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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 167 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 76 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(x1))) -> 1(3(2(x1))) 1(1(4(2(3(x1))))) -> 1(5(4(4(x1)))) 2(2(1(3(5(x1))))) -> 1(2(0(5(x1)))) 3(0(0(2(1(x1))))) -> 0(5(4(0(x1)))) 4(3(2(5(0(x1))))) -> 3(4(1(5(5(x1))))) 0(0(0(3(2(5(x1)))))) -> 2(0(5(5(4(5(x1)))))) 2(2(1(1(3(2(x1)))))) -> 2(1(4(0(0(x1))))) 3(3(1(1(2(2(x1)))))) -> 3(3(0(0(2(x1))))) 0(2(5(1(0(4(2(2(x1)))))))) -> 0(0(3(5(1(5(4(x1))))))) 4(0(5(3(5(1(3(5(x1)))))))) -> 4(0(0(1(3(0(1(5(x1)))))))) 4(4(4(3(5(1(4(0(x1)))))))) -> 3(0(2(2(2(2(3(2(x1)))))))) 0(0(5(3(2(2(5(0(3(x1))))))))) -> 2(5(5(4(2(2(5(0(3(x1))))))))) 3(2(2(0(0(0(0(3(1(0(5(x1))))))))))) -> 0(4(3(0(4(4(2(4(1(4(0(5(x1)))))))))))) 4(3(3(3(5(0(0(3(2(4(4(1(2(x1))))))))))))) -> 4(4(5(2(2(0(5(0(1(4(3(0(x1)))))))))))) 4(4(3(3(1(2(2(5(3(5(3(2(3(x1))))))))))))) -> 4(1(4(0(0(2(5(4(4(2(0(3(x1)))))))))))) 0(0(5(0(1(4(4(3(5(2(0(0(3(3(x1)))))))))))))) -> 2(1(2(1(2(0(4(0(2(2(4(3(3(5(4(x1))))))))))))))) 3(0(1(5(5(1(0(4(0(0(2(1(0(3(x1)))))))))))))) -> 3(4(0(1(2(5(2(2(0(3(0(4(5(1(x1)))))))))))))) 5(5(0(5(4(4(4(3(4(0(5(4(3(3(x1)))))))))))))) -> 2(4(5(0(2(0(3(0(2(5(3(1(3(3(x1)))))))))))))) 1(0(4(3(2(1(1(1(2(4(4(5(5(0(1(x1))))))))))))))) -> 1(0(2(4(5(5(0(1(1(4(4(5(0(1(x1)))))))))))))) 3(4(1(1(4(4(0(4(4(2(4(1(0(0(5(3(2(x1))))))))))))))))) -> 3(5(5(5(0(1(3(2(4(2(0(3(5(3(0(x1))))))))))))))) 0(2(1(3(5(3(4(1(1(4(4(0(4(3(4(1(0(2(x1)))))))))))))))))) -> 1(5(2(2(0(3(2(3(4(2(0(1(1(1(3(2(1(x1))))))))))))))))) 3(3(0(0(1(2(3(5(3(0(5(2(0(0(2(4(4(1(x1)))))))))))))))))) -> 2(2(4(1(4(4(2(5(2(2(5(1(4(2(5(2(0(4(1(x1))))))))))))))))))) 3(3(0(5(2(3(1(3(0(0(3(1(5(2(2(1(2(2(x1)))))))))))))))))) -> 1(0(3(0(4(2(4(3(2(0(4(2(1(5(5(2(2(x1))))))))))))))))) 4(5(4(0(1(1(5(5(4(5(3(2(1(3(2(4(4(2(x1)))))))))))))))))) -> 3(4(5(5(3(0(4(3(3(3(0(5(3(2(2(5(0(x1))))))))))))))))) 0(0(4(1(2(3(3(5(5(2(0(3(1(2(2(4(0(1(5(x1))))))))))))))))))) -> 2(2(0(1(0(1(5(0(1(0(0(5(0(1(1(4(5(x1))))))))))))))))) 0(5(0(3(2(3(2(3(0(1(5(5(5(3(4(0(0(2(2(x1))))))))))))))))))) -> 2(2(5(4(0(0(1(5(5(3(2(2(5(0(0(5(4(0(x1)))))))))))))))))) 4(0(3(4(1(3(2(0(0(0(2(1(0(1(1(3(1(5(1(x1))))))))))))))))))) -> 0(3(5(3(4(5(1(0(0(3(1(0(2(4(1(3(3(0(x1)))))))))))))))))) 4(4(5(4(5(4(3(1(2(2(0(2(5(4(2(0(4(1(2(x1))))))))))))))))))) -> 0(3(5(5(2(0(3(4(2(4(5(0(2(1(3(2(2(1(x1)))))))))))))))))) 3(3(3(5(2(0(0(3(3(1(2(5(2(1(3(1(0(5(2(2(x1)))))))))))))))))))) -> 3(4(5(0(5(3(1(3(5(5(5(4(0(4(1(2(5(5(0(x1))))))))))))))))))) 2(5(4(3(4(4(4(4(2(1(2(2(1(5(5(2(5(0(1(1(1(x1))))))))))))))))))))) -> 5(4(5(4(1(3(4(3(0(3(3(1(1(4(4(2(1(2(0(1(0(x1))))))))))))))))))))) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0(x_1)) = 17 + x_1 POL(1(x_1)) = 13 + x_1 POL(2(x_1)) = 13 + x_1 POL(3(x_1)) = 13 + x_1 POL(4(x_1)) = 12 + x_1 POL(5(x_1)) = 17 + 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(x1))) -> 1(3(2(x1))) 1(1(4(2(3(x1))))) -> 1(5(4(4(x1)))) 2(2(1(3(5(x1))))) -> 1(2(0(5(x1)))) 3(0(0(2(1(x1))))) -> 0(5(4(0(x1)))) 0(0(0(3(2(5(x1)))))) -> 2(0(5(5(4(5(x1)))))) 2(2(1(1(3(2(x1)))))) -> 2(1(4(0(0(x1))))) 3(3(1(1(2(2(x1)))))) -> 3(3(0(0(2(x1))))) 0(2(5(1(0(4(2(2(x1)))))))) -> 0(0(3(5(1(5(4(x1))))))) 0(0(5(3(2(2(5(0(3(x1))))))))) -> 2(5(5(4(2(2(5(0(3(x1))))))))) 4(3(3(3(5(0(0(3(2(4(4(1(2(x1))))))))))))) -> 4(4(5(2(2(0(5(0(1(4(3(0(x1)))))))))))) 4(4(3(3(1(2(2(5(3(5(3(2(3(x1))))))))))))) -> 4(1(4(0(0(2(5(4(4(2(0(3(x1)))))))))))) 0(0(5(0(1(4(4(3(5(2(0(0(3(3(x1)))))))))))))) -> 2(1(2(1(2(0(4(0(2(2(4(3(3(5(4(x1))))))))))))))) 3(0(1(5(5(1(0(4(0(0(2(1(0(3(x1)))))))))))))) -> 3(4(0(1(2(5(2(2(0(3(0(4(5(1(x1)))))))))))))) 1(0(4(3(2(1(1(1(2(4(4(5(5(0(1(x1))))))))))))))) -> 1(0(2(4(5(5(0(1(1(4(4(5(0(1(x1)))))))))))))) 3(4(1(1(4(4(0(4(4(2(4(1(0(0(5(3(2(x1))))))))))))))))) -> 3(5(5(5(0(1(3(2(4(2(0(3(5(3(0(x1))))))))))))))) 0(2(1(3(5(3(4(1(1(4(4(0(4(3(4(1(0(2(x1)))))))))))))))))) -> 1(5(2(2(0(3(2(3(4(2(0(1(1(1(3(2(1(x1))))))))))))))))) 3(3(0(0(1(2(3(5(3(0(5(2(0(0(2(4(4(1(x1)))))))))))))))))) -> 2(2(4(1(4(4(2(5(2(2(5(1(4(2(5(2(0(4(1(x1))))))))))))))))))) 3(3(0(5(2(3(1(3(0(0(3(1(5(2(2(1(2(2(x1)))))))))))))))))) -> 1(0(3(0(4(2(4(3(2(0(4(2(1(5(5(2(2(x1))))))))))))))))) 4(5(4(0(1(1(5(5(4(5(3(2(1(3(2(4(4(2(x1)))))))))))))))))) -> 3(4(5(5(3(0(4(3(3(3(0(5(3(2(2(5(0(x1))))))))))))))))) 0(0(4(1(2(3(3(5(5(2(0(3(1(2(2(4(0(1(5(x1))))))))))))))))))) -> 2(2(0(1(0(1(5(0(1(0(0(5(0(1(1(4(5(x1))))))))))))))))) 0(5(0(3(2(3(2(3(0(1(5(5(5(3(4(0(0(2(2(x1))))))))))))))))))) -> 2(2(5(4(0(0(1(5(5(3(2(2(5(0(0(5(4(0(x1)))))))))))))))))) 4(0(3(4(1(3(2(0(0(0(2(1(0(1(1(3(1(5(1(x1))))))))))))))))))) -> 0(3(5(3(4(5(1(0(0(3(1(0(2(4(1(3(3(0(x1)))))))))))))))))) 4(4(5(4(5(4(3(1(2(2(0(2(5(4(2(0(4(1(2(x1))))))))))))))))))) -> 0(3(5(5(2(0(3(4(2(4(5(0(2(1(3(2(2(1(x1)))))))))))))))))) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 4(3(2(5(0(x1))))) -> 3(4(1(5(5(x1))))) 4(0(5(3(5(1(3(5(x1)))))))) -> 4(0(0(1(3(0(1(5(x1)))))))) 4(4(4(3(5(1(4(0(x1)))))))) -> 3(0(2(2(2(2(3(2(x1)))))))) 3(2(2(0(0(0(0(3(1(0(5(x1))))))))))) -> 0(4(3(0(4(4(2(4(1(4(0(5(x1)))))))))))) 5(5(0(5(4(4(4(3(4(0(5(4(3(3(x1)))))))))))))) -> 2(4(5(0(2(0(3(0(2(5(3(1(3(3(x1)))))))))))))) 3(3(3(5(2(0(0(3(3(1(2(5(2(1(3(1(0(5(2(2(x1)))))))))))))))))))) -> 3(4(5(0(5(3(1(3(5(5(5(4(0(4(1(2(5(5(0(x1))))))))))))))))))) 2(5(4(3(4(4(4(4(2(1(2(2(1(5(5(2(5(0(1(1(1(x1))))))))))))))))))))) -> 5(4(5(4(1(3(4(3(0(3(3(1(1(4(4(2(1(2(0(1(0(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: 4^1(3(2(5(0(x1))))) -> 3^1(4(1(5(5(x1))))) 4^1(3(2(5(0(x1))))) -> 4^1(1(5(5(x1)))) 4^1(3(2(5(0(x1))))) -> 5^1(5(x1)) 4^1(3(2(5(0(x1))))) -> 5^1(x1) 4^1(0(5(3(5(1(3(5(x1)))))))) -> 4^1(0(0(1(3(0(1(5(x1)))))))) 4^1(0(5(3(5(1(3(5(x1)))))))) -> 3^1(0(1(5(x1)))) 4^1(4(4(3(5(1(4(0(x1)))))))) -> 3^1(0(2(2(2(2(3(2(x1)))))))) 4^1(4(4(3(5(1(4(0(x1)))))))) -> 2^1(2(2(2(3(2(x1)))))) 4^1(4(4(3(5(1(4(0(x1)))))))) -> 2^1(2(2(3(2(x1))))) 4^1(4(4(3(5(1(4(0(x1)))))))) -> 2^1(2(3(2(x1)))) 4^1(4(4(3(5(1(4(0(x1)))))))) -> 2^1(3(2(x1))) 4^1(4(4(3(5(1(4(0(x1)))))))) -> 3^1(2(x1)) 4^1(4(4(3(5(1(4(0(x1)))))))) -> 2^1(x1) 3^1(2(2(0(0(0(0(3(1(0(5(x1))))))))))) -> 4^1(3(0(4(4(2(4(1(4(0(5(x1))))))))))) 3^1(2(2(0(0(0(0(3(1(0(5(x1))))))))))) -> 3^1(0(4(4(2(4(1(4(0(5(x1)))))))))) 3^1(2(2(0(0(0(0(3(1(0(5(x1))))))))))) -> 4^1(4(2(4(1(4(0(5(x1)))))))) 3^1(2(2(0(0(0(0(3(1(0(5(x1))))))))))) -> 4^1(2(4(1(4(0(5(x1))))))) 3^1(2(2(0(0(0(0(3(1(0(5(x1))))))))))) -> 2^1(4(1(4(0(5(x1)))))) 3^1(2(2(0(0(0(0(3(1(0(5(x1))))))))))) -> 4^1(1(4(0(5(x1))))) 3^1(2(2(0(0(0(0(3(1(0(5(x1))))))))))) -> 4^1(0(5(x1))) 5^1(5(0(5(4(4(4(3(4(0(5(4(3(3(x1)))))))))))))) -> 2^1(4(5(0(2(0(3(0(2(5(3(1(3(3(x1)))))))))))))) 5^1(5(0(5(4(4(4(3(4(0(5(4(3(3(x1)))))))))))))) -> 4^1(5(0(2(0(3(0(2(5(3(1(3(3(x1))))))))))))) 5^1(5(0(5(4(4(4(3(4(0(5(4(3(3(x1)))))))))))))) -> 5^1(0(2(0(3(0(2(5(3(1(3(3(x1)))))))))))) 5^1(5(0(5(4(4(4(3(4(0(5(4(3(3(x1)))))))))))))) -> 2^1(0(3(0(2(5(3(1(3(3(x1)))))))))) 5^1(5(0(5(4(4(4(3(4(0(5(4(3(3(x1)))))))))))))) -> 3^1(0(2(5(3(1(3(3(x1)))))))) 5^1(5(0(5(4(4(4(3(4(0(5(4(3(3(x1)))))))))))))) -> 2^1(5(3(1(3(3(x1)))))) 5^1(5(0(5(4(4(4(3(4(0(5(4(3(3(x1)))))))))))))) -> 5^1(3(1(3(3(x1))))) 5^1(5(0(5(4(4(4(3(4(0(5(4(3(3(x1)))))))))))))) -> 3^1(1(3(3(x1)))) 3^1(3(3(5(2(0(0(3(3(1(2(5(2(1(3(1(0(5(2(2(x1)))))))))))))))))))) -> 3^1(4(5(0(5(3(1(3(5(5(5(4(0(4(1(2(5(5(0(x1))))))))))))))))))) 3^1(3(3(5(2(0(0(3(3(1(2(5(2(1(3(1(0(5(2(2(x1)))))))))))))))))))) -> 4^1(5(0(5(3(1(3(5(5(5(4(0(4(1(2(5(5(0(x1)))))))))))))))))) 3^1(3(3(5(2(0(0(3(3(1(2(5(2(1(3(1(0(5(2(2(x1)))))))))))))))))))) -> 5^1(0(5(3(1(3(5(5(5(4(0(4(1(2(5(5(0(x1))))))))))))))))) 3^1(3(3(5(2(0(0(3(3(1(2(5(2(1(3(1(0(5(2(2(x1)))))))))))))))))))) -> 5^1(3(1(3(5(5(5(4(0(4(1(2(5(5(0(x1))))))))))))))) 3^1(3(3(5(2(0(0(3(3(1(2(5(2(1(3(1(0(5(2(2(x1)))))))))))))))))))) -> 3^1(1(3(5(5(5(4(0(4(1(2(5(5(0(x1)))))))))))))) 3^1(3(3(5(2(0(0(3(3(1(2(5(2(1(3(1(0(5(2(2(x1)))))))))))))))))))) -> 3^1(5(5(5(4(0(4(1(2(5(5(0(x1)))))))))))) 3^1(3(3(5(2(0(0(3(3(1(2(5(2(1(3(1(0(5(2(2(x1)))))))))))))))))))) -> 5^1(5(5(4(0(4(1(2(5(5(0(x1))))))))))) 3^1(3(3(5(2(0(0(3(3(1(2(5(2(1(3(1(0(5(2(2(x1)))))))))))))))))))) -> 5^1(5(4(0(4(1(2(5(5(0(x1)))))))))) 3^1(3(3(5(2(0(0(3(3(1(2(5(2(1(3(1(0(5(2(2(x1)))))))))))))))))))) -> 5^1(4(0(4(1(2(5(5(0(x1))))))))) 3^1(3(3(5(2(0(0(3(3(1(2(5(2(1(3(1(0(5(2(2(x1)))))))))))))))))))) -> 4^1(0(4(1(2(5(5(0(x1)))))))) 3^1(3(3(5(2(0(0(3(3(1(2(5(2(1(3(1(0(5(2(2(x1)))))))))))))))))))) -> 4^1(1(2(5(5(0(x1)))))) 3^1(3(3(5(2(0(0(3(3(1(2(5(2(1(3(1(0(5(2(2(x1)))))))))))))))))))) -> 2^1(5(5(0(x1)))) 3^1(3(3(5(2(0(0(3(3(1(2(5(2(1(3(1(0(5(2(2(x1)))))))))))))))))))) -> 5^1(5(0(x1))) 3^1(3(3(5(2(0(0(3(3(1(2(5(2(1(3(1(0(5(2(2(x1)))))))))))))))))))) -> 5^1(0(x1)) 2^1(5(4(3(4(4(4(4(2(1(2(2(1(5(5(2(5(0(1(1(1(x1))))))))))))))))))))) -> 5^1(4(5(4(1(3(4(3(0(3(3(1(1(4(4(2(1(2(0(1(0(x1))))))))))))))))))))) 2^1(5(4(3(4(4(4(4(2(1(2(2(1(5(5(2(5(0(1(1(1(x1))))))))))))))))))))) -> 4^1(5(4(1(3(4(3(0(3(3(1(1(4(4(2(1(2(0(1(0(x1)))))))))))))))))))) 2^1(5(4(3(4(4(4(4(2(1(2(2(1(5(5(2(5(0(1(1(1(x1))))))))))))))))))))) -> 5^1(4(1(3(4(3(0(3(3(1(1(4(4(2(1(2(0(1(0(x1))))))))))))))))))) 2^1(5(4(3(4(4(4(4(2(1(2(2(1(5(5(2(5(0(1(1(1(x1))))))))))))))))))))) -> 4^1(1(3(4(3(0(3(3(1(1(4(4(2(1(2(0(1(0(x1)))))))))))))))))) 2^1(5(4(3(4(4(4(4(2(1(2(2(1(5(5(2(5(0(1(1(1(x1))))))))))))))))))))) -> 3^1(4(3(0(3(3(1(1(4(4(2(1(2(0(1(0(x1)))))))))))))))) 2^1(5(4(3(4(4(4(4(2(1(2(2(1(5(5(2(5(0(1(1(1(x1))))))))))))))))))))) -> 4^1(3(0(3(3(1(1(4(4(2(1(2(0(1(0(x1))))))))))))))) 2^1(5(4(3(4(4(4(4(2(1(2(2(1(5(5(2(5(0(1(1(1(x1))))))))))))))))))))) -> 3^1(0(3(3(1(1(4(4(2(1(2(0(1(0(x1)))))))))))))) 2^1(5(4(3(4(4(4(4(2(1(2(2(1(5(5(2(5(0(1(1(1(x1))))))))))))))))))))) -> 3^1(3(1(1(4(4(2(1(2(0(1(0(x1)))))))))))) 2^1(5(4(3(4(4(4(4(2(1(2(2(1(5(5(2(5(0(1(1(1(x1))))))))))))))))))))) -> 3^1(1(1(4(4(2(1(2(0(1(0(x1))))))))))) 2^1(5(4(3(4(4(4(4(2(1(2(2(1(5(5(2(5(0(1(1(1(x1))))))))))))))))))))) -> 4^1(4(2(1(2(0(1(0(x1)))))))) 2^1(5(4(3(4(4(4(4(2(1(2(2(1(5(5(2(5(0(1(1(1(x1))))))))))))))))))))) -> 4^1(2(1(2(0(1(0(x1))))))) 2^1(5(4(3(4(4(4(4(2(1(2(2(1(5(5(2(5(0(1(1(1(x1))))))))))))))))))))) -> 2^1(1(2(0(1(0(x1)))))) 2^1(5(4(3(4(4(4(4(2(1(2(2(1(5(5(2(5(0(1(1(1(x1))))))))))))))))))))) -> 2^1(0(1(0(x1)))) The TRS R consists of the following rules: 4(3(2(5(0(x1))))) -> 3(4(1(5(5(x1))))) 4(0(5(3(5(1(3(5(x1)))))))) -> 4(0(0(1(3(0(1(5(x1)))))))) 4(4(4(3(5(1(4(0(x1)))))))) -> 3(0(2(2(2(2(3(2(x1)))))))) 3(2(2(0(0(0(0(3(1(0(5(x1))))))))))) -> 0(4(3(0(4(4(2(4(1(4(0(5(x1)))))))))))) 5(5(0(5(4(4(4(3(4(0(5(4(3(3(x1)))))))))))))) -> 2(4(5(0(2(0(3(0(2(5(3(1(3(3(x1)))))))))))))) 3(3(3(5(2(0(0(3(3(1(2(5(2(1(3(1(0(5(2(2(x1)))))))))))))))))))) -> 3(4(5(0(5(3(1(3(5(5(5(4(0(4(1(2(5(5(0(x1))))))))))))))))))) 2(5(4(3(4(4(4(4(2(1(2(2(1(5(5(2(5(0(1(1(1(x1))))))))))))))))))))) -> 5(4(5(4(1(3(4(3(0(3(3(1(1(4(4(2(1(2(0(1(0(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 55 less nodes. ---------------------------------------- (6) TRUE