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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 84 ms] (4) QTRS (5) DependencyPairsProof [EQUIVALENT, 108 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) AND (9) QDP (10) QDPOrderProof [EQUIVALENT, 64 ms] (11) QDP (12) PisEmptyProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) QDP (17) QDPSizeChangeProof [EQUIVALENT, 0 ms] (18) YES (19) QDP (20) QDPSizeChangeProof [EQUIVALENT, 0 ms] (21) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(5(0(x1))) -> 3(4(3(3(2(0(4(4(4(0(x1)))))))))) 0(2(5(0(x1)))) -> 0(2(1(1(5(2(1(4(1(3(x1)))))))))) 0(5(0(5(x1)))) -> 3(2(0(2(0(4(4(1(5(4(x1)))))))))) 2(5(5(3(x1)))) -> 4(3(0(2(1(0(1(3(4(3(x1)))))))))) 5(5(5(0(x1)))) -> 4(1(2(1(2(3(5(0(1(3(x1)))))))))) 0(5(1(5(0(x1))))) -> 3(4(0(1(4(5(2(2(3(1(x1)))))))))) 3(0(0(5(3(x1))))) -> 1(3(4(3(5(2(4(1(3(3(x1)))))))))) 3(5(5(0(0(x1))))) -> 1(4(1(0(0(4(4(0(4(1(x1)))))))))) 0(2(5(1(5(0(x1)))))) -> 2(1(0(1(5(2(4(0(2(0(x1)))))))))) 0(2(5(5(1(4(x1)))))) -> 4(0(1(1(1(1(0(4(1(5(x1)))))))))) 0(5(3(1(2(5(x1)))))) -> 0(2(3(1(1(2(4(4(5(5(x1)))))))))) 0(5(5(1(5(1(x1)))))) -> 0(5(1(1(3(3(4(2(1(0(x1)))))))))) 4(5(5(4(2(0(x1)))))) -> 2(4(0(1(3(4(4(4(1(0(x1)))))))))) 5(0(0(3(5(2(x1)))))) -> 4(4(2(3(0(1(2(0(5(2(x1)))))))))) 5(1(5(0(2(5(x1)))))) -> 5(0(0(1(4(2(3(2(1(5(x1)))))))))) 5(2(0(2(5(5(x1)))))) -> 5(5(4(4(4(5(4(4(1(4(x1)))))))))) 5(5(0(2(5(0(x1)))))) -> 2(0(5(0(2(1(0(0(3(0(x1)))))))))) 5(5(0(3(4(5(x1)))))) -> 2(0(5(5(2(1(3(2(3(2(x1)))))))))) 5(5(3(5(0(5(x1)))))) -> 5(4(4(3(5(1(3(3(4(5(x1)))))))))) 0(4(4(0(0(5(1(x1))))))) -> 1(3(2(0(4(1(5(1(1(2(x1)))))))))) 0(4(4(2(5(5(5(x1))))))) -> 0(2(4(5(5(4(2(0(1(1(x1)))))))))) 1(0(2(5(2(0(0(x1))))))) -> 3(1(4(4(0(3(0(1(2(2(x1)))))))))) 1(2(0(4(2(5(0(x1))))))) -> 4(2(4(0(3(2(2(4(1(0(x1)))))))))) 1(2(5(5(0(3(3(x1))))))) -> 3(4(1(2(0(3(3(1(0(3(x1)))))))))) 1(5(5(3(3(3(4(x1))))))) -> 1(5(1(0(0(2(2(2(3(5(x1)))))))))) 2(5(4(5(2(5(1(x1))))))) -> 4(3(2(1(4(2(2(4(5(2(x1)))))))))) 3(2(3(5(1(5(2(x1))))))) -> 2(0(3(2(3(2(1(5(5(1(x1)))))))))) 3(3(4(2(5(5(2(x1))))))) -> 1(2(3(3(4(4(1(4(0(1(x1)))))))))) 3(5(0(5(5(5(0(x1))))))) -> 0(0(3(0(3(5(0(3(2(0(x1)))))))))) 4(3(1(2(5(2(4(x1))))))) -> 2(3(1(1(4(3(4(4(2(4(x1)))))))))) 4(5(5(3(1(0(5(x1))))))) -> 1(1(5(2(0(3(3(3(2(1(x1)))))))))) 5(0(5(3(1(0(5(x1))))))) -> 5(2(4(4(2(1(3(5(1(5(x1)))))))))) 5(0(5(3(5(1(5(x1))))))) -> 5(1(1(2(4(0(0(3(2(5(x1)))))))))) 5(1(5(3(3(0(5(x1))))))) -> 4(4(3(2(2(2(5(0(1(1(x1)))))))))) 5(2(0(2(5(3(3(x1))))))) -> 5(2(0(5(1(1(3(2(0(3(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: 0(5(0(x1))) -> 0(4(4(4(0(2(3(3(4(3(x1)))))))))) 0(5(2(0(x1)))) -> 3(1(4(1(2(5(1(1(2(0(x1)))))))))) 5(0(5(0(x1)))) -> 4(5(1(4(4(0(2(0(2(3(x1)))))))))) 3(5(5(2(x1)))) -> 3(4(3(1(0(1(2(0(3(4(x1)))))))))) 0(5(5(5(x1)))) -> 3(1(0(5(3(2(1(2(1(4(x1)))))))))) 0(5(1(5(0(x1))))) -> 1(3(2(2(5(4(1(0(4(3(x1)))))))))) 3(5(0(0(3(x1))))) -> 3(3(1(4(2(5(3(4(3(1(x1)))))))))) 0(0(5(5(3(x1))))) -> 1(4(0(4(4(0(0(1(4(1(x1)))))))))) 0(5(1(5(2(0(x1)))))) -> 0(2(0(4(2(5(1(0(1(2(x1)))))))))) 4(1(5(5(2(0(x1)))))) -> 5(1(4(0(1(1(1(1(0(4(x1)))))))))) 5(2(1(3(5(0(x1)))))) -> 5(5(4(4(2(1(1(3(2(0(x1)))))))))) 1(5(1(5(5(0(x1)))))) -> 0(1(2(4(3(3(1(1(5(0(x1)))))))))) 0(2(4(5(5(4(x1)))))) -> 0(1(4(4(4(3(1(0(4(2(x1)))))))))) 2(5(3(0(0(5(x1)))))) -> 2(5(0(2(1(0(3(2(4(4(x1)))))))))) 5(2(0(5(1(5(x1)))))) -> 5(1(2(3(2(4(1(0(0(5(x1)))))))))) 5(5(2(0(2(5(x1)))))) -> 4(1(4(4(5(4(4(4(5(5(x1)))))))))) 0(5(2(0(5(5(x1)))))) -> 0(3(0(0(1(2(0(5(0(2(x1)))))))))) 5(4(3(0(5(5(x1)))))) -> 2(3(2(3(1(2(5(5(0(2(x1)))))))))) 5(0(5(3(5(5(x1)))))) -> 5(4(3(3(1(5(3(4(4(5(x1)))))))))) 1(5(0(0(4(4(0(x1))))))) -> 2(1(1(5(1(4(0(2(3(1(x1)))))))))) 5(5(5(2(4(4(0(x1))))))) -> 1(1(0(2(4(5(5(4(2(0(x1)))))))))) 0(0(2(5(2(0(1(x1))))))) -> 2(2(1(0(3(0(4(4(1(3(x1)))))))))) 0(5(2(4(0(2(1(x1))))))) -> 0(1(4(2(2(3(0(4(2(4(x1)))))))))) 3(3(0(5(5(2(1(x1))))))) -> 3(0(1(3(3(0(2(1(4(3(x1)))))))))) 4(3(3(3(5(5(1(x1))))))) -> 5(3(2(2(2(0(0(1(5(1(x1)))))))))) 1(5(2(5(4(5(2(x1))))))) -> 2(5(4(2(2(4(1(2(3(4(x1)))))))))) 2(5(1(5(3(2(3(x1))))))) -> 1(5(5(1(2(3(2(3(0(2(x1)))))))))) 2(5(5(2(4(3(3(x1))))))) -> 1(0(4(1(4(4(3(3(2(1(x1)))))))))) 0(5(5(5(0(5(3(x1))))))) -> 0(2(3(0(5(3(0(3(0(0(x1)))))))))) 4(2(5(2(1(3(4(x1))))))) -> 4(2(4(4(3(4(1(1(3(2(x1)))))))))) 5(0(1(3(5(5(4(x1))))))) -> 1(2(3(3(3(0(2(5(1(1(x1)))))))))) 5(0(1(3(5(0(5(x1))))))) -> 5(1(5(3(1(2(4(4(2(5(x1)))))))))) 5(1(5(3(5(0(5(x1))))))) -> 5(2(3(0(0(4(2(1(1(5(x1)))))))))) 5(0(3(3(5(1(5(x1))))))) -> 1(1(0(5(2(2(2(3(4(4(x1)))))))))) 3(3(5(2(0(2(5(x1))))))) -> 3(0(2(3(1(1(5(0(2(5(x1)))))))))) Q is empty. ---------------------------------------- (3) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0(x_1)) = x_1 POL(1(x_1)) = x_1 POL(2(x_1)) = x_1 POL(3(x_1)) = x_1 POL(4(x_1)) = x_1 POL(5(x_1)) = 1 + x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 0(5(0(x1))) -> 0(4(4(4(0(2(3(3(4(3(x1)))))))))) 5(0(5(0(x1)))) -> 4(5(1(4(4(0(2(0(2(3(x1)))))))))) 3(5(5(2(x1)))) -> 3(4(3(1(0(1(2(0(3(4(x1)))))))))) 0(5(5(5(x1)))) -> 3(1(0(5(3(2(1(2(1(4(x1)))))))))) 0(5(1(5(0(x1))))) -> 1(3(2(2(5(4(1(0(4(3(x1)))))))))) 0(0(5(5(3(x1))))) -> 1(4(0(4(4(0(0(1(4(1(x1)))))))))) 0(5(1(5(2(0(x1)))))) -> 0(2(0(4(2(5(1(0(1(2(x1)))))))))) 4(1(5(5(2(0(x1)))))) -> 5(1(4(0(1(1(1(1(0(4(x1)))))))))) 1(5(1(5(5(0(x1)))))) -> 0(1(2(4(3(3(1(1(5(0(x1)))))))))) 0(2(4(5(5(4(x1)))))) -> 0(1(4(4(4(3(1(0(4(2(x1)))))))))) 2(5(3(0(0(5(x1)))))) -> 2(5(0(2(1(0(3(2(4(4(x1)))))))))) 5(2(0(5(1(5(x1)))))) -> 5(1(2(3(2(4(1(0(0(5(x1)))))))))) 0(5(2(0(5(5(x1)))))) -> 0(3(0(0(1(2(0(5(0(2(x1)))))))))) 5(4(3(0(5(5(x1)))))) -> 2(3(2(3(1(2(5(5(0(2(x1)))))))))) 5(0(5(3(5(5(x1)))))) -> 5(4(3(3(1(5(3(4(4(5(x1)))))))))) 5(5(5(2(4(4(0(x1))))))) -> 1(1(0(2(4(5(5(4(2(0(x1)))))))))) 0(0(2(5(2(0(1(x1))))))) -> 2(2(1(0(3(0(4(4(1(3(x1)))))))))) 0(5(2(4(0(2(1(x1))))))) -> 0(1(4(2(2(3(0(4(2(4(x1)))))))))) 3(3(0(5(5(2(1(x1))))))) -> 3(0(1(3(3(0(2(1(4(3(x1)))))))))) 1(5(2(5(4(5(2(x1))))))) -> 2(5(4(2(2(4(1(2(3(4(x1)))))))))) 2(5(5(2(4(3(3(x1))))))) -> 1(0(4(1(4(4(3(3(2(1(x1)))))))))) 0(5(5(5(0(5(3(x1))))))) -> 0(2(3(0(5(3(0(3(0(0(x1)))))))))) 4(2(5(2(1(3(4(x1))))))) -> 4(2(4(4(3(4(1(1(3(2(x1)))))))))) 5(0(1(3(5(5(4(x1))))))) -> 1(2(3(3(3(0(2(5(1(1(x1)))))))))) 5(1(5(3(5(0(5(x1))))))) -> 5(2(3(0(0(4(2(1(1(5(x1)))))))))) 5(0(3(3(5(1(5(x1))))))) -> 1(1(0(5(2(2(2(3(4(4(x1)))))))))) ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(5(2(0(x1)))) -> 3(1(4(1(2(5(1(1(2(0(x1)))))))))) 3(5(0(0(3(x1))))) -> 3(3(1(4(2(5(3(4(3(1(x1)))))))))) 5(2(1(3(5(0(x1)))))) -> 5(5(4(4(2(1(1(3(2(0(x1)))))))))) 5(5(2(0(2(5(x1)))))) -> 4(1(4(4(5(4(4(4(5(5(x1)))))))))) 1(5(0(0(4(4(0(x1))))))) -> 2(1(1(5(1(4(0(2(3(1(x1)))))))))) 4(3(3(3(5(5(1(x1))))))) -> 5(3(2(2(2(0(0(1(5(1(x1)))))))))) 2(5(1(5(3(2(3(x1))))))) -> 1(5(5(1(2(3(2(3(0(2(x1)))))))))) 5(0(1(3(5(0(5(x1))))))) -> 5(1(5(3(1(2(4(4(2(5(x1)))))))))) 3(3(5(2(0(2(5(x1))))))) -> 3(0(2(3(1(1(5(0(2(5(x1)))))))))) Q is empty. ---------------------------------------- (5) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: 0^1(5(2(0(x1)))) -> 3^1(1(4(1(2(5(1(1(2(0(x1)))))))))) 0^1(5(2(0(x1)))) -> 1^1(4(1(2(5(1(1(2(0(x1))))))))) 0^1(5(2(0(x1)))) -> 4^1(1(2(5(1(1(2(0(x1)))))))) 0^1(5(2(0(x1)))) -> 1^1(2(5(1(1(2(0(x1))))))) 0^1(5(2(0(x1)))) -> 2^1(5(1(1(2(0(x1)))))) 0^1(5(2(0(x1)))) -> 5^1(1(1(2(0(x1))))) 0^1(5(2(0(x1)))) -> 1^1(1(2(0(x1)))) 0^1(5(2(0(x1)))) -> 1^1(2(0(x1))) 3^1(5(0(0(3(x1))))) -> 3^1(3(1(4(2(5(3(4(3(1(x1)))))))))) 3^1(5(0(0(3(x1))))) -> 3^1(1(4(2(5(3(4(3(1(x1))))))))) 3^1(5(0(0(3(x1))))) -> 1^1(4(2(5(3(4(3(1(x1)))))))) 3^1(5(0(0(3(x1))))) -> 4^1(2(5(3(4(3(1(x1))))))) 3^1(5(0(0(3(x1))))) -> 2^1(5(3(4(3(1(x1)))))) 3^1(5(0(0(3(x1))))) -> 5^1(3(4(3(1(x1))))) 3^1(5(0(0(3(x1))))) -> 3^1(4(3(1(x1)))) 3^1(5(0(0(3(x1))))) -> 4^1(3(1(x1))) 3^1(5(0(0(3(x1))))) -> 3^1(1(x1)) 3^1(5(0(0(3(x1))))) -> 1^1(x1) 5^1(2(1(3(5(0(x1)))))) -> 5^1(5(4(4(2(1(1(3(2(0(x1)))))))))) 5^1(2(1(3(5(0(x1)))))) -> 5^1(4(4(2(1(1(3(2(0(x1))))))))) 5^1(2(1(3(5(0(x1)))))) -> 4^1(4(2(1(1(3(2(0(x1)))))))) 5^1(2(1(3(5(0(x1)))))) -> 4^1(2(1(1(3(2(0(x1))))))) 5^1(2(1(3(5(0(x1)))))) -> 2^1(1(1(3(2(0(x1)))))) 5^1(2(1(3(5(0(x1)))))) -> 1^1(1(3(2(0(x1))))) 5^1(2(1(3(5(0(x1)))))) -> 1^1(3(2(0(x1)))) 5^1(2(1(3(5(0(x1)))))) -> 3^1(2(0(x1))) 5^1(2(1(3(5(0(x1)))))) -> 2^1(0(x1)) 5^1(5(2(0(2(5(x1)))))) -> 4^1(1(4(4(5(4(4(4(5(5(x1)))))))))) 5^1(5(2(0(2(5(x1)))))) -> 1^1(4(4(5(4(4(4(5(5(x1))))))))) 5^1(5(2(0(2(5(x1)))))) -> 4^1(4(5(4(4(4(5(5(x1)))))))) 5^1(5(2(0(2(5(x1)))))) -> 4^1(5(4(4(4(5(5(x1))))))) 5^1(5(2(0(2(5(x1)))))) -> 5^1(4(4(4(5(5(x1)))))) 5^1(5(2(0(2(5(x1)))))) -> 4^1(4(4(5(5(x1))))) 5^1(5(2(0(2(5(x1)))))) -> 4^1(4(5(5(x1)))) 5^1(5(2(0(2(5(x1)))))) -> 4^1(5(5(x1))) 5^1(5(2(0(2(5(x1)))))) -> 5^1(5(x1)) 1^1(5(0(0(4(4(0(x1))))))) -> 2^1(1(1(5(1(4(0(2(3(1(x1)))))))))) 1^1(5(0(0(4(4(0(x1))))))) -> 1^1(1(5(1(4(0(2(3(1(x1))))))))) 1^1(5(0(0(4(4(0(x1))))))) -> 1^1(5(1(4(0(2(3(1(x1)))))))) 1^1(5(0(0(4(4(0(x1))))))) -> 5^1(1(4(0(2(3(1(x1))))))) 1^1(5(0(0(4(4(0(x1))))))) -> 1^1(4(0(2(3(1(x1)))))) 1^1(5(0(0(4(4(0(x1))))))) -> 4^1(0(2(3(1(x1))))) 1^1(5(0(0(4(4(0(x1))))))) -> 0^1(2(3(1(x1)))) 1^1(5(0(0(4(4(0(x1))))))) -> 2^1(3(1(x1))) 1^1(5(0(0(4(4(0(x1))))))) -> 3^1(1(x1)) 1^1(5(0(0(4(4(0(x1))))))) -> 1^1(x1) 4^1(3(3(3(5(5(1(x1))))))) -> 5^1(3(2(2(2(0(0(1(5(1(x1)))))))))) 4^1(3(3(3(5(5(1(x1))))))) -> 3^1(2(2(2(0(0(1(5(1(x1))))))))) 4^1(3(3(3(5(5(1(x1))))))) -> 2^1(2(2(0(0(1(5(1(x1)))))))) 4^1(3(3(3(5(5(1(x1))))))) -> 2^1(2(0(0(1(5(1(x1))))))) 4^1(3(3(3(5(5(1(x1))))))) -> 2^1(0(0(1(5(1(x1)))))) 4^1(3(3(3(5(5(1(x1))))))) -> 0^1(0(1(5(1(x1))))) 4^1(3(3(3(5(5(1(x1))))))) -> 0^1(1(5(1(x1)))) 4^1(3(3(3(5(5(1(x1))))))) -> 1^1(5(1(x1))) 2^1(5(1(5(3(2(3(x1))))))) -> 1^1(5(5(1(2(3(2(3(0(2(x1)))))))))) 2^1(5(1(5(3(2(3(x1))))))) -> 5^1(5(1(2(3(2(3(0(2(x1))))))))) 2^1(5(1(5(3(2(3(x1))))))) -> 5^1(1(2(3(2(3(0(2(x1)))))))) 2^1(5(1(5(3(2(3(x1))))))) -> 1^1(2(3(2(3(0(2(x1))))))) 2^1(5(1(5(3(2(3(x1))))))) -> 2^1(3(2(3(0(2(x1)))))) 2^1(5(1(5(3(2(3(x1))))))) -> 3^1(2(3(0(2(x1))))) 2^1(5(1(5(3(2(3(x1))))))) -> 2^1(3(0(2(x1)))) 2^1(5(1(5(3(2(3(x1))))))) -> 3^1(0(2(x1))) 2^1(5(1(5(3(2(3(x1))))))) -> 0^1(2(x1)) 2^1(5(1(5(3(2(3(x1))))))) -> 2^1(x1) 5^1(0(1(3(5(0(5(x1))))))) -> 5^1(1(5(3(1(2(4(4(2(5(x1)))))))))) 5^1(0(1(3(5(0(5(x1))))))) -> 1^1(5(3(1(2(4(4(2(5(x1))))))))) 5^1(0(1(3(5(0(5(x1))))))) -> 5^1(3(1(2(4(4(2(5(x1)))))))) 5^1(0(1(3(5(0(5(x1))))))) -> 3^1(1(2(4(4(2(5(x1))))))) 5^1(0(1(3(5(0(5(x1))))))) -> 1^1(2(4(4(2(5(x1)))))) 5^1(0(1(3(5(0(5(x1))))))) -> 2^1(4(4(2(5(x1))))) 5^1(0(1(3(5(0(5(x1))))))) -> 4^1(4(2(5(x1)))) 5^1(0(1(3(5(0(5(x1))))))) -> 4^1(2(5(x1))) 5^1(0(1(3(5(0(5(x1))))))) -> 2^1(5(x1)) 3^1(3(5(2(0(2(5(x1))))))) -> 3^1(0(2(3(1(1(5(0(2(5(x1)))))))))) 3^1(3(5(2(0(2(5(x1))))))) -> 0^1(2(3(1(1(5(0(2(5(x1))))))))) 3^1(3(5(2(0(2(5(x1))))))) -> 2^1(3(1(1(5(0(2(5(x1)))))))) 3^1(3(5(2(0(2(5(x1))))))) -> 3^1(1(1(5(0(2(5(x1))))))) 3^1(3(5(2(0(2(5(x1))))))) -> 1^1(1(5(0(2(5(x1)))))) 3^1(3(5(2(0(2(5(x1))))))) -> 1^1(5(0(2(5(x1))))) 3^1(3(5(2(0(2(5(x1))))))) -> 5^1(0(2(5(x1)))) The TRS R consists of the following rules: 0(5(2(0(x1)))) -> 3(1(4(1(2(5(1(1(2(0(x1)))))))))) 3(5(0(0(3(x1))))) -> 3(3(1(4(2(5(3(4(3(1(x1)))))))))) 5(2(1(3(5(0(x1)))))) -> 5(5(4(4(2(1(1(3(2(0(x1)))))))))) 5(5(2(0(2(5(x1)))))) -> 4(1(4(4(5(4(4(4(5(5(x1)))))))))) 1(5(0(0(4(4(0(x1))))))) -> 2(1(1(5(1(4(0(2(3(1(x1)))))))))) 4(3(3(3(5(5(1(x1))))))) -> 5(3(2(2(2(0(0(1(5(1(x1)))))))))) 2(5(1(5(3(2(3(x1))))))) -> 1(5(5(1(2(3(2(3(0(2(x1)))))))))) 5(0(1(3(5(0(5(x1))))))) -> 5(1(5(3(1(2(4(4(2(5(x1)))))))))) 3(3(5(2(0(2(5(x1))))))) -> 3(0(2(3(1(1(5(0(2(5(x1)))))))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (7) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 75 less nodes. ---------------------------------------- (8) Complex Obligation (AND) ---------------------------------------- (9) Obligation: Q DP problem: The TRS P consists of the following rules: 2^1(5(1(5(3(2(3(x1))))))) -> 2^1(3(0(2(x1)))) 2^1(5(1(5(3(2(3(x1))))))) -> 2^1(3(2(3(0(2(x1)))))) 2^1(5(1(5(3(2(3(x1))))))) -> 2^1(x1) The TRS R consists of the following rules: 0(5(2(0(x1)))) -> 3(1(4(1(2(5(1(1(2(0(x1)))))))))) 3(5(0(0(3(x1))))) -> 3(3(1(4(2(5(3(4(3(1(x1)))))))))) 5(2(1(3(5(0(x1)))))) -> 5(5(4(4(2(1(1(3(2(0(x1)))))))))) 5(5(2(0(2(5(x1)))))) -> 4(1(4(4(5(4(4(4(5(5(x1)))))))))) 1(5(0(0(4(4(0(x1))))))) -> 2(1(1(5(1(4(0(2(3(1(x1)))))))))) 4(3(3(3(5(5(1(x1))))))) -> 5(3(2(2(2(0(0(1(5(1(x1)))))))))) 2(5(1(5(3(2(3(x1))))))) -> 1(5(5(1(2(3(2(3(0(2(x1)))))))))) 5(0(1(3(5(0(5(x1))))))) -> 5(1(5(3(1(2(4(4(2(5(x1)))))))))) 3(3(5(2(0(2(5(x1))))))) -> 3(0(2(3(1(1(5(0(2(5(x1)))))))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (10) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. 2^1(5(1(5(3(2(3(x1))))))) -> 2^1(3(0(2(x1)))) 2^1(5(1(5(3(2(3(x1))))))) -> 2^1(3(2(3(0(2(x1)))))) 2^1(5(1(5(3(2(3(x1))))))) -> 2^1(x1) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial interpretation [POLO]: POL(0(x_1)) = x_1 POL(1(x_1)) = x_1 POL(2(x_1)) = x_1 POL(2^1(x_1)) = x_1 POL(3(x_1)) = x_1 POL(4(x_1)) = x_1 POL(5(x_1)) = 1 + x_1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 2(5(1(5(3(2(3(x1))))))) -> 1(5(5(1(2(3(2(3(0(2(x1)))))))))) 1(5(0(0(4(4(0(x1))))))) -> 2(1(1(5(1(4(0(2(3(1(x1)))))))))) 0(5(2(0(x1)))) -> 3(1(4(1(2(5(1(1(2(0(x1)))))))))) 3(3(5(2(0(2(5(x1))))))) -> 3(0(2(3(1(1(5(0(2(5(x1)))))))))) 3(5(0(0(3(x1))))) -> 3(3(1(4(2(5(3(4(3(1(x1)))))))))) 5(5(2(0(2(5(x1)))))) -> 4(1(4(4(5(4(4(4(5(5(x1)))))))))) 4(3(3(3(5(5(1(x1))))))) -> 5(3(2(2(2(0(0(1(5(1(x1)))))))))) 5(2(1(3(5(0(x1)))))) -> 5(5(4(4(2(1(1(3(2(0(x1)))))))))) 5(0(1(3(5(0(5(x1))))))) -> 5(1(5(3(1(2(4(4(2(5(x1)))))))))) ---------------------------------------- (11) Obligation: Q DP problem: P is empty. The TRS R consists of the following rules: 0(5(2(0(x1)))) -> 3(1(4(1(2(5(1(1(2(0(x1)))))))))) 3(5(0(0(3(x1))))) -> 3(3(1(4(2(5(3(4(3(1(x1)))))))))) 5(2(1(3(5(0(x1)))))) -> 5(5(4(4(2(1(1(3(2(0(x1)))))))))) 5(5(2(0(2(5(x1)))))) -> 4(1(4(4(5(4(4(4(5(5(x1)))))))))) 1(5(0(0(4(4(0(x1))))))) -> 2(1(1(5(1(4(0(2(3(1(x1)))))))))) 4(3(3(3(5(5(1(x1))))))) -> 5(3(2(2(2(0(0(1(5(1(x1)))))))))) 2(5(1(5(3(2(3(x1))))))) -> 1(5(5(1(2(3(2(3(0(2(x1)))))))))) 5(0(1(3(5(0(5(x1))))))) -> 5(1(5(3(1(2(4(4(2(5(x1)))))))))) 3(3(5(2(0(2(5(x1))))))) -> 3(0(2(3(1(1(5(0(2(5(x1)))))))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (12) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (13) YES ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: 1^1(5(0(0(4(4(0(x1))))))) -> 1^1(x1) The TRS R consists of the following rules: 0(5(2(0(x1)))) -> 3(1(4(1(2(5(1(1(2(0(x1)))))))))) 3(5(0(0(3(x1))))) -> 3(3(1(4(2(5(3(4(3(1(x1)))))))))) 5(2(1(3(5(0(x1)))))) -> 5(5(4(4(2(1(1(3(2(0(x1)))))))))) 5(5(2(0(2(5(x1)))))) -> 4(1(4(4(5(4(4(4(5(5(x1)))))))))) 1(5(0(0(4(4(0(x1))))))) -> 2(1(1(5(1(4(0(2(3(1(x1)))))))))) 4(3(3(3(5(5(1(x1))))))) -> 5(3(2(2(2(0(0(1(5(1(x1)))))))))) 2(5(1(5(3(2(3(x1))))))) -> 1(5(5(1(2(3(2(3(0(2(x1)))))))))) 5(0(1(3(5(0(5(x1))))))) -> 5(1(5(3(1(2(4(4(2(5(x1)))))))))) 3(3(5(2(0(2(5(x1))))))) -> 3(0(2(3(1(1(5(0(2(5(x1)))))))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (15) 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. ---------------------------------------- (16) Obligation: Q DP problem: The TRS P consists of the following rules: 1^1(5(0(0(4(4(0(x1))))))) -> 1^1(x1) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (17) 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: *1^1(5(0(0(4(4(0(x1))))))) -> 1^1(x1) The graph contains the following edges 1 > 1 ---------------------------------------- (18) YES ---------------------------------------- (19) Obligation: Q DP problem: The TRS P consists of the following rules: 5^1(5(2(0(2(5(x1)))))) -> 5^1(5(x1)) The TRS R consists of the following rules: 0(5(2(0(x1)))) -> 3(1(4(1(2(5(1(1(2(0(x1)))))))))) 3(5(0(0(3(x1))))) -> 3(3(1(4(2(5(3(4(3(1(x1)))))))))) 5(2(1(3(5(0(x1)))))) -> 5(5(4(4(2(1(1(3(2(0(x1)))))))))) 5(5(2(0(2(5(x1)))))) -> 4(1(4(4(5(4(4(4(5(5(x1)))))))))) 1(5(0(0(4(4(0(x1))))))) -> 2(1(1(5(1(4(0(2(3(1(x1)))))))))) 4(3(3(3(5(5(1(x1))))))) -> 5(3(2(2(2(0(0(1(5(1(x1)))))))))) 2(5(1(5(3(2(3(x1))))))) -> 1(5(5(1(2(3(2(3(0(2(x1)))))))))) 5(0(1(3(5(0(5(x1))))))) -> 5(1(5(3(1(2(4(4(2(5(x1)))))))))) 3(3(5(2(0(2(5(x1))))))) -> 3(0(2(3(1(1(5(0(2(5(x1)))))))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (20) 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: *5^1(5(2(0(2(5(x1)))))) -> 5^1(5(x1)) The graph contains the following edges 1 > 1 ---------------------------------------- (21) YES