YES proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 139 ms] (4) QTRS (5) Overlay + Local Confluence [EQUIVALENT, 0 ms] (6) QTRS (7) DependencyPairsProof [EQUIVALENT, 30 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) QDP (11) UsableRulesProof [EQUIVALENT, 0 ms] (12) QDP (13) QReductionProof [EQUIVALENT, 0 ms] (14) QDP (15) QDPSizeChangeProof [EQUIVALENT, 0 ms] (16) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(0(1(2(1(x1))))) -> 1(0(1(1(0(x1))))) 0(3(1(2(4(x1))))) -> 0(5(1(4(x1)))) 4(3(0(1(1(x1))))) -> 5(4(5(2(x1)))) 4(3(4(2(2(x1))))) -> 0(5(0(2(x1)))) 2(1(1(4(0(2(x1)))))) -> 1(1(5(3(5(x1))))) 2(2(4(1(3(4(2(x1))))))) -> 2(3(2(4(3(5(x1)))))) 4(0(5(4(2(4(0(x1))))))) -> 0(1(2(2(2(2(0(x1))))))) 4(2(4(2(5(1(0(1(5(x1))))))))) -> 5(0(3(1(0(2(2(5(x1)))))))) 5(2(4(5(0(3(0(2(3(x1))))))))) -> 5(3(2(1(0(1(3(4(0(x1))))))))) 1(4(1(3(2(3(3(1(2(1(x1)))))))))) -> 1(5(3(2(4(5(1(3(4(x1))))))))) 4(0(4(2(3(4(5(1(1(5(1(x1))))))))))) -> 5(4(3(2(2(3(2(1(2(4(0(x1))))))))))) 4(5(2(2(5(4(4(3(4(5(4(x1))))))))))) -> 5(2(3(2(5(0(0(0(5(4(x1)))))))))) 4(3(4(4(0(3(0(3(2(3(2(1(x1)))))))))))) -> 5(5(5(5(5(0(2(2(4(4(2(0(x1)))))))))))) 5(3(4(4(3(3(5(2(5(2(1(1(4(2(x1)))))))))))))) -> 5(4(0(5(2(5(5(3(1(0(3(3(5(x1))))))))))))) 0(5(1(0(3(3(2(5(5(4(0(5(5(5(2(x1))))))))))))))) -> 1(0(1(3(4(5(3(3(3(2(2(3(3(5(2(x1))))))))))))))) 4(5(3(2(1(1(5(2(2(3(4(3(2(3(1(x1))))))))))))))) -> 0(1(3(5(0(1(3(4(0(3(5(4(3(1(x1)))))))))))))) 5(0(1(0(1(1(5(1(1(5(5(2(1(1(0(x1))))))))))))))) -> 5(1(5(1(1(1(3(0(3(3(3(3(1(0(x1)))))))))))))) 5(3(0(4(4(1(1(5(3(4(1(1(2(3(2(x1))))))))))))))) -> 5(4(1(4(0(2(1(2(2(5(3(5(3(4(4(x1))))))))))))))) 2(4(1(0(2(3(2(3(5(3(1(2(3(1(1(4(x1)))))))))))))))) -> 2(2(2(1(4(5(0(1(0(3(1(3(5(1(2(x1))))))))))))))) 0(1(1(3(2(2(0(0(0(5(0(2(4(3(3(0(1(x1))))))))))))))))) -> 5(4(1(1(0(5(2(0(2(3(3(3(0(5(0(1(x1)))))))))))))))) 2(1(1(5(3(1(3(4(3(5(3(3(2(4(3(1(4(x1))))))))))))))))) -> 1(0(1(0(0(2(1(3(2(2(0(3(0(5(2(4(x1)))))))))))))))) 2(4(3(5(0(2(5(5(1(5(0(4(4(4(1(4(3(x1))))))))))))))))) -> 2(0(5(2(2(0(5(4(1(3(2(4(1(4(1(1(0(x1))))))))))))))))) 0(4(5(4(5(0(2(3(1(2(4(5(3(5(0(4(3(3(2(x1))))))))))))))))))) -> 1(0(2(4(5(5(2(2(4(2(1(1(4(0(1(2(3(0(2(5(x1)))))))))))))))))))) 4(5(1(0(2(0(5(4(5(4(4(2(5(5(2(3(5(4(2(3(x1)))))))))))))))))))) -> 0(5(5(2(0(5(2(4(2(5(2(5(2(0(1(5(2(3(3(0(x1)))))))))))))))))))) 3(1(2(4(3(4(3(2(0(3(2(3(4(3(4(5(4(3(4(1(1(x1))))))))))))))))))))) -> 3(4(0(0(2(4(5(0(0(4(3(5(4(3(0(3(2(2(1(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: 1(2(1(0(0(x1))))) -> 0(1(1(0(1(x1))))) 4(2(1(3(0(x1))))) -> 4(1(5(0(x1)))) 1(1(0(3(4(x1))))) -> 2(5(4(5(x1)))) 2(2(4(3(4(x1))))) -> 2(0(5(0(x1)))) 2(0(4(1(1(2(x1)))))) -> 5(3(5(1(1(x1))))) 2(4(3(1(4(2(2(x1))))))) -> 5(3(4(2(3(2(x1)))))) 0(4(2(4(5(0(4(x1))))))) -> 0(2(2(2(2(1(0(x1))))))) 5(1(0(1(5(2(4(2(4(x1))))))))) -> 5(2(2(0(1(3(0(5(x1)))))))) 3(2(0(3(0(5(4(2(5(x1))))))))) -> 0(4(3(1(0(1(2(3(5(x1))))))))) 1(2(1(3(3(2(3(1(4(1(x1)))))))))) -> 4(3(1(5(4(2(3(5(1(x1))))))))) 1(5(1(1(5(4(3(2(4(0(4(x1))))))))))) -> 0(4(2(1(2(3(2(2(3(4(5(x1))))))))))) 4(5(4(3(4(4(5(2(2(5(4(x1))))))))))) -> 4(5(0(0(0(5(2(3(2(5(x1)))))))))) 1(2(3(2(3(0(3(0(4(4(3(4(x1)))))))))))) -> 0(2(4(4(2(2(0(5(5(5(5(5(x1)))))))))))) 2(4(1(1(2(5(2(5(3(3(4(4(3(5(x1)))))))))))))) -> 5(3(3(0(1(3(5(5(2(5(0(4(5(x1))))))))))))) 2(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) -> 2(5(3(3(2(2(3(3(3(5(4(3(1(0(1(x1))))))))))))))) 1(3(2(3(4(3(2(2(5(1(1(2(3(5(4(x1))))))))))))))) -> 1(3(4(5(3(0(4(3(1(0(5(3(1(0(x1)))))))))))))) 0(1(1(2(5(5(1(1(5(1(1(0(1(0(5(x1))))))))))))))) -> 0(1(3(3(3(3(0(3(1(1(1(5(1(5(x1)))))))))))))) 2(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) -> 4(4(3(5(3(5(2(2(1(2(0(4(1(4(5(x1))))))))))))))) 4(1(1(3(2(1(3(5(3(2(3(2(0(1(4(2(x1)))))))))))))))) -> 2(1(5(3(1(3(0(1(0(5(4(1(2(2(2(x1))))))))))))))) 1(0(3(3(4(2(0(5(0(0(0(2(2(3(1(1(0(x1))))))))))))))))) -> 1(0(5(0(3(3(3(2(0(2(5(0(1(1(4(5(x1)))))))))))))))) 4(1(3(4(2(3(3(5(3(4(3(1(3(5(1(1(2(x1))))))))))))))))) -> 4(2(5(0(3(0(2(2(3(1(2(0(0(1(0(1(x1)))))))))))))))) 3(4(1(4(4(4(0(5(1(5(5(2(0(5(3(4(2(x1))))))))))))))))) -> 0(1(1(4(1(4(2(3(1(4(5(0(2(2(5(0(2(x1))))))))))))))))) 2(3(3(4(0(5(3(5(4(2(1(3(2(0(5(4(5(4(0(x1))))))))))))))))))) -> 5(2(0(3(2(1(0(4(1(1(2(4(2(2(5(5(4(2(0(1(x1)))))))))))))))))))) 3(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) -> 0(3(3(2(5(1(0(2(5(2(5(2(4(2(5(0(2(5(5(0(x1)))))))))))))))))))) 1(1(4(3(4(5(4(3(4(3(2(3(0(2(3(4(3(4(2(1(3(x1))))))))))))))))))))) -> 1(1(2(2(3(0(3(4(5(3(4(0(0(5(4(2(0(0(4(3(x1)))))))))))))))))))) Q is empty. ---------------------------------------- (3) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0(x_1)) = 9 + x_1 POL(1(x_1)) = 7 + x_1 POL(2(x_1)) = 7 + x_1 POL(3(x_1)) = 9 + x_1 POL(4(x_1)) = 8 + x_1 POL(5(x_1)) = 8 + x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 4(2(1(3(0(x1))))) -> 4(1(5(0(x1)))) 1(1(0(3(4(x1))))) -> 2(5(4(5(x1)))) 2(2(4(3(4(x1))))) -> 2(0(5(0(x1)))) 2(0(4(1(1(2(x1)))))) -> 5(3(5(1(1(x1))))) 2(4(3(1(4(2(2(x1))))))) -> 5(3(4(2(3(2(x1)))))) 0(4(2(4(5(0(4(x1))))))) -> 0(2(2(2(2(1(0(x1))))))) 5(1(0(1(5(2(4(2(4(x1))))))))) -> 5(2(2(0(1(3(0(5(x1)))))))) 3(2(0(3(0(5(4(2(5(x1))))))))) -> 0(4(3(1(0(1(2(3(5(x1))))))))) 1(2(1(3(3(2(3(1(4(1(x1)))))))))) -> 4(3(1(5(4(2(3(5(1(x1))))))))) 4(5(4(3(4(4(5(2(2(5(4(x1))))))))))) -> 4(5(0(0(0(5(2(3(2(5(x1)))))))))) 1(2(3(2(3(0(3(0(4(4(3(4(x1)))))))))))) -> 0(2(4(4(2(2(0(5(5(5(5(5(x1)))))))))))) 2(4(1(1(2(5(2(5(3(3(4(4(3(5(x1)))))))))))))) -> 5(3(3(0(1(3(5(5(2(5(0(4(5(x1))))))))))))) 1(3(2(3(4(3(2(2(5(1(1(2(3(5(4(x1))))))))))))))) -> 1(3(4(5(3(0(4(3(1(0(5(3(1(0(x1)))))))))))))) 0(1(1(2(5(5(1(1(5(1(1(0(1(0(5(x1))))))))))))))) -> 0(1(3(3(3(3(0(3(1(1(1(5(1(5(x1)))))))))))))) 4(1(1(3(2(1(3(5(3(2(3(2(0(1(4(2(x1)))))))))))))))) -> 2(1(5(3(1(3(0(1(0(5(4(1(2(2(2(x1))))))))))))))) 1(0(3(3(4(2(0(5(0(0(0(2(2(3(1(1(0(x1))))))))))))))))) -> 1(0(5(0(3(3(3(2(0(2(5(0(1(1(4(5(x1)))))))))))))))) 4(1(3(4(2(3(3(5(3(4(3(1(3(5(1(1(2(x1))))))))))))))))) -> 4(2(5(0(3(0(2(2(3(1(2(0(0(1(0(1(x1)))))))))))))))) 3(4(1(4(4(4(0(5(1(5(5(2(0(5(3(4(2(x1))))))))))))))))) -> 0(1(1(4(1(4(2(3(1(4(5(0(2(2(5(0(2(x1))))))))))))))))) 2(3(3(4(0(5(3(5(4(2(1(3(2(0(5(4(5(4(0(x1))))))))))))))))))) -> 5(2(0(3(2(1(0(4(1(1(2(4(2(2(5(5(4(2(0(1(x1)))))))))))))))))))) 1(1(4(3(4(5(4(3(4(3(2(3(0(2(3(4(3(4(2(1(3(x1))))))))))))))))))))) -> 1(1(2(2(3(0(3(4(5(3(4(0(0(5(4(2(0(0(4(3(x1)))))))))))))))))))) ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 1(2(1(0(0(x1))))) -> 0(1(1(0(1(x1))))) 1(5(1(1(5(4(3(2(4(0(4(x1))))))))))) -> 0(4(2(1(2(3(2(2(3(4(5(x1))))))))))) 2(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) -> 2(5(3(3(2(2(3(3(3(5(4(3(1(0(1(x1))))))))))))))) 2(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) -> 4(4(3(5(3(5(2(2(1(2(0(4(1(4(5(x1))))))))))))))) 3(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) -> 0(3(3(2(5(1(0(2(5(2(5(2(4(2(5(0(2(5(5(0(x1)))))))))))))))))))) Q is empty. ---------------------------------------- (5) Overlay + Local Confluence (EQUIVALENT) The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. ---------------------------------------- (6) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 1(2(1(0(0(x1))))) -> 0(1(1(0(1(x1))))) 1(5(1(1(5(4(3(2(4(0(4(x1))))))))))) -> 0(4(2(1(2(3(2(2(3(4(5(x1))))))))))) 2(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) -> 2(5(3(3(2(2(3(3(3(5(4(3(1(0(1(x1))))))))))))))) 2(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) -> 4(4(3(5(3(5(2(2(1(2(0(4(1(4(5(x1))))))))))))))) 3(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) -> 0(3(3(2(5(1(0(2(5(2(5(2(4(2(5(0(2(5(5(0(x1)))))))))))))))))))) The set Q consists of the following terms: 1(2(1(0(0(x0))))) 1(5(1(1(5(4(3(2(4(0(4(x0))))))))))) 2(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x0))))))))))))))) 2(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x0))))))))))))))) 3(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x0)))))))))))))))))))) ---------------------------------------- (7) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (8) Obligation: Q DP problem: The TRS P consists of the following rules: 1^1(2(1(0(0(x1))))) -> 1^1(1(0(1(x1)))) 1^1(2(1(0(0(x1))))) -> 1^1(0(1(x1))) 1^1(2(1(0(0(x1))))) -> 1^1(x1) 1^1(5(1(1(5(4(3(2(4(0(4(x1))))))))))) -> 2^1(1(2(3(2(2(3(4(5(x1))))))))) 1^1(5(1(1(5(4(3(2(4(0(4(x1))))))))))) -> 1^1(2(3(2(2(3(4(5(x1)))))))) 1^1(5(1(1(5(4(3(2(4(0(4(x1))))))))))) -> 2^1(3(2(2(3(4(5(x1))))))) 1^1(5(1(1(5(4(3(2(4(0(4(x1))))))))))) -> 3^1(2(2(3(4(5(x1)))))) 1^1(5(1(1(5(4(3(2(4(0(4(x1))))))))))) -> 2^1(2(3(4(5(x1))))) 1^1(5(1(1(5(4(3(2(4(0(4(x1))))))))))) -> 2^1(3(4(5(x1)))) 1^1(5(1(1(5(4(3(2(4(0(4(x1))))))))))) -> 3^1(4(5(x1))) 2^1(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) -> 2^1(5(3(3(2(2(3(3(3(5(4(3(1(0(1(x1))))))))))))))) 2^1(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) -> 3^1(3(2(2(3(3(3(5(4(3(1(0(1(x1))))))))))))) 2^1(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) -> 3^1(2(2(3(3(3(5(4(3(1(0(1(x1)))))))))))) 2^1(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) -> 2^1(2(3(3(3(5(4(3(1(0(1(x1))))))))))) 2^1(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) -> 2^1(3(3(3(5(4(3(1(0(1(x1)))))))))) 2^1(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) -> 3^1(3(3(5(4(3(1(0(1(x1))))))))) 2^1(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) -> 3^1(3(5(4(3(1(0(1(x1)))))))) 2^1(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) -> 3^1(5(4(3(1(0(1(x1))))))) 2^1(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) -> 3^1(1(0(1(x1)))) 2^1(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) -> 1^1(0(1(x1))) 2^1(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) -> 1^1(x1) 2^1(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) -> 3^1(5(3(5(2(2(1(2(0(4(1(4(5(x1))))))))))))) 2^1(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) -> 3^1(5(2(2(1(2(0(4(1(4(5(x1))))))))))) 2^1(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) -> 2^1(2(1(2(0(4(1(4(5(x1))))))))) 2^1(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) -> 2^1(1(2(0(4(1(4(5(x1)))))))) 2^1(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) -> 1^1(2(0(4(1(4(5(x1))))))) 2^1(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) -> 2^1(0(4(1(4(5(x1)))))) 2^1(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) -> 1^1(4(5(x1))) 3^1(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) -> 3^1(3(2(5(1(0(2(5(2(5(2(4(2(5(0(2(5(5(0(x1))))))))))))))))))) 3^1(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) -> 3^1(2(5(1(0(2(5(2(5(2(4(2(5(0(2(5(5(0(x1)))))))))))))))))) 3^1(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) -> 2^1(5(1(0(2(5(2(5(2(4(2(5(0(2(5(5(0(x1))))))))))))))))) 3^1(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) -> 1^1(0(2(5(2(5(2(4(2(5(0(2(5(5(0(x1))))))))))))))) 3^1(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) -> 2^1(5(2(5(2(4(2(5(0(2(5(5(0(x1))))))))))))) 3^1(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) -> 2^1(5(2(4(2(5(0(2(5(5(0(x1))))))))))) 3^1(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) -> 2^1(4(2(5(0(2(5(5(0(x1))))))))) 3^1(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) -> 2^1(5(0(2(5(5(0(x1))))))) 3^1(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) -> 2^1(5(5(0(x1)))) The TRS R consists of the following rules: 1(2(1(0(0(x1))))) -> 0(1(1(0(1(x1))))) 1(5(1(1(5(4(3(2(4(0(4(x1))))))))))) -> 0(4(2(1(2(3(2(2(3(4(5(x1))))))))))) 2(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) -> 2(5(3(3(2(2(3(3(3(5(4(3(1(0(1(x1))))))))))))))) 2(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) -> 4(4(3(5(3(5(2(2(1(2(0(4(1(4(5(x1))))))))))))))) 3(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) -> 0(3(3(2(5(1(0(2(5(2(5(2(4(2(5(0(2(5(5(0(x1)))))))))))))))))))) The set Q consists of the following terms: 1(2(1(0(0(x0))))) 1(5(1(1(5(4(3(2(4(0(4(x0))))))))))) 2(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x0))))))))))))))) 2(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x0))))))))))))))) 3(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x0)))))))))))))))))))) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (9) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 36 less nodes. ---------------------------------------- (10) Obligation: Q DP problem: The TRS P consists of the following rules: 1^1(2(1(0(0(x1))))) -> 1^1(x1) The TRS R consists of the following rules: 1(2(1(0(0(x1))))) -> 0(1(1(0(1(x1))))) 1(5(1(1(5(4(3(2(4(0(4(x1))))))))))) -> 0(4(2(1(2(3(2(2(3(4(5(x1))))))))))) 2(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x1))))))))))))))) -> 2(5(3(3(2(2(3(3(3(5(4(3(1(0(1(x1))))))))))))))) 2(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x1))))))))))))))) -> 4(4(3(5(3(5(2(2(1(2(0(4(1(4(5(x1))))))))))))))) 3(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x1)))))))))))))))))))) -> 0(3(3(2(5(1(0(2(5(2(5(2(4(2(5(0(2(5(5(0(x1)))))))))))))))))))) The set Q consists of the following terms: 1(2(1(0(0(x0))))) 1(5(1(1(5(4(3(2(4(0(4(x0))))))))))) 2(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x0))))))))))))))) 2(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x0))))))))))))))) 3(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x0)))))))))))))))))))) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (11) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (12) Obligation: Q DP problem: The TRS P consists of the following rules: 1^1(2(1(0(0(x1))))) -> 1^1(x1) R is empty. The set Q consists of the following terms: 1(2(1(0(0(x0))))) 1(5(1(1(5(4(3(2(4(0(4(x0))))))))))) 2(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x0))))))))))))))) 2(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x0))))))))))))))) 3(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x0)))))))))))))))))))) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (13) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 3(2(4(5(3(2(5(5(2(4(4(5(4(5(0(2(0(1(5(4(x0)))))))))))))))))))) ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: 1^1(2(1(0(0(x1))))) -> 1^1(x1) R is empty. The set Q consists of the following terms: 1(2(1(0(0(x0))))) 1(5(1(1(5(4(3(2(4(0(4(x0))))))))))) 2(5(5(5(0(4(5(5(2(3(3(0(1(5(0(x0))))))))))))))) 2(3(2(1(1(4(3(5(1(1(4(4(0(3(5(x0))))))))))))))) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (15) 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(2(1(0(0(x1))))) -> 1^1(x1) The graph contains the following edges 1 > 1 ---------------------------------------- (16) YES