> 1^1(2(x1)) 34.62/10.07 The remaining pairs can at least be oriented weakly. 34.62/10.07 Used ordering: Polynomial interpretation [POLO]: 34.62/10.07 34.62/10.07 POL(0(x_1)) = 1 + x_1 34.62/10.07 POL(1(x_1)) = 1 + x_1 34.62/10.07 POL(1^1(x_1)) = 1 + x_1 34.62/10.07 POL(2(x_1)) = x_1 34.62/10.07 POL(2^1(x_1)) = x_1 34.62/10.07 POL(3(x_1)) = x_1 34.62/10.07 POL(4(x_1)) = x_1 34.62/10.07 POL(5(x_1)) = x_1 34.62/10.07 34.62/10.07 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 34.62/10.07 34.62/10.07 2(1(0(x1))) -> 1(3(0(3(2(x1))))) 34.62/10.07 2(0(0(0(x1)))) -> 2(0(3(0(0(3(x1)))))) 34.62/10.07 2(2(0(0(x1)))) -> 2(3(0(0(3(2(x1)))))) 34.62/10.07 2(1(1(0(x1)))) -> 1(1(3(0(3(2(x1)))))) 34.62/10.07 2(2(1(0(x1)))) -> 2(1(3(0(3(2(x1)))))) 34.62/10.07 2(5(1(0(x1)))) -> 5(3(0(1(4(2(x1)))))) 34.62/10.07 2(5(1(0(x1)))) -> 5(0(1(3(4(2(x1)))))) 34.62/10.07 2(5(1(0(x1)))) -> 2(5(0(3(1(3(x1)))))) 34.62/10.07 2(1(2(0(x1)))) -> 1(3(0(3(2(2(x1)))))) 34.62/10.07 2(5(2(0(x1)))) -> 5(0(3(2(2(x1))))) 34.62/10.07 2(1(0(2(x1)))) -> 1(3(0(3(2(2(x1)))))) 34.62/10.07 2(1(0(5(x1)))) -> 2(1(3(5(0(3(x1)))))) 34.62/10.07 2(1(0(5(x1)))) -> 1(5(0(3(2(4(x1)))))) 34.62/10.07 2(5(1(0(0(x1))))) -> 2(5(3(0(1(0(x1)))))) 34.62/10.07 2(1(1(1(0(x1))))) -> 2(1(1(3(0(1(x1)))))) 34.62/10.07 2(5(3(1(0(x1))))) -> 2(5(1(4(0(3(x1)))))) 34.62/10.07 2(4(4(1(0(x1))))) -> 2(4(4(4(0(1(x1)))))) 34.62/10.07 2(5(1(0(2(x1))))) -> 2(5(3(0(1(2(x1)))))) 34.62/10.07 2(1(5(0(2(x1))))) -> 2(2(5(1(3(0(x1)))))) 34.62/10.07 2(1(1(2(2(x1))))) -> 2(1(3(1(2(2(x1)))))) 34.62/10.07 2(5(1(5(2(x1))))) -> 2(5(5(1(4(2(x1)))))) 34.62/10.07 1(0(0(0(0(x1))))) -> 3(0(0(1(0(0(x1)))))) 34.62/10.07 1(0(5(1(0(x1))))) -> 3(5(0(1(1(0(x1)))))) 34.62/10.07 1(2(1(0(x1)))) -> 3(1(4(2(1(0(x1)))))) 34.62/10.07 1(5(1(0(x1)))) -> 1(3(5(3(0(1(x1)))))) 34.62/10.07 1(5(5(1(x1)))) -> 3(5(1(3(5(1(x1)))))) 34.62/10.07 1(2(4(2(0(x1))))) -> 0(3(4(2(1(2(x1)))))) 34.62/10.07 1(2(0(4(0(x1))))) -> 0(2(1(4(0(3(x1)))))) 34.62/10.07 1(2(4(0(2(x1))))) -> 0(3(2(4(1(2(x1)))))) 34.62/10.07 1(2(0(4(5(x1))))) -> 0(2(5(3(1(4(x1)))))) 34.62/10.07 5(2(1(0(x1)))) -> 1(3(5(0(2(x1))))) 34.62/10.07 5(5(1(0(x1)))) -> 5(5(0(3(1(x1))))) 34.62/10.07 5(4(1(1(x1)))) -> 1(3(1(4(5(x1))))) 34.62/10.07 5(1(5(1(x1)))) -> 1(3(5(5(3(1(x1)))))) 34.62/10.07 5(1(0(2(x1)))) -> 5(0(3(1(2(x1))))) 34.62/10.07 5(2(1(0(0(x1))))) -> 1(3(5(0(0(2(x1)))))) 34.62/10.07 5(4(0(1(0(x1))))) -> 1(3(0(0(4(5(x1)))))) 34.62/10.07 5(1(2(1(0(x1))))) -> 1(3(5(0(1(2(x1)))))) 34.62/10.07 5(2(4(1(0(x1))))) -> 1(5(0(3(4(2(x1)))))) 34.62/10.07 5(1(0(0(1(x1))))) -> 1(3(0(0(1(5(x1)))))) 34.62/10.07 5(4(1(0(1(x1))))) -> 5(0(1(4(4(1(x1)))))) 34.62/10.07 5(1(0(4(1(x1))))) -> 5(0(3(1(4(1(x1)))))) 34.62/10.07 5(4(1(5(5(x1))))) -> 5(5(3(1(4(5(x1)))))) 34.62/10.07 5(0(5(1(0(x1))))) -> 0(5(0(1(5(3(x1)))))) 34.62/10.07 5(1(0(5(0(x1))))) -> 1(3(5(0(5(0(x1)))))) 34.62/10.07 5(4(1(0(5(x1))))) -> 5(5(3(0(1(4(x1)))))) 34.62/10.07 5(1(0(1(5(x1))))) -> 5(0(1(5(1(3(x1)))))) 34.62/10.07 34.62/10.07 34.62/10.07 ---------------------------------------- 34.62/10.07 34.62/10.07 (14) 34.62/10.07 Obligation: 34.62/10.07 Q DP problem: 34.62/10.07 P is empty. 34.62/10.07 The TRS R consists of the following rules: 34.62/10.07 34.62/10.07 2(1(0(x1))) -> 1(3(0(3(2(x1))))) 34.62/10.07 2(0(0(0(x1)))) -> 2(0(3(0(0(3(x1)))))) 34.62/10.07 2(2(0(0(x1)))) -> 2(3(0(0(3(2(x1)))))) 34.62/10.07 2(1(1(0(x1)))) -> 1(1(3(0(3(2(x1)))))) 34.62/10.07 1(2(1(0(x1)))) -> 3(1(4(2(1(0(x1)))))) 34.62/10.07 2(2(1(0(x1)))) -> 2(1(3(0(3(2(x1)))))) 34.62/10.07 5(2(1(0(x1)))) -> 1(3(5(0(2(x1))))) 34.62/10.07 1(5(1(0(x1)))) -> 1(3(5(3(0(1(x1)))))) 34.62/10.07 2(5(1(0(x1)))) -> 5(3(0(1(4(2(x1)))))) 34.62/10.07 2(5(1(0(x1)))) -> 5(0(1(3(4(2(x1)))))) 34.62/10.07 2(5(1(0(x1)))) -> 2(5(0(3(1(3(x1)))))) 34.62/10.07 5(5(1(0(x1)))) -> 5(5(0(3(1(x1))))) 34.62/10.07 2(1(2(0(x1)))) -> 1(3(0(3(2(2(x1)))))) 34.62/10.07 2(5(2(0(x1)))) -> 5(0(3(2(2(x1))))) 34.62/10.07 5(4(1(1(x1)))) -> 1(3(1(4(5(x1))))) 34.62/10.07 5(1(5(1(x1)))) -> 1(3(5(5(3(1(x1)))))) 34.62/10.07 1(5(5(1(x1)))) -> 3(5(1(3(5(1(x1)))))) 34.62/10.07 2(1(0(2(x1)))) -> 1(3(0(3(2(2(x1)))))) 34.62/10.07 5(1(0(2(x1)))) -> 5(0(3(1(2(x1))))) 34.62/10.07 2(1(0(5(x1)))) -> 2(1(3(5(0(3(x1)))))) 34.62/10.07 2(1(0(5(x1)))) -> 1(5(0(3(2(4(x1)))))) 34.62/10.07 1(0(0(0(0(x1))))) -> 3(0(0(1(0(0(x1)))))) 34.62/10.07 5(2(1(0(0(x1))))) -> 1(3(5(0(0(2(x1)))))) 34.62/10.07 2(5(1(0(0(x1))))) -> 2(5(3(0(1(0(x1)))))) 34.62/10.07 5(4(0(1(0(x1))))) -> 1(3(0(0(4(5(x1)))))) 34.62/10.07 2(1(1(1(0(x1))))) -> 2(1(1(3(0(1(x1)))))) 34.62/10.07 5(1(2(1(0(x1))))) -> 1(3(5(0(1(2(x1)))))) 34.62/10.07 2(5(3(1(0(x1))))) -> 2(5(1(4(0(3(x1)))))) 34.62/10.07 5(2(4(1(0(x1))))) -> 1(5(0(3(4(2(x1)))))) 34.62/10.07 2(4(4(1(0(x1))))) -> 2(4(4(4(0(1(x1)))))) 34.62/10.07 1(0(5(1(0(x1))))) -> 3(5(0(1(1(0(x1)))))) 34.62/10.07 5(0(5(1(0(x1))))) -> 0(5(0(1(5(3(x1)))))) 34.62/10.07 1(2(4(2(0(x1))))) -> 0(3(4(2(1(2(x1)))))) 34.62/10.07 1(2(0(4(0(x1))))) -> 0(2(1(4(0(3(x1)))))) 34.62/10.07 5(1(0(5(0(x1))))) -> 1(3(5(0(5(0(x1)))))) 34.62/10.07 5(1(0(0(1(x1))))) -> 1(3(0(0(1(5(x1)))))) 34.62/10.07 5(4(1(0(1(x1))))) -> 5(0(1(4(4(1(x1)))))) 34.62/10.07 5(1(0(4(1(x1))))) -> 5(0(3(1(4(1(x1)))))) 34.62/10.07 2(5(1(0(2(x1))))) -> 2(5(3(0(1(2(x1)))))) 34.62/10.07 1(2(4(0(2(x1))))) -> 0(3(2(4(1(2(x1)))))) 34.62/10.07 2(1(5(0(2(x1))))) -> 2(2(5(1(3(0(x1)))))) 34.62/10.07 2(1(1(2(2(x1))))) -> 2(1(3(1(2(2(x1)))))) 34.62/10.07 2(5(1(5(2(x1))))) -> 2(5(5(1(4(2(x1)))))) 34.62/10.07 5(4(1(0(5(x1))))) -> 5(5(3(0(1(4(x1)))))) 34.62/10.07 5(1(0(1(5(x1))))) -> 5(0(1(5(1(3(x1)))))) 34.62/10.07 1(2(0(4(5(x1))))) -> 0(2(5(3(1(4(x1)))))) 34.62/10.07 5(4(1(5(5(x1))))) -> 5(5(3(1(4(5(x1)))))) 34.62/10.07 34.62/10.07 Q is empty. 34.62/10.07 We have to consider all minimal (P,Q,R)-chains. 34.62/10.07 ---------------------------------------- 34.62/10.07 34.62/10.07 (15) PisEmptyProof (EQUIVALENT) 34.62/10.07 The TRS P is empty. Hence, there is no (P,Q,R) chain. 34.62/10.07 ---------------------------------------- 34.62/10.07 34.62/10.07 (16) 34.62/10.07 YES 34.62/10.07 34.62/10.07 ---------------------------------------- 34.62/10.07 34.62/10.07 (17) 34.62/10.07 Obligation: 34.62/10.07 Q DP problem: 34.62/10.07 The TRS P consists of the following rules: 34.62/10.07 34.62/10.07 5^1(4(0(1(0(x1))))) -> 5^1(x1) 34.62/10.07 5^1(4(1(1(x1)))) -> 5^1(x1) 34.62/10.07 5^1(1(0(0(1(x1))))) -> 5^1(x1) 34.62/10.07 34.62/10.07 The TRS R consists of the following rules: 34.62/10.07 34.62/10.07 2(1(0(x1))) -> 1(3(0(3(2(x1))))) 34.62/10.07 2(0(0(0(x1)))) -> 2(0(3(0(0(3(x1)))))) 34.62/10.07 2(2(0(0(x1)))) -> 2(3(0(0(3(2(x1)))))) 34.62/10.07 2(1(1(0(x1)))) -> 1(1(3(0(3(2(x1)))))) 34.62/10.07 1(2(1(0(x1)))) -> 3(1(4(2(1(0(x1)))))) 34.62/10.07 2(2(1(0(x1)))) -> 2(1(3(0(3(2(x1)))))) 34.62/10.07 5(2(1(0(x1)))) -> 1(3(5(0(2(x1))))) 34.62/10.07 1(5(1(0(x1)))) -> 1(3(5(3(0(1(x1)))))) 34.62/10.07 2(5(1(0(x1)))) -> 5(3(0(1(4(2(x1)))))) 34.62/10.07 2(5(1(0(x1)))) -> 5(0(1(3(4(2(x1)))))) 34.62/10.07 2(5(1(0(x1)))) -> 2(5(0(3(1(3(x1)))))) 34.62/10.07 5(5(1(0(x1)))) -> 5(5(0(3(1(x1))))) 34.62/10.07 2(1(2(0(x1)))) -> 1(3(0(3(2(2(x1)))))) 34.62/10.07 2(5(2(0(x1)))) -> 5(0(3(2(2(x1))))) 34.62/10.07 5(4(1(1(x1)))) -> 1(3(1(4(5(x1))))) 34.62/10.07 5(1(5(1(x1)))) -> 1(3(5(5(3(1(x1)))))) 34.62/10.07 1(5(5(1(x1)))) -> 3(5(1(3(5(1(x1)))))) 34.62/10.07 2(1(0(2(x1)))) -> 1(3(0(3(2(2(x1)))))) 34.62/10.07 5(1(0(2(x1)))) -> 5(0(3(1(2(x1))))) 34.62/10.07 2(1(0(5(x1)))) -> 2(1(3(5(0(3(x1)))))) 34.62/10.07 2(1(0(5(x1)))) -> 1(5(0(3(2(4(x1)))))) 34.62/10.07 1(0(0(0(0(x1))))) -> 3(0(0(1(0(0(x1)))))) 34.62/10.07 5(2(1(0(0(x1))))) -> 1(3(5(0(0(2(x1)))))) 34.62/10.07 2(5(1(0(0(x1))))) -> 2(5(3(0(1(0(x1)))))) 34.62/10.07 5(4(0(1(0(x1))))) -> 1(3(0(0(4(5(x1)))))) 34.62/10.07 2(1(1(1(0(x1))))) -> 2(1(1(3(0(1(x1)))))) 34.62/10.07 5(1(2(1(0(x1))))) -> 1(3(5(0(1(2(x1)))))) 34.62/10.07 2(5(3(1(0(x1))))) -> 2(5(1(4(0(3(x1)))))) 34.62/10.07 5(2(4(1(0(x1))))) -> 1(5(0(3(4(2(x1)))))) 34.62/10.07 2(4(4(1(0(x1))))) -> 2(4(4(4(0(1(x1)))))) 34.62/10.07 1(0(5(1(0(x1))))) -> 3(5(0(1(1(0(x1)))))) 34.62/10.07 5(0(5(1(0(x1))))) -> 0(5(0(1(5(3(x1)))))) 34.62/10.07 1(2(4(2(0(x1))))) -> 0(3(4(2(1(2(x1)))))) 34.62/10.07 1(2(0(4(0(x1))))) -> 0(2(1(4(0(3(x1)))))) 34.62/10.07 5(1(0(5(0(x1))))) -> 1(3(5(0(5(0(x1)))))) 34.62/10.07 5(1(0(0(1(x1))))) -> 1(3(0(0(1(5(x1)))))) 34.62/10.07 5(4(1(0(1(x1))))) -> 5(0(1(4(4(1(x1)))))) 34.62/10.07 5(1(0(4(1(x1))))) -> 5(0(3(1(4(1(x1)))))) 34.62/10.07 2(5(1(0(2(x1))))) -> 2(5(3(0(1(2(x1)))))) 34.62/10.07 1(2(4(0(2(x1))))) -> 0(3(2(4(1(2(x1)))))) 34.62/10.07 2(1(5(0(2(x1))))) -> 2(2(5(1(3(0(x1)))))) 34.62/10.07 2(1(1(2(2(x1))))) -> 2(1(3(1(2(2(x1)))))) 34.62/10.07 2(5(1(5(2(x1))))) -> 2(5(5(1(4(2(x1)))))) 34.62/10.07 5(4(1(0(5(x1))))) -> 5(5(3(0(1(4(x1)))))) 34.62/10.07 5(1(0(1(5(x1))))) -> 5(0(1(5(1(3(x1)))))) 34.62/10.07 1(2(0(4(5(x1))))) -> 0(2(5(3(1(4(x1)))))) 34.62/10.07 5(4(1(5(5(x1))))) -> 5(5(3(1(4(5(x1)))))) 34.62/10.07 34.62/10.07 Q is empty. 34.62/10.07 We have to consider all minimal (P,Q,R)-chains. 34.62/10.07 ---------------------------------------- 34.62/10.07 34.62/10.07 (18) UsableRulesProof (EQUIVALENT) 34.62/10.07 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. 34.62/10.07 ---------------------------------------- 34.62/10.07 34.62/10.07 (19) 34.62/10.07 Obligation: 34.62/10.07 Q DP problem: 34.62/10.07 The TRS P consists of the following rules: 34.62/10.07 34.62/10.07 5^1(4(0(1(0(x1))))) -> 5^1(x1) 34.62/10.07 5^1(4(1(1(x1)))) -> 5^1(x1) 34.62/10.07 5^1(1(0(0(1(x1))))) -> 5^1(x1) 34.62/10.07 34.62/10.07 R is empty. 34.62/10.07 Q is empty. 34.62/10.07 We have to consider all minimal (P,Q,R)-chains. 34.62/10.07 ---------------------------------------- 34.62/10.07 34.62/10.07 (20) QDPSizeChangeProof (EQUIVALENT) 34.62/10.07 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. 34.62/10.07 34.62/10.07 From the DPs we obtained the following set of size-change graphs: 34.62/10.07 *5^1(4(0(1(0(x1))))) -> 5^1(x1) 34.62/10.07 The graph contains the following edges 1 > 1 34.62/10.07 34.62/10.07 34.62/10.07 *5^1(4(1(1(x1)))) -> 5^1(x1) 34.62/10.07 The graph contains the following edges 1 > 1 34.62/10.07 34.62/10.07 34.62/10.07 *5^1(1(0(0(1(x1))))) -> 5^1(x1) 34.62/10.07 The graph contains the following edges 1 > 1 34.62/10.07 34.62/10.07 34.62/10.07 ---------------------------------------- 34.62/10.07 34.62/10.07 (21) 34.62/10.07 YES 34.72/11.42 EOF