16.43/5.02 YES 16.75/5.05 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 16.75/5.05 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 16.75/5.05 16.75/5.05 16.75/5.05 Termination w.r.t. Q of the given QTRS could be proven: 16.75/5.05 16.75/5.05 (0) QTRS 16.75/5.05 (1) QTRS Reverse [EQUIVALENT, 0 ms] 16.75/5.05 (2) QTRS 16.75/5.05 (3) DependencyPairsProof [EQUIVALENT, 130 ms] 16.75/5.05 (4) QDP 16.75/5.05 (5) DependencyGraphProof [EQUIVALENT, 0 ms] 16.75/5.05 (6) QDP 16.75/5.05 (7) UsableRulesProof [EQUIVALENT, 2 ms] 16.75/5.05 (8) QDP 16.75/5.05 (9) QDPOrderProof [EQUIVALENT, 55 ms] 16.75/5.05 (10) QDP 16.75/5.05 (11) PisEmptyProof [EQUIVALENT, 0 ms] 16.75/5.05 (12) YES 16.75/5.05 16.75/5.05 16.75/5.05 ---------------------------------------- 16.75/5.05 16.75/5.05 (0) 16.75/5.05 Obligation: 16.75/5.05 Q restricted rewrite system: 16.75/5.05 The TRS R consists of the following rules: 16.75/5.05 16.75/5.05 0(1(2(x1))) -> 0(2(1(1(x1)))) 16.75/5.05 0(1(2(x1))) -> 0(2(1(3(x1)))) 16.75/5.05 0(1(2(x1))) -> 0(2(1(1(3(x1))))) 16.75/5.05 0(1(2(x1))) -> 0(2(4(1(1(x1))))) 16.75/5.05 0(1(2(x1))) -> 0(2(4(1(3(x1))))) 16.75/5.05 0(3(2(x1))) -> 0(2(1(3(x1)))) 16.75/5.05 0(3(2(x1))) -> 0(2(1(1(3(x1))))) 16.75/5.05 0(3(2(x1))) -> 0(0(2(1(1(3(x1)))))) 16.75/5.05 0(1(1(2(x1)))) -> 0(2(1(1(3(x1))))) 16.75/5.05 0(1(2(2(x1)))) -> 0(2(2(1(1(x1))))) 16.75/5.05 0(1(2(2(x1)))) -> 0(2(3(2(1(x1))))) 16.75/5.05 0(1(3(2(x1)))) -> 0(2(1(4(3(x1))))) 16.75/5.05 0(1(3(2(x1)))) -> 3(3(0(2(1(x1))))) 16.75/5.05 0(1(3(2(x1)))) -> 3(0(2(4(1(1(x1)))))) 16.75/5.05 0(1(5(2(x1)))) -> 5(0(2(1(1(x1))))) 16.75/5.05 0(3(2(2(x1)))) -> 0(0(2(3(2(x1))))) 16.75/5.05 0(3(2(2(x1)))) -> 0(0(4(2(3(2(x1)))))) 16.75/5.05 2(0(3(2(x1)))) -> 1(3(0(2(2(x1))))) 16.75/5.05 3(0(3(2(x1)))) -> 3(0(0(0(2(3(x1)))))) 16.75/5.05 3(1(5(2(x1)))) -> 0(2(1(5(3(x1))))) 16.75/5.05 3(3(5(2(x1)))) -> 3(0(2(1(5(3(x1)))))) 16.75/5.05 4(5(2(2(x1)))) -> 2(5(0(2(4(1(x1)))))) 16.75/5.05 5(0(1(2(x1)))) -> 0(2(1(3(4(5(x1)))))) 16.75/5.05 5(0(2(2(x1)))) -> 2(0(2(4(1(5(x1)))))) 16.75/5.05 5(0(3(2(x1)))) -> 3(0(2(1(4(5(x1)))))) 16.75/5.05 5(1(2(2(x1)))) -> 0(2(1(5(2(1(x1)))))) 16.75/5.05 5(1(2(2(x1)))) -> 2(1(5(2(1(1(x1)))))) 16.75/5.05 0(1(2(4(2(x1))))) -> 4(2(1(0(2(1(x1)))))) 16.75/5.05 0(1(2(5(2(x1))))) -> 2(0(2(1(5(4(x1)))))) 16.75/5.05 0(1(5(1(2(x1))))) -> 1(0(2(1(1(5(x1)))))) 16.75/5.05 0(1(5(5(2(x1))))) -> 0(2(1(5(5(1(x1)))))) 16.75/5.05 0(5(0(1(2(x1))))) -> 0(4(5(0(2(1(x1)))))) 16.75/5.05 0(5(0(3(2(x1))))) -> 0(5(0(0(2(3(x1)))))) 16.75/5.05 0(5(5(2(2(x1))))) -> 0(2(1(5(5(2(x1)))))) 16.75/5.05 3(0(3(1(2(x1))))) -> 1(3(3(0(2(3(x1)))))) 16.75/5.05 3(0(3(4(2(x1))))) -> 1(3(4(0(2(3(x1)))))) 16.75/5.05 4(5(1(4(2(x1))))) -> 2(4(4(4(1(5(x1)))))) 16.75/5.05 5(0(1(3(2(x1))))) -> 3(5(0(2(1(3(x1)))))) 16.75/5.05 5(0(3(1(2(x1))))) -> 0(2(1(4(3(5(x1)))))) 16.75/5.05 5(0(4(2(2(x1))))) -> 2(0(2(4(1(5(x1)))))) 16.75/5.05 5(1(0(3(2(x1))))) -> 0(4(3(5(1(2(x1)))))) 16.75/5.05 5(1(0(3(2(x1))))) -> 5(3(1(1(0(2(x1)))))) 16.75/5.05 5(1(0(5(2(x1))))) -> 3(5(5(0(2(1(x1)))))) 16.75/5.05 5(1(0(5(2(x1))))) -> 5(0(2(1(5(5(x1)))))) 16.75/5.05 5(2(0(3(2(x1))))) -> 0(2(5(2(3(1(x1)))))) 16.75/5.05 16.75/5.05 Q is empty. 16.75/5.05 16.75/5.05 ---------------------------------------- 16.75/5.05 16.75/5.05 (1) QTRS Reverse (EQUIVALENT) 16.75/5.05 We applied the QTRS Reverse Processor [REVERSE]. 16.75/5.05 ---------------------------------------- 16.75/5.05 16.75/5.05 (2) 16.75/5.05 Obligation: 16.75/5.05 Q restricted rewrite system: 16.75/5.05 The TRS R consists of the following rules: 16.75/5.05 16.75/5.05 2(1(0(x1))) -> 1(1(2(0(x1)))) 16.75/5.05 2(1(0(x1))) -> 3(1(2(0(x1)))) 16.75/5.05 2(1(0(x1))) -> 3(1(1(2(0(x1))))) 16.75/5.05 2(1(0(x1))) -> 1(1(4(2(0(x1))))) 16.75/5.05 2(1(0(x1))) -> 3(1(4(2(0(x1))))) 16.75/5.05 2(3(0(x1))) -> 3(1(2(0(x1)))) 16.75/5.05 2(3(0(x1))) -> 3(1(1(2(0(x1))))) 16.75/5.05 2(3(0(x1))) -> 3(1(1(2(0(0(x1)))))) 16.75/5.05 2(1(1(0(x1)))) -> 3(1(1(2(0(x1))))) 16.75/5.05 2(2(1(0(x1)))) -> 1(1(2(2(0(x1))))) 16.75/5.05 2(2(1(0(x1)))) -> 1(2(3(2(0(x1))))) 16.75/5.05 2(3(1(0(x1)))) -> 3(4(1(2(0(x1))))) 16.75/5.05 2(3(1(0(x1)))) -> 1(2(0(3(3(x1))))) 16.75/5.05 2(3(1(0(x1)))) -> 1(1(4(2(0(3(x1)))))) 16.75/5.05 2(5(1(0(x1)))) -> 1(1(2(0(5(x1))))) 16.75/5.05 2(2(3(0(x1)))) -> 2(3(2(0(0(x1))))) 16.75/5.05 2(2(3(0(x1)))) -> 2(3(2(4(0(0(x1)))))) 16.75/5.05 2(3(0(2(x1)))) -> 2(2(0(3(1(x1))))) 16.75/5.05 2(3(0(3(x1)))) -> 3(2(0(0(0(3(x1)))))) 16.75/5.05 2(5(1(3(x1)))) -> 3(5(1(2(0(x1))))) 16.75/5.05 2(5(3(3(x1)))) -> 3(5(1(2(0(3(x1)))))) 16.75/5.05 2(2(5(4(x1)))) -> 1(4(2(0(5(2(x1)))))) 16.75/5.05 2(1(0(5(x1)))) -> 5(4(3(1(2(0(x1)))))) 16.75/5.05 2(2(0(5(x1)))) -> 5(1(4(2(0(2(x1)))))) 16.75/5.05 2(3(0(5(x1)))) -> 5(4(1(2(0(3(x1)))))) 16.75/5.05 2(2(1(5(x1)))) -> 1(2(5(1(2(0(x1)))))) 16.75/5.05 2(2(1(5(x1)))) -> 1(1(2(5(1(2(x1)))))) 16.75/5.05 2(4(2(1(0(x1))))) -> 1(2(0(1(2(4(x1)))))) 16.75/5.05 2(5(2(1(0(x1))))) -> 4(5(1(2(0(2(x1)))))) 16.75/5.05 2(1(5(1(0(x1))))) -> 5(1(1(2(0(1(x1)))))) 16.75/5.05 2(5(5(1(0(x1))))) -> 1(5(5(1(2(0(x1)))))) 16.75/5.05 2(1(0(5(0(x1))))) -> 1(2(0(5(4(0(x1)))))) 16.75/5.05 2(3(0(5(0(x1))))) -> 3(2(0(0(5(0(x1)))))) 16.75/5.05 2(2(5(5(0(x1))))) -> 2(5(5(1(2(0(x1)))))) 16.75/5.05 2(1(3(0(3(x1))))) -> 3(2(0(3(3(1(x1)))))) 16.75/5.05 2(4(3(0(3(x1))))) -> 3(2(0(4(3(1(x1)))))) 16.75/5.05 2(4(1(5(4(x1))))) -> 5(1(4(4(4(2(x1)))))) 16.75/5.05 2(3(1(0(5(x1))))) -> 3(1(2(0(5(3(x1)))))) 16.75/5.05 2(1(3(0(5(x1))))) -> 5(3(4(1(2(0(x1)))))) 16.75/5.05 2(2(4(0(5(x1))))) -> 5(1(4(2(0(2(x1)))))) 16.75/5.05 2(3(0(1(5(x1))))) -> 2(1(5(3(4(0(x1)))))) 16.75/5.05 2(3(0(1(5(x1))))) -> 2(0(1(1(3(5(x1)))))) 16.75/5.05 2(5(0(1(5(x1))))) -> 1(2(0(5(5(3(x1)))))) 16.75/5.05 2(5(0(1(5(x1))))) -> 5(5(1(2(0(5(x1)))))) 16.75/5.05 2(3(0(2(5(x1))))) -> 1(3(2(5(2(0(x1)))))) 16.75/5.05 16.75/5.05 Q is empty. 16.75/5.05 16.75/5.05 ---------------------------------------- 16.75/5.05 16.75/5.05 (3) DependencyPairsProof (EQUIVALENT) 16.75/5.05 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 16.75/5.05 ---------------------------------------- 16.75/5.05 16.75/5.05 (4) 16.75/5.05 Obligation: 16.75/5.05 Q DP problem: 16.75/5.05 The TRS P consists of the following rules: 16.75/5.05 16.75/5.05 2^1(1(0(x1))) -> 2^1(0(x1)) 16.75/5.05 2^1(3(0(x1))) -> 2^1(0(x1)) 16.75/5.05 2^1(3(0(x1))) -> 2^1(0(0(x1))) 16.75/5.05 2^1(1(1(0(x1)))) -> 2^1(0(x1)) 16.75/5.05 2^1(2(1(0(x1)))) -> 2^1(2(0(x1))) 16.75/5.05 2^1(2(1(0(x1)))) -> 2^1(0(x1)) 16.75/5.05 2^1(2(1(0(x1)))) -> 2^1(3(2(0(x1)))) 16.75/5.05 2^1(3(1(0(x1)))) -> 2^1(0(x1)) 16.75/5.05 2^1(3(1(0(x1)))) -> 2^1(0(3(3(x1)))) 16.75/5.05 2^1(3(1(0(x1)))) -> 2^1(0(3(x1))) 16.75/5.05 2^1(5(1(0(x1)))) -> 2^1(0(5(x1))) 16.75/5.05 2^1(2(3(0(x1)))) -> 2^1(3(2(0(0(x1))))) 16.75/5.05 2^1(2(3(0(x1)))) -> 2^1(0(0(x1))) 16.75/5.05 2^1(2(3(0(x1)))) -> 2^1(3(2(4(0(0(x1)))))) 16.75/5.05 2^1(2(3(0(x1)))) -> 2^1(4(0(0(x1)))) 16.75/5.05 2^1(3(0(2(x1)))) -> 2^1(2(0(3(1(x1))))) 16.75/5.05 2^1(3(0(2(x1)))) -> 2^1(0(3(1(x1)))) 16.75/5.05 2^1(3(0(3(x1)))) -> 2^1(0(0(0(3(x1))))) 16.75/5.05 2^1(5(1(3(x1)))) -> 2^1(0(x1)) 16.75/5.05 2^1(5(3(3(x1)))) -> 2^1(0(3(x1))) 16.75/5.05 2^1(2(5(4(x1)))) -> 2^1(0(5(2(x1)))) 16.75/5.05 2^1(2(5(4(x1)))) -> 2^1(x1) 16.75/5.05 2^1(1(0(5(x1)))) -> 2^1(0(x1)) 16.75/5.05 2^1(2(0(5(x1)))) -> 2^1(0(2(x1))) 16.75/5.05 2^1(2(0(5(x1)))) -> 2^1(x1) 16.75/5.05 2^1(3(0(5(x1)))) -> 2^1(0(3(x1))) 16.75/5.05 2^1(2(1(5(x1)))) -> 2^1(5(1(2(0(x1))))) 16.75/5.05 2^1(2(1(5(x1)))) -> 2^1(0(x1)) 16.75/5.05 2^1(2(1(5(x1)))) -> 2^1(5(1(2(x1)))) 16.75/5.05 2^1(2(1(5(x1)))) -> 2^1(x1) 16.75/5.05 2^1(4(2(1(0(x1))))) -> 2^1(0(1(2(4(x1))))) 16.75/5.05 2^1(4(2(1(0(x1))))) -> 2^1(4(x1)) 16.75/5.05 2^1(5(2(1(0(x1))))) -> 2^1(0(2(x1))) 16.75/5.05 2^1(5(2(1(0(x1))))) -> 2^1(x1) 16.75/5.05 2^1(1(5(1(0(x1))))) -> 2^1(0(1(x1))) 16.75/5.05 2^1(5(5(1(0(x1))))) -> 2^1(0(x1)) 16.75/5.05 2^1(1(0(5(0(x1))))) -> 2^1(0(5(4(0(x1))))) 16.75/5.05 2^1(3(0(5(0(x1))))) -> 2^1(0(0(5(0(x1))))) 16.75/5.05 2^1(2(5(5(0(x1))))) -> 2^1(5(5(1(2(0(x1)))))) 16.75/5.05 2^1(2(5(5(0(x1))))) -> 2^1(0(x1)) 16.75/5.05 2^1(1(3(0(3(x1))))) -> 2^1(0(3(3(1(x1))))) 16.75/5.05 2^1(4(3(0(3(x1))))) -> 2^1(0(4(3(1(x1))))) 16.75/5.05 2^1(4(1(5(4(x1))))) -> 2^1(x1) 16.75/5.05 2^1(3(1(0(5(x1))))) -> 2^1(0(5(3(x1)))) 16.75/5.05 2^1(1(3(0(5(x1))))) -> 2^1(0(x1)) 16.75/5.05 2^1(2(4(0(5(x1))))) -> 2^1(0(2(x1))) 16.75/5.05 2^1(2(4(0(5(x1))))) -> 2^1(x1) 16.75/5.05 2^1(3(0(1(5(x1))))) -> 2^1(1(5(3(4(0(x1)))))) 16.75/5.05 2^1(3(0(1(5(x1))))) -> 2^1(0(1(1(3(5(x1)))))) 16.75/5.05 2^1(5(0(1(5(x1))))) -> 2^1(0(5(5(3(x1))))) 16.75/5.05 2^1(5(0(1(5(x1))))) -> 2^1(0(5(x1))) 16.75/5.05 2^1(3(0(2(5(x1))))) -> 2^1(5(2(0(x1)))) 16.75/5.05 2^1(3(0(2(5(x1))))) -> 2^1(0(x1)) 16.75/5.05 16.75/5.05 The TRS R consists of the following rules: 16.75/5.05 16.75/5.05 2(1(0(x1))) -> 1(1(2(0(x1)))) 16.75/5.05 2(1(0(x1))) -> 3(1(2(0(x1)))) 16.75/5.05 2(1(0(x1))) -> 3(1(1(2(0(x1))))) 16.75/5.05 2(1(0(x1))) -> 1(1(4(2(0(x1))))) 16.75/5.05 2(1(0(x1))) -> 3(1(4(2(0(x1))))) 16.75/5.05 2(3(0(x1))) -> 3(1(2(0(x1)))) 16.75/5.05 2(3(0(x1))) -> 3(1(1(2(0(x1))))) 16.75/5.05 2(3(0(x1))) -> 3(1(1(2(0(0(x1)))))) 16.75/5.05 2(1(1(0(x1)))) -> 3(1(1(2(0(x1))))) 16.75/5.05 2(2(1(0(x1)))) -> 1(1(2(2(0(x1))))) 16.75/5.05 2(2(1(0(x1)))) -> 1(2(3(2(0(x1))))) 16.75/5.05 2(3(1(0(x1)))) -> 3(4(1(2(0(x1))))) 16.75/5.05 2(3(1(0(x1)))) -> 1(2(0(3(3(x1))))) 16.75/5.05 2(3(1(0(x1)))) -> 1(1(4(2(0(3(x1)))))) 16.75/5.05 2(5(1(0(x1)))) -> 1(1(2(0(5(x1))))) 16.75/5.05 2(2(3(0(x1)))) -> 2(3(2(0(0(x1))))) 16.75/5.05 2(2(3(0(x1)))) -> 2(3(2(4(0(0(x1)))))) 16.75/5.05 2(3(0(2(x1)))) -> 2(2(0(3(1(x1))))) 16.75/5.05 2(3(0(3(x1)))) -> 3(2(0(0(0(3(x1)))))) 16.75/5.05 2(5(1(3(x1)))) -> 3(5(1(2(0(x1))))) 16.75/5.05 2(5(3(3(x1)))) -> 3(5(1(2(0(3(x1)))))) 16.75/5.05 2(2(5(4(x1)))) -> 1(4(2(0(5(2(x1)))))) 16.75/5.05 2(1(0(5(x1)))) -> 5(4(3(1(2(0(x1)))))) 16.75/5.05 2(2(0(5(x1)))) -> 5(1(4(2(0(2(x1)))))) 16.75/5.05 2(3(0(5(x1)))) -> 5(4(1(2(0(3(x1)))))) 16.75/5.05 2(2(1(5(x1)))) -> 1(2(5(1(2(0(x1)))))) 16.75/5.05 2(2(1(5(x1)))) -> 1(1(2(5(1(2(x1)))))) 16.75/5.05 2(4(2(1(0(x1))))) -> 1(2(0(1(2(4(x1)))))) 16.75/5.05 2(5(2(1(0(x1))))) -> 4(5(1(2(0(2(x1)))))) 16.75/5.05 2(1(5(1(0(x1))))) -> 5(1(1(2(0(1(x1)))))) 16.75/5.05 2(5(5(1(0(x1))))) -> 1(5(5(1(2(0(x1)))))) 16.75/5.05 2(1(0(5(0(x1))))) -> 1(2(0(5(4(0(x1)))))) 16.75/5.05 2(3(0(5(0(x1))))) -> 3(2(0(0(5(0(x1)))))) 16.75/5.05 2(2(5(5(0(x1))))) -> 2(5(5(1(2(0(x1)))))) 16.75/5.05 2(1(3(0(3(x1))))) -> 3(2(0(3(3(1(x1)))))) 16.75/5.05 2(4(3(0(3(x1))))) -> 3(2(0(4(3(1(x1)))))) 16.75/5.05 2(4(1(5(4(x1))))) -> 5(1(4(4(4(2(x1)))))) 16.75/5.05 2(3(1(0(5(x1))))) -> 3(1(2(0(5(3(x1)))))) 16.75/5.05 2(1(3(0(5(x1))))) -> 5(3(4(1(2(0(x1)))))) 16.75/5.05 2(2(4(0(5(x1))))) -> 5(1(4(2(0(2(x1)))))) 16.75/5.05 2(3(0(1(5(x1))))) -> 2(1(5(3(4(0(x1)))))) 16.75/5.05 2(3(0(1(5(x1))))) -> 2(0(1(1(3(5(x1)))))) 16.75/5.05 2(5(0(1(5(x1))))) -> 1(2(0(5(5(3(x1)))))) 16.75/5.05 2(5(0(1(5(x1))))) -> 5(5(1(2(0(5(x1)))))) 16.75/5.05 2(3(0(2(5(x1))))) -> 1(3(2(5(2(0(x1)))))) 16.75/5.05 16.75/5.05 Q is empty. 16.75/5.05 We have to consider all minimal (P,Q,R)-chains. 16.75/5.05 ---------------------------------------- 16.75/5.05 16.75/5.05 (5) DependencyGraphProof (EQUIVALENT) 16.75/5.05 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 45 less nodes. 16.75/5.05 ---------------------------------------- 16.75/5.05 16.75/5.05 (6) 16.75/5.05 Obligation: 16.75/5.05 Q DP problem: 16.75/5.05 The TRS P consists of the following rules: 16.75/5.05 16.75/5.05 2^1(2(1(0(x1)))) -> 2^1(2(0(x1))) 16.75/5.05 2^1(2(0(5(x1)))) -> 2^1(x1) 16.75/5.05 2^1(2(5(4(x1)))) -> 2^1(x1) 16.75/5.05 2^1(2(1(5(x1)))) -> 2^1(x1) 16.75/5.05 2^1(4(2(1(0(x1))))) -> 2^1(4(x1)) 16.75/5.05 2^1(4(1(5(4(x1))))) -> 2^1(x1) 16.75/5.05 2^1(5(2(1(0(x1))))) -> 2^1(x1) 16.75/5.05 2^1(2(4(0(5(x1))))) -> 2^1(x1) 16.75/5.05 16.75/5.05 The TRS R consists of the following rules: 16.75/5.05 16.75/5.05 2(1(0(x1))) -> 1(1(2(0(x1)))) 16.75/5.05 2(1(0(x1))) -> 3(1(2(0(x1)))) 16.75/5.05 2(1(0(x1))) -> 3(1(1(2(0(x1))))) 16.75/5.05 2(1(0(x1))) -> 1(1(4(2(0(x1))))) 16.75/5.05 2(1(0(x1))) -> 3(1(4(2(0(x1))))) 16.75/5.05 2(3(0(x1))) -> 3(1(2(0(x1)))) 16.75/5.05 2(3(0(x1))) -> 3(1(1(2(0(x1))))) 16.75/5.05 2(3(0(x1))) -> 3(1(1(2(0(0(x1)))))) 16.75/5.05 2(1(1(0(x1)))) -> 3(1(1(2(0(x1))))) 16.75/5.05 2(2(1(0(x1)))) -> 1(1(2(2(0(x1))))) 16.75/5.05 2(2(1(0(x1)))) -> 1(2(3(2(0(x1))))) 16.75/5.05 2(3(1(0(x1)))) -> 3(4(1(2(0(x1))))) 16.75/5.05 2(3(1(0(x1)))) -> 1(2(0(3(3(x1))))) 16.75/5.05 2(3(1(0(x1)))) -> 1(1(4(2(0(3(x1)))))) 16.75/5.05 2(5(1(0(x1)))) -> 1(1(2(0(5(x1))))) 16.75/5.05 2(2(3(0(x1)))) -> 2(3(2(0(0(x1))))) 16.75/5.05 2(2(3(0(x1)))) -> 2(3(2(4(0(0(x1)))))) 16.75/5.05 2(3(0(2(x1)))) -> 2(2(0(3(1(x1))))) 16.75/5.05 2(3(0(3(x1)))) -> 3(2(0(0(0(3(x1)))))) 16.75/5.05 2(5(1(3(x1)))) -> 3(5(1(2(0(x1))))) 16.75/5.05 2(5(3(3(x1)))) -> 3(5(1(2(0(3(x1)))))) 16.75/5.05 2(2(5(4(x1)))) -> 1(4(2(0(5(2(x1)))))) 16.75/5.05 2(1(0(5(x1)))) -> 5(4(3(1(2(0(x1)))))) 16.75/5.05 2(2(0(5(x1)))) -> 5(1(4(2(0(2(x1)))))) 16.75/5.05 2(3(0(5(x1)))) -> 5(4(1(2(0(3(x1)))))) 16.75/5.05 2(2(1(5(x1)))) -> 1(2(5(1(2(0(x1)))))) 16.75/5.05 2(2(1(5(x1)))) -> 1(1(2(5(1(2(x1)))))) 16.75/5.05 2(4(2(1(0(x1))))) -> 1(2(0(1(2(4(x1)))))) 16.75/5.05 2(5(2(1(0(x1))))) -> 4(5(1(2(0(2(x1)))))) 16.75/5.05 2(1(5(1(0(x1))))) -> 5(1(1(2(0(1(x1)))))) 16.75/5.05 2(5(5(1(0(x1))))) -> 1(5(5(1(2(0(x1)))))) 16.75/5.05 2(1(0(5(0(x1))))) -> 1(2(0(5(4(0(x1)))))) 16.75/5.05 2(3(0(5(0(x1))))) -> 3(2(0(0(5(0(x1)))))) 16.75/5.05 2(2(5(5(0(x1))))) -> 2(5(5(1(2(0(x1)))))) 16.75/5.05 2(1(3(0(3(x1))))) -> 3(2(0(3(3(1(x1)))))) 16.75/5.05 2(4(3(0(3(x1))))) -> 3(2(0(4(3(1(x1)))))) 16.75/5.05 2(4(1(5(4(x1))))) -> 5(1(4(4(4(2(x1)))))) 16.75/5.05 2(3(1(0(5(x1))))) -> 3(1(2(0(5(3(x1)))))) 16.75/5.05 2(1(3(0(5(x1))))) -> 5(3(4(1(2(0(x1)))))) 16.75/5.05 2(2(4(0(5(x1))))) -> 5(1(4(2(0(2(x1)))))) 16.75/5.05 2(3(0(1(5(x1))))) -> 2(1(5(3(4(0(x1)))))) 16.75/5.05 2(3(0(1(5(x1))))) -> 2(0(1(1(3(5(x1)))))) 16.75/5.05 2(5(0(1(5(x1))))) -> 1(2(0(5(5(3(x1)))))) 16.75/5.05 2(5(0(1(5(x1))))) -> 5(5(1(2(0(5(x1)))))) 16.75/5.05 2(3(0(2(5(x1))))) -> 1(3(2(5(2(0(x1)))))) 16.75/5.05 16.75/5.05 Q is empty. 16.75/5.05 We have to consider all minimal (P,Q,R)-chains. 16.75/5.05 ---------------------------------------- 16.75/5.05 16.75/5.05 (7) UsableRulesProof (EQUIVALENT) 16.75/5.05 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. 16.75/5.05 ---------------------------------------- 16.75/5.05 16.75/5.05 (8) 16.75/5.05 Obligation: 16.75/5.05 Q DP problem: 16.75/5.05 The TRS P consists of the following rules: 16.75/5.05 16.75/5.05 2^1(2(1(0(x1)))) -> 2^1(2(0(x1))) 16.75/5.05 2^1(2(0(5(x1)))) -> 2^1(x1) 16.75/5.05 2^1(2(5(4(x1)))) -> 2^1(x1) 16.75/5.05 2^1(2(1(5(x1)))) -> 2^1(x1) 16.75/5.05 2^1(4(2(1(0(x1))))) -> 2^1(4(x1)) 16.75/5.05 2^1(4(1(5(4(x1))))) -> 2^1(x1) 16.75/5.05 2^1(5(2(1(0(x1))))) -> 2^1(x1) 16.75/5.05 2^1(2(4(0(5(x1))))) -> 2^1(x1) 16.75/5.05 16.75/5.05 R is empty. 16.75/5.05 Q is empty. 16.75/5.05 We have to consider all minimal (P,Q,R)-chains. 16.75/5.05 ---------------------------------------- 16.75/5.05 16.75/5.05 (9) QDPOrderProof (EQUIVALENT) 16.75/5.05 We use the reduction pair processor [LPAR04,JAR06]. 16.75/5.05 16.75/5.05 16.75/5.05 The following pairs can be oriented strictly and are deleted. 16.75/5.05 16.75/5.05 2^1(2(1(0(x1)))) -> 2^1(2(0(x1))) 16.75/5.05 2^1(2(0(5(x1)))) -> 2^1(x1) 16.75/5.05 2^1(2(5(4(x1)))) -> 2^1(x1) 16.75/5.05 2^1(2(1(5(x1)))) -> 2^1(x1) 16.75/5.05 2^1(4(2(1(0(x1))))) -> 2^1(4(x1)) 16.75/5.05 2^1(4(1(5(4(x1))))) -> 2^1(x1) 16.75/5.05 2^1(5(2(1(0(x1))))) -> 2^1(x1) 16.75/5.05 2^1(2(4(0(5(x1))))) -> 2^1(x1) 16.75/5.05 The remaining pairs can at least be oriented weakly. 16.75/5.05 Used ordering: Polynomial interpretation [POLO]: 16.75/5.05 16.75/5.05 POL(0(x_1)) = 1 + x_1 16.75/5.05 POL(1(x_1)) = 1 + x_1 16.75/5.05 POL(2(x_1)) = 1 + x_1 16.75/5.05 POL(2^1(x_1)) = x_1 16.75/5.05 POL(4(x_1)) = 1 + x_1 16.75/5.05 POL(5(x_1)) = 1 + x_1 16.75/5.05 16.75/5.05 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 16.75/5.05 none 16.75/5.05 16.75/5.05 16.75/5.05 ---------------------------------------- 16.75/5.05 16.75/5.05 (10) 16.75/5.05 Obligation: 16.75/5.05 Q DP problem: 16.75/5.05 P is empty. 16.75/5.05 R is empty. 16.75/5.05 Q is empty. 16.75/5.05 We have to consider all minimal (P,Q,R)-chains. 16.75/5.05 ---------------------------------------- 16.75/5.05 16.75/5.05 (11) PisEmptyProof (EQUIVALENT) 16.75/5.05 The TRS P is empty. Hence, there is no (P,Q,R) chain. 16.75/5.05 ---------------------------------------- 16.75/5.05 16.75/5.05 (12) 16.75/5.05 YES 16.79/5.10 EOF