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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 165 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 66 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(1(2(1(x1)))) -> 3(3(2(x1))) 1(2(3(2(x1)))) -> 3(4(4(2(x1)))) 0(5(1(4(1(x1))))) -> 4(1(4(3(x1)))) 4(3(0(2(2(x1))))) -> 4(1(1(4(5(x1))))) 5(5(1(5(2(x1))))) -> 4(3(5(2(x1)))) 0(4(3(3(4(4(1(x1))))))) -> 4(4(2(5(0(2(2(x1))))))) 2(0(4(1(2(2(1(3(x1)))))))) -> 2(3(0(4(5(5(1(1(x1)))))))) 1(2(2(1(5(2(1(2(1(x1))))))))) -> 1(2(2(2(0(2(4(4(3(x1))))))))) 4(5(1(4(3(4(3(5(4(3(x1)))))))))) -> 4(3(2(0(2(4(3(2(3(x1))))))))) 1(4(1(2(5(3(4(3(3(2(2(x1))))))))))) -> 0(0(1(4(0(4(5(2(3(0(4(x1))))))))))) 4(5(1(3(2(2(5(4(3(5(4(x1))))))))))) -> 4(0(1(1(5(3(5(4(2(2(4(x1))))))))))) 5(1(4(0(1(5(5(3(3(0(3(2(x1)))))))))))) -> 3(0(5(3(2(0(1(0(4(1(2(x1))))))))))) 0(4(1(1(3(3(2(5(4(2(2(1(3(x1))))))))))))) -> 1(0(1(4(3(4(4(2(3(4(2(2(1(x1))))))))))))) 5(1(3(3(5(3(1(3(2(1(2(0(4(x1))))))))))))) -> 2(2(3(2(3(2(3(5(2(5(1(4(x1)))))))))))) 0(5(3(5(3(3(3(3(4(5(5(5(4(4(x1)))))))))))))) -> 0(1(0(3(2(3(4(0(5(5(2(4(0(x1))))))))))))) 1(5(0(1(0(4(4(2(2(3(4(1(4(1(x1)))))))))))))) -> 3(4(5(0(0(4(0(3(5(0(4(1(5(4(x1)))))))))))))) 5(1(0(5(2(2(2(3(3(2(5(1(5(1(x1)))))))))))))) -> 5(4(0(1(4(3(2(2(3(3(3(5(5(1(x1)))))))))))))) 5(1(2(2(4(0(2(4(2(5(2(1(4(0(5(x1))))))))))))))) -> 0(1(4(3(0(5(3(4(3(3(1(4(1(5(x1)))))))))))))) 0(4(0(3(2(0(2(1(2(0(0(2(4(2(3(4(x1)))))))))))))))) -> 3(2(1(3(3(4(5(5(4(0(3(2(1(2(3(x1))))))))))))))) 1(5(1(3(3(3(0(4(0(2(3(1(5(1(4(2(x1)))))))))))))))) -> 3(4(0(5(0(4(4(0(2(1(3(1(4(0(4(2(x1)))))))))))))))) 5(1(1(0(0(3(2(5(0(3(4(2(1(2(5(1(x1)))))))))))))))) -> 5(1(3(2(1(0(1(0(5(5(3(1(1(4(1(0(x1)))))))))))))))) 3(2(1(2(4(2(1(1(3(3(3(5(2(2(0(4(4(x1))))))))))))))))) -> 3(2(3(1(0(2(0(2(4(5(5(4(1(0(0(2(5(0(x1)))))))))))))))))) 4(1(0(3(0(4(3(2(2(1(3(2(2(4(0(2(4(x1))))))))))))))))) -> 4(2(5(3(3(3(1(2(4(5(3(5(3(5(1(3(4(x1))))))))))))))))) 3(0(0(2(1(1(3(5(1(2(2(2(5(1(0(0(0(1(x1)))))))))))))))))) -> 3(5(1(0(4(0(1(2(2(5(0(3(4(3(5(5(4(3(x1)))))))))))))))))) 4(5(5(1(5(3(5(3(2(0(4(4(2(1(0(3(5(3(x1)))))))))))))))))) -> 4(4(0(5(1(3(5(5(3(4(4(0(0(4(3(0(0(0(x1)))))))))))))))))) 4(5(5(2(5(1(0(2(1(0(1(4(4(4(2(1(5(1(x1)))))))))))))))))) -> 4(3(2(3(0(2(5(3(4(1(4(4(5(1(1(4(0(x1))))))))))))))))) 0(2(0(2(2(0(1(1(2(4(1(1(0(3(3(2(1(4(1(4(x1)))))))))))))))))))) -> 2(2(3(3(0(2(1(3(5(3(4(4(1(2(4(4(4(4(4(0(x1)))))))))))))))))))) 3(4(1(1(0(3(4(0(5(5(5(5(3(5(2(3(2(3(1(3(x1)))))))))))))))))))) -> 3(4(3(4(3(1(0(1(1(4(5(5(2(3(2(3(0(2(3(x1))))))))))))))))))) 4(5(2(3(5(4(5(0(5(1(2(3(0(1(1(0(3(5(0(3(0(x1))))))))))))))))))))) -> 4(4(3(1(4(4(0(5(3(5(2(1(4(2(4(1(0(2(4(5(0(x1))))))))))))))))))))) 5(1(4(0(0(3(4(2(3(0(3(5(4(0(4(2(4(0(0(5(0(x1))))))))))))))))))))) -> 5(0(3(2(2(0(4(1(1(5(3(0(1(5(0(1(3(2(2(3(x1)))))))))))))))))))) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0(x_1)) = 61 + x_1 POL(1(x_1)) = 70 + x_1 POL(2(x_1)) = 76 + x_1 POL(3(x_1)) = 77 + x_1 POL(4(x_1)) = 64 + x_1 POL(5(x_1)) = 60 + 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(1(x1)))) -> 3(3(2(x1))) 1(2(3(2(x1)))) -> 3(4(4(2(x1)))) 0(5(1(4(1(x1))))) -> 4(1(4(3(x1)))) 4(3(0(2(2(x1))))) -> 4(1(1(4(5(x1))))) 5(5(1(5(2(x1))))) -> 4(3(5(2(x1)))) 2(0(4(1(2(2(1(3(x1)))))))) -> 2(3(0(4(5(5(1(1(x1)))))))) 1(2(2(1(5(2(1(2(1(x1))))))))) -> 1(2(2(2(0(2(4(4(3(x1))))))))) 4(5(1(4(3(4(3(5(4(3(x1)))))))))) -> 4(3(2(0(2(4(3(2(3(x1))))))))) 1(4(1(2(5(3(4(3(3(2(2(x1))))))))))) -> 0(0(1(4(0(4(5(2(3(0(4(x1))))))))))) 4(5(1(3(2(2(5(4(3(5(4(x1))))))))))) -> 4(0(1(1(5(3(5(4(2(2(4(x1))))))))))) 5(1(4(0(1(5(5(3(3(0(3(2(x1)))))))))))) -> 3(0(5(3(2(0(1(0(4(1(2(x1))))))))))) 0(4(1(1(3(3(2(5(4(2(2(1(3(x1))))))))))))) -> 1(0(1(4(3(4(4(2(3(4(2(2(1(x1))))))))))))) 5(1(3(3(5(3(1(3(2(1(2(0(4(x1))))))))))))) -> 2(2(3(2(3(2(3(5(2(5(1(4(x1)))))))))))) 0(5(3(5(3(3(3(3(4(5(5(5(4(4(x1)))))))))))))) -> 0(1(0(3(2(3(4(0(5(5(2(4(0(x1))))))))))))) 1(5(0(1(0(4(4(2(2(3(4(1(4(1(x1)))))))))))))) -> 3(4(5(0(0(4(0(3(5(0(4(1(5(4(x1)))))))))))))) 5(1(2(2(4(0(2(4(2(5(2(1(4(0(5(x1))))))))))))))) -> 0(1(4(3(0(5(3(4(3(3(1(4(1(5(x1)))))))))))))) 0(4(0(3(2(0(2(1(2(0(0(2(4(2(3(4(x1)))))))))))))))) -> 3(2(1(3(3(4(5(5(4(0(3(2(1(2(3(x1))))))))))))))) 1(5(1(3(3(3(0(4(0(2(3(1(5(1(4(2(x1)))))))))))))))) -> 3(4(0(5(0(4(4(0(2(1(3(1(4(0(4(2(x1)))))))))))))))) 5(1(1(0(0(3(2(5(0(3(4(2(1(2(5(1(x1)))))))))))))))) -> 5(1(3(2(1(0(1(0(5(5(3(1(1(4(1(0(x1)))))))))))))))) 4(1(0(3(0(4(3(2(2(1(3(2(2(4(0(2(4(x1))))))))))))))))) -> 4(2(5(3(3(3(1(2(4(5(3(5(3(5(1(3(4(x1))))))))))))))))) 3(0(0(2(1(1(3(5(1(2(2(2(5(1(0(0(0(1(x1)))))))))))))))))) -> 3(5(1(0(4(0(1(2(2(5(0(3(4(3(5(5(4(3(x1)))))))))))))))))) 4(5(5(1(5(3(5(3(2(0(4(4(2(1(0(3(5(3(x1)))))))))))))))))) -> 4(4(0(5(1(3(5(5(3(4(4(0(0(4(3(0(0(0(x1)))))))))))))))))) 4(5(5(2(5(1(0(2(1(0(1(4(4(4(2(1(5(1(x1)))))))))))))))))) -> 4(3(2(3(0(2(5(3(4(1(4(4(5(1(1(4(0(x1))))))))))))))))) 0(2(0(2(2(0(1(1(2(4(1(1(0(3(3(2(1(4(1(4(x1)))))))))))))))))))) -> 2(2(3(3(0(2(1(3(5(3(4(4(1(2(4(4(4(4(4(0(x1)))))))))))))))))))) 3(4(1(1(0(3(4(0(5(5(5(5(3(5(2(3(2(3(1(3(x1)))))))))))))))))))) -> 3(4(3(4(3(1(0(1(1(4(5(5(2(3(2(3(0(2(3(x1))))))))))))))))))) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(4(3(3(4(4(1(x1))))))) -> 4(4(2(5(0(2(2(x1))))))) 5(1(0(5(2(2(2(3(3(2(5(1(5(1(x1)))))))))))))) -> 5(4(0(1(4(3(2(2(3(3(3(5(5(1(x1)))))))))))))) 3(2(1(2(4(2(1(1(3(3(3(5(2(2(0(4(4(x1))))))))))))))))) -> 3(2(3(1(0(2(0(2(4(5(5(4(1(0(0(2(5(0(x1)))))))))))))))))) 4(5(2(3(5(4(5(0(5(1(2(3(0(1(1(0(3(5(0(3(0(x1))))))))))))))))))))) -> 4(4(3(1(4(4(0(5(3(5(2(1(4(2(4(1(0(2(4(5(0(x1))))))))))))))))))))) 5(1(4(0(0(3(4(2(3(0(3(5(4(0(4(2(4(0(0(5(0(x1))))))))))))))))))))) -> 5(0(3(2(2(0(4(1(1(5(3(0(1(5(0(1(3(2(2(3(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: 0^1(4(3(3(4(4(1(x1))))))) -> 4^1(4(2(5(0(2(2(x1))))))) 0^1(4(3(3(4(4(1(x1))))))) -> 4^1(2(5(0(2(2(x1)))))) 0^1(4(3(3(4(4(1(x1))))))) -> 5^1(0(2(2(x1)))) 0^1(4(3(3(4(4(1(x1))))))) -> 0^1(2(2(x1))) 5^1(1(0(5(2(2(2(3(3(2(5(1(5(1(x1)))))))))))))) -> 5^1(4(0(1(4(3(2(2(3(3(3(5(5(1(x1)))))))))))))) 5^1(1(0(5(2(2(2(3(3(2(5(1(5(1(x1)))))))))))))) -> 4^1(0(1(4(3(2(2(3(3(3(5(5(1(x1))))))))))))) 5^1(1(0(5(2(2(2(3(3(2(5(1(5(1(x1)))))))))))))) -> 0^1(1(4(3(2(2(3(3(3(5(5(1(x1)))))))))))) 5^1(1(0(5(2(2(2(3(3(2(5(1(5(1(x1)))))))))))))) -> 4^1(3(2(2(3(3(3(5(5(1(x1)))))))))) 5^1(1(0(5(2(2(2(3(3(2(5(1(5(1(x1)))))))))))))) -> 3^1(2(2(3(3(3(5(5(1(x1))))))))) 5^1(1(0(5(2(2(2(3(3(2(5(1(5(1(x1)))))))))))))) -> 3^1(3(3(5(5(1(x1)))))) 5^1(1(0(5(2(2(2(3(3(2(5(1(5(1(x1)))))))))))))) -> 3^1(3(5(5(1(x1))))) 5^1(1(0(5(2(2(2(3(3(2(5(1(5(1(x1)))))))))))))) -> 3^1(5(5(1(x1)))) 5^1(1(0(5(2(2(2(3(3(2(5(1(5(1(x1)))))))))))))) -> 5^1(5(1(x1))) 3^1(2(1(2(4(2(1(1(3(3(3(5(2(2(0(4(4(x1))))))))))))))))) -> 3^1(2(3(1(0(2(0(2(4(5(5(4(1(0(0(2(5(0(x1)))))))))))))))))) 3^1(2(1(2(4(2(1(1(3(3(3(5(2(2(0(4(4(x1))))))))))))))))) -> 3^1(1(0(2(0(2(4(5(5(4(1(0(0(2(5(0(x1)))))))))))))))) 3^1(2(1(2(4(2(1(1(3(3(3(5(2(2(0(4(4(x1))))))))))))))))) -> 0^1(2(0(2(4(5(5(4(1(0(0(2(5(0(x1)))))))))))))) 3^1(2(1(2(4(2(1(1(3(3(3(5(2(2(0(4(4(x1))))))))))))))))) -> 0^1(2(4(5(5(4(1(0(0(2(5(0(x1)))))))))))) 3^1(2(1(2(4(2(1(1(3(3(3(5(2(2(0(4(4(x1))))))))))))))))) -> 4^1(5(5(4(1(0(0(2(5(0(x1)))))))))) 3^1(2(1(2(4(2(1(1(3(3(3(5(2(2(0(4(4(x1))))))))))))))))) -> 5^1(5(4(1(0(0(2(5(0(x1))))))))) 3^1(2(1(2(4(2(1(1(3(3(3(5(2(2(0(4(4(x1))))))))))))))))) -> 5^1(4(1(0(0(2(5(0(x1)))))))) 3^1(2(1(2(4(2(1(1(3(3(3(5(2(2(0(4(4(x1))))))))))))))))) -> 4^1(1(0(0(2(5(0(x1))))))) 3^1(2(1(2(4(2(1(1(3(3(3(5(2(2(0(4(4(x1))))))))))))))))) -> 0^1(0(2(5(0(x1))))) 3^1(2(1(2(4(2(1(1(3(3(3(5(2(2(0(4(4(x1))))))))))))))))) -> 0^1(2(5(0(x1)))) 3^1(2(1(2(4(2(1(1(3(3(3(5(2(2(0(4(4(x1))))))))))))))))) -> 5^1(0(x1)) 3^1(2(1(2(4(2(1(1(3(3(3(5(2(2(0(4(4(x1))))))))))))))))) -> 0^1(x1) 4^1(5(2(3(5(4(5(0(5(1(2(3(0(1(1(0(3(5(0(3(0(x1))))))))))))))))))))) -> 4^1(4(3(1(4(4(0(5(3(5(2(1(4(2(4(1(0(2(4(5(0(x1))))))))))))))))))))) 4^1(5(2(3(5(4(5(0(5(1(2(3(0(1(1(0(3(5(0(3(0(x1))))))))))))))))))))) -> 4^1(3(1(4(4(0(5(3(5(2(1(4(2(4(1(0(2(4(5(0(x1)))))))))))))))))))) 4^1(5(2(3(5(4(5(0(5(1(2(3(0(1(1(0(3(5(0(3(0(x1))))))))))))))))))))) -> 3^1(1(4(4(0(5(3(5(2(1(4(2(4(1(0(2(4(5(0(x1))))))))))))))))))) 4^1(5(2(3(5(4(5(0(5(1(2(3(0(1(1(0(3(5(0(3(0(x1))))))))))))))))))))) -> 4^1(4(0(5(3(5(2(1(4(2(4(1(0(2(4(5(0(x1))))))))))))))))) 4^1(5(2(3(5(4(5(0(5(1(2(3(0(1(1(0(3(5(0(3(0(x1))))))))))))))))))))) -> 4^1(0(5(3(5(2(1(4(2(4(1(0(2(4(5(0(x1)))))))))))))))) 4^1(5(2(3(5(4(5(0(5(1(2(3(0(1(1(0(3(5(0(3(0(x1))))))))))))))))))))) -> 0^1(5(3(5(2(1(4(2(4(1(0(2(4(5(0(x1))))))))))))))) 4^1(5(2(3(5(4(5(0(5(1(2(3(0(1(1(0(3(5(0(3(0(x1))))))))))))))))))))) -> 5^1(3(5(2(1(4(2(4(1(0(2(4(5(0(x1)))))))))))))) 4^1(5(2(3(5(4(5(0(5(1(2(3(0(1(1(0(3(5(0(3(0(x1))))))))))))))))))))) -> 3^1(5(2(1(4(2(4(1(0(2(4(5(0(x1))))))))))))) 4^1(5(2(3(5(4(5(0(5(1(2(3(0(1(1(0(3(5(0(3(0(x1))))))))))))))))))))) -> 5^1(2(1(4(2(4(1(0(2(4(5(0(x1)))))))))))) 4^1(5(2(3(5(4(5(0(5(1(2(3(0(1(1(0(3(5(0(3(0(x1))))))))))))))))))))) -> 4^1(2(4(1(0(2(4(5(0(x1))))))))) 4^1(5(2(3(5(4(5(0(5(1(2(3(0(1(1(0(3(5(0(3(0(x1))))))))))))))))))))) -> 4^1(1(0(2(4(5(0(x1))))))) 4^1(5(2(3(5(4(5(0(5(1(2(3(0(1(1(0(3(5(0(3(0(x1))))))))))))))))))))) -> 0^1(2(4(5(0(x1))))) 4^1(5(2(3(5(4(5(0(5(1(2(3(0(1(1(0(3(5(0(3(0(x1))))))))))))))))))))) -> 4^1(5(0(x1))) 4^1(5(2(3(5(4(5(0(5(1(2(3(0(1(1(0(3(5(0(3(0(x1))))))))))))))))))))) -> 5^1(0(x1)) 5^1(1(4(0(0(3(4(2(3(0(3(5(4(0(4(2(4(0(0(5(0(x1))))))))))))))))))))) -> 5^1(0(3(2(2(0(4(1(1(5(3(0(1(5(0(1(3(2(2(3(x1)))))))))))))))))))) 5^1(1(4(0(0(3(4(2(3(0(3(5(4(0(4(2(4(0(0(5(0(x1))))))))))))))))))))) -> 0^1(3(2(2(0(4(1(1(5(3(0(1(5(0(1(3(2(2(3(x1))))))))))))))))))) 5^1(1(4(0(0(3(4(2(3(0(3(5(4(0(4(2(4(0(0(5(0(x1))))))))))))))))))))) -> 3^1(2(2(0(4(1(1(5(3(0(1(5(0(1(3(2(2(3(x1)))))))))))))))))) 5^1(1(4(0(0(3(4(2(3(0(3(5(4(0(4(2(4(0(0(5(0(x1))))))))))))))))))))) -> 0^1(4(1(1(5(3(0(1(5(0(1(3(2(2(3(x1))))))))))))))) 5^1(1(4(0(0(3(4(2(3(0(3(5(4(0(4(2(4(0(0(5(0(x1))))))))))))))))))))) -> 4^1(1(1(5(3(0(1(5(0(1(3(2(2(3(x1)))))))))))))) 5^1(1(4(0(0(3(4(2(3(0(3(5(4(0(4(2(4(0(0(5(0(x1))))))))))))))))))))) -> 5^1(3(0(1(5(0(1(3(2(2(3(x1))))))))))) 5^1(1(4(0(0(3(4(2(3(0(3(5(4(0(4(2(4(0(0(5(0(x1))))))))))))))))))))) -> 3^1(0(1(5(0(1(3(2(2(3(x1)))))))))) 5^1(1(4(0(0(3(4(2(3(0(3(5(4(0(4(2(4(0(0(5(0(x1))))))))))))))))))))) -> 0^1(1(5(0(1(3(2(2(3(x1))))))))) 5^1(1(4(0(0(3(4(2(3(0(3(5(4(0(4(2(4(0(0(5(0(x1))))))))))))))))))))) -> 5^1(0(1(3(2(2(3(x1))))))) 5^1(1(4(0(0(3(4(2(3(0(3(5(4(0(4(2(4(0(0(5(0(x1))))))))))))))))))))) -> 0^1(1(3(2(2(3(x1)))))) 5^1(1(4(0(0(3(4(2(3(0(3(5(4(0(4(2(4(0(0(5(0(x1))))))))))))))))))))) -> 3^1(2(2(3(x1)))) 5^1(1(4(0(0(3(4(2(3(0(3(5(4(0(4(2(4(0(0(5(0(x1))))))))))))))))))))) -> 3^1(x1) The TRS R consists of the following rules: 0(4(3(3(4(4(1(x1))))))) -> 4(4(2(5(0(2(2(x1))))))) 5(1(0(5(2(2(2(3(3(2(5(1(5(1(x1)))))))))))))) -> 5(4(0(1(4(3(2(2(3(3(3(5(5(1(x1)))))))))))))) 3(2(1(2(4(2(1(1(3(3(3(5(2(2(0(4(4(x1))))))))))))))))) -> 3(2(3(1(0(2(0(2(4(5(5(4(1(0(0(2(5(0(x1)))))))))))))))))) 4(5(2(3(5(4(5(0(5(1(2(3(0(1(1(0(3(5(0(3(0(x1))))))))))))))))))))) -> 4(4(3(1(4(4(0(5(3(5(2(1(4(2(4(1(0(2(4(5(0(x1))))))))))))))))))))) 5(1(4(0(0(3(4(2(3(0(3(5(4(0(4(2(4(0(0(5(0(x1))))))))))))))))))))) -> 5(0(3(2(2(0(4(1(1(5(3(0(1(5(0(1(3(2(2(3(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 51 less nodes. ---------------------------------------- (6) TRUE