YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 255 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 1 ms] (6) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) DependencyGraphProof [EQUIVALENT, 0 ms] (11) QDP (12) UsableRulesProof [EQUIVALENT, 0 ms] (13) QDP (14) QDPSizeChangeProof [EQUIVALENT, 1 ms] (15) YES (16) QDP (17) UsableRulesProof [EQUIVALENT, 0 ms] (18) QDP (19) DependencyGraphProof [EQUIVALENT, 0 ms] (20) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(0(1(2(x1)))) -> 0(2(0(1(1(x1))))) 0(1(2(2(x1)))) -> 0(2(2(1(0(x1))))) 0(1(2(2(x1)))) -> 1(0(2(2(0(x1))))) 0(1(2(3(x1)))) -> 0(2(0(1(3(x1))))) 0(1(2(3(x1)))) -> 0(2(3(3(1(x1))))) 0(2(1(2(x1)))) -> 0(2(2(0(1(x1))))) 0(2(1(2(x1)))) -> 0(2(2(2(1(x1))))) 0(2(1(2(x1)))) -> 0(2(2(2(1(1(x1)))))) 0(3(2(2(x1)))) -> 3(4(0(2(2(x1))))) 0(3(2(2(x1)))) -> 0(2(2(2(2(3(x1)))))) 0(4(1(2(x1)))) -> 0(2(2(1(4(x1))))) 0(4(1(2(x1)))) -> 4(0(2(0(1(x1))))) 0(5(0(1(x1)))) -> 0(2(0(2(5(1(x1)))))) 0(5(0(5(x1)))) -> 0(2(0(5(5(x1))))) 0(5(2(1(x1)))) -> 0(2(2(5(1(x1))))) 0(5(2(5(x1)))) -> 0(2(2(5(5(x1))))) 0(5(4(2(x1)))) -> 4(0(2(2(0(5(x1)))))) 2(1(0(3(x1)))) -> 4(0(2(2(3(1(x1)))))) 2(1(0(4(x1)))) -> 1(4(0(2(2(2(x1)))))) 2(5(4(2(x1)))) -> 4(0(2(2(5(x1))))) 0(0(1(0(4(x1))))) -> 0(0(2(0(1(4(x1)))))) 0(0(5(4(2(x1))))) -> 0(4(0(0(2(5(x1)))))) 0(1(0(1(2(x1))))) -> 0(2(0(1(4(1(x1)))))) 0(1(2(0(3(x1))))) -> 0(2(0(4(1(3(x1)))))) 0(1(2(2(2(x1))))) -> 0(2(2(2(1(2(x1)))))) 0(1(2(3(2(x1))))) -> 1(3(4(0(2(2(x1)))))) 0(1(3(2(3(x1))))) -> 0(0(2(3(3(1(x1)))))) 0(1(3(4(2(x1))))) -> 0(2(3(4(1(1(x1)))))) 0(2(1(0(1(x1))))) -> 0(0(2(0(1(1(x1)))))) 0(2(1(2(2(x1))))) -> 0(2(0(2(2(1(x1)))))) 0(2(3(0(5(x1))))) -> 0(2(0(0(5(3(x1)))))) 0(3(0(1(3(x1))))) -> 0(0(4(3(1(3(x1)))))) 0(3(0(4(1(x1))))) -> 0(0(1(4(4(3(x1)))))) 0(3(2(0(4(x1))))) -> 4(0(0(2(3(4(x1)))))) 0(4(5(2(3(x1))))) -> 0(2(2(3(4(5(x1)))))) 0(5(0(0(3(x1))))) -> 0(2(0(3(0(5(x1)))))) 0(5(0(1(2(x1))))) -> 0(0(2(0(1(5(x1)))))) 0(5(1(4(2(x1))))) -> 0(2(0(1(4(5(x1)))))) 0(5(2(5(1(x1))))) -> 0(2(0(5(5(1(x1)))))) 2(1(0(0(4(x1))))) -> 1(4(4(0(0(2(x1)))))) 2(5(0(0(3(x1))))) -> 0(2(0(0(5(3(x1)))))) 2(5(3(0(1(x1))))) -> 5(0(2(2(3(1(x1)))))) 5(0(1(2(2(x1))))) -> 5(1(0(2(0(2(x1)))))) 5(2(0(1(2(x1))))) -> 1(5(4(0(2(2(x1)))))) 5(2(1(0(1(x1))))) -> 0(2(3(1(5(1(x1)))))) 5(2(3(0(1(x1))))) -> 1(5(0(2(2(3(x1)))))) 5(3(0(4(1(x1))))) -> 4(5(0(2(3(1(x1)))))) Q is empty. ---------------------------------------- (1) QTRS Reverse (EQUIVALENT) We applied the QTRS Reverse Processor [REVERSE]. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 2(1(0(0(x1)))) -> 1(1(0(2(0(x1))))) 2(2(1(0(x1)))) -> 0(1(2(2(0(x1))))) 2(2(1(0(x1)))) -> 0(2(2(0(1(x1))))) 3(2(1(0(x1)))) -> 3(1(0(2(0(x1))))) 3(2(1(0(x1)))) -> 1(3(3(2(0(x1))))) 2(1(2(0(x1)))) -> 1(0(2(2(0(x1))))) 2(1(2(0(x1)))) -> 1(2(2(2(0(x1))))) 2(1(2(0(x1)))) -> 1(1(2(2(2(0(x1)))))) 2(2(3(0(x1)))) -> 2(2(0(4(3(x1))))) 2(2(3(0(x1)))) -> 3(2(2(2(2(0(x1)))))) 2(1(4(0(x1)))) -> 4(1(2(2(0(x1))))) 2(1(4(0(x1)))) -> 1(0(2(0(4(x1))))) 1(0(5(0(x1)))) -> 1(5(2(0(2(0(x1)))))) 5(0(5(0(x1)))) -> 5(5(0(2(0(x1))))) 1(2(5(0(x1)))) -> 1(5(2(2(0(x1))))) 5(2(5(0(x1)))) -> 5(5(2(2(0(x1))))) 2(4(5(0(x1)))) -> 5(0(2(2(0(4(x1)))))) 3(0(1(2(x1)))) -> 1(3(2(2(0(4(x1)))))) 4(0(1(2(x1)))) -> 2(2(2(0(4(1(x1)))))) 2(4(5(2(x1)))) -> 5(2(2(0(4(x1))))) 4(0(1(0(0(x1))))) -> 4(1(0(2(0(0(x1)))))) 2(4(5(0(0(x1))))) -> 5(2(0(0(4(0(x1)))))) 2(1(0(1(0(x1))))) -> 1(4(1(0(2(0(x1)))))) 3(0(2(1(0(x1))))) -> 3(1(4(0(2(0(x1)))))) 2(2(2(1(0(x1))))) -> 2(1(2(2(2(0(x1)))))) 2(3(2(1(0(x1))))) -> 2(2(0(4(3(1(x1)))))) 3(2(3(1(0(x1))))) -> 1(3(3(2(0(0(x1)))))) 2(4(3(1(0(x1))))) -> 1(1(4(3(2(0(x1)))))) 1(0(1(2(0(x1))))) -> 1(1(0(2(0(0(x1)))))) 2(2(1(2(0(x1))))) -> 1(2(2(0(2(0(x1)))))) 5(0(3(2(0(x1))))) -> 3(5(0(0(2(0(x1)))))) 3(1(0(3(0(x1))))) -> 3(1(3(4(0(0(x1)))))) 1(4(0(3(0(x1))))) -> 3(4(4(1(0(0(x1)))))) 4(0(2(3(0(x1))))) -> 4(3(2(0(0(4(x1)))))) 3(2(5(4(0(x1))))) -> 5(4(3(2(2(0(x1)))))) 3(0(0(5(0(x1))))) -> 5(0(3(0(2(0(x1)))))) 2(1(0(5(0(x1))))) -> 5(1(0(2(0(0(x1)))))) 2(4(1(5(0(x1))))) -> 5(4(1(0(2(0(x1)))))) 1(5(2(5(0(x1))))) -> 1(5(5(0(2(0(x1)))))) 4(0(0(1(2(x1))))) -> 2(0(0(4(4(1(x1)))))) 3(0(0(5(2(x1))))) -> 3(5(0(0(2(0(x1)))))) 1(0(3(5(2(x1))))) -> 1(3(2(2(0(5(x1)))))) 2(2(1(0(5(x1))))) -> 2(0(2(0(1(5(x1)))))) 2(1(0(2(5(x1))))) -> 2(2(0(4(5(1(x1)))))) 1(0(1(2(5(x1))))) -> 1(5(1(3(2(0(x1)))))) 1(0(3(2(5(x1))))) -> 3(2(2(0(5(1(x1)))))) 1(4(0(3(5(x1))))) -> 1(3(2(0(5(4(x1)))))) Q is empty. ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: 2^1(1(0(0(x1)))) -> 1^1(1(0(2(0(x1))))) 2^1(1(0(0(x1)))) -> 1^1(0(2(0(x1)))) 2^1(1(0(0(x1)))) -> 2^1(0(x1)) 2^1(2(1(0(x1)))) -> 1^1(2(2(0(x1)))) 2^1(2(1(0(x1)))) -> 2^1(2(0(x1))) 2^1(2(1(0(x1)))) -> 2^1(0(x1)) 2^1(2(1(0(x1)))) -> 2^1(2(0(1(x1)))) 2^1(2(1(0(x1)))) -> 2^1(0(1(x1))) 2^1(2(1(0(x1)))) -> 1^1(x1) 3^1(2(1(0(x1)))) -> 3^1(1(0(2(0(x1))))) 3^1(2(1(0(x1)))) -> 1^1(0(2(0(x1)))) 3^1(2(1(0(x1)))) -> 2^1(0(x1)) 3^1(2(1(0(x1)))) -> 1^1(3(3(2(0(x1))))) 3^1(2(1(0(x1)))) -> 3^1(3(2(0(x1)))) 3^1(2(1(0(x1)))) -> 3^1(2(0(x1))) 2^1(1(2(0(x1)))) -> 1^1(0(2(2(0(x1))))) 2^1(1(2(0(x1)))) -> 2^1(2(0(x1))) 2^1(1(2(0(x1)))) -> 1^1(2(2(2(0(x1))))) 2^1(1(2(0(x1)))) -> 2^1(2(2(0(x1)))) 2^1(1(2(0(x1)))) -> 1^1(1(2(2(2(0(x1)))))) 2^1(2(3(0(x1)))) -> 2^1(2(0(4(3(x1))))) 2^1(2(3(0(x1)))) -> 2^1(0(4(3(x1)))) 2^1(2(3(0(x1)))) -> 4^1(3(x1)) 2^1(2(3(0(x1)))) -> 3^1(x1) 2^1(2(3(0(x1)))) -> 3^1(2(2(2(2(0(x1)))))) 2^1(2(3(0(x1)))) -> 2^1(2(2(2(0(x1))))) 2^1(2(3(0(x1)))) -> 2^1(2(2(0(x1)))) 2^1(2(3(0(x1)))) -> 2^1(2(0(x1))) 2^1(2(3(0(x1)))) -> 2^1(0(x1)) 2^1(1(4(0(x1)))) -> 4^1(1(2(2(0(x1))))) 2^1(1(4(0(x1)))) -> 1^1(2(2(0(x1)))) 2^1(1(4(0(x1)))) -> 2^1(2(0(x1))) 2^1(1(4(0(x1)))) -> 2^1(0(x1)) 2^1(1(4(0(x1)))) -> 1^1(0(2(0(4(x1))))) 2^1(1(4(0(x1)))) -> 2^1(0(4(x1))) 2^1(1(4(0(x1)))) -> 4^1(x1) 1^1(0(5(0(x1)))) -> 1^1(5(2(0(2(0(x1)))))) 1^1(0(5(0(x1)))) -> 5^1(2(0(2(0(x1))))) 1^1(0(5(0(x1)))) -> 2^1(0(2(0(x1)))) 1^1(0(5(0(x1)))) -> 2^1(0(x1)) 5^1(0(5(0(x1)))) -> 5^1(5(0(2(0(x1))))) 5^1(0(5(0(x1)))) -> 5^1(0(2(0(x1)))) 5^1(0(5(0(x1)))) -> 2^1(0(x1)) 1^1(2(5(0(x1)))) -> 1^1(5(2(2(0(x1))))) 1^1(2(5(0(x1)))) -> 5^1(2(2(0(x1)))) 1^1(2(5(0(x1)))) -> 2^1(2(0(x1))) 1^1(2(5(0(x1)))) -> 2^1(0(x1)) 5^1(2(5(0(x1)))) -> 5^1(5(2(2(0(x1))))) 5^1(2(5(0(x1)))) -> 5^1(2(2(0(x1)))) 5^1(2(5(0(x1)))) -> 2^1(2(0(x1))) 5^1(2(5(0(x1)))) -> 2^1(0(x1)) 2^1(4(5(0(x1)))) -> 5^1(0(2(2(0(4(x1)))))) 2^1(4(5(0(x1)))) -> 2^1(2(0(4(x1)))) 2^1(4(5(0(x1)))) -> 2^1(0(4(x1))) 2^1(4(5(0(x1)))) -> 4^1(x1) 3^1(0(1(2(x1)))) -> 1^1(3(2(2(0(4(x1)))))) 3^1(0(1(2(x1)))) -> 3^1(2(2(0(4(x1))))) 3^1(0(1(2(x1)))) -> 2^1(2(0(4(x1)))) 3^1(0(1(2(x1)))) -> 2^1(0(4(x1))) 3^1(0(1(2(x1)))) -> 4^1(x1) 4^1(0(1(2(x1)))) -> 2^1(2(2(0(4(1(x1)))))) 4^1(0(1(2(x1)))) -> 2^1(2(0(4(1(x1))))) 4^1(0(1(2(x1)))) -> 2^1(0(4(1(x1)))) 4^1(0(1(2(x1)))) -> 4^1(1(x1)) 4^1(0(1(2(x1)))) -> 1^1(x1) 2^1(4(5(2(x1)))) -> 5^1(2(2(0(4(x1))))) 2^1(4(5(2(x1)))) -> 2^1(2(0(4(x1)))) 2^1(4(5(2(x1)))) -> 2^1(0(4(x1))) 2^1(4(5(2(x1)))) -> 4^1(x1) 4^1(0(1(0(0(x1))))) -> 4^1(1(0(2(0(0(x1)))))) 4^1(0(1(0(0(x1))))) -> 1^1(0(2(0(0(x1))))) 4^1(0(1(0(0(x1))))) -> 2^1(0(0(x1))) 2^1(4(5(0(0(x1))))) -> 5^1(2(0(0(4(0(x1)))))) 2^1(4(5(0(0(x1))))) -> 2^1(0(0(4(0(x1))))) 2^1(4(5(0(0(x1))))) -> 4^1(0(x1)) 2^1(1(0(1(0(x1))))) -> 1^1(4(1(0(2(0(x1)))))) 2^1(1(0(1(0(x1))))) -> 4^1(1(0(2(0(x1))))) 2^1(1(0(1(0(x1))))) -> 1^1(0(2(0(x1)))) 2^1(1(0(1(0(x1))))) -> 2^1(0(x1)) 3^1(0(2(1(0(x1))))) -> 3^1(1(4(0(2(0(x1)))))) 3^1(0(2(1(0(x1))))) -> 1^1(4(0(2(0(x1))))) 3^1(0(2(1(0(x1))))) -> 4^1(0(2(0(x1)))) 3^1(0(2(1(0(x1))))) -> 2^1(0(x1)) 2^1(2(2(1(0(x1))))) -> 2^1(1(2(2(2(0(x1)))))) 2^1(2(2(1(0(x1))))) -> 1^1(2(2(2(0(x1))))) 2^1(2(2(1(0(x1))))) -> 2^1(2(2(0(x1)))) 2^1(2(2(1(0(x1))))) -> 2^1(2(0(x1))) 2^1(2(2(1(0(x1))))) -> 2^1(0(x1)) 2^1(3(2(1(0(x1))))) -> 2^1(2(0(4(3(1(x1)))))) 2^1(3(2(1(0(x1))))) -> 2^1(0(4(3(1(x1))))) 2^1(3(2(1(0(x1))))) -> 4^1(3(1(x1))) 2^1(3(2(1(0(x1))))) -> 3^1(1(x1)) 2^1(3(2(1(0(x1))))) -> 1^1(x1) 3^1(2(3(1(0(x1))))) -> 1^1(3(3(2(0(0(x1)))))) 3^1(2(3(1(0(x1))))) -> 3^1(3(2(0(0(x1))))) 3^1(2(3(1(0(x1))))) -> 3^1(2(0(0(x1)))) 3^1(2(3(1(0(x1))))) -> 2^1(0(0(x1))) 2^1(4(3(1(0(x1))))) -> 1^1(1(4(3(2(0(x1)))))) 2^1(4(3(1(0(x1))))) -> 1^1(4(3(2(0(x1))))) 2^1(4(3(1(0(x1))))) -> 4^1(3(2(0(x1)))) 2^1(4(3(1(0(x1))))) -> 3^1(2(0(x1))) 2^1(4(3(1(0(x1))))) -> 2^1(0(x1)) 1^1(0(1(2(0(x1))))) -> 1^1(1(0(2(0(0(x1)))))) 1^1(0(1(2(0(x1))))) -> 1^1(0(2(0(0(x1))))) 1^1(0(1(2(0(x1))))) -> 2^1(0(0(x1))) 2^1(2(1(2(0(x1))))) -> 1^1(2(2(0(2(0(x1)))))) 2^1(2(1(2(0(x1))))) -> 2^1(2(0(2(0(x1))))) 2^1(2(1(2(0(x1))))) -> 2^1(0(2(0(x1)))) 5^1(0(3(2(0(x1))))) -> 3^1(5(0(0(2(0(x1)))))) 5^1(0(3(2(0(x1))))) -> 5^1(0(0(2(0(x1))))) 3^1(1(0(3(0(x1))))) -> 3^1(1(3(4(0(0(x1)))))) 3^1(1(0(3(0(x1))))) -> 1^1(3(4(0(0(x1))))) 3^1(1(0(3(0(x1))))) -> 3^1(4(0(0(x1)))) 3^1(1(0(3(0(x1))))) -> 4^1(0(0(x1))) 1^1(4(0(3(0(x1))))) -> 3^1(4(4(1(0(0(x1)))))) 1^1(4(0(3(0(x1))))) -> 4^1(4(1(0(0(x1))))) 1^1(4(0(3(0(x1))))) -> 4^1(1(0(0(x1)))) 1^1(4(0(3(0(x1))))) -> 1^1(0(0(x1))) 4^1(0(2(3(0(x1))))) -> 4^1(3(2(0(0(4(x1)))))) 4^1(0(2(3(0(x1))))) -> 3^1(2(0(0(4(x1))))) 4^1(0(2(3(0(x1))))) -> 2^1(0(0(4(x1)))) 4^1(0(2(3(0(x1))))) -> 4^1(x1) 3^1(2(5(4(0(x1))))) -> 5^1(4(3(2(2(0(x1)))))) 3^1(2(5(4(0(x1))))) -> 4^1(3(2(2(0(x1))))) 3^1(2(5(4(0(x1))))) -> 3^1(2(2(0(x1)))) 3^1(2(5(4(0(x1))))) -> 2^1(2(0(x1))) 3^1(2(5(4(0(x1))))) -> 2^1(0(x1)) 3^1(0(0(5(0(x1))))) -> 5^1(0(3(0(2(0(x1)))))) 3^1(0(0(5(0(x1))))) -> 3^1(0(2(0(x1)))) 3^1(0(0(5(0(x1))))) -> 2^1(0(x1)) 2^1(1(0(5(0(x1))))) -> 5^1(1(0(2(0(0(x1)))))) 2^1(1(0(5(0(x1))))) -> 1^1(0(2(0(0(x1))))) 2^1(1(0(5(0(x1))))) -> 2^1(0(0(x1))) 2^1(4(1(5(0(x1))))) -> 5^1(4(1(0(2(0(x1)))))) 2^1(4(1(5(0(x1))))) -> 4^1(1(0(2(0(x1))))) 2^1(4(1(5(0(x1))))) -> 1^1(0(2(0(x1)))) 2^1(4(1(5(0(x1))))) -> 2^1(0(x1)) 1^1(5(2(5(0(x1))))) -> 1^1(5(5(0(2(0(x1)))))) 1^1(5(2(5(0(x1))))) -> 5^1(5(0(2(0(x1))))) 1^1(5(2(5(0(x1))))) -> 5^1(0(2(0(x1)))) 1^1(5(2(5(0(x1))))) -> 2^1(0(x1)) 4^1(0(0(1(2(x1))))) -> 2^1(0(0(4(4(1(x1)))))) 4^1(0(0(1(2(x1))))) -> 4^1(4(1(x1))) 4^1(0(0(1(2(x1))))) -> 4^1(1(x1)) 4^1(0(0(1(2(x1))))) -> 1^1(x1) 3^1(0(0(5(2(x1))))) -> 3^1(5(0(0(2(0(x1)))))) 3^1(0(0(5(2(x1))))) -> 5^1(0(0(2(0(x1))))) 3^1(0(0(5(2(x1))))) -> 2^1(0(x1)) 1^1(0(3(5(2(x1))))) -> 1^1(3(2(2(0(5(x1)))))) 1^1(0(3(5(2(x1))))) -> 3^1(2(2(0(5(x1))))) 1^1(0(3(5(2(x1))))) -> 2^1(2(0(5(x1)))) 1^1(0(3(5(2(x1))))) -> 2^1(0(5(x1))) 1^1(0(3(5(2(x1))))) -> 5^1(x1) 2^1(2(1(0(5(x1))))) -> 2^1(0(2(0(1(5(x1)))))) 2^1(2(1(0(5(x1))))) -> 2^1(0(1(5(x1)))) 2^1(2(1(0(5(x1))))) -> 1^1(5(x1)) 2^1(1(0(2(5(x1))))) -> 2^1(2(0(4(5(1(x1)))))) 2^1(1(0(2(5(x1))))) -> 2^1(0(4(5(1(x1))))) 2^1(1(0(2(5(x1))))) -> 4^1(5(1(x1))) 2^1(1(0(2(5(x1))))) -> 5^1(1(x1)) 2^1(1(0(2(5(x1))))) -> 1^1(x1) 1^1(0(1(2(5(x1))))) -> 1^1(5(1(3(2(0(x1)))))) 1^1(0(1(2(5(x1))))) -> 5^1(1(3(2(0(x1))))) 1^1(0(1(2(5(x1))))) -> 1^1(3(2(0(x1)))) 1^1(0(1(2(5(x1))))) -> 3^1(2(0(x1))) 1^1(0(1(2(5(x1))))) -> 2^1(0(x1)) 1^1(0(3(2(5(x1))))) -> 3^1(2(2(0(5(1(x1)))))) 1^1(0(3(2(5(x1))))) -> 2^1(2(0(5(1(x1))))) 1^1(0(3(2(5(x1))))) -> 2^1(0(5(1(x1)))) 1^1(0(3(2(5(x1))))) -> 5^1(1(x1)) 1^1(0(3(2(5(x1))))) -> 1^1(x1) 1^1(4(0(3(5(x1))))) -> 1^1(3(2(0(5(4(x1)))))) 1^1(4(0(3(5(x1))))) -> 3^1(2(0(5(4(x1))))) 1^1(4(0(3(5(x1))))) -> 2^1(0(5(4(x1)))) 1^1(4(0(3(5(x1))))) -> 5^1(4(x1)) 1^1(4(0(3(5(x1))))) -> 4^1(x1) The TRS R consists of the following rules: 2(1(0(0(x1)))) -> 1(1(0(2(0(x1))))) 2(2(1(0(x1)))) -> 0(1(2(2(0(x1))))) 2(2(1(0(x1)))) -> 0(2(2(0(1(x1))))) 3(2(1(0(x1)))) -> 3(1(0(2(0(x1))))) 3(2(1(0(x1)))) -> 1(3(3(2(0(x1))))) 2(1(2(0(x1)))) -> 1(0(2(2(0(x1))))) 2(1(2(0(x1)))) -> 1(2(2(2(0(x1))))) 2(1(2(0(x1)))) -> 1(1(2(2(2(0(x1)))))) 2(2(3(0(x1)))) -> 2(2(0(4(3(x1))))) 2(2(3(0(x1)))) -> 3(2(2(2(2(0(x1)))))) 2(1(4(0(x1)))) -> 4(1(2(2(0(x1))))) 2(1(4(0(x1)))) -> 1(0(2(0(4(x1))))) 1(0(5(0(x1)))) -> 1(5(2(0(2(0(x1)))))) 5(0(5(0(x1)))) -> 5(5(0(2(0(x1))))) 1(2(5(0(x1)))) -> 1(5(2(2(0(x1))))) 5(2(5(0(x1)))) -> 5(5(2(2(0(x1))))) 2(4(5(0(x1)))) -> 5(0(2(2(0(4(x1)))))) 3(0(1(2(x1)))) -> 1(3(2(2(0(4(x1)))))) 4(0(1(2(x1)))) -> 2(2(2(0(4(1(x1)))))) 2(4(5(2(x1)))) -> 5(2(2(0(4(x1))))) 4(0(1(0(0(x1))))) -> 4(1(0(2(0(0(x1)))))) 2(4(5(0(0(x1))))) -> 5(2(0(0(4(0(x1)))))) 2(1(0(1(0(x1))))) -> 1(4(1(0(2(0(x1)))))) 3(0(2(1(0(x1))))) -> 3(1(4(0(2(0(x1)))))) 2(2(2(1(0(x1))))) -> 2(1(2(2(2(0(x1)))))) 2(3(2(1(0(x1))))) -> 2(2(0(4(3(1(x1)))))) 3(2(3(1(0(x1))))) -> 1(3(3(2(0(0(x1)))))) 2(4(3(1(0(x1))))) -> 1(1(4(3(2(0(x1)))))) 1(0(1(2(0(x1))))) -> 1(1(0(2(0(0(x1)))))) 2(2(1(2(0(x1))))) -> 1(2(2(0(2(0(x1)))))) 5(0(3(2(0(x1))))) -> 3(5(0(0(2(0(x1)))))) 3(1(0(3(0(x1))))) -> 3(1(3(4(0(0(x1)))))) 1(4(0(3(0(x1))))) -> 3(4(4(1(0(0(x1)))))) 4(0(2(3(0(x1))))) -> 4(3(2(0(0(4(x1)))))) 3(2(5(4(0(x1))))) -> 5(4(3(2(2(0(x1)))))) 3(0(0(5(0(x1))))) -> 5(0(3(0(2(0(x1)))))) 2(1(0(5(0(x1))))) -> 5(1(0(2(0(0(x1)))))) 2(4(1(5(0(x1))))) -> 5(4(1(0(2(0(x1)))))) 1(5(2(5(0(x1))))) -> 1(5(5(0(2(0(x1)))))) 4(0(0(1(2(x1))))) -> 2(0(0(4(4(1(x1)))))) 3(0(0(5(2(x1))))) -> 3(5(0(0(2(0(x1)))))) 1(0(3(5(2(x1))))) -> 1(3(2(2(0(5(x1)))))) 2(2(1(0(5(x1))))) -> 2(0(2(0(1(5(x1)))))) 2(1(0(2(5(x1))))) -> 2(2(0(4(5(1(x1)))))) 1(0(1(2(5(x1))))) -> 1(5(1(3(2(0(x1)))))) 1(0(3(2(5(x1))))) -> 3(2(2(0(5(1(x1)))))) 1(4(0(3(5(x1))))) -> 1(3(2(0(5(4(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 166 less nodes. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (7) Obligation: Q DP problem: The TRS P consists of the following rules: 1^1(0(3(2(5(x1))))) -> 1^1(x1) 1^1(4(0(3(5(x1))))) -> 4^1(x1) 4^1(0(1(2(x1)))) -> 4^1(1(x1)) 4^1(0(1(2(x1)))) -> 1^1(x1) 4^1(0(2(3(0(x1))))) -> 4^1(x1) 4^1(0(0(1(2(x1))))) -> 4^1(4(1(x1))) 4^1(0(0(1(2(x1))))) -> 4^1(1(x1)) 4^1(0(0(1(2(x1))))) -> 1^1(x1) The TRS R consists of the following rules: 2(1(0(0(x1)))) -> 1(1(0(2(0(x1))))) 2(2(1(0(x1)))) -> 0(1(2(2(0(x1))))) 2(2(1(0(x1)))) -> 0(2(2(0(1(x1))))) 3(2(1(0(x1)))) -> 3(1(0(2(0(x1))))) 3(2(1(0(x1)))) -> 1(3(3(2(0(x1))))) 2(1(2(0(x1)))) -> 1(0(2(2(0(x1))))) 2(1(2(0(x1)))) -> 1(2(2(2(0(x1))))) 2(1(2(0(x1)))) -> 1(1(2(2(2(0(x1)))))) 2(2(3(0(x1)))) -> 2(2(0(4(3(x1))))) 2(2(3(0(x1)))) -> 3(2(2(2(2(0(x1)))))) 2(1(4(0(x1)))) -> 4(1(2(2(0(x1))))) 2(1(4(0(x1)))) -> 1(0(2(0(4(x1))))) 1(0(5(0(x1)))) -> 1(5(2(0(2(0(x1)))))) 5(0(5(0(x1)))) -> 5(5(0(2(0(x1))))) 1(2(5(0(x1)))) -> 1(5(2(2(0(x1))))) 5(2(5(0(x1)))) -> 5(5(2(2(0(x1))))) 2(4(5(0(x1)))) -> 5(0(2(2(0(4(x1)))))) 3(0(1(2(x1)))) -> 1(3(2(2(0(4(x1)))))) 4(0(1(2(x1)))) -> 2(2(2(0(4(1(x1)))))) 2(4(5(2(x1)))) -> 5(2(2(0(4(x1))))) 4(0(1(0(0(x1))))) -> 4(1(0(2(0(0(x1)))))) 2(4(5(0(0(x1))))) -> 5(2(0(0(4(0(x1)))))) 2(1(0(1(0(x1))))) -> 1(4(1(0(2(0(x1)))))) 3(0(2(1(0(x1))))) -> 3(1(4(0(2(0(x1)))))) 2(2(2(1(0(x1))))) -> 2(1(2(2(2(0(x1)))))) 2(3(2(1(0(x1))))) -> 2(2(0(4(3(1(x1)))))) 3(2(3(1(0(x1))))) -> 1(3(3(2(0(0(x1)))))) 2(4(3(1(0(x1))))) -> 1(1(4(3(2(0(x1)))))) 1(0(1(2(0(x1))))) -> 1(1(0(2(0(0(x1)))))) 2(2(1(2(0(x1))))) -> 1(2(2(0(2(0(x1)))))) 5(0(3(2(0(x1))))) -> 3(5(0(0(2(0(x1)))))) 3(1(0(3(0(x1))))) -> 3(1(3(4(0(0(x1)))))) 1(4(0(3(0(x1))))) -> 3(4(4(1(0(0(x1)))))) 4(0(2(3(0(x1))))) -> 4(3(2(0(0(4(x1)))))) 3(2(5(4(0(x1))))) -> 5(4(3(2(2(0(x1)))))) 3(0(0(5(0(x1))))) -> 5(0(3(0(2(0(x1)))))) 2(1(0(5(0(x1))))) -> 5(1(0(2(0(0(x1)))))) 2(4(1(5(0(x1))))) -> 5(4(1(0(2(0(x1)))))) 1(5(2(5(0(x1))))) -> 1(5(5(0(2(0(x1)))))) 4(0(0(1(2(x1))))) -> 2(0(0(4(4(1(x1)))))) 3(0(0(5(2(x1))))) -> 3(5(0(0(2(0(x1)))))) 1(0(3(5(2(x1))))) -> 1(3(2(2(0(5(x1)))))) 2(2(1(0(5(x1))))) -> 2(0(2(0(1(5(x1)))))) 2(1(0(2(5(x1))))) -> 2(2(0(4(5(1(x1)))))) 1(0(1(2(5(x1))))) -> 1(5(1(3(2(0(x1)))))) 1(0(3(2(5(x1))))) -> 3(2(2(0(5(1(x1)))))) 1(4(0(3(5(x1))))) -> 1(3(2(0(5(4(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (8) UsableRulesProof (EQUIVALENT) 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. ---------------------------------------- (9) Obligation: Q DP problem: The TRS P consists of the following rules: 1^1(0(3(2(5(x1))))) -> 1^1(x1) 1^1(4(0(3(5(x1))))) -> 4^1(x1) 4^1(0(1(2(x1)))) -> 4^1(1(x1)) 4^1(0(1(2(x1)))) -> 1^1(x1) 4^1(0(2(3(0(x1))))) -> 4^1(x1) 4^1(0(0(1(2(x1))))) -> 4^1(4(1(x1))) 4^1(0(0(1(2(x1))))) -> 4^1(1(x1)) 4^1(0(0(1(2(x1))))) -> 1^1(x1) The TRS R consists of the following rules: 1(0(5(0(x1)))) -> 1(5(2(0(2(0(x1)))))) 1(2(5(0(x1)))) -> 1(5(2(2(0(x1))))) 1(0(1(2(0(x1))))) -> 1(1(0(2(0(0(x1)))))) 1(4(0(3(0(x1))))) -> 3(4(4(1(0(0(x1)))))) 1(5(2(5(0(x1))))) -> 1(5(5(0(2(0(x1)))))) 1(0(3(5(2(x1))))) -> 1(3(2(2(0(5(x1)))))) 1(0(1(2(5(x1))))) -> 1(5(1(3(2(0(x1)))))) 1(0(3(2(5(x1))))) -> 3(2(2(0(5(1(x1)))))) 1(4(0(3(5(x1))))) -> 1(3(2(0(5(4(x1)))))) 4(0(1(2(x1)))) -> 2(2(2(0(4(1(x1)))))) 4(0(1(0(0(x1))))) -> 4(1(0(2(0(0(x1)))))) 4(0(2(3(0(x1))))) -> 4(3(2(0(0(4(x1)))))) 4(0(0(1(2(x1))))) -> 2(0(0(4(4(1(x1)))))) 5(0(5(0(x1)))) -> 5(5(0(2(0(x1))))) 5(2(5(0(x1)))) -> 5(5(2(2(0(x1))))) 5(0(3(2(0(x1))))) -> 3(5(0(0(2(0(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (10) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes. ---------------------------------------- (11) Obligation: Q DP problem: The TRS P consists of the following rules: 1^1(4(0(3(5(x1))))) -> 4^1(x1) 4^1(0(1(2(x1)))) -> 1^1(x1) 1^1(0(3(2(5(x1))))) -> 1^1(x1) 4^1(0(2(3(0(x1))))) -> 4^1(x1) 4^1(0(0(1(2(x1))))) -> 1^1(x1) The TRS R consists of the following rules: 1(0(5(0(x1)))) -> 1(5(2(0(2(0(x1)))))) 1(2(5(0(x1)))) -> 1(5(2(2(0(x1))))) 1(0(1(2(0(x1))))) -> 1(1(0(2(0(0(x1)))))) 1(4(0(3(0(x1))))) -> 3(4(4(1(0(0(x1)))))) 1(5(2(5(0(x1))))) -> 1(5(5(0(2(0(x1)))))) 1(0(3(5(2(x1))))) -> 1(3(2(2(0(5(x1)))))) 1(0(1(2(5(x1))))) -> 1(5(1(3(2(0(x1)))))) 1(0(3(2(5(x1))))) -> 3(2(2(0(5(1(x1)))))) 1(4(0(3(5(x1))))) -> 1(3(2(0(5(4(x1)))))) 4(0(1(2(x1)))) -> 2(2(2(0(4(1(x1)))))) 4(0(1(0(0(x1))))) -> 4(1(0(2(0(0(x1)))))) 4(0(2(3(0(x1))))) -> 4(3(2(0(0(4(x1)))))) 4(0(0(1(2(x1))))) -> 2(0(0(4(4(1(x1)))))) 5(0(5(0(x1)))) -> 5(5(0(2(0(x1))))) 5(2(5(0(x1)))) -> 5(5(2(2(0(x1))))) 5(0(3(2(0(x1))))) -> 3(5(0(0(2(0(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (12) UsableRulesProof (EQUIVALENT) We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (13) Obligation: Q DP problem: The TRS P consists of the following rules: 1^1(4(0(3(5(x1))))) -> 4^1(x1) 4^1(0(1(2(x1)))) -> 1^1(x1) 1^1(0(3(2(5(x1))))) -> 1^1(x1) 4^1(0(2(3(0(x1))))) -> 4^1(x1) 4^1(0(0(1(2(x1))))) -> 1^1(x1) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (14) QDPSizeChangeProof (EQUIVALENT) 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. From the DPs we obtained the following set of size-change graphs: *4^1(0(2(3(0(x1))))) -> 4^1(x1) The graph contains the following edges 1 > 1 *1^1(0(3(2(5(x1))))) -> 1^1(x1) The graph contains the following edges 1 > 1 *1^1(4(0(3(5(x1))))) -> 4^1(x1) The graph contains the following edges 1 > 1 *4^1(0(1(2(x1)))) -> 1^1(x1) The graph contains the following edges 1 > 1 *4^1(0(0(1(2(x1))))) -> 1^1(x1) The graph contains the following edges 1 > 1 ---------------------------------------- (15) YES ---------------------------------------- (16) Obligation: Q DP problem: The TRS P consists of the following rules: 3^1(1(0(3(0(x1))))) -> 3^1(4(0(0(x1)))) 3^1(1(0(3(0(x1))))) -> 3^1(1(3(4(0(0(x1)))))) The TRS R consists of the following rules: 2(1(0(0(x1)))) -> 1(1(0(2(0(x1))))) 2(2(1(0(x1)))) -> 0(1(2(2(0(x1))))) 2(2(1(0(x1)))) -> 0(2(2(0(1(x1))))) 3(2(1(0(x1)))) -> 3(1(0(2(0(x1))))) 3(2(1(0(x1)))) -> 1(3(3(2(0(x1))))) 2(1(2(0(x1)))) -> 1(0(2(2(0(x1))))) 2(1(2(0(x1)))) -> 1(2(2(2(0(x1))))) 2(1(2(0(x1)))) -> 1(1(2(2(2(0(x1)))))) 2(2(3(0(x1)))) -> 2(2(0(4(3(x1))))) 2(2(3(0(x1)))) -> 3(2(2(2(2(0(x1)))))) 2(1(4(0(x1)))) -> 4(1(2(2(0(x1))))) 2(1(4(0(x1)))) -> 1(0(2(0(4(x1))))) 1(0(5(0(x1)))) -> 1(5(2(0(2(0(x1)))))) 5(0(5(0(x1)))) -> 5(5(0(2(0(x1))))) 1(2(5(0(x1)))) -> 1(5(2(2(0(x1))))) 5(2(5(0(x1)))) -> 5(5(2(2(0(x1))))) 2(4(5(0(x1)))) -> 5(0(2(2(0(4(x1)))))) 3(0(1(2(x1)))) -> 1(3(2(2(0(4(x1)))))) 4(0(1(2(x1)))) -> 2(2(2(0(4(1(x1)))))) 2(4(5(2(x1)))) -> 5(2(2(0(4(x1))))) 4(0(1(0(0(x1))))) -> 4(1(0(2(0(0(x1)))))) 2(4(5(0(0(x1))))) -> 5(2(0(0(4(0(x1)))))) 2(1(0(1(0(x1))))) -> 1(4(1(0(2(0(x1)))))) 3(0(2(1(0(x1))))) -> 3(1(4(0(2(0(x1)))))) 2(2(2(1(0(x1))))) -> 2(1(2(2(2(0(x1)))))) 2(3(2(1(0(x1))))) -> 2(2(0(4(3(1(x1)))))) 3(2(3(1(0(x1))))) -> 1(3(3(2(0(0(x1)))))) 2(4(3(1(0(x1))))) -> 1(1(4(3(2(0(x1)))))) 1(0(1(2(0(x1))))) -> 1(1(0(2(0(0(x1)))))) 2(2(1(2(0(x1))))) -> 1(2(2(0(2(0(x1)))))) 5(0(3(2(0(x1))))) -> 3(5(0(0(2(0(x1)))))) 3(1(0(3(0(x1))))) -> 3(1(3(4(0(0(x1)))))) 1(4(0(3(0(x1))))) -> 3(4(4(1(0(0(x1)))))) 4(0(2(3(0(x1))))) -> 4(3(2(0(0(4(x1)))))) 3(2(5(4(0(x1))))) -> 5(4(3(2(2(0(x1)))))) 3(0(0(5(0(x1))))) -> 5(0(3(0(2(0(x1)))))) 2(1(0(5(0(x1))))) -> 5(1(0(2(0(0(x1)))))) 2(4(1(5(0(x1))))) -> 5(4(1(0(2(0(x1)))))) 1(5(2(5(0(x1))))) -> 1(5(5(0(2(0(x1)))))) 4(0(0(1(2(x1))))) -> 2(0(0(4(4(1(x1)))))) 3(0(0(5(2(x1))))) -> 3(5(0(0(2(0(x1)))))) 1(0(3(5(2(x1))))) -> 1(3(2(2(0(5(x1)))))) 2(2(1(0(5(x1))))) -> 2(0(2(0(1(5(x1)))))) 2(1(0(2(5(x1))))) -> 2(2(0(4(5(1(x1)))))) 1(0(1(2(5(x1))))) -> 1(5(1(3(2(0(x1)))))) 1(0(3(2(5(x1))))) -> 3(2(2(0(5(1(x1)))))) 1(4(0(3(5(x1))))) -> 1(3(2(0(5(4(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (17) UsableRulesProof (EQUIVALENT) 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. ---------------------------------------- (18) Obligation: Q DP problem: The TRS P consists of the following rules: 3^1(1(0(3(0(x1))))) -> 3^1(4(0(0(x1)))) 3^1(1(0(3(0(x1))))) -> 3^1(1(3(4(0(0(x1)))))) The TRS R consists of the following rules: 4(0(0(1(2(x1))))) -> 2(0(0(4(4(1(x1)))))) 3(2(1(0(x1)))) -> 3(1(0(2(0(x1))))) 3(2(1(0(x1)))) -> 1(3(3(2(0(x1))))) 3(0(1(2(x1)))) -> 1(3(2(2(0(4(x1)))))) 3(0(2(1(0(x1))))) -> 3(1(4(0(2(0(x1)))))) 3(2(3(1(0(x1))))) -> 1(3(3(2(0(0(x1)))))) 3(1(0(3(0(x1))))) -> 3(1(3(4(0(0(x1)))))) 3(2(5(4(0(x1))))) -> 5(4(3(2(2(0(x1)))))) 3(0(0(5(0(x1))))) -> 5(0(3(0(2(0(x1)))))) 3(0(0(5(2(x1))))) -> 3(5(0(0(2(0(x1)))))) 1(0(5(0(x1)))) -> 1(5(2(0(2(0(x1)))))) 1(2(5(0(x1)))) -> 1(5(2(2(0(x1))))) 1(0(1(2(0(x1))))) -> 1(1(0(2(0(0(x1)))))) 1(4(0(3(0(x1))))) -> 3(4(4(1(0(0(x1)))))) 1(5(2(5(0(x1))))) -> 1(5(5(0(2(0(x1)))))) 1(0(3(5(2(x1))))) -> 1(3(2(2(0(5(x1)))))) 1(0(1(2(5(x1))))) -> 1(5(1(3(2(0(x1)))))) 1(0(3(2(5(x1))))) -> 3(2(2(0(5(1(x1)))))) 1(4(0(3(5(x1))))) -> 1(3(2(0(5(4(x1)))))) 4(0(1(2(x1)))) -> 2(2(2(0(4(1(x1)))))) 4(0(1(0(0(x1))))) -> 4(1(0(2(0(0(x1)))))) 4(0(2(3(0(x1))))) -> 4(3(2(0(0(4(x1)))))) 5(0(5(0(x1)))) -> 5(5(0(2(0(x1))))) 5(2(5(0(x1)))) -> 5(5(2(2(0(x1))))) 5(0(3(2(0(x1))))) -> 3(5(0(0(2(0(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (19) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes. ---------------------------------------- (20) TRUE