51.79/14.45 YES 52.03/14.50 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 52.03/14.50 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 52.03/14.50 52.03/14.50 52.03/14.50 Termination w.r.t. Q of the given QTRS could be proven: 52.03/14.50 52.03/14.50 (0) QTRS 52.03/14.50 (1) DependencyPairsProof [EQUIVALENT, 280 ms] 52.03/14.50 (2) QDP 52.03/14.50 (3) DependencyGraphProof [EQUIVALENT, 4 ms] 52.03/14.50 (4) AND 52.03/14.50 (5) QDP 52.03/14.50 (6) UsableRulesProof [EQUIVALENT, 0 ms] 52.03/14.50 (7) QDP 52.03/14.50 (8) MRRProof [EQUIVALENT, 0 ms] 52.03/14.50 (9) QDP 52.03/14.50 (10) QDPOrderProof [EQUIVALENT, 2 ms] 52.03/14.50 (11) QDP 52.03/14.50 (12) PisEmptyProof [EQUIVALENT, 0 ms] 52.03/14.50 (13) YES 52.03/14.50 (14) QDP 52.03/14.50 (15) QDPOrderProof [EQUIVALENT, 59 ms] 52.03/14.50 (16) QDP 52.03/14.50 (17) QDPOrderProof [EQUIVALENT, 91 ms] 52.03/14.50 (18) QDP 52.03/14.50 (19) QDPOrderProof [EQUIVALENT, 61 ms] 52.03/14.50 (20) QDP 52.03/14.50 (21) PisEmptyProof [EQUIVALENT, 0 ms] 52.03/14.50 (22) YES 52.03/14.50 52.03/14.50 52.03/14.50 ---------------------------------------- 52.03/14.50 52.03/14.50 (0) 52.03/14.50 Obligation: 52.03/14.50 Q restricted rewrite system: 52.03/14.50 The TRS R consists of the following rules: 52.03/14.50 52.03/14.50 0(1(2(x1))) -> 1(3(0(2(x1)))) 52.03/14.50 0(1(2(x1))) -> 3(1(2(0(x1)))) 52.03/14.50 0(1(2(x1))) -> 2(0(4(1(3(x1))))) 52.03/14.50 0(1(2(x1))) -> 3(0(2(1(3(x1))))) 52.03/14.50 0(1(2(x1))) -> 3(3(0(2(1(x1))))) 52.03/14.50 0(1(2(x1))) -> 1(3(3(0(2(3(x1)))))) 52.03/14.50 0(1(2(x1))) -> 2(0(4(3(1(3(x1)))))) 52.03/14.50 0(1(2(x1))) -> 3(0(1(3(1(2(x1)))))) 52.03/14.50 0(1(2(x1))) -> 3(0(5(3(1(2(x1)))))) 52.03/14.50 0(5(2(x1))) -> 1(5(0(2(x1)))) 52.03/14.50 0(5(2(x1))) -> 3(5(0(2(x1)))) 52.03/14.50 0(5(2(x1))) -> 3(0(5(1(2(x1))))) 52.03/14.50 0(5(2(x1))) -> 3(5(3(0(2(x1))))) 52.03/14.50 0(5(2(x1))) -> 1(5(5(3(0(2(x1)))))) 52.03/14.50 0(5(2(x1))) -> 5(0(4(3(1(2(x1)))))) 52.03/14.50 1(5(2(x1))) -> 5(3(1(2(x1)))) 52.03/14.50 1(5(2(x1))) -> 1(5(3(1(2(x1))))) 52.03/14.50 0(0(5(4(x1)))) -> 5(3(0(4(0(0(x1)))))) 52.03/14.50 0(1(0(1(x1)))) -> 1(1(3(0(3(0(x1)))))) 52.03/14.50 0(1(1(2(x1)))) -> 2(1(1(3(0(2(x1)))))) 52.03/14.50 0(1(2(5(x1)))) -> 3(0(5(1(2(x1))))) 52.03/14.50 0(1(2(5(x1)))) -> 3(5(3(1(2(0(x1)))))) 52.03/14.50 0(1(3(2(x1)))) -> 3(0(2(1(3(5(x1)))))) 52.03/14.50 0(1(4(2(x1)))) -> 3(0(0(4(1(2(x1)))))) 52.03/14.50 0(1(4(2(x1)))) -> 4(0(1(3(1(2(x1)))))) 52.03/14.50 0(5(2(4(x1)))) -> 5(5(3(0(2(4(x1)))))) 52.03/14.50 0(5(2(5(x1)))) -> 0(2(3(5(5(x1))))) 52.03/14.50 0(5(3(2(x1)))) -> 3(0(2(1(5(x1))))) 52.03/14.50 0(5(3(2(x1)))) -> 5(3(1(4(0(2(x1)))))) 52.03/14.50 1(0(3(4(x1)))) -> 1(0(4(1(3(x1))))) 52.03/14.50 1(5(3(2(x1)))) -> 3(1(3(5(2(x1))))) 52.03/14.50 5(0(1(2(x1)))) -> 1(5(0(4(1(2(x1)))))) 52.03/14.50 5(1(0(2(x1)))) -> 3(1(5(0(2(x1))))) 52.03/14.50 5(1(1(2(x1)))) -> 1(3(5(2(1(3(x1)))))) 52.03/14.50 5(1(5(2(x1)))) -> 1(3(5(5(2(x1))))) 52.03/14.50 0(0(1(3(2(x1))))) -> 0(3(1(2(0(3(x1)))))) 52.03/14.50 0(0(5(3(2(x1))))) -> 0(3(0(3(2(5(x1)))))) 52.03/14.50 0(1(0(3(4(x1))))) -> 4(0(1(3(0(3(x1)))))) 52.03/14.50 0(1(0(5(4(x1))))) -> 3(0(4(0(1(5(x1)))))) 52.03/14.50 0(1(2(4(4(x1))))) -> 0(4(1(2(1(4(x1)))))) 52.03/14.50 0(1(5(3(2(x1))))) -> 1(2(3(0(4(5(x1)))))) 52.03/14.50 0(5(4(2(4(x1))))) -> 4(5(3(0(4(2(x1)))))) 52.03/14.50 0(5(4(4(2(x1))))) -> 0(4(4(3(5(2(x1)))))) 52.03/14.50 1(0(1(3(4(x1))))) -> 4(1(1(3(0(3(x1)))))) 52.03/14.50 1(0(3(4(5(x1))))) -> 2(1(3(0(4(5(x1)))))) 52.03/14.50 1(1(3(4(2(x1))))) -> 2(5(1(3(1(4(x1)))))) 52.03/14.50 1(5(3(2(2(x1))))) -> 2(3(5(3(1(2(x1)))))) 52.03/14.50 5(0(5(2(4(x1))))) -> 5(1(5(0(4(2(x1)))))) 52.03/14.50 5(1(0(5(2(x1))))) -> 1(5(0(2(3(5(x1)))))) 52.03/14.50 5(1(1(2(1(x1))))) -> 3(1(2(5(1(1(x1)))))) 52.03/14.50 52.03/14.50 Q is empty. 52.03/14.50 52.03/14.50 ---------------------------------------- 52.03/14.50 52.03/14.50 (1) DependencyPairsProof (EQUIVALENT) 52.03/14.50 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 52.03/14.50 ---------------------------------------- 52.03/14.50 52.03/14.50 (2) 52.03/14.50 Obligation: 52.03/14.50 Q DP problem: 52.03/14.50 The TRS P consists of the following rules: 52.03/14.50 52.03/14.50 0^1(1(2(x1))) -> 1^1(3(0(2(x1)))) 52.03/14.50 0^1(1(2(x1))) -> 0^1(2(x1)) 52.03/14.50 0^1(1(2(x1))) -> 1^1(2(0(x1))) 52.03/14.50 0^1(1(2(x1))) -> 0^1(x1) 52.03/14.50 0^1(1(2(x1))) -> 0^1(4(1(3(x1)))) 52.03/14.50 0^1(1(2(x1))) -> 1^1(3(x1)) 52.03/14.50 0^1(1(2(x1))) -> 0^1(2(1(3(x1)))) 52.03/14.50 0^1(1(2(x1))) -> 0^1(2(1(x1))) 52.03/14.50 0^1(1(2(x1))) -> 1^1(x1) 52.03/14.50 0^1(1(2(x1))) -> 1^1(3(3(0(2(3(x1)))))) 52.03/14.50 0^1(1(2(x1))) -> 0^1(2(3(x1))) 52.03/14.50 0^1(1(2(x1))) -> 0^1(4(3(1(3(x1))))) 52.03/14.50 0^1(1(2(x1))) -> 0^1(1(3(1(2(x1))))) 52.03/14.50 0^1(1(2(x1))) -> 1^1(3(1(2(x1)))) 52.03/14.50 0^1(1(2(x1))) -> 0^1(5(3(1(2(x1))))) 52.03/14.50 0^1(1(2(x1))) -> 5^1(3(1(2(x1)))) 52.03/14.50 0^1(5(2(x1))) -> 1^1(5(0(2(x1)))) 52.03/14.50 0^1(5(2(x1))) -> 5^1(0(2(x1))) 52.03/14.50 0^1(5(2(x1))) -> 0^1(2(x1)) 52.03/14.50 0^1(5(2(x1))) -> 0^1(5(1(2(x1)))) 52.03/14.50 0^1(5(2(x1))) -> 5^1(1(2(x1))) 52.03/14.50 0^1(5(2(x1))) -> 1^1(2(x1)) 52.03/14.50 0^1(5(2(x1))) -> 5^1(3(0(2(x1)))) 52.03/14.50 0^1(5(2(x1))) -> 1^1(5(5(3(0(2(x1)))))) 52.03/14.50 0^1(5(2(x1))) -> 5^1(5(3(0(2(x1))))) 52.03/14.50 0^1(5(2(x1))) -> 5^1(0(4(3(1(2(x1)))))) 52.03/14.50 0^1(5(2(x1))) -> 0^1(4(3(1(2(x1))))) 52.03/14.50 1^1(5(2(x1))) -> 5^1(3(1(2(x1)))) 52.03/14.50 1^1(5(2(x1))) -> 1^1(2(x1)) 52.03/14.50 1^1(5(2(x1))) -> 1^1(5(3(1(2(x1))))) 52.03/14.50 0^1(0(5(4(x1)))) -> 5^1(3(0(4(0(0(x1)))))) 52.03/14.50 0^1(0(5(4(x1)))) -> 0^1(4(0(0(x1)))) 52.03/14.50 0^1(0(5(4(x1)))) -> 0^1(0(x1)) 52.03/14.50 0^1(0(5(4(x1)))) -> 0^1(x1) 52.03/14.50 0^1(1(0(1(x1)))) -> 1^1(1(3(0(3(0(x1)))))) 52.03/14.50 0^1(1(0(1(x1)))) -> 1^1(3(0(3(0(x1))))) 52.03/14.50 0^1(1(0(1(x1)))) -> 0^1(3(0(x1))) 52.03/14.50 0^1(1(0(1(x1)))) -> 0^1(x1) 52.03/14.50 0^1(1(1(2(x1)))) -> 1^1(1(3(0(2(x1))))) 52.03/14.50 0^1(1(1(2(x1)))) -> 1^1(3(0(2(x1)))) 52.03/14.50 0^1(1(1(2(x1)))) -> 0^1(2(x1)) 52.03/14.50 0^1(1(2(5(x1)))) -> 0^1(5(1(2(x1)))) 52.03/14.50 0^1(1(2(5(x1)))) -> 5^1(1(2(x1))) 52.03/14.50 0^1(1(2(5(x1)))) -> 1^1(2(x1)) 52.03/14.50 0^1(1(2(5(x1)))) -> 5^1(3(1(2(0(x1))))) 52.03/14.50 0^1(1(2(5(x1)))) -> 1^1(2(0(x1))) 52.03/14.50 0^1(1(2(5(x1)))) -> 0^1(x1) 52.03/14.50 0^1(1(3(2(x1)))) -> 0^1(2(1(3(5(x1))))) 52.03/14.50 0^1(1(3(2(x1)))) -> 1^1(3(5(x1))) 52.03/14.50 0^1(1(3(2(x1)))) -> 5^1(x1) 52.03/14.50 0^1(1(4(2(x1)))) -> 0^1(0(4(1(2(x1))))) 52.03/14.50 0^1(1(4(2(x1)))) -> 0^1(4(1(2(x1)))) 52.03/14.50 0^1(1(4(2(x1)))) -> 1^1(2(x1)) 52.03/14.50 0^1(1(4(2(x1)))) -> 0^1(1(3(1(2(x1))))) 52.03/14.50 0^1(1(4(2(x1)))) -> 1^1(3(1(2(x1)))) 52.03/14.50 0^1(5(2(4(x1)))) -> 5^1(5(3(0(2(4(x1)))))) 52.03/14.50 0^1(5(2(4(x1)))) -> 5^1(3(0(2(4(x1))))) 52.03/14.50 0^1(5(2(4(x1)))) -> 0^1(2(4(x1))) 52.03/14.50 0^1(5(2(5(x1)))) -> 0^1(2(3(5(5(x1))))) 52.03/14.50 0^1(5(2(5(x1)))) -> 5^1(5(x1)) 52.03/14.50 0^1(5(3(2(x1)))) -> 0^1(2(1(5(x1)))) 52.03/14.50 0^1(5(3(2(x1)))) -> 1^1(5(x1)) 52.03/14.50 0^1(5(3(2(x1)))) -> 5^1(x1) 52.03/14.50 0^1(5(3(2(x1)))) -> 5^1(3(1(4(0(2(x1)))))) 52.03/14.50 0^1(5(3(2(x1)))) -> 1^1(4(0(2(x1)))) 52.03/14.50 0^1(5(3(2(x1)))) -> 0^1(2(x1)) 52.03/14.50 1^1(0(3(4(x1)))) -> 1^1(0(4(1(3(x1))))) 52.03/14.50 1^1(0(3(4(x1)))) -> 0^1(4(1(3(x1)))) 52.03/14.50 1^1(0(3(4(x1)))) -> 1^1(3(x1)) 52.03/14.50 1^1(5(3(2(x1)))) -> 1^1(3(5(2(x1)))) 52.03/14.50 1^1(5(3(2(x1)))) -> 5^1(2(x1)) 52.03/14.50 5^1(0(1(2(x1)))) -> 1^1(5(0(4(1(2(x1)))))) 52.03/14.50 5^1(0(1(2(x1)))) -> 5^1(0(4(1(2(x1))))) 52.03/14.50 5^1(0(1(2(x1)))) -> 0^1(4(1(2(x1)))) 52.03/14.50 5^1(1(0(2(x1)))) -> 1^1(5(0(2(x1)))) 52.03/14.50 5^1(1(0(2(x1)))) -> 5^1(0(2(x1))) 52.03/14.50 5^1(1(1(2(x1)))) -> 1^1(3(5(2(1(3(x1)))))) 52.03/14.50 5^1(1(1(2(x1)))) -> 5^1(2(1(3(x1)))) 52.03/14.50 5^1(1(1(2(x1)))) -> 1^1(3(x1)) 52.03/14.50 5^1(1(5(2(x1)))) -> 1^1(3(5(5(2(x1))))) 52.03/14.50 5^1(1(5(2(x1)))) -> 5^1(5(2(x1))) 52.03/14.50 0^1(0(1(3(2(x1))))) -> 0^1(3(1(2(0(3(x1)))))) 52.03/14.50 0^1(0(1(3(2(x1))))) -> 1^1(2(0(3(x1)))) 52.03/14.50 0^1(0(1(3(2(x1))))) -> 0^1(3(x1)) 52.03/14.50 0^1(0(5(3(2(x1))))) -> 0^1(3(0(3(2(5(x1)))))) 52.03/14.50 0^1(0(5(3(2(x1))))) -> 0^1(3(2(5(x1)))) 52.03/14.50 0^1(0(5(3(2(x1))))) -> 5^1(x1) 52.03/14.50 0^1(1(0(3(4(x1))))) -> 0^1(1(3(0(3(x1))))) 52.03/14.50 0^1(1(0(3(4(x1))))) -> 1^1(3(0(3(x1)))) 52.03/14.50 0^1(1(0(3(4(x1))))) -> 0^1(3(x1)) 52.03/14.50 0^1(1(0(5(4(x1))))) -> 0^1(4(0(1(5(x1))))) 52.03/14.50 0^1(1(0(5(4(x1))))) -> 0^1(1(5(x1))) 52.03/14.50 0^1(1(0(5(4(x1))))) -> 1^1(5(x1)) 52.03/14.50 0^1(1(0(5(4(x1))))) -> 5^1(x1) 52.03/14.50 0^1(1(2(4(4(x1))))) -> 0^1(4(1(2(1(4(x1)))))) 52.03/14.50 0^1(1(2(4(4(x1))))) -> 1^1(2(1(4(x1)))) 52.03/14.50 0^1(1(2(4(4(x1))))) -> 1^1(4(x1)) 52.03/14.50 0^1(1(5(3(2(x1))))) -> 1^1(2(3(0(4(5(x1)))))) 52.03/14.50 0^1(1(5(3(2(x1))))) -> 0^1(4(5(x1))) 52.03/14.50 0^1(1(5(3(2(x1))))) -> 5^1(x1) 52.03/14.50 0^1(5(4(2(4(x1))))) -> 5^1(3(0(4(2(x1))))) 52.03/14.50 0^1(5(4(2(4(x1))))) -> 0^1(4(2(x1))) 52.03/14.50 0^1(5(4(4(2(x1))))) -> 0^1(4(4(3(5(2(x1)))))) 52.03/14.50 0^1(5(4(4(2(x1))))) -> 5^1(2(x1)) 52.03/14.50 1^1(0(1(3(4(x1))))) -> 1^1(1(3(0(3(x1))))) 52.03/14.50 1^1(0(1(3(4(x1))))) -> 1^1(3(0(3(x1)))) 52.03/14.50 1^1(0(1(3(4(x1))))) -> 0^1(3(x1)) 52.03/14.50 1^1(0(3(4(5(x1))))) -> 1^1(3(0(4(5(x1))))) 52.03/14.50 1^1(0(3(4(5(x1))))) -> 0^1(4(5(x1))) 52.03/14.50 1^1(1(3(4(2(x1))))) -> 5^1(1(3(1(4(x1))))) 52.03/14.50 1^1(1(3(4(2(x1))))) -> 1^1(3(1(4(x1)))) 52.03/14.50 1^1(1(3(4(2(x1))))) -> 1^1(4(x1)) 52.03/14.50 1^1(5(3(2(2(x1))))) -> 5^1(3(1(2(x1)))) 52.03/14.50 1^1(5(3(2(2(x1))))) -> 1^1(2(x1)) 52.03/14.50 5^1(0(5(2(4(x1))))) -> 5^1(1(5(0(4(2(x1)))))) 52.03/14.50 5^1(0(5(2(4(x1))))) -> 1^1(5(0(4(2(x1))))) 52.03/14.50 5^1(0(5(2(4(x1))))) -> 5^1(0(4(2(x1)))) 52.03/14.50 5^1(0(5(2(4(x1))))) -> 0^1(4(2(x1))) 52.03/14.50 5^1(1(0(5(2(x1))))) -> 1^1(5(0(2(3(5(x1)))))) 52.03/14.50 5^1(1(0(5(2(x1))))) -> 5^1(0(2(3(5(x1))))) 52.03/14.50 5^1(1(0(5(2(x1))))) -> 0^1(2(3(5(x1)))) 52.03/14.50 5^1(1(0(5(2(x1))))) -> 5^1(x1) 52.03/14.50 5^1(1(1(2(1(x1))))) -> 1^1(2(5(1(1(x1))))) 52.03/14.50 5^1(1(1(2(1(x1))))) -> 5^1(1(1(x1))) 52.03/14.50 5^1(1(1(2(1(x1))))) -> 1^1(1(x1)) 52.03/14.51 52.03/14.51 The TRS R consists of the following rules: 52.03/14.51 52.03/14.51 0(1(2(x1))) -> 1(3(0(2(x1)))) 52.03/14.51 0(1(2(x1))) -> 3(1(2(0(x1)))) 52.03/14.51 0(1(2(x1))) -> 2(0(4(1(3(x1))))) 52.03/14.51 0(1(2(x1))) -> 3(0(2(1(3(x1))))) 52.03/14.51 0(1(2(x1))) -> 3(3(0(2(1(x1))))) 52.03/14.51 0(1(2(x1))) -> 1(3(3(0(2(3(x1)))))) 52.03/14.51 0(1(2(x1))) -> 2(0(4(3(1(3(x1)))))) 52.03/14.51 0(1(2(x1))) -> 3(0(1(3(1(2(x1)))))) 52.03/14.51 0(1(2(x1))) -> 3(0(5(3(1(2(x1)))))) 52.03/14.51 0(5(2(x1))) -> 1(5(0(2(x1)))) 52.03/14.51 0(5(2(x1))) -> 3(5(0(2(x1)))) 52.03/14.51 0(5(2(x1))) -> 3(0(5(1(2(x1))))) 52.03/14.51 0(5(2(x1))) -> 3(5(3(0(2(x1))))) 52.03/14.51 0(5(2(x1))) -> 1(5(5(3(0(2(x1)))))) 52.03/14.51 0(5(2(x1))) -> 5(0(4(3(1(2(x1)))))) 52.03/14.51 1(5(2(x1))) -> 5(3(1(2(x1)))) 52.03/14.51 1(5(2(x1))) -> 1(5(3(1(2(x1))))) 52.03/14.51 0(0(5(4(x1)))) -> 5(3(0(4(0(0(x1)))))) 52.03/14.51 0(1(0(1(x1)))) -> 1(1(3(0(3(0(x1)))))) 52.03/14.51 0(1(1(2(x1)))) -> 2(1(1(3(0(2(x1)))))) 52.03/14.51 0(1(2(5(x1)))) -> 3(0(5(1(2(x1))))) 52.03/14.51 0(1(2(5(x1)))) -> 3(5(3(1(2(0(x1)))))) 52.03/14.51 0(1(3(2(x1)))) -> 3(0(2(1(3(5(x1)))))) 52.03/14.51 0(1(4(2(x1)))) -> 3(0(0(4(1(2(x1)))))) 52.03/14.51 0(1(4(2(x1)))) -> 4(0(1(3(1(2(x1)))))) 52.03/14.51 0(5(2(4(x1)))) -> 5(5(3(0(2(4(x1)))))) 52.03/14.51 0(5(2(5(x1)))) -> 0(2(3(5(5(x1))))) 52.03/14.51 0(5(3(2(x1)))) -> 3(0(2(1(5(x1))))) 52.03/14.51 0(5(3(2(x1)))) -> 5(3(1(4(0(2(x1)))))) 52.03/14.51 1(0(3(4(x1)))) -> 1(0(4(1(3(x1))))) 52.03/14.51 1(5(3(2(x1)))) -> 3(1(3(5(2(x1))))) 52.03/14.51 5(0(1(2(x1)))) -> 1(5(0(4(1(2(x1)))))) 52.03/14.51 5(1(0(2(x1)))) -> 3(1(5(0(2(x1))))) 52.03/14.51 5(1(1(2(x1)))) -> 1(3(5(2(1(3(x1)))))) 52.03/14.51 5(1(5(2(x1)))) -> 1(3(5(5(2(x1))))) 52.03/14.51 0(0(1(3(2(x1))))) -> 0(3(1(2(0(3(x1)))))) 52.03/14.51 0(0(5(3(2(x1))))) -> 0(3(0(3(2(5(x1)))))) 52.03/14.51 0(1(0(3(4(x1))))) -> 4(0(1(3(0(3(x1)))))) 52.03/14.51 0(1(0(5(4(x1))))) -> 3(0(4(0(1(5(x1)))))) 52.03/14.51 0(1(2(4(4(x1))))) -> 0(4(1(2(1(4(x1)))))) 52.03/14.51 0(1(5(3(2(x1))))) -> 1(2(3(0(4(5(x1)))))) 52.03/14.51 0(5(4(2(4(x1))))) -> 4(5(3(0(4(2(x1)))))) 52.03/14.51 0(5(4(4(2(x1))))) -> 0(4(4(3(5(2(x1)))))) 52.03/14.51 1(0(1(3(4(x1))))) -> 4(1(1(3(0(3(x1)))))) 52.03/14.51 1(0(3(4(5(x1))))) -> 2(1(3(0(4(5(x1)))))) 52.03/14.51 1(1(3(4(2(x1))))) -> 2(5(1(3(1(4(x1)))))) 52.03/14.51 1(5(3(2(2(x1))))) -> 2(3(5(3(1(2(x1)))))) 52.03/14.51 5(0(5(2(4(x1))))) -> 5(1(5(0(4(2(x1)))))) 52.03/14.51 5(1(0(5(2(x1))))) -> 1(5(0(2(3(5(x1)))))) 52.03/14.51 5(1(1(2(1(x1))))) -> 3(1(2(5(1(1(x1)))))) 52.03/14.51 52.03/14.51 Q is empty. 52.03/14.51 We have to consider all minimal (P,Q,R)-chains. 52.03/14.51 ---------------------------------------- 52.03/14.51 52.03/14.51 (3) DependencyGraphProof (EQUIVALENT) 52.03/14.51 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 117 less nodes. 52.03/14.51 ---------------------------------------- 52.03/14.51 52.03/14.51 (4) 52.03/14.51 Complex Obligation (AND) 52.03/14.51 52.03/14.51 ---------------------------------------- 52.03/14.51 52.03/14.51 (5) 52.03/14.51 Obligation: 52.03/14.51 Q DP problem: 52.03/14.51 The TRS P consists of the following rules: 52.03/14.51 52.03/14.51 5^1(1(1(2(1(x1))))) -> 5^1(1(1(x1))) 52.03/14.51 5^1(1(0(5(2(x1))))) -> 5^1(x1) 52.03/14.51 52.03/14.51 The TRS R consists of the following rules: 52.03/14.51 52.03/14.51 0(1(2(x1))) -> 1(3(0(2(x1)))) 52.03/14.51 0(1(2(x1))) -> 3(1(2(0(x1)))) 52.03/14.51 0(1(2(x1))) -> 2(0(4(1(3(x1))))) 52.03/14.51 0(1(2(x1))) -> 3(0(2(1(3(x1))))) 52.03/14.51 0(1(2(x1))) -> 3(3(0(2(1(x1))))) 52.03/14.51 0(1(2(x1))) -> 1(3(3(0(2(3(x1)))))) 52.03/14.51 0(1(2(x1))) -> 2(0(4(3(1(3(x1)))))) 52.03/14.51 0(1(2(x1))) -> 3(0(1(3(1(2(x1)))))) 52.03/14.51 0(1(2(x1))) -> 3(0(5(3(1(2(x1)))))) 52.03/14.51 0(5(2(x1))) -> 1(5(0(2(x1)))) 52.03/14.51 0(5(2(x1))) -> 3(5(0(2(x1)))) 52.03/14.51 0(5(2(x1))) -> 3(0(5(1(2(x1))))) 52.03/14.51 0(5(2(x1))) -> 3(5(3(0(2(x1))))) 52.03/14.51 0(5(2(x1))) -> 1(5(5(3(0(2(x1)))))) 52.03/14.51 0(5(2(x1))) -> 5(0(4(3(1(2(x1)))))) 52.03/14.51 1(5(2(x1))) -> 5(3(1(2(x1)))) 52.03/14.51 1(5(2(x1))) -> 1(5(3(1(2(x1))))) 52.03/14.51 0(0(5(4(x1)))) -> 5(3(0(4(0(0(x1)))))) 52.03/14.51 0(1(0(1(x1)))) -> 1(1(3(0(3(0(x1)))))) 52.03/14.51 0(1(1(2(x1)))) -> 2(1(1(3(0(2(x1)))))) 52.03/14.51 0(1(2(5(x1)))) -> 3(0(5(1(2(x1))))) 52.03/14.51 0(1(2(5(x1)))) -> 3(5(3(1(2(0(x1)))))) 52.03/14.51 0(1(3(2(x1)))) -> 3(0(2(1(3(5(x1)))))) 52.03/14.51 0(1(4(2(x1)))) -> 3(0(0(4(1(2(x1)))))) 52.03/14.51 0(1(4(2(x1)))) -> 4(0(1(3(1(2(x1)))))) 52.03/14.51 0(5(2(4(x1)))) -> 5(5(3(0(2(4(x1)))))) 52.03/14.51 0(5(2(5(x1)))) -> 0(2(3(5(5(x1))))) 52.03/14.51 0(5(3(2(x1)))) -> 3(0(2(1(5(x1))))) 52.03/14.51 0(5(3(2(x1)))) -> 5(3(1(4(0(2(x1)))))) 52.03/14.51 1(0(3(4(x1)))) -> 1(0(4(1(3(x1))))) 52.03/14.51 1(5(3(2(x1)))) -> 3(1(3(5(2(x1))))) 52.03/14.51 5(0(1(2(x1)))) -> 1(5(0(4(1(2(x1)))))) 52.03/14.51 5(1(0(2(x1)))) -> 3(1(5(0(2(x1))))) 52.03/14.51 5(1(1(2(x1)))) -> 1(3(5(2(1(3(x1)))))) 52.03/14.51 5(1(5(2(x1)))) -> 1(3(5(5(2(x1))))) 52.03/14.51 0(0(1(3(2(x1))))) -> 0(3(1(2(0(3(x1)))))) 52.03/14.51 0(0(5(3(2(x1))))) -> 0(3(0(3(2(5(x1)))))) 52.03/14.51 0(1(0(3(4(x1))))) -> 4(0(1(3(0(3(x1)))))) 52.03/14.51 0(1(0(5(4(x1))))) -> 3(0(4(0(1(5(x1)))))) 52.03/14.51 0(1(2(4(4(x1))))) -> 0(4(1(2(1(4(x1)))))) 52.03/14.51 0(1(5(3(2(x1))))) -> 1(2(3(0(4(5(x1)))))) 52.03/14.51 0(5(4(2(4(x1))))) -> 4(5(3(0(4(2(x1)))))) 52.03/14.51 0(5(4(4(2(x1))))) -> 0(4(4(3(5(2(x1)))))) 52.03/14.51 1(0(1(3(4(x1))))) -> 4(1(1(3(0(3(x1)))))) 52.03/14.51 1(0(3(4(5(x1))))) -> 2(1(3(0(4(5(x1)))))) 52.03/14.51 1(1(3(4(2(x1))))) -> 2(5(1(3(1(4(x1)))))) 52.03/14.51 1(5(3(2(2(x1))))) -> 2(3(5(3(1(2(x1)))))) 52.03/14.51 5(0(5(2(4(x1))))) -> 5(1(5(0(4(2(x1)))))) 52.03/14.51 5(1(0(5(2(x1))))) -> 1(5(0(2(3(5(x1)))))) 52.03/14.51 5(1(1(2(1(x1))))) -> 3(1(2(5(1(1(x1)))))) 52.03/14.51 52.03/14.51 Q is empty. 52.03/14.51 We have to consider all minimal (P,Q,R)-chains. 52.03/14.51 ---------------------------------------- 52.03/14.51 52.03/14.51 (6) UsableRulesProof (EQUIVALENT) 52.03/14.51 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. 52.03/14.51 ---------------------------------------- 52.03/14.51 52.03/14.51 (7) 52.03/14.51 Obligation: 52.03/14.51 Q DP problem: 52.03/14.51 The TRS P consists of the following rules: 52.03/14.51 52.03/14.51 5^1(1(1(2(1(x1))))) -> 5^1(1(1(x1))) 52.03/14.51 5^1(1(0(5(2(x1))))) -> 5^1(x1) 52.03/14.51 52.03/14.51 The TRS R consists of the following rules: 52.03/14.51 52.03/14.51 1(5(2(x1))) -> 5(3(1(2(x1)))) 52.03/14.51 1(5(2(x1))) -> 1(5(3(1(2(x1))))) 52.03/14.51 1(0(3(4(x1)))) -> 1(0(4(1(3(x1))))) 52.03/14.51 1(5(3(2(x1)))) -> 3(1(3(5(2(x1))))) 52.03/14.51 1(0(1(3(4(x1))))) -> 4(1(1(3(0(3(x1)))))) 52.03/14.51 1(0(3(4(5(x1))))) -> 2(1(3(0(4(5(x1)))))) 52.03/14.51 1(1(3(4(2(x1))))) -> 2(5(1(3(1(4(x1)))))) 52.03/14.51 1(5(3(2(2(x1))))) -> 2(3(5(3(1(2(x1)))))) 52.03/14.51 5(0(1(2(x1)))) -> 1(5(0(4(1(2(x1)))))) 52.03/14.51 5(1(0(2(x1)))) -> 3(1(5(0(2(x1))))) 52.03/14.51 5(1(1(2(x1)))) -> 1(3(5(2(1(3(x1)))))) 52.03/14.51 5(1(5(2(x1)))) -> 1(3(5(5(2(x1))))) 52.03/14.51 5(0(5(2(4(x1))))) -> 5(1(5(0(4(2(x1)))))) 52.03/14.51 5(1(0(5(2(x1))))) -> 1(5(0(2(3(5(x1)))))) 52.03/14.51 5(1(1(2(1(x1))))) -> 3(1(2(5(1(1(x1)))))) 52.03/14.51 52.03/14.51 Q is empty. 52.03/14.51 We have to consider all minimal (P,Q,R)-chains. 52.03/14.51 ---------------------------------------- 52.03/14.51 52.03/14.51 (8) MRRProof (EQUIVALENT) 52.03/14.51 By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. 52.03/14.51 52.03/14.51 Strictly oriented dependency pairs: 52.03/14.51 52.03/14.51 5^1(1(0(5(2(x1))))) -> 5^1(x1) 52.03/14.51 52.03/14.51 52.03/14.51 Used ordering: Polynomial interpretation [POLO]: 52.03/14.51 52.03/14.51 POL(0(x_1)) = 2 + 2*x_1 52.03/14.51 POL(1(x_1)) = x_1 52.03/14.51 POL(2(x_1)) = x_1 52.03/14.51 POL(3(x_1)) = x_1 52.03/14.51 POL(4(x_1)) = x_1 52.03/14.51 POL(5(x_1)) = x_1 52.03/14.51 POL(5^1(x_1)) = 2*x_1 52.03/14.51 52.03/14.51 52.03/14.51 ---------------------------------------- 52.03/14.51 52.03/14.51 (9) 52.03/14.51 Obligation: 52.03/14.51 Q DP problem: 52.03/14.51 The TRS P consists of the following rules: 52.03/14.51 52.03/14.51 5^1(1(1(2(1(x1))))) -> 5^1(1(1(x1))) 52.03/14.51 52.03/14.51 The TRS R consists of the following rules: 52.03/14.51 52.03/14.51 1(5(2(x1))) -> 5(3(1(2(x1)))) 52.03/14.51 1(5(2(x1))) -> 1(5(3(1(2(x1))))) 52.03/14.51 1(0(3(4(x1)))) -> 1(0(4(1(3(x1))))) 52.03/14.51 1(5(3(2(x1)))) -> 3(1(3(5(2(x1))))) 52.03/14.51 1(0(1(3(4(x1))))) -> 4(1(1(3(0(3(x1)))))) 52.03/14.51 1(0(3(4(5(x1))))) -> 2(1(3(0(4(5(x1)))))) 52.03/14.51 1(1(3(4(2(x1))))) -> 2(5(1(3(1(4(x1)))))) 52.03/14.51 1(5(3(2(2(x1))))) -> 2(3(5(3(1(2(x1)))))) 52.03/14.51 5(0(1(2(x1)))) -> 1(5(0(4(1(2(x1)))))) 52.03/14.51 5(1(0(2(x1)))) -> 3(1(5(0(2(x1))))) 52.03/14.51 5(1(1(2(x1)))) -> 1(3(5(2(1(3(x1)))))) 52.03/14.51 5(1(5(2(x1)))) -> 1(3(5(5(2(x1))))) 52.03/14.51 5(0(5(2(4(x1))))) -> 5(1(5(0(4(2(x1)))))) 52.03/14.51 5(1(0(5(2(x1))))) -> 1(5(0(2(3(5(x1)))))) 52.03/14.51 5(1(1(2(1(x1))))) -> 3(1(2(5(1(1(x1)))))) 52.03/14.51 52.03/14.51 Q is empty. 52.03/14.51 We have to consider all minimal (P,Q,R)-chains. 52.03/14.51 ---------------------------------------- 52.03/14.51 52.03/14.51 (10) QDPOrderProof (EQUIVALENT) 52.03/14.51 We use the reduction pair processor [LPAR04,JAR06]. 52.03/14.51 52.03/14.51 52.03/14.51 The following pairs can be oriented strictly and are deleted. 52.03/14.51 52.03/14.51 5^1(1(1(2(1(x1))))) -> 5^1(1(1(x1))) 52.03/14.51 The remaining pairs can at least be oriented weakly. 52.03/14.51 Used ordering: Polynomial interpretation [POLO]: 52.03/14.51 52.03/14.51 POL(0(x_1)) = 1 52.03/14.51 POL(1(x_1)) = 1 + x_1 52.03/14.51 POL(2(x_1)) = x_1 52.03/14.51 POL(3(x_1)) = 0 52.03/14.51 POL(4(x_1)) = 0 52.03/14.51 POL(5(x_1)) = 1 52.03/14.51 POL(5^1(x_1)) = x_1 52.03/14.51 52.03/14.51 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 52.03/14.51 52.03/14.51 1(5(2(x1))) -> 5(3(1(2(x1)))) 52.03/14.51 1(5(2(x1))) -> 1(5(3(1(2(x1))))) 52.03/14.51 1(0(3(4(x1)))) -> 1(0(4(1(3(x1))))) 52.03/14.51 1(5(3(2(x1)))) -> 3(1(3(5(2(x1))))) 52.03/14.51 1(0(1(3(4(x1))))) -> 4(1(1(3(0(3(x1)))))) 52.03/14.51 1(0(3(4(5(x1))))) -> 2(1(3(0(4(5(x1)))))) 52.03/14.52 1(1(3(4(2(x1))))) -> 2(5(1(3(1(4(x1)))))) 52.03/14.52 1(5(3(2(2(x1))))) -> 2(3(5(3(1(2(x1)))))) 52.03/14.52 52.03/14.52 52.03/14.52 ---------------------------------------- 52.03/14.52 52.03/14.52 (11) 52.03/14.52 Obligation: 52.03/14.52 Q DP problem: 52.03/14.52 P is empty. 52.03/14.52 The TRS R consists of the following rules: 52.03/14.52 52.03/14.52 1(5(2(x1))) -> 5(3(1(2(x1)))) 52.03/14.52 1(5(2(x1))) -> 1(5(3(1(2(x1))))) 52.03/14.52 1(0(3(4(x1)))) -> 1(0(4(1(3(x1))))) 52.03/14.52 1(5(3(2(x1)))) -> 3(1(3(5(2(x1))))) 52.03/14.52 1(0(1(3(4(x1))))) -> 4(1(1(3(0(3(x1)))))) 52.03/14.52 1(0(3(4(5(x1))))) -> 2(1(3(0(4(5(x1)))))) 52.03/14.52 1(1(3(4(2(x1))))) -> 2(5(1(3(1(4(x1)))))) 52.03/14.52 1(5(3(2(2(x1))))) -> 2(3(5(3(1(2(x1)))))) 52.03/14.52 5(0(1(2(x1)))) -> 1(5(0(4(1(2(x1)))))) 52.03/14.52 5(1(0(2(x1)))) -> 3(1(5(0(2(x1))))) 52.03/14.52 5(1(1(2(x1)))) -> 1(3(5(2(1(3(x1)))))) 52.03/14.52 5(1(5(2(x1)))) -> 1(3(5(5(2(x1))))) 52.03/14.52 5(0(5(2(4(x1))))) -> 5(1(5(0(4(2(x1)))))) 52.03/14.52 5(1(0(5(2(x1))))) -> 1(5(0(2(3(5(x1)))))) 52.03/14.52 5(1(1(2(1(x1))))) -> 3(1(2(5(1(1(x1)))))) 52.03/14.52 52.03/14.52 Q is empty. 52.03/14.52 We have to consider all minimal (P,Q,R)-chains. 52.03/14.52 ---------------------------------------- 52.03/14.52 52.03/14.52 (12) PisEmptyProof (EQUIVALENT) 52.03/14.52 The TRS P is empty. Hence, there is no (P,Q,R) chain. 52.03/14.52 ---------------------------------------- 52.03/14.52 52.03/14.52 (13) 52.03/14.52 YES 52.03/14.52 52.03/14.52 ---------------------------------------- 52.03/14.52 52.03/14.52 (14) 52.03/14.52 Obligation: 52.03/14.52 Q DP problem: 52.03/14.52 The TRS P consists of the following rules: 52.03/14.52 52.03/14.52 0^1(0(5(4(x1)))) -> 0^1(0(x1)) 52.03/14.52 0^1(1(2(x1))) -> 0^1(x1) 52.03/14.52 0^1(0(5(4(x1)))) -> 0^1(x1) 52.03/14.52 0^1(1(0(1(x1)))) -> 0^1(x1) 52.03/14.52 0^1(1(2(5(x1)))) -> 0^1(x1) 52.03/14.52 0^1(1(0(5(4(x1))))) -> 0^1(1(5(x1))) 52.03/14.52 52.03/14.52 The TRS R consists of the following rules: 52.03/14.52 52.03/14.52 0(1(2(x1))) -> 1(3(0(2(x1)))) 52.03/14.52 0(1(2(x1))) -> 3(1(2(0(x1)))) 52.03/14.52 0(1(2(x1))) -> 2(0(4(1(3(x1))))) 52.03/14.52 0(1(2(x1))) -> 3(0(2(1(3(x1))))) 52.03/14.52 0(1(2(x1))) -> 3(3(0(2(1(x1))))) 52.03/14.52 0(1(2(x1))) -> 1(3(3(0(2(3(x1)))))) 52.03/14.52 0(1(2(x1))) -> 2(0(4(3(1(3(x1)))))) 52.03/14.52 0(1(2(x1))) -> 3(0(1(3(1(2(x1)))))) 52.03/14.52 0(1(2(x1))) -> 3(0(5(3(1(2(x1)))))) 52.03/14.52 0(5(2(x1))) -> 1(5(0(2(x1)))) 52.03/14.52 0(5(2(x1))) -> 3(5(0(2(x1)))) 52.03/14.52 0(5(2(x1))) -> 3(0(5(1(2(x1))))) 52.03/14.52 0(5(2(x1))) -> 3(5(3(0(2(x1))))) 52.03/14.52 0(5(2(x1))) -> 1(5(5(3(0(2(x1)))))) 52.03/14.52 0(5(2(x1))) -> 5(0(4(3(1(2(x1)))))) 52.03/14.52 1(5(2(x1))) -> 5(3(1(2(x1)))) 52.03/14.52 1(5(2(x1))) -> 1(5(3(1(2(x1))))) 52.03/14.52 0(0(5(4(x1)))) -> 5(3(0(4(0(0(x1)))))) 52.03/14.52 0(1(0(1(x1)))) -> 1(1(3(0(3(0(x1)))))) 52.03/14.52 0(1(1(2(x1)))) -> 2(1(1(3(0(2(x1)))))) 52.03/14.52 0(1(2(5(x1)))) -> 3(0(5(1(2(x1))))) 52.03/14.52 0(1(2(5(x1)))) -> 3(5(3(1(2(0(x1)))))) 52.03/14.52 0(1(3(2(x1)))) -> 3(0(2(1(3(5(x1)))))) 52.03/14.52 0(1(4(2(x1)))) -> 3(0(0(4(1(2(x1)))))) 52.03/14.52 0(1(4(2(x1)))) -> 4(0(1(3(1(2(x1)))))) 52.03/14.52 0(5(2(4(x1)))) -> 5(5(3(0(2(4(x1)))))) 52.03/14.52 0(5(2(5(x1)))) -> 0(2(3(5(5(x1))))) 52.03/14.52 0(5(3(2(x1)))) -> 3(0(2(1(5(x1))))) 52.03/14.52 0(5(3(2(x1)))) -> 5(3(1(4(0(2(x1)))))) 52.03/14.52 1(0(3(4(x1)))) -> 1(0(4(1(3(x1))))) 52.03/14.52 1(5(3(2(x1)))) -> 3(1(3(5(2(x1))))) 52.03/14.52 5(0(1(2(x1)))) -> 1(5(0(4(1(2(x1)))))) 52.03/14.52 5(1(0(2(x1)))) -> 3(1(5(0(2(x1))))) 52.03/14.52 5(1(1(2(x1)))) -> 1(3(5(2(1(3(x1)))))) 52.03/14.52 5(1(5(2(x1)))) -> 1(3(5(5(2(x1))))) 52.03/14.52 0(0(1(3(2(x1))))) -> 0(3(1(2(0(3(x1)))))) 52.03/14.52 0(0(5(3(2(x1))))) -> 0(3(0(3(2(5(x1)))))) 52.03/14.52 0(1(0(3(4(x1))))) -> 4(0(1(3(0(3(x1)))))) 52.03/14.52 0(1(0(5(4(x1))))) -> 3(0(4(0(1(5(x1)))))) 52.03/14.52 0(1(2(4(4(x1))))) -> 0(4(1(2(1(4(x1)))))) 52.03/14.52 0(1(5(3(2(x1))))) -> 1(2(3(0(4(5(x1)))))) 52.03/14.52 0(5(4(2(4(x1))))) -> 4(5(3(0(4(2(x1)))))) 52.03/14.52 0(5(4(4(2(x1))))) -> 0(4(4(3(5(2(x1)))))) 52.03/14.52 1(0(1(3(4(x1))))) -> 4(1(1(3(0(3(x1)))))) 52.03/14.52 1(0(3(4(5(x1))))) -> 2(1(3(0(4(5(x1)))))) 52.03/14.52 1(1(3(4(2(x1))))) -> 2(5(1(3(1(4(x1)))))) 52.03/14.52 1(5(3(2(2(x1))))) -> 2(3(5(3(1(2(x1)))))) 52.03/14.52 5(0(5(2(4(x1))))) -> 5(1(5(0(4(2(x1)))))) 52.03/14.52 5(1(0(5(2(x1))))) -> 1(5(0(2(3(5(x1)))))) 52.03/14.52 5(1(1(2(1(x1))))) -> 3(1(2(5(1(1(x1)))))) 52.03/14.52 52.03/14.52 Q is empty. 52.03/14.52 We have to consider all minimal (P,Q,R)-chains. 52.03/14.52 ---------------------------------------- 52.03/14.52 52.03/14.52 (15) QDPOrderProof (EQUIVALENT) 52.03/14.52 We use the reduction pair processor [LPAR04,JAR06]. 52.03/14.52 52.03/14.52 52.03/14.52 The following pairs can be oriented strictly and are deleted. 52.03/14.52 52.03/14.52 0^1(0(5(4(x1)))) -> 0^1(x1) 52.03/14.52 0^1(1(0(1(x1)))) -> 0^1(x1) 52.03/14.52 0^1(1(0(5(4(x1))))) -> 0^1(1(5(x1))) 52.03/14.52 The remaining pairs can at least be oriented weakly. 52.03/14.52 Used ordering: Polynomial interpretation [POLO]: 52.03/14.52 52.03/14.52 POL(0(x_1)) = 1 + x_1 52.03/14.52 POL(0^1(x_1)) = x_1 52.03/14.52 POL(1(x_1)) = x_1 52.03/14.52 POL(2(x_1)) = x_1 52.03/14.52 POL(3(x_1)) = 0 52.03/14.52 POL(4(x_1)) = x_1 52.03/14.52 POL(5(x_1)) = x_1 52.03/14.52 52.03/14.52 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 52.03/14.52 52.03/14.52 0(1(2(x1))) -> 1(3(0(2(x1)))) 52.03/14.52 0(1(2(x1))) -> 3(1(2(0(x1)))) 52.03/14.52 0(1(2(x1))) -> 2(0(4(1(3(x1))))) 52.03/14.52 0(1(2(x1))) -> 3(0(2(1(3(x1))))) 52.03/14.52 0(1(2(x1))) -> 3(3(0(2(1(x1))))) 52.03/14.52 0(1(2(x1))) -> 1(3(3(0(2(3(x1)))))) 52.03/14.52 0(1(2(x1))) -> 2(0(4(3(1(3(x1)))))) 52.03/14.52 0(1(2(x1))) -> 3(0(1(3(1(2(x1)))))) 52.03/14.52 0(1(2(x1))) -> 3(0(5(3(1(2(x1)))))) 52.03/14.52 0(5(2(x1))) -> 1(5(0(2(x1)))) 52.03/14.52 0(5(2(x1))) -> 3(5(0(2(x1)))) 52.03/14.52 0(5(2(x1))) -> 3(0(5(1(2(x1))))) 52.03/14.52 0(5(2(x1))) -> 3(5(3(0(2(x1))))) 52.03/14.52 0(5(2(x1))) -> 1(5(5(3(0(2(x1)))))) 52.03/14.52 0(5(2(x1))) -> 5(0(4(3(1(2(x1)))))) 52.03/14.52 0(0(5(4(x1)))) -> 5(3(0(4(0(0(x1)))))) 52.03/14.52 0(1(0(1(x1)))) -> 1(1(3(0(3(0(x1)))))) 52.03/14.52 0(1(1(2(x1)))) -> 2(1(1(3(0(2(x1)))))) 52.03/14.52 0(1(2(5(x1)))) -> 3(0(5(1(2(x1))))) 52.03/14.52 0(1(2(5(x1)))) -> 3(5(3(1(2(0(x1)))))) 52.03/14.52 0(1(3(2(x1)))) -> 3(0(2(1(3(5(x1)))))) 52.03/14.52 0(1(4(2(x1)))) -> 3(0(0(4(1(2(x1)))))) 52.03/14.52 0(1(4(2(x1)))) -> 4(0(1(3(1(2(x1)))))) 52.03/14.52 0(5(2(4(x1)))) -> 5(5(3(0(2(4(x1)))))) 52.03/14.52 0(5(2(5(x1)))) -> 0(2(3(5(5(x1))))) 52.03/14.52 0(5(3(2(x1)))) -> 3(0(2(1(5(x1))))) 52.03/14.52 0(5(3(2(x1)))) -> 5(3(1(4(0(2(x1)))))) 52.03/14.52 0(0(1(3(2(x1))))) -> 0(3(1(2(0(3(x1)))))) 52.03/14.52 0(0(5(3(2(x1))))) -> 0(3(0(3(2(5(x1)))))) 52.03/14.52 0(1(0(3(4(x1))))) -> 4(0(1(3(0(3(x1)))))) 52.03/14.52 0(1(0(5(4(x1))))) -> 3(0(4(0(1(5(x1)))))) 52.03/14.52 0(1(2(4(4(x1))))) -> 0(4(1(2(1(4(x1)))))) 52.03/14.52 0(1(5(3(2(x1))))) -> 1(2(3(0(4(5(x1)))))) 52.03/14.52 0(5(4(2(4(x1))))) -> 4(5(3(0(4(2(x1)))))) 52.03/14.52 0(5(4(4(2(x1))))) -> 0(4(4(3(5(2(x1)))))) 52.03/14.52 5(0(1(2(x1)))) -> 1(5(0(4(1(2(x1)))))) 52.03/14.52 5(1(0(2(x1)))) -> 3(1(5(0(2(x1))))) 52.03/14.52 5(1(1(2(x1)))) -> 1(3(5(2(1(3(x1)))))) 52.03/14.52 5(1(5(2(x1)))) -> 1(3(5(5(2(x1))))) 52.03/14.52 5(0(5(2(4(x1))))) -> 5(1(5(0(4(2(x1)))))) 52.03/14.52 5(1(0(5(2(x1))))) -> 1(5(0(2(3(5(x1)))))) 52.03/14.52 5(1(1(2(1(x1))))) -> 3(1(2(5(1(1(x1)))))) 52.03/14.52 1(5(2(x1))) -> 5(3(1(2(x1)))) 52.03/14.52 1(5(2(x1))) -> 1(5(3(1(2(x1))))) 52.03/14.52 1(0(3(4(x1)))) -> 1(0(4(1(3(x1))))) 52.03/14.52 1(5(3(2(x1)))) -> 3(1(3(5(2(x1))))) 52.03/14.52 1(0(1(3(4(x1))))) -> 4(1(1(3(0(3(x1)))))) 52.03/14.52 1(0(3(4(5(x1))))) -> 2(1(3(0(4(5(x1)))))) 52.03/14.52 1(1(3(4(2(x1))))) -> 2(5(1(3(1(4(x1)))))) 52.03/14.52 1(5(3(2(2(x1))))) -> 2(3(5(3(1(2(x1)))))) 52.03/14.52 52.03/14.52 52.03/14.52 ---------------------------------------- 52.03/14.52 52.03/14.52 (16) 52.03/14.52 Obligation: 52.03/14.52 Q DP problem: 52.03/14.52 The TRS P consists of the following rules: 52.03/14.52 52.03/14.52 0^1(0(5(4(x1)))) -> 0^1(0(x1)) 52.03/14.52 0^1(1(2(x1))) -> 0^1(x1) 52.03/14.52 0^1(1(2(5(x1)))) -> 0^1(x1) 52.03/14.52 52.03/14.52 The TRS R consists of the following rules: 52.03/14.52 52.03/14.52 0(1(2(x1))) -> 1(3(0(2(x1)))) 52.03/14.52 0(1(2(x1))) -> 3(1(2(0(x1)))) 52.03/14.52 0(1(2(x1))) -> 2(0(4(1(3(x1))))) 52.03/14.52 0(1(2(x1))) -> 3(0(2(1(3(x1))))) 52.03/14.52 0(1(2(x1))) -> 3(3(0(2(1(x1))))) 52.03/14.52 0(1(2(x1))) -> 1(3(3(0(2(3(x1)))))) 52.03/14.52 0(1(2(x1))) -> 2(0(4(3(1(3(x1)))))) 52.03/14.52 0(1(2(x1))) -> 3(0(1(3(1(2(x1)))))) 52.03/14.52 0(1(2(x1))) -> 3(0(5(3(1(2(x1)))))) 52.03/14.52 0(5(2(x1))) -> 1(5(0(2(x1)))) 52.03/14.52 0(5(2(x1))) -> 3(5(0(2(x1)))) 52.03/14.52 0(5(2(x1))) -> 3(0(5(1(2(x1))))) 52.03/14.52 0(5(2(x1))) -> 3(5(3(0(2(x1))))) 52.03/14.52 0(5(2(x1))) -> 1(5(5(3(0(2(x1)))))) 52.03/14.52 0(5(2(x1))) -> 5(0(4(3(1(2(x1)))))) 52.03/14.52 1(5(2(x1))) -> 5(3(1(2(x1)))) 52.03/14.52 1(5(2(x1))) -> 1(5(3(1(2(x1))))) 52.03/14.52 0(0(5(4(x1)))) -> 5(3(0(4(0(0(x1)))))) 52.03/14.52 0(1(0(1(x1)))) -> 1(1(3(0(3(0(x1)))))) 52.03/14.52 0(1(1(2(x1)))) -> 2(1(1(3(0(2(x1)))))) 52.03/14.52 0(1(2(5(x1)))) -> 3(0(5(1(2(x1))))) 52.03/14.52 0(1(2(5(x1)))) -> 3(5(3(1(2(0(x1)))))) 52.03/14.52 0(1(3(2(x1)))) -> 3(0(2(1(3(5(x1)))))) 52.03/14.52 0(1(4(2(x1)))) -> 3(0(0(4(1(2(x1)))))) 52.03/14.52 0(1(4(2(x1)))) -> 4(0(1(3(1(2(x1)))))) 52.03/14.52 0(5(2(4(x1)))) -> 5(5(3(0(2(4(x1)))))) 52.03/14.52 0(5(2(5(x1)))) -> 0(2(3(5(5(x1))))) 52.03/14.52 0(5(3(2(x1)))) -> 3(0(2(1(5(x1))))) 52.03/14.52 0(5(3(2(x1)))) -> 5(3(1(4(0(2(x1)))))) 52.03/14.52 1(0(3(4(x1)))) -> 1(0(4(1(3(x1))))) 52.03/14.52 1(5(3(2(x1)))) -> 3(1(3(5(2(x1))))) 52.03/14.52 5(0(1(2(x1)))) -> 1(5(0(4(1(2(x1)))))) 52.03/14.52 5(1(0(2(x1)))) -> 3(1(5(0(2(x1))))) 52.03/14.52 5(1(1(2(x1)))) -> 1(3(5(2(1(3(x1)))))) 52.03/14.52 5(1(5(2(x1)))) -> 1(3(5(5(2(x1))))) 52.03/14.52 0(0(1(3(2(x1))))) -> 0(3(1(2(0(3(x1)))))) 52.03/14.52 0(0(5(3(2(x1))))) -> 0(3(0(3(2(5(x1)))))) 52.03/14.52 0(1(0(3(4(x1))))) -> 4(0(1(3(0(3(x1)))))) 52.03/14.52 0(1(0(5(4(x1))))) -> 3(0(4(0(1(5(x1)))))) 52.03/14.52 0(1(2(4(4(x1))))) -> 0(4(1(2(1(4(x1)))))) 52.03/14.52 0(1(5(3(2(x1))))) -> 1(2(3(0(4(5(x1)))))) 52.03/14.52 0(5(4(2(4(x1))))) -> 4(5(3(0(4(2(x1)))))) 52.03/14.52 0(5(4(4(2(x1))))) -> 0(4(4(3(5(2(x1)))))) 52.03/14.53 1(0(1(3(4(x1))))) -> 4(1(1(3(0(3(x1)))))) 52.03/14.53 1(0(3(4(5(x1))))) -> 2(1(3(0(4(5(x1)))))) 52.03/14.53 1(1(3(4(2(x1))))) -> 2(5(1(3(1(4(x1)))))) 52.03/14.53 1(5(3(2(2(x1))))) -> 2(3(5(3(1(2(x1)))))) 52.03/14.53 5(0(5(2(4(x1))))) -> 5(1(5(0(4(2(x1)))))) 52.03/14.53 5(1(0(5(2(x1))))) -> 1(5(0(2(3(5(x1)))))) 52.03/14.53 5(1(1(2(1(x1))))) -> 3(1(2(5(1(1(x1)))))) 52.03/14.53 52.03/14.53 Q is empty. 52.03/14.53 We have to consider all minimal (P,Q,R)-chains. 52.03/14.53 ---------------------------------------- 52.03/14.53 52.03/14.53 (17) QDPOrderProof (EQUIVALENT) 52.03/14.53 We use the reduction pair processor [LPAR04,JAR06]. 52.03/14.53 52.03/14.53 52.03/14.53 The following pairs can be oriented strictly and are deleted. 52.03/14.53 52.03/14.53 0^1(1(2(x1))) -> 0^1(x1) 52.03/14.53 0^1(1(2(5(x1)))) -> 0^1(x1) 52.03/14.53 The remaining pairs can at least be oriented weakly. 52.03/14.53 Used ordering: Polynomial interpretation [POLO]: 52.03/14.53 52.03/14.53 POL(0(x_1)) = 1 + x_1 52.03/14.53 POL(0^1(x_1)) = x_1 52.03/14.53 POL(1(x_1)) = x_1 52.03/14.53 POL(2(x_1)) = 1 + x_1 52.03/14.53 POL(3(x_1)) = 0 52.03/14.53 POL(4(x_1)) = x_1 52.03/14.53 POL(5(x_1)) = x_1 52.03/14.53 52.03/14.53 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 52.03/14.53 52.03/14.53 0(1(2(x1))) -> 1(3(0(2(x1)))) 52.03/14.53 0(1(2(x1))) -> 3(1(2(0(x1)))) 52.03/14.53 0(1(2(x1))) -> 2(0(4(1(3(x1))))) 52.03/14.53 0(1(2(x1))) -> 3(0(2(1(3(x1))))) 52.03/14.53 0(1(2(x1))) -> 3(3(0(2(1(x1))))) 52.03/14.53 0(1(2(x1))) -> 1(3(3(0(2(3(x1)))))) 52.03/14.53 0(1(2(x1))) -> 2(0(4(3(1(3(x1)))))) 52.03/14.53 0(1(2(x1))) -> 3(0(1(3(1(2(x1)))))) 52.03/14.53 0(1(2(x1))) -> 3(0(5(3(1(2(x1)))))) 52.03/14.53 0(5(2(x1))) -> 1(5(0(2(x1)))) 52.03/14.53 0(5(2(x1))) -> 3(5(0(2(x1)))) 52.03/14.53 0(5(2(x1))) -> 3(0(5(1(2(x1))))) 52.03/14.53 0(5(2(x1))) -> 3(5(3(0(2(x1))))) 52.03/14.53 0(5(2(x1))) -> 1(5(5(3(0(2(x1)))))) 52.03/14.53 0(5(2(x1))) -> 5(0(4(3(1(2(x1)))))) 52.03/14.53 0(0(5(4(x1)))) -> 5(3(0(4(0(0(x1)))))) 52.03/14.53 0(1(0(1(x1)))) -> 1(1(3(0(3(0(x1)))))) 52.03/14.53 0(1(1(2(x1)))) -> 2(1(1(3(0(2(x1)))))) 52.03/14.53 0(1(2(5(x1)))) -> 3(0(5(1(2(x1))))) 52.03/14.53 0(1(2(5(x1)))) -> 3(5(3(1(2(0(x1)))))) 52.03/14.53 0(1(3(2(x1)))) -> 3(0(2(1(3(5(x1)))))) 52.03/14.53 0(1(4(2(x1)))) -> 3(0(0(4(1(2(x1)))))) 52.03/14.53 0(1(4(2(x1)))) -> 4(0(1(3(1(2(x1)))))) 52.03/14.53 0(5(2(4(x1)))) -> 5(5(3(0(2(4(x1)))))) 52.03/14.53 0(5(2(5(x1)))) -> 0(2(3(5(5(x1))))) 52.03/14.53 0(5(3(2(x1)))) -> 3(0(2(1(5(x1))))) 52.03/14.53 0(5(3(2(x1)))) -> 5(3(1(4(0(2(x1)))))) 52.03/14.53 0(0(1(3(2(x1))))) -> 0(3(1(2(0(3(x1)))))) 52.03/14.53 0(0(5(3(2(x1))))) -> 0(3(0(3(2(5(x1)))))) 52.03/14.53 0(1(0(3(4(x1))))) -> 4(0(1(3(0(3(x1)))))) 52.03/14.53 0(1(0(5(4(x1))))) -> 3(0(4(0(1(5(x1)))))) 52.03/14.53 0(1(2(4(4(x1))))) -> 0(4(1(2(1(4(x1)))))) 52.03/14.53 0(1(5(3(2(x1))))) -> 1(2(3(0(4(5(x1)))))) 52.03/14.53 0(5(4(2(4(x1))))) -> 4(5(3(0(4(2(x1)))))) 52.03/14.53 0(5(4(4(2(x1))))) -> 0(4(4(3(5(2(x1)))))) 52.03/14.53 52.03/14.53 52.03/14.53 ---------------------------------------- 52.03/14.53 52.03/14.53 (18) 52.03/14.53 Obligation: 52.03/14.53 Q DP problem: 52.03/14.53 The TRS P consists of the following rules: 52.03/14.53 52.03/14.53 0^1(0(5(4(x1)))) -> 0^1(0(x1)) 52.03/14.53 52.03/14.53 The TRS R consists of the following rules: 52.03/14.53 52.03/14.53 0(1(2(x1))) -> 1(3(0(2(x1)))) 52.03/14.53 0(1(2(x1))) -> 3(1(2(0(x1)))) 52.03/14.53 0(1(2(x1))) -> 2(0(4(1(3(x1))))) 52.03/14.53 0(1(2(x1))) -> 3(0(2(1(3(x1))))) 52.03/14.53 0(1(2(x1))) -> 3(3(0(2(1(x1))))) 52.03/14.53 0(1(2(x1))) -> 1(3(3(0(2(3(x1)))))) 52.03/14.53 0(1(2(x1))) -> 2(0(4(3(1(3(x1)))))) 52.03/14.53 0(1(2(x1))) -> 3(0(1(3(1(2(x1)))))) 52.03/14.53 0(1(2(x1))) -> 3(0(5(3(1(2(x1)))))) 52.03/14.53 0(5(2(x1))) -> 1(5(0(2(x1)))) 52.03/14.53 0(5(2(x1))) -> 3(5(0(2(x1)))) 52.03/14.53 0(5(2(x1))) -> 3(0(5(1(2(x1))))) 52.03/14.53 0(5(2(x1))) -> 3(5(3(0(2(x1))))) 52.03/14.53 0(5(2(x1))) -> 1(5(5(3(0(2(x1)))))) 52.03/14.53 0(5(2(x1))) -> 5(0(4(3(1(2(x1)))))) 52.03/14.53 1(5(2(x1))) -> 5(3(1(2(x1)))) 52.03/14.53 1(5(2(x1))) -> 1(5(3(1(2(x1))))) 52.03/14.53 0(0(5(4(x1)))) -> 5(3(0(4(0(0(x1)))))) 52.03/14.53 0(1(0(1(x1)))) -> 1(1(3(0(3(0(x1)))))) 52.03/14.53 0(1(1(2(x1)))) -> 2(1(1(3(0(2(x1)))))) 52.03/14.53 0(1(2(5(x1)))) -> 3(0(5(1(2(x1))))) 52.03/14.53 0(1(2(5(x1)))) -> 3(5(3(1(2(0(x1)))))) 52.03/14.53 0(1(3(2(x1)))) -> 3(0(2(1(3(5(x1)))))) 52.03/14.53 0(1(4(2(x1)))) -> 3(0(0(4(1(2(x1)))))) 52.03/14.53 0(1(4(2(x1)))) -> 4(0(1(3(1(2(x1)))))) 52.03/14.53 0(5(2(4(x1)))) -> 5(5(3(0(2(4(x1)))))) 52.03/14.53 0(5(2(5(x1)))) -> 0(2(3(5(5(x1))))) 52.03/14.53 0(5(3(2(x1)))) -> 3(0(2(1(5(x1))))) 52.03/14.53 0(5(3(2(x1)))) -> 5(3(1(4(0(2(x1)))))) 52.03/14.53 1(0(3(4(x1)))) -> 1(0(4(1(3(x1))))) 52.03/14.53 1(5(3(2(x1)))) -> 3(1(3(5(2(x1))))) 52.03/14.53 5(0(1(2(x1)))) -> 1(5(0(4(1(2(x1)))))) 52.03/14.53 5(1(0(2(x1)))) -> 3(1(5(0(2(x1))))) 52.03/14.53 5(1(1(2(x1)))) -> 1(3(5(2(1(3(x1)))))) 52.03/14.53 5(1(5(2(x1)))) -> 1(3(5(5(2(x1))))) 52.03/14.53 0(0(1(3(2(x1))))) -> 0(3(1(2(0(3(x1)))))) 52.03/14.53 0(0(5(3(2(x1))))) -> 0(3(0(3(2(5(x1)))))) 52.03/14.53 0(1(0(3(4(x1))))) -> 4(0(1(3(0(3(x1)))))) 52.03/14.53 0(1(0(5(4(x1))))) -> 3(0(4(0(1(5(x1)))))) 52.03/14.53 0(1(2(4(4(x1))))) -> 0(4(1(2(1(4(x1)))))) 52.03/14.53 0(1(5(3(2(x1))))) -> 1(2(3(0(4(5(x1)))))) 52.03/14.53 0(5(4(2(4(x1))))) -> 4(5(3(0(4(2(x1)))))) 52.03/14.53 0(5(4(4(2(x1))))) -> 0(4(4(3(5(2(x1)))))) 52.03/14.53 1(0(1(3(4(x1))))) -> 4(1(1(3(0(3(x1)))))) 52.03/14.53 1(0(3(4(5(x1))))) -> 2(1(3(0(4(5(x1)))))) 52.03/14.53 1(1(3(4(2(x1))))) -> 2(5(1(3(1(4(x1)))))) 52.03/14.53 1(5(3(2(2(x1))))) -> 2(3(5(3(1(2(x1)))))) 52.03/14.53 5(0(5(2(4(x1))))) -> 5(1(5(0(4(2(x1)))))) 52.03/14.53 5(1(0(5(2(x1))))) -> 1(5(0(2(3(5(x1)))))) 52.03/14.53 5(1(1(2(1(x1))))) -> 3(1(2(5(1(1(x1)))))) 52.03/14.53 52.03/14.53 Q is empty. 52.03/14.53 We have to consider all minimal (P,Q,R)-chains. 52.03/14.53 ---------------------------------------- 52.03/14.53 52.03/14.53 (19) QDPOrderProof (EQUIVALENT) 52.03/14.53 We use the reduction pair processor [LPAR04,JAR06]. 52.03/14.53 52.03/14.53 52.03/14.53 The following pairs can be oriented strictly and are deleted. 52.03/14.53 52.03/14.53 0^1(0(5(4(x1)))) -> 0^1(0(x1)) 52.03/14.53 The remaining pairs can at least be oriented weakly. 52.03/14.53 Used ordering: Polynomial interpretation [POLO]: 52.03/14.53 52.03/14.53 POL(0(x_1)) = 1 + x_1 52.03/14.53 POL(0^1(x_1)) = x_1 52.03/14.53 POL(1(x_1)) = 0 52.03/14.53 POL(2(x_1)) = 0 52.03/14.53 POL(3(x_1)) = 0 52.03/14.53 POL(4(x_1)) = x_1 52.03/14.53 POL(5(x_1)) = 1 + x_1 52.03/14.53 52.03/14.53 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 52.03/14.53 52.03/14.53 0(1(2(x1))) -> 1(3(0(2(x1)))) 52.03/14.53 0(1(2(x1))) -> 3(1(2(0(x1)))) 52.03/14.53 0(1(2(x1))) -> 2(0(4(1(3(x1))))) 52.03/14.53 0(1(2(x1))) -> 3(0(2(1(3(x1))))) 52.03/14.53 0(1(2(x1))) -> 3(3(0(2(1(x1))))) 52.03/14.53 0(1(2(x1))) -> 1(3(3(0(2(3(x1)))))) 52.03/14.53 0(1(2(x1))) -> 2(0(4(3(1(3(x1)))))) 52.03/14.53 0(1(2(x1))) -> 3(0(1(3(1(2(x1)))))) 52.03/14.53 0(1(2(x1))) -> 3(0(5(3(1(2(x1)))))) 52.03/14.53 0(5(2(x1))) -> 1(5(0(2(x1)))) 52.03/14.53 0(5(2(x1))) -> 3(5(0(2(x1)))) 52.03/14.53 0(5(2(x1))) -> 3(0(5(1(2(x1))))) 52.03/14.53 0(5(2(x1))) -> 3(5(3(0(2(x1))))) 52.03/14.53 0(5(2(x1))) -> 1(5(5(3(0(2(x1)))))) 52.03/14.53 0(5(2(x1))) -> 5(0(4(3(1(2(x1)))))) 52.03/14.53 0(0(5(4(x1)))) -> 5(3(0(4(0(0(x1)))))) 52.03/14.53 0(1(0(1(x1)))) -> 1(1(3(0(3(0(x1)))))) 52.03/14.53 0(1(1(2(x1)))) -> 2(1(1(3(0(2(x1)))))) 52.03/14.53 0(1(2(5(x1)))) -> 3(0(5(1(2(x1))))) 52.03/14.53 0(1(2(5(x1)))) -> 3(5(3(1(2(0(x1)))))) 52.03/14.53 0(1(3(2(x1)))) -> 3(0(2(1(3(5(x1)))))) 52.03/14.53 0(1(4(2(x1)))) -> 3(0(0(4(1(2(x1)))))) 52.03/14.53 0(1(4(2(x1)))) -> 4(0(1(3(1(2(x1)))))) 52.03/14.53 0(5(2(4(x1)))) -> 5(5(3(0(2(4(x1)))))) 52.03/14.53 0(5(2(5(x1)))) -> 0(2(3(5(5(x1))))) 52.03/14.53 0(5(3(2(x1)))) -> 3(0(2(1(5(x1))))) 52.03/14.53 0(5(3(2(x1)))) -> 5(3(1(4(0(2(x1)))))) 52.03/14.53 0(0(1(3(2(x1))))) -> 0(3(1(2(0(3(x1)))))) 52.03/14.53 0(0(5(3(2(x1))))) -> 0(3(0(3(2(5(x1)))))) 52.03/14.53 0(1(0(3(4(x1))))) -> 4(0(1(3(0(3(x1)))))) 52.03/14.53 0(1(0(5(4(x1))))) -> 3(0(4(0(1(5(x1)))))) 52.03/14.53 0(1(2(4(4(x1))))) -> 0(4(1(2(1(4(x1)))))) 52.03/14.53 0(1(5(3(2(x1))))) -> 1(2(3(0(4(5(x1)))))) 52.03/14.53 0(5(4(2(4(x1))))) -> 4(5(3(0(4(2(x1)))))) 52.03/14.53 0(5(4(4(2(x1))))) -> 0(4(4(3(5(2(x1)))))) 52.03/14.53 52.03/14.53 52.03/14.53 ---------------------------------------- 52.03/14.53 52.03/14.53 (20) 52.03/14.53 Obligation: 52.03/14.53 Q DP problem: 52.03/14.53 P is empty. 52.03/14.53 The TRS R consists of the following rules: 52.03/14.53 52.03/14.53 0(1(2(x1))) -> 1(3(0(2(x1)))) 52.03/14.53 0(1(2(x1))) -> 3(1(2(0(x1)))) 52.03/14.53 0(1(2(x1))) -> 2(0(4(1(3(x1))))) 52.03/14.53 0(1(2(x1))) -> 3(0(2(1(3(x1))))) 52.03/14.53 0(1(2(x1))) -> 3(3(0(2(1(x1))))) 52.03/14.53 0(1(2(x1))) -> 1(3(3(0(2(3(x1)))))) 52.03/14.53 0(1(2(x1))) -> 2(0(4(3(1(3(x1)))))) 52.03/14.53 0(1(2(x1))) -> 3(0(1(3(1(2(x1)))))) 52.03/14.53 0(1(2(x1))) -> 3(0(5(3(1(2(x1)))))) 52.03/14.53 0(5(2(x1))) -> 1(5(0(2(x1)))) 52.03/14.53 0(5(2(x1))) -> 3(5(0(2(x1)))) 52.03/14.53 0(5(2(x1))) -> 3(0(5(1(2(x1))))) 52.03/14.53 0(5(2(x1))) -> 3(5(3(0(2(x1))))) 52.03/14.53 0(5(2(x1))) -> 1(5(5(3(0(2(x1)))))) 52.03/14.53 0(5(2(x1))) -> 5(0(4(3(1(2(x1)))))) 52.03/14.53 1(5(2(x1))) -> 5(3(1(2(x1)))) 52.03/14.53 1(5(2(x1))) -> 1(5(3(1(2(x1))))) 52.03/14.53 0(0(5(4(x1)))) -> 5(3(0(4(0(0(x1)))))) 52.03/14.53 0(1(0(1(x1)))) -> 1(1(3(0(3(0(x1)))))) 52.03/14.53 0(1(1(2(x1)))) -> 2(1(1(3(0(2(x1)))))) 52.03/14.53 0(1(2(5(x1)))) -> 3(0(5(1(2(x1))))) 52.03/14.53 0(1(2(5(x1)))) -> 3(5(3(1(2(0(x1)))))) 52.03/14.53 0(1(3(2(x1)))) -> 3(0(2(1(3(5(x1)))))) 52.03/14.53 0(1(4(2(x1)))) -> 3(0(0(4(1(2(x1)))))) 52.03/14.53 0(1(4(2(x1)))) -> 4(0(1(3(1(2(x1)))))) 52.03/14.53 0(5(2(4(x1)))) -> 5(5(3(0(2(4(x1)))))) 52.03/14.53 0(5(2(5(x1)))) -> 0(2(3(5(5(x1))))) 52.03/14.53 0(5(3(2(x1)))) -> 3(0(2(1(5(x1))))) 52.03/14.53 0(5(3(2(x1)))) -> 5(3(1(4(0(2(x1)))))) 52.03/14.53 1(0(3(4(x1)))) -> 1(0(4(1(3(x1))))) 52.03/14.53 1(5(3(2(x1)))) -> 3(1(3(5(2(x1))))) 52.03/14.53 5(0(1(2(x1)))) -> 1(5(0(4(1(2(x1)))))) 52.03/14.53 5(1(0(2(x1)))) -> 3(1(5(0(2(x1))))) 52.03/14.53 5(1(1(2(x1)))) -> 1(3(5(2(1(3(x1)))))) 52.03/14.53 5(1(5(2(x1)))) -> 1(3(5(5(2(x1))))) 52.03/14.53 0(0(1(3(2(x1))))) -> 0(3(1(2(0(3(x1)))))) 52.03/14.53 0(0(5(3(2(x1))))) -> 0(3(0(3(2(5(x1)))))) 52.03/14.53 0(1(0(3(4(x1))))) -> 4(0(1(3(0(3(x1)))))) 52.03/14.53 0(1(0(5(4(x1))))) -> 3(0(4(0(1(5(x1)))))) 52.03/14.53 0(1(2(4(4(x1))))) -> 0(4(1(2(1(4(x1)))))) 52.03/14.53 0(1(5(3(2(x1))))) -> 1(2(3(0(4(5(x1)))))) 52.03/14.53 0(5(4(2(4(x1))))) -> 4(5(3(0(4(2(x1)))))) 52.03/14.53 0(5(4(4(2(x1))))) -> 0(4(4(3(5(2(x1)))))) 52.03/14.53 1(0(1(3(4(x1))))) -> 4(1(1(3(0(3(x1)))))) 52.03/14.53 1(0(3(4(5(x1))))) -> 2(1(3(0(4(5(x1)))))) 52.03/14.53 1(1(3(4(2(x1))))) -> 2(5(1(3(1(4(x1)))))) 52.03/14.53 1(5(3(2(2(x1))))) -> 2(3(5(3(1(2(x1)))))) 52.03/14.53 5(0(5(2(4(x1))))) -> 5(1(5(0(4(2(x1)))))) 52.03/14.53 5(1(0(5(2(x1))))) -> 1(5(0(2(3(5(x1)))))) 52.03/14.53 5(1(1(2(1(x1))))) -> 3(1(2(5(1(1(x1)))))) 52.03/14.53 52.03/14.53 Q is empty. 52.03/14.53 We have to consider all minimal (P,Q,R)-chains. 52.03/14.53 ---------------------------------------- 52.03/14.53 52.03/14.53 (21) PisEmptyProof (EQUIVALENT) 52.03/14.53 The TRS P is empty. Hence, there is no (P,Q,R) chain. 52.03/14.53 ---------------------------------------- 52.03/14.53 52.03/14.53 (22) 52.03/14.53 YES 52.33/14.63 EOF