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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 113 ms] (4) QTRS (5) Overlay + Local Confluence [EQUIVALENT, 0 ms] (6) QTRS (7) DependencyPairsProof [EQUIVALENT, 18 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(1(2(3(x1)))) -> 4(4(2(3(x1)))) 5(3(3(5(4(x1))))) -> 5(1(0(2(x1)))) 2(3(1(5(0(5(x1)))))) -> 2(1(3(5(0(5(x1)))))) 5(3(3(5(5(4(x1)))))) -> 4(2(4(3(2(x1))))) 5(1(4(5(1(1(5(x1))))))) -> 1(4(0(2(3(2(5(x1))))))) 3(3(4(3(1(3(0(5(x1)))))))) -> 3(5(2(4(5(0(5(2(x1)))))))) 3(1(2(2(2(1(3(1(3(x1))))))))) -> 1(4(3(1(5(0(2(2(x1)))))))) 3(4(2(0(5(2(3(5(3(x1))))))))) -> 3(5(4(4(2(2(0(5(1(x1))))))))) 5(5(1(3(3(5(4(0(0(x1))))))))) -> 3(1(0(1(4(2(4(3(x1)))))))) 3(0(2(5(1(5(0(1(5(0(x1)))))))))) -> 1(2(2(0(0(4(3(4(4(x1))))))))) 3(5(5(4(4(4(2(0(0(3(x1)))))))))) -> 1(1(2(3(2(3(4(1(x1)))))))) 3(0(4(3(3(5(0(4(4(0(4(2(x1)))))))))))) -> 3(4(5(5(3(2(0(5(1(4(2(x1))))))))))) 5(2(0(4(5(0(2(1(1(1(2(0(x1)))))))))))) -> 3(0(0(2(2(4(5(1(3(1(0(x1))))))))))) 5(5(4(3(3(4(5(4(5(0(0(4(5(x1))))))))))))) -> 5(0(1(0(3(1(4(1(2(3(1(x1))))))))))) 5(2(1(3(0(2(2(4(5(2(2(0(0(1(x1)))))))))))))) -> 3(4(5(1(4(3(3(5(0(3(0(1(x1)))))))))))) 3(1(5(2(5(5(3(3(4(4(5(2(3(2(4(x1))))))))))))))) -> 3(0(5(4(4(4(2(0(0(1(4(3(2(4(x1)))))))))))))) 4(5(5(4(3(4(4(2(4(2(4(3(3(3(3(x1))))))))))))))) -> 4(5(0(0(4(4(5(4(4(3(4(0(0(0(x1)))))))))))))) 0(1(2(4(3(1(1(4(1(5(0(2(5(3(2(4(3(x1))))))))))))))))) -> 4(2(2(1(3(1(3(0(4(5(1(2(2(5(5(4(1(x1))))))))))))))))) 2(4(3(0(4(2(0(0(2(5(1(0(2(0(0(4(4(x1))))))))))))))))) -> 5(4(1(2(1(2(1(0(2(0(4(3(1(0(0(2(x1)))))))))))))))) 3(3(3(1(0(2(1(1(5(2(4(0(0(4(5(2(2(0(2(x1))))))))))))))))))) -> 3(2(2(3(1(5(5(5(3(0(3(1(4(3(2(3(1(x1))))))))))))))))) 5(3(2(2(5(2(1(3(0(2(4(3(2(5(3(3(0(5(4(x1))))))))))))))))))) -> 1(3(0(3(3(4(5(5(0(5(5(4(0(2(1(1(0(0(2(x1))))))))))))))))))) 5(4(5(5(5(2(0(1(2(1(0(1(2(1(5(3(1(3(1(x1))))))))))))))))))) -> 0(0(3(5(3(0(2(0(1(4(0(5(4(3(0(2(4(1(x1)))))))))))))))))) 4(0(4(0(5(1(0(3(2(5(3(1(3(0(2(5(3(5(0(0(x1)))))))))))))))))))) -> 1(5(3(5(2(0(5(4(4(5(0(1(4(4(3(1(3(2(5(1(x1)))))))))))))))))))) 5(4(2(1(3(2(5(4(2(2(0(0(5(5(1(0(5(1(3(0(x1)))))))))))))))))))) -> 4(4(2(4(0(1(3(2(5(1(3(4(4(0(0(1(1(1(2(0(x1)))))))))))))))))))) 3(0(4(5(4(1(4(3(5(5(3(5(4(0(1(4(3(5(0(3(2(x1))))))))))))))))))))) -> 1(2(4(1(1(2(5(4(2(4(0(4(2(5(1(4(2(1(3(1(2(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: 3(2(1(0(x1)))) -> 3(2(4(4(x1)))) 4(5(3(3(5(x1))))) -> 2(0(1(5(x1)))) 5(0(5(1(3(2(x1)))))) -> 5(0(5(3(1(2(x1)))))) 4(5(5(3(3(5(x1)))))) -> 2(3(4(2(4(x1))))) 5(1(1(5(4(1(5(x1))))))) -> 5(2(3(2(0(4(1(x1))))))) 5(0(3(1(3(4(3(3(x1)))))))) -> 2(5(0(5(4(2(5(3(x1)))))))) 3(1(3(1(2(2(2(1(3(x1))))))))) -> 2(2(0(5(1(3(4(1(x1)))))))) 3(5(3(2(5(0(2(4(3(x1))))))))) -> 1(5(0(2(2(4(4(5(3(x1))))))))) 0(0(4(5(3(3(1(5(5(x1))))))))) -> 3(4(2(4(1(0(1(3(x1)))))))) 0(5(1(0(5(1(5(2(0(3(x1)))))))))) -> 4(4(3(4(0(0(2(2(1(x1))))))))) 3(0(0(2(4(4(4(5(5(3(x1)))))))))) -> 1(4(3(2(3(2(1(1(x1)))))))) 2(4(0(4(4(0(5(3(3(4(0(3(x1)))))))))))) -> 2(4(1(5(0(2(3(5(5(4(3(x1))))))))))) 0(2(1(1(1(2(0(5(4(0(2(5(x1)))))))))))) -> 0(1(3(1(5(4(2(2(0(0(3(x1))))))))))) 5(4(0(0(5(4(5(4(3(3(4(5(5(x1))))))))))))) -> 1(3(2(1(4(1(3(0(1(0(5(x1))))))))))) 1(0(0(2(2(5(4(2(2(0(3(1(2(5(x1)))))))))))))) -> 1(0(3(0(5(3(3(4(1(5(4(3(x1)))))))))))) 4(2(3(2(5(4(4(3(3(5(5(2(5(1(3(x1))))))))))))))) -> 4(2(3(4(1(0(0(2(4(4(4(5(0(3(x1)))))))))))))) 3(3(3(3(4(2(4(2(4(4(3(4(5(5(4(x1))))))))))))))) -> 0(0(0(4(3(4(4(5(4(4(0(0(5(4(x1)))))))))))))) 3(4(2(3(5(2(0(5(1(4(1(1(3(4(2(1(0(x1))))))))))))))))) -> 1(4(5(5(2(2(1(5(4(0(3(1(3(1(2(2(4(x1))))))))))))))))) 4(4(0(0(2(0(1(5(2(0(0(2(4(0(3(4(2(x1))))))))))))))))) -> 2(0(0(1(3(4(0(2(0(1(2(1(2(1(4(5(x1)))))))))))))))) 2(0(2(2(5(4(0(0(4(2(5(1(1(2(0(1(3(3(3(x1))))))))))))))))))) -> 1(3(2(3(4(1(3(0(3(5(5(5(1(3(2(2(3(x1))))))))))))))))) 4(5(0(3(3(5(2(3(4(2(0(3(1(2(5(2(2(3(5(x1))))))))))))))))))) -> 2(0(0(1(1(2(0(4(5(5(0(5(5(4(3(3(0(3(1(x1))))))))))))))))))) 1(3(1(3(5(1(2(1(0(1(2(1(0(2(5(5(5(4(5(x1))))))))))))))))))) -> 1(4(2(0(3(4(5(0(4(1(0(2(0(3(5(3(0(0(x1)))))))))))))))))) 0(0(5(3(5(2(0(3(1(3(5(2(3(0(1(5(0(4(0(4(x1)))))))))))))))))))) -> 1(5(2(3(1(3(4(4(1(0(5(4(4(5(0(2(5(3(5(1(x1)))))))))))))))))))) 0(3(1(5(0(1(5(5(0(0(2(2(4(5(2(3(1(2(4(5(x1)))))))))))))))))))) -> 0(2(1(1(1(0(0(4(4(3(1(5(2(3(1(0(4(2(4(4(x1)))))))))))))))))))) 2(3(0(5(3(4(1(0(4(5(3(5(5(3(4(1(4(5(4(0(3(x1))))))))))))))))))))) -> 2(1(3(1(2(4(1(5(2(4(0(4(2(4(5(2(1(1(4(2(1(x1))))))))))))))))))))) Q is empty. ---------------------------------------- (3) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0(x_1)) = 5 + x_1 POL(1(x_1)) = 4 + x_1 POL(2(x_1)) = 4 + x_1 POL(3(x_1)) = 6 + x_1 POL(4(x_1)) = 4 + x_1 POL(5(x_1)) = 6 + x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 3(2(1(0(x1)))) -> 3(2(4(4(x1)))) 4(5(3(3(5(x1))))) -> 2(0(1(5(x1)))) 4(5(5(3(3(5(x1)))))) -> 2(3(4(2(4(x1))))) 5(1(1(5(4(1(5(x1))))))) -> 5(2(3(2(0(4(1(x1))))))) 5(0(3(1(3(4(3(3(x1)))))))) -> 2(5(0(5(4(2(5(3(x1)))))))) 3(1(3(1(2(2(2(1(3(x1))))))))) -> 2(2(0(5(1(3(4(1(x1)))))))) 3(5(3(2(5(0(2(4(3(x1))))))))) -> 1(5(0(2(2(4(4(5(3(x1))))))))) 0(0(4(5(3(3(1(5(5(x1))))))))) -> 3(4(2(4(1(0(1(3(x1)))))))) 0(5(1(0(5(1(5(2(0(3(x1)))))))))) -> 4(4(3(4(0(0(2(2(1(x1))))))))) 3(0(0(2(4(4(4(5(5(3(x1)))))))))) -> 1(4(3(2(3(2(1(1(x1)))))))) 2(4(0(4(4(0(5(3(3(4(0(3(x1)))))))))))) -> 2(4(1(5(0(2(3(5(5(4(3(x1))))))))))) 0(2(1(1(1(2(0(5(4(0(2(5(x1)))))))))))) -> 0(1(3(1(5(4(2(2(0(0(3(x1))))))))))) 5(4(0(0(5(4(5(4(3(3(4(5(5(x1))))))))))))) -> 1(3(2(1(4(1(3(0(1(0(5(x1))))))))))) 1(0(0(2(2(5(4(2(2(0(3(1(2(5(x1)))))))))))))) -> 1(0(3(0(5(3(3(4(1(5(4(3(x1)))))))))))) 4(2(3(2(5(4(4(3(3(5(5(2(5(1(3(x1))))))))))))))) -> 4(2(3(4(1(0(0(2(4(4(4(5(0(3(x1)))))))))))))) 3(3(3(3(4(2(4(2(4(4(3(4(5(5(4(x1))))))))))))))) -> 0(0(0(4(3(4(4(5(4(4(0(0(5(4(x1)))))))))))))) 3(4(2(3(5(2(0(5(1(4(1(1(3(4(2(1(0(x1))))))))))))))))) -> 1(4(5(5(2(2(1(5(4(0(3(1(3(1(2(2(4(x1))))))))))))))))) 4(4(0(0(2(0(1(5(2(0(0(2(4(0(3(4(2(x1))))))))))))))))) -> 2(0(0(1(3(4(0(2(0(1(2(1(2(1(4(5(x1)))))))))))))))) 2(0(2(2(5(4(0(0(4(2(5(1(1(2(0(1(3(3(3(x1))))))))))))))))))) -> 1(3(2(3(4(1(3(0(3(5(5(5(1(3(2(2(3(x1))))))))))))))))) 4(5(0(3(3(5(2(3(4(2(0(3(1(2(5(2(2(3(5(x1))))))))))))))))))) -> 2(0(0(1(1(2(0(4(5(5(0(5(5(4(3(3(0(3(1(x1))))))))))))))))))) 1(3(1(3(5(1(2(1(0(1(2(1(0(2(5(5(5(4(5(x1))))))))))))))))))) -> 1(4(2(0(3(4(5(0(4(1(0(2(0(3(5(3(0(0(x1)))))))))))))))))) 0(0(5(3(5(2(0(3(1(3(5(2(3(0(1(5(0(4(0(4(x1)))))))))))))))))))) -> 1(5(2(3(1(3(4(4(1(0(5(4(4(5(0(2(5(3(5(1(x1)))))))))))))))))))) 0(3(1(5(0(1(5(5(0(0(2(2(4(5(2(3(1(2(4(5(x1)))))))))))))))))))) -> 0(2(1(1(1(0(0(4(4(3(1(5(2(3(1(0(4(2(4(4(x1)))))))))))))))))))) 2(3(0(5(3(4(1(0(4(5(3(5(5(3(4(1(4(5(4(0(3(x1))))))))))))))))))))) -> 2(1(3(1(2(4(1(5(2(4(0(4(2(4(5(2(1(1(4(2(1(x1))))))))))))))))))))) ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 5(0(5(1(3(2(x1)))))) -> 5(0(5(3(1(2(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: 5(0(5(1(3(2(x1)))))) -> 5(0(5(3(1(2(x1)))))) The set Q consists of the following terms: 5(0(5(1(3(2(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: 5^1(0(5(1(3(2(x1)))))) -> 5^1(0(5(3(1(2(x1)))))) 5^1(0(5(1(3(2(x1)))))) -> 5^1(3(1(2(x1)))) The TRS R consists of the following rules: 5(0(5(1(3(2(x1)))))) -> 5(0(5(3(1(2(x1)))))) The set Q consists of the following terms: 5(0(5(1(3(2(x0)))))) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (9) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes. ---------------------------------------- (10) TRUE