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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 111 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 1 ms] (6) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(1(2(2(x1)))) -> 0(1(0(1(x1)))) 0(0(0(0(0(x1))))) -> 2(0(2(0(x1)))) 3(2(3(4(0(4(1(0(x1)))))))) -> 2(3(2(2(5(3(1(x1))))))) 1(4(1(0(4(1(2(2(5(3(x1)))))))))) -> 1(4(4(2(5(3(1(5(2(x1))))))))) 1(4(2(3(1(4(0(2(1(1(x1)))))))))) -> 1(1(4(1(0(4(2(3(2(1(x1)))))))))) 2(5(1(2(4(5(1(3(1(5(x1)))))))))) -> 2(4(3(0(3(4(2(4(5(x1))))))))) 2(3(2(4(3(2(3(4(4(0(0(0(x1)))))))))))) -> 2(4(2(0(2(2(2(2(4(3(0(x1))))))))))) 5(0(2(4(2(4(1(4(4(5(1(4(x1)))))))))))) -> 1(4(5(3(3(2(3(2(3(3(4(x1))))))))))) 1(0(1(3(5(5(1(2(5(2(3(5(4(x1))))))))))))) -> 1(0(3(4(1(2(3(4(5(3(3(5(4(x1))))))))))))) 1(4(4(2(5(3(1(5(1(2(1(5(0(x1))))))))))))) -> 3(5(2(5(4(1(5(2(4(1(3(2(x1)))))))))))) 4(1(2(5(1(1(0(0(5(4(1(3(1(x1))))))))))))) -> 4(1(1(2(2(3(5(1(4(2(3(1(x1)))))))))))) 5(5(3(2(0(3(4(0(0(3(1(4(3(x1))))))))))))) -> 5(5(3(5(0(3(1(5(2(3(1(3(4(x1))))))))))))) 0(5(3(2(4(0(2(1(2(3(3(4(3(3(x1)))))))))))))) -> 0(5(4(4(5(3(4(0(1(0(1(3(1(2(3(x1))))))))))))))) 4(3(0(5(5(2(5(2(3(5(3(0(2(2(4(x1))))))))))))))) -> 4(3(3(4(3(1(4(1(5(0(0(5(1(5(3(4(x1)))))))))))))))) 3(1(5(4(1(2(0(0(1(0(0(0(2(0(4(5(x1)))))))))))))))) -> 3(0(2(2(3(3(1(3(2(2(1(2(2(5(5(x1))))))))))))))) 4(1(1(0(5(4(2(0(4(0(5(1(2(0(3(1(x1)))))))))))))))) -> 2(4(3(1(4(1(4(0(1(1(0(5(4(0(5(0(x1)))))))))))))))) 4(0(2(2(4(4(1(1(1(1(0(4(1(5(1(2(0(1(x1)))))))))))))))))) -> 4(5(5(2(5(0(2(1(5(2(4(1(1(1(5(3(2(x1))))))))))))))))) 4(4(4(4(3(1(1(3(3(4(2(2(4(0(3(5(4(2(5(2(3(x1))))))))))))))))))))) -> 2(5(5(4(0(1(5(3(3(5(0(1(5(1(5(4(2(4(2(2(x1)))))))))))))))))))) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0(x_1)) = 13 + x_1 POL(1(x_1)) = 9 + x_1 POL(2(x_1)) = 14 + x_1 POL(3(x_1)) = 11 + x_1 POL(4(x_1)) = 8 + x_1 POL(5(x_1)) = 10 + x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 0(1(2(2(x1)))) -> 0(1(0(1(x1)))) 0(0(0(0(0(x1))))) -> 2(0(2(0(x1)))) 3(2(3(4(0(4(1(0(x1)))))))) -> 2(3(2(2(5(3(1(x1))))))) 1(4(1(0(4(1(2(2(5(3(x1)))))))))) -> 1(4(4(2(5(3(1(5(2(x1))))))))) 2(5(1(2(4(5(1(3(1(5(x1)))))))))) -> 2(4(3(0(3(4(2(4(5(x1))))))))) 2(3(2(4(3(2(3(4(4(0(0(0(x1)))))))))))) -> 2(4(2(0(2(2(2(2(4(3(0(x1))))))))))) 5(0(2(4(2(4(1(4(4(5(1(4(x1)))))))))))) -> 1(4(5(3(3(2(3(2(3(3(4(x1))))))))))) 1(0(1(3(5(5(1(2(5(2(3(5(4(x1))))))))))))) -> 1(0(3(4(1(2(3(4(5(3(3(5(4(x1))))))))))))) 1(4(4(2(5(3(1(5(1(2(1(5(0(x1))))))))))))) -> 3(5(2(5(4(1(5(2(4(1(3(2(x1)))))))))))) 4(1(2(5(1(1(0(0(5(4(1(3(1(x1))))))))))))) -> 4(1(1(2(2(3(5(1(4(2(3(1(x1)))))))))))) 5(5(3(2(0(3(4(0(0(3(1(4(3(x1))))))))))))) -> 5(5(3(5(0(3(1(5(2(3(1(3(4(x1))))))))))))) 0(5(3(2(4(0(2(1(2(3(3(4(3(3(x1)))))))))))))) -> 0(5(4(4(5(3(4(0(1(0(1(3(1(2(3(x1))))))))))))))) 4(3(0(5(5(2(5(2(3(5(3(0(2(2(4(x1))))))))))))))) -> 4(3(3(4(3(1(4(1(5(0(0(5(1(5(3(4(x1)))))))))))))))) 3(1(5(4(1(2(0(0(1(0(0(0(2(0(4(5(x1)))))))))))))))) -> 3(0(2(2(3(3(1(3(2(2(1(2(2(5(5(x1))))))))))))))) 4(1(1(0(5(4(2(0(4(0(5(1(2(0(3(1(x1)))))))))))))))) -> 2(4(3(1(4(1(4(0(1(1(0(5(4(0(5(0(x1)))))))))))))))) 4(0(2(2(4(4(1(1(1(1(0(4(1(5(1(2(0(1(x1)))))))))))))))))) -> 4(5(5(2(5(0(2(1(5(2(4(1(1(1(5(3(2(x1))))))))))))))))) 4(4(4(4(3(1(1(3(3(4(2(2(4(0(3(5(4(2(5(2(3(x1))))))))))))))))))))) -> 2(5(5(4(0(1(5(3(3(5(0(1(5(1(5(4(2(4(2(2(x1)))))))))))))))))))) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 1(4(2(3(1(4(0(2(1(1(x1)))))))))) -> 1(1(4(1(0(4(2(3(2(1(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: 1^1(4(2(3(1(4(0(2(1(1(x1)))))))))) -> 1^1(1(4(1(0(4(2(3(2(1(x1)))))))))) 1^1(4(2(3(1(4(0(2(1(1(x1)))))))))) -> 1^1(4(1(0(4(2(3(2(1(x1))))))))) 1^1(4(2(3(1(4(0(2(1(1(x1)))))))))) -> 1^1(0(4(2(3(2(1(x1))))))) The TRS R consists of the following rules: 1(4(2(3(1(4(0(2(1(1(x1)))))))))) -> 1(1(4(1(0(4(2(3(2(1(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 0 SCCs with 3 less nodes. ---------------------------------------- (6) TRUE