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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 170 ms] (2) QTRS (3) Overlay + Local Confluence [EQUIVALENT, 0 ms] (4) QTRS (5) DependencyPairsProof [EQUIVALENT, 5 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 3 ms] (8) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(x1) -> 1(x1) 0(0(x1)) -> 0(x1) 3(4(5(x1))) -> 4(3(5(x1))) 2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(x1))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))) -> 1(1(0(1(0(0(0(1(0(1(0(1(1(0(0(0(1(1(1(0(0(1(1(0(0(1(1(0(1(1(0(1(0(1(1(1(1(1(1(1(1(1(1(0(0(1(0(0(1(1(1(1(0(1(1(0(1(1(1(0(0(0(1(0(1(0(0(0(1(0(0(0(0(0(0(1(1(1(1(1(1(0(0(0(1(0(0(1(1(1(1(1(1(1(0(0(1(0(1(1(0(0(1(0(1(1(1(0(1(0(0(0(0(0(0(0(1(0(0(0(0(0(0(0(0(1(0(1(0(1(1(0(0(0(1(1(1(1(0(0(0(0(0(0(1(1(0(1(1(1(1(1(1(1(0(0(1(0(1(1(0(1(1(0(0(0(0(0(0(0(0(0(0(1(1(0(0(0(1(1(1(1(1(0(0(1(1(1(0(0(1(0(0(0(1(1(0(0(1(0(0(0(1(0(0(0(0(0(0(1(1(1(1(1(0(0(1(1(0(1(1(1(0(0(1(0(1(1(1(1(0(0(1(0(0(1(0(1(0(1(1(0(1(0(1(0(1(0(1(1(1(1(0(0(1(1(0(1(1(0(0(1(0(1(0(1(1(0(1(0(1(0(0(1(0(1(0(1(1(0(0(1(0(1(1(0(0(0(0(1(0(1(0(1(1(0(1(0(1(0(x1)))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))) 1(0(1(0(1(1(1(1(0(1(0(0(1(0(1(0(0(0(0(1(1(0(1(1(0(0(0(1(1(1(1(1(1(0(1(0(0(1(0(1(1(0(1(0(1(1(1(1(0(1(0(1(1(0(0(0(1(0(0(0(1(0(0(1(0(1(1(0(1(1(0(1(0(1(1(0(0(0(0(1(1(1(0(0(1(0(0(1(1(1(1(1(1(0(1(1(0(1(0(1(0(1(0(1(1(0(0(0(0(1(1(1(0(1(1(0(1(0(0(1(1(1(1(0(0(0(0(1(1(0(1(1(0(0(0(1(1(0(1(0(1(1(1(0(1(1(1(0(1(0(0(1(1(1(1(0(0(0(0(0(0(1(0(0(1(0(0(0(0(0(0(0(1(0(0(1(0(1(0(1(1(0(0(0(1(1(0(0(0(1(1(1(0(0(1(0(0(0(0(1(1(1(0(1(0(0(0(0(1(0(0(1(1(1(0(0(1(1(0(1(1(0(1(1(0(1(1(0(0(0(0(1(1(0(1(1(1(0(0(0(1(0(1(1(1(0(1(0(1(1(1(0(0(0(0(0(0(1(0(0(1(0(0(1(1(0(0(0(1(1(0(1(0(1(0(0(1(0(1(1(1(1(1(0(0(1(1(0(1(1(1(1(0(0(0(0(1(0(0(1(0(1(1(0(x1)))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))) -> 2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(x1))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0(x_1)) = 23 + x_1 POL(1(x_1)) = x_1 POL(2(x_1)) = 15 + x_1 POL(3(x_1)) = x_1 POL(4(x_1)) = x_1 POL(5(x_1)) = x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 0(x1) -> 1(x1) 0(0(x1)) -> 0(x1) 2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(x1))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))) -> 1(1(0(1(0(0(0(1(0(1(0(1(1(0(0(0(1(1(1(0(0(1(1(0(0(1(1(0(1(1(0(1(0(1(1(1(1(1(1(1(1(1(1(0(0(1(0(0(1(1(1(1(0(1(1(0(1(1(1(0(0(0(1(0(1(0(0(0(1(0(0(0(0(0(0(1(1(1(1(1(1(0(0(0(1(0(0(1(1(1(1(1(1(1(0(0(1(0(1(1(0(0(1(0(1(1(1(0(1(0(0(0(0(0(0(0(1(0(0(0(0(0(0(0(0(1(0(1(0(1(1(0(0(0(1(1(1(1(0(0(0(0(0(0(1(1(0(1(1(1(1(1(1(1(0(0(1(0(1(1(0(1(1(0(0(0(0(0(0(0(0(0(0(1(1(0(0(0(1(1(1(1(1(0(0(1(1(1(0(0(1(0(0(0(1(1(0(0(1(0(0(0(1(0(0(0(0(0(0(1(1(1(1(1(0(0(1(1(0(1(1(1(0(0(1(0(1(1(1(1(0(0(1(0(0(1(0(1(0(1(1(0(1(0(1(0(1(0(1(1(1(1(0(0(1(1(0(1(1(0(0(1(0(1(0(1(1(0(1(0(1(0(0(1(0(1(0(1(1(0(0(1(0(1(1(0(0(0(0(1(0(1(0(1(1(0(1(0(1(0(x1)))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))) 1(0(1(0(1(1(1(1(0(1(0(0(1(0(1(0(0(0(0(1(1(0(1(1(0(0(0(1(1(1(1(1(1(0(1(0(0(1(0(1(1(0(1(0(1(1(1(1(0(1(0(1(1(0(0(0(1(0(0(0(1(0(0(1(0(1(1(0(1(1(0(1(0(1(1(0(0(0(0(1(1(1(0(0(1(0(0(1(1(1(1(1(1(0(1(1(0(1(0(1(0(1(0(1(1(0(0(0(0(1(1(1(0(1(1(0(1(0(0(1(1(1(1(0(0(0(0(1(1(0(1(1(0(0(0(1(1(0(1(0(1(1(1(0(1(1(1(0(1(0(0(1(1(1(1(0(0(0(0(0(0(1(0(0(1(0(0(0(0(0(0(0(1(0(0(1(0(1(0(1(1(0(0(0(1(1(0(0(0(1(1(1(0(0(1(0(0(0(0(1(1(1(0(1(0(0(0(0(1(0(0(1(1(1(0(0(1(1(0(1(1(0(1(1(0(1(1(0(0(0(0(1(1(0(1(1(1(0(0(0(1(0(1(1(1(0(1(0(1(1(1(0(0(0(0(0(0(1(0(0(1(0(0(1(1(0(0(0(1(1(0(1(0(1(0(0(1(0(1(1(1(1(1(0(0(1(1(0(1(1(1(1(0(0(0(0(1(0(0(1(0(1(1(0(x1)))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))) -> 2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(x1))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 3(4(5(x1))) -> 4(3(5(x1))) Q is empty. ---------------------------------------- (3) Overlay + Local Confluence (EQUIVALENT) The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 3(4(5(x1))) -> 4(3(5(x1))) The set Q consists of the following terms: 3(4(5(x0))) ---------------------------------------- (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: 3^1(4(5(x1))) -> 3^1(5(x1)) The TRS R consists of the following rules: 3(4(5(x1))) -> 4(3(5(x1))) The set Q consists of the following terms: 3(4(5(x0))) 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 1 less node. ---------------------------------------- (8) TRUE