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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 99 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 0 ms] (4) QTRS (5) DependencyPairsProof [EQUIVALENT, 9 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(0(1(2(x1)))) -> 2(3(0(x1))) 4(2(5(2(x1)))) -> 1(4(0(x1))) 4(5(3(4(x1)))) -> 4(4(2(4(x1)))) 1(3(5(5(2(x1))))) -> 1(3(2(3(x1)))) 5(2(2(1(2(x1))))) -> 0(0(2(3(x1)))) 2(2(5(3(2(2(x1)))))) -> 5(1(1(0(3(x1))))) 2(5(1(2(1(1(x1)))))) -> 5(2(1(2(4(1(x1)))))) 3(4(1(4(2(4(x1)))))) -> 1(3(3(3(x1)))) 3(5(2(2(4(5(x1)))))) -> 3(2(4(3(0(x1))))) 5(2(1(0(1(5(x1)))))) -> 5(4(2(4(5(1(x1)))))) 1(3(5(4(1(2(2(x1))))))) -> 3(3(3(4(4(0(x1)))))) 4(5(4(3(0(5(1(x1))))))) -> 4(3(3(5(4(1(x1)))))) 2(1(5(2(1(3(4(4(x1)))))))) -> 4(0(3(4(0(1(2(x1))))))) 5(4(0(2(2(4(0(4(x1)))))))) -> 3(1(5(1(3(0(4(x1))))))) 3(4(2(1(1(2(2(5(4(x1))))))))) -> 3(3(3(1(3(3(4(x1))))))) 5(4(4(5(0(1(4(5(4(x1))))))))) -> 1(5(5(0(4(1(4(5(4(x1))))))))) 5(2(1(3(1(5(2(5(4(4(x1)))))))))) -> 5(3(4(5(0(1(4(0(3(x1))))))))) 2(4(1(2(5(2(4(1(3(2(0(3(x1)))))))))))) -> 4(3(0(4(2(3(4(3(4(2(0(x1))))))))))) 0(2(3(5(4(2(2(1(0(3(3(5(0(x1))))))))))))) -> 3(3(1(2(3(0(4(0(0(0(2(0(x1)))))))))))) 2(1(0(2(1(4(0(0(2(0(0(0(5(2(x1)))))))))))))) -> 2(2(4(2(1(4(3(0(5(1(3(3(0(x1))))))))))))) 4(5(0(3(1(3(2(2(5(2(2(4(1(3(2(x1))))))))))))))) -> 4(5(3(2(1(4(5(0(0(0(4(5(4(0(0(x1))))))))))))))) 5(2(2(5(2(4(4(1(2(0(1(1(0(1(1(x1))))))))))))))) -> 5(0(5(4(3(2(1(0(3(3(5(0(4(1(x1)))))))))))))) 5(4(2(0(3(3(0(0(4(0(3(2(0(5(1(x1))))))))))))))) -> 5(0(1(0(0(2(1(1(0(3(2(2(1(3(x1)))))))))))))) 5(4(1(5(1(5(4(4(2(2(0(4(3(1(5(4(4(3(1(x1))))))))))))))))))) -> 3(0(4(5(1(1(3(5(3(4(4(4(5(1(4(3(3(1(x1)))))))))))))))))) 2(0(3(0(2(2(2(0(1(4(2(1(0(4(4(3(3(1(4(4(x1)))))))))))))))))))) -> 5(2(4(1(1(4(5(1(0(1(2(0(3(0(1(2(3(4(3(1(x1)))))))))))))))))))) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0(x_1)) = 22 + x_1 POL(1(x_1)) = 14 + x_1 POL(2(x_1)) = 19 + x_1 POL(3(x_1)) = 21 + x_1 POL(4(x_1)) = 14 + x_1 POL(5(x_1)) = 13 + x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 0(0(1(2(x1)))) -> 2(3(0(x1))) 4(2(5(2(x1)))) -> 1(4(0(x1))) 4(5(3(4(x1)))) -> 4(4(2(4(x1)))) 1(3(5(5(2(x1))))) -> 1(3(2(3(x1)))) 2(2(5(3(2(2(x1)))))) -> 5(1(1(0(3(x1))))) 3(4(1(4(2(4(x1)))))) -> 1(3(3(3(x1)))) 3(5(2(2(4(5(x1)))))) -> 3(2(4(3(0(x1))))) 5(2(1(0(1(5(x1)))))) -> 5(4(2(4(5(1(x1)))))) 1(3(5(4(1(2(2(x1))))))) -> 3(3(3(4(4(0(x1)))))) 4(5(4(3(0(5(1(x1))))))) -> 4(3(3(5(4(1(x1)))))) 2(1(5(2(1(3(4(4(x1)))))))) -> 4(0(3(4(0(1(2(x1))))))) 5(4(0(2(2(4(0(4(x1)))))))) -> 3(1(5(1(3(0(4(x1))))))) 3(4(2(1(1(2(2(5(4(x1))))))))) -> 3(3(3(1(3(3(4(x1))))))) 2(4(1(2(5(2(4(1(3(2(0(3(x1)))))))))))) -> 4(3(0(4(2(3(4(3(4(2(0(x1))))))))))) 0(2(3(5(4(2(2(1(0(3(3(5(0(x1))))))))))))) -> 3(3(1(2(3(0(4(0(0(0(2(0(x1)))))))))))) 2(1(0(2(1(4(0(0(2(0(0(0(5(2(x1)))))))))))))) -> 2(2(4(2(1(4(3(0(5(1(3(3(0(x1))))))))))))) 4(5(0(3(1(3(2(2(5(2(2(4(1(3(2(x1))))))))))))))) -> 4(5(3(2(1(4(5(0(0(0(4(5(4(0(0(x1))))))))))))))) 5(2(2(5(2(4(4(1(2(0(1(1(0(1(1(x1))))))))))))))) -> 5(0(5(4(3(2(1(0(3(3(5(0(4(1(x1)))))))))))))) 5(4(2(0(3(3(0(0(4(0(3(2(0(5(1(x1))))))))))))))) -> 5(0(1(0(0(2(1(1(0(3(2(2(1(3(x1)))))))))))))) 5(4(1(5(1(5(4(4(2(2(0(4(3(1(5(4(4(3(1(x1))))))))))))))))))) -> 3(0(4(5(1(1(3(5(3(4(4(4(5(1(4(3(3(1(x1)))))))))))))))))) 2(0(3(0(2(2(2(0(1(4(2(1(0(4(4(3(3(1(4(4(x1)))))))))))))))))))) -> 5(2(4(1(1(4(5(1(0(1(2(0(3(0(1(2(3(4(3(1(x1)))))))))))))))))))) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 5(2(2(1(2(x1))))) -> 0(0(2(3(x1)))) 2(5(1(2(1(1(x1)))))) -> 5(2(1(2(4(1(x1)))))) 5(4(4(5(0(1(4(5(4(x1))))))))) -> 1(5(5(0(4(1(4(5(4(x1))))))))) 5(2(1(3(1(5(2(5(4(4(x1)))))))))) -> 5(3(4(5(0(1(4(0(3(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: 5(2(2(1(2(x1))))) -> 0(0(2(3(x1)))) 5(2(1(3(1(5(2(5(4(4(x1)))))))))) -> 5(3(4(5(0(1(4(0(3(x1))))))))) ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 2(5(1(2(1(1(x1)))))) -> 5(2(1(2(4(1(x1)))))) 5(4(4(5(0(1(4(5(4(x1))))))))) -> 1(5(5(0(4(1(4(5(4(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: 2^1(5(1(2(1(1(x1)))))) -> 5^1(2(1(2(4(1(x1)))))) 2^1(5(1(2(1(1(x1)))))) -> 2^1(1(2(4(1(x1))))) 2^1(5(1(2(1(1(x1)))))) -> 2^1(4(1(x1))) 5^1(4(4(5(0(1(4(5(4(x1))))))))) -> 5^1(5(0(4(1(4(5(4(x1)))))))) 5^1(4(4(5(0(1(4(5(4(x1))))))))) -> 5^1(0(4(1(4(5(4(x1))))))) The TRS R consists of the following rules: 2(5(1(2(1(1(x1)))))) -> 5(2(1(2(4(1(x1)))))) 5(4(4(5(0(1(4(5(4(x1))))))))) -> 1(5(5(0(4(1(4(5(4(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 0 SCCs with 5 less nodes. ---------------------------------------- (8) TRUE