/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 147 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 18 ms]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(0(2(x1)))) -> 3(4(4(x1))) 2(0(0(3(x1)))) -> 3(4(1(x1))) 5(3(5(0(1(x1))))) -> 5(2(3(0(x1)))) 0(0(1(2(2(3(x1)))))) -> 4(0(2(2(3(x1))))) 0(5(3(1(4(3(x1)))))) -> 1(4(0(5(2(x1))))) 2(0(4(3(5(3(3(1(x1)))))))) -> 2(1(2(2(2(0(3(1(x1)))))))) 2(4(0(2(3(0(0(2(x1)))))))) -> 2(4(0(3(4(4(2(x1))))))) 2(1(4(0(4(1(5(0(2(x1))))))))) -> 3(3(2(2(3(0(3(3(4(x1))))))))) 5(0(4(0(0(1(3(5(0(x1))))))))) -> 5(0(0(0(3(5(5(3(2(x1))))))))) 5(3(5(4(4(2(2(2(1(x1))))))))) -> 5(0(3(0(0(4(5(2(1(x1))))))))) 2(0(4(2(3(3(3(5(4(2(x1)))))))))) -> 3(4(4(2(0(2(3(3(5(2(x1)))))))))) 4(4(3(2(0(1(1(4(0(2(x1)))))))))) -> 2(3(4(0(0(1(2(1(3(4(x1)))))))))) 1(5(5(2(4(2(4(0(3(3(0(x1))))))))))) -> 1(3(5(4(5(5(5(3(2(4(2(x1))))))))))) 2(5(5(0(1(1(5(1(4(2(3(3(x1)))))))))))) -> 4(2(3(0(3(0(0(2(3(2(3(x1))))))))))) 3(0(2(4(5(2(1(2(0(2(5(1(x1)))))))))))) -> 3(1(3(0(5(0(2(2(4(0(4(x1))))))))))) 3(3(3(4(3(0(0(4(4(5(0(5(x1)))))))))))) -> 3(1(5(5(0(3(1(0(5(2(5(x1))))))))))) 0(5(3(5(1(0(5(1(2(4(4(5(0(x1))))))))))))) -> 0(3(1(0(4(5(5(2(5(5(2(4(0(x1))))))))))))) 2(5(2(4(1(5(3(3(1(0(2(5(3(x1))))))))))))) -> 2(3(1(3(3(5(1(2(0(3(2(3(0(x1))))))))))))) 2(5(5(3(5(3(1(3(1(2(5(0(0(x1))))))))))))) -> 2(5(5(5(0(4(5(5(1(1(3(2(2(x1))))))))))))) 3(5(3(3(3(2(3(0(2(4(2(1(3(x1))))))))))))) -> 5(1(3(1(4(0(0(1(5(0(4(5(x1)))))))))))) 2(0(3(0(2(3(1(0(0(3(0(3(5(3(x1)))))))))))))) -> 3(4(0(2(1(4(4(4(0(5(5(5(2(x1))))))))))))) 3(5(3(3(1(2(4(0(4(3(5(4(2(1(x1)))))))))))))) -> 5(2(0(1(4(4(3(3(3(5(3(4(1(2(x1)))))))))))))) 3(5(5(0(2(2(0(0(3(5(1(3(3(5(3(5(x1)))))))))))))))) -> 3(1(4(2(0(3(4(0(4(1(5(4(2(5(x1)))))))))))))) 5(5(3(5(0(5(3(5(2(3(1(3(3(4(0(2(x1)))))))))))))))) -> 5(2(4(2(4(4(0(5(5(1(3(4(0(4(0(x1))))))))))))))) 2(2(1(3(4(0(1(1(4(2(2(2(0(4(4(2(4(4(x1)))))))))))))))))) -> 4(3(4(4(5(5(0(1(0(1(3(2(5(5(1(1(4(4(x1)))))))))))))))))) 0(3(0(2(2(2(5(5(2(4(4(3(5(1(0(0(0(3(3(x1))))))))))))))))))) -> 3(3(1(3(4(5(5(2(0(4(2(3(1(1(5(1(2(3(x1)))))))))))))))))) 3(3(4(3(3(4(3(0(2(5(3(1(4(5(2(5(2(4(3(5(x1)))))))))))))))))))) -> 5(2(5(0(5(0(1(1(1(3(2(4(4(0(3(2(4(1(5(x1))))))))))))))))))) 2(1(5(5(3(0(1(3(3(3(1(2(0(5(2(0(3(5(2(2(2(x1))))))))))))))))))))) -> 3(0(3(3(4(0(1(4(5(1(3(3(3(4(1(2(1(3(2(1(x1)))))))))))))))))))) 5(1(3(4(0(3(0(2(5(3(0(2(2(0(1(2(3(3(4(1(1(x1))))))))))))))))))))) -> 5(0(1(1(4(2(0(1(1(2(1(4(0(4(3(1(2(0(0(2(x1)))))))))))))))))))) 5(4(0(0(3(2(5(3(0(3(0(4(5(0(4(4(5(1(1(1(3(x1))))))))))))))))))))) -> 5(2(5(1(0(3(5(3(2(1(0(0(5(5(4(5(4(4(2(2(2(x1))))))))))))))))))))) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0(x_1)) = 130 + x_1 POL(1(x_1)) = 129 + x_1 POL(2(x_1)) = 120 + x_1 POL(3(x_1)) = 132 + x_1 POL(4(x_1)) = 151 + x_1 POL(5(x_1)) = 122 + x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 0(1(0(2(x1)))) -> 3(4(4(x1))) 2(0(0(3(x1)))) -> 3(4(1(x1))) 5(3(5(0(1(x1))))) -> 5(2(3(0(x1)))) 0(0(1(2(2(3(x1)))))) -> 4(0(2(2(3(x1))))) 0(5(3(1(4(3(x1)))))) -> 1(4(0(5(2(x1))))) 2(0(4(3(5(3(3(1(x1)))))))) -> 2(1(2(2(2(0(3(1(x1)))))))) 2(4(0(2(3(0(0(2(x1)))))))) -> 2(4(0(3(4(4(2(x1))))))) 2(1(4(0(4(1(5(0(2(x1))))))))) -> 3(3(2(2(3(0(3(3(4(x1))))))))) 5(0(4(0(0(1(3(5(0(x1))))))))) -> 5(0(0(0(3(5(5(3(2(x1))))))))) 5(3(5(4(4(2(2(2(1(x1))))))))) -> 5(0(3(0(0(4(5(2(1(x1))))))))) 4(4(3(2(0(1(1(4(0(2(x1)))))))))) -> 2(3(4(0(0(1(2(1(3(4(x1)))))))))) 1(5(5(2(4(2(4(0(3(3(0(x1))))))))))) -> 1(3(5(4(5(5(5(3(2(4(2(x1))))))))))) 2(5(5(0(1(1(5(1(4(2(3(3(x1)))))))))))) -> 4(2(3(0(3(0(0(2(3(2(3(x1))))))))))) 3(0(2(4(5(2(1(2(0(2(5(1(x1)))))))))))) -> 3(1(3(0(5(0(2(2(4(0(4(x1))))))))))) 3(3(3(4(3(0(0(4(4(5(0(5(x1)))))))))))) -> 3(1(5(5(0(3(1(0(5(2(5(x1))))))))))) 0(5(3(5(1(0(5(1(2(4(4(5(0(x1))))))))))))) -> 0(3(1(0(4(5(5(2(5(5(2(4(0(x1))))))))))))) 2(5(2(4(1(5(3(3(1(0(2(5(3(x1))))))))))))) -> 2(3(1(3(3(5(1(2(0(3(2(3(0(x1))))))))))))) 2(5(5(3(5(3(1(3(1(2(5(0(0(x1))))))))))))) -> 2(5(5(5(0(4(5(5(1(1(3(2(2(x1))))))))))))) 3(5(3(3(3(2(3(0(2(4(2(1(3(x1))))))))))))) -> 5(1(3(1(4(0(0(1(5(0(4(5(x1)))))))))))) 2(0(3(0(2(3(1(0(0(3(0(3(5(3(x1)))))))))))))) -> 3(4(0(2(1(4(4(4(0(5(5(5(2(x1))))))))))))) 3(5(5(0(2(2(0(0(3(5(1(3(3(5(3(5(x1)))))))))))))))) -> 3(1(4(2(0(3(4(0(4(1(5(4(2(5(x1)))))))))))))) 5(5(3(5(0(5(3(5(2(3(1(3(3(4(0(2(x1)))))))))))))))) -> 5(2(4(2(4(4(0(5(5(1(3(4(0(4(0(x1))))))))))))))) 2(2(1(3(4(0(1(1(4(2(2(2(0(4(4(2(4(4(x1)))))))))))))))))) -> 4(3(4(4(5(5(0(1(0(1(3(2(5(5(1(1(4(4(x1)))))))))))))))))) 0(3(0(2(2(2(5(5(2(4(4(3(5(1(0(0(0(3(3(x1))))))))))))))))))) -> 3(3(1(3(4(5(5(2(0(4(2(3(1(1(5(1(2(3(x1)))))))))))))))))) 3(3(4(3(3(4(3(0(2(5(3(1(4(5(2(5(2(4(3(5(x1)))))))))))))))))))) -> 5(2(5(0(5(0(1(1(1(3(2(4(4(0(3(2(4(1(5(x1))))))))))))))))))) 2(1(5(5(3(0(1(3(3(3(1(2(0(5(2(0(3(5(2(2(2(x1))))))))))))))))))))) -> 3(0(3(3(4(0(1(4(5(1(3(3(3(4(1(2(1(3(2(1(x1)))))))))))))))))))) 5(1(3(4(0(3(0(2(5(3(0(2(2(0(1(2(3(3(4(1(1(x1))))))))))))))))))))) -> 5(0(1(1(4(2(0(1(1(2(1(4(0(4(3(1(2(0(0(2(x1)))))))))))))))))))) 5(4(0(0(3(2(5(3(0(3(0(4(5(0(4(4(5(1(1(1(3(x1))))))))))))))))))))) -> 5(2(5(1(0(3(5(3(2(1(0(0(5(5(4(5(4(4(2(2(2(x1))))))))))))))))))))) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 2(0(4(2(3(3(3(5(4(2(x1)))))))))) -> 3(4(4(2(0(2(3(3(5(2(x1)))))))))) 3(5(3(3(1(2(4(0(4(3(5(4(2(1(x1)))))))))))))) -> 5(2(0(1(4(4(3(3(3(5(3(4(1(2(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: 2^1(0(4(2(3(3(3(5(4(2(x1)))))))))) -> 3^1(4(4(2(0(2(3(3(5(2(x1)))))))))) 2^1(0(4(2(3(3(3(5(4(2(x1)))))))))) -> 2^1(0(2(3(3(5(2(x1))))))) 2^1(0(4(2(3(3(3(5(4(2(x1)))))))))) -> 2^1(3(3(5(2(x1))))) 2^1(0(4(2(3(3(3(5(4(2(x1)))))))))) -> 3^1(3(5(2(x1)))) 2^1(0(4(2(3(3(3(5(4(2(x1)))))))))) -> 3^1(5(2(x1))) 3^1(5(3(3(1(2(4(0(4(3(5(4(2(1(x1)))))))))))))) -> 2^1(0(1(4(4(3(3(3(5(3(4(1(2(x1))))))))))))) 3^1(5(3(3(1(2(4(0(4(3(5(4(2(1(x1)))))))))))))) -> 3^1(3(3(5(3(4(1(2(x1)))))))) 3^1(5(3(3(1(2(4(0(4(3(5(4(2(1(x1)))))))))))))) -> 3^1(3(5(3(4(1(2(x1))))))) 3^1(5(3(3(1(2(4(0(4(3(5(4(2(1(x1)))))))))))))) -> 3^1(5(3(4(1(2(x1)))))) 3^1(5(3(3(1(2(4(0(4(3(5(4(2(1(x1)))))))))))))) -> 3^1(4(1(2(x1)))) 3^1(5(3(3(1(2(4(0(4(3(5(4(2(1(x1)))))))))))))) -> 2^1(x1) The TRS R consists of the following rules: 2(0(4(2(3(3(3(5(4(2(x1)))))))))) -> 3(4(4(2(0(2(3(3(5(2(x1)))))))))) 3(5(3(3(1(2(4(0(4(3(5(4(2(1(x1)))))))))))))) -> 5(2(0(1(4(4(3(3(3(5(3(4(1(2(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 11 less nodes. ---------------------------------------- (6) TRUE