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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) QDPSizeChangeProof [EQUIVALENT, 0 ms] (9) YES (10) QDP (11) QDPOrderProof [EQUIVALENT, 20 ms] (12) QDP (13) PisEmptyProof [EQUIVALENT, 0 ms] (14) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(a, h(x)) -> f(g(x), h(x)) h(g(x)) -> h(a) g(h(x)) -> g(x) h(h(x)) -> x Q is empty. ---------------------------------------- (1) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: F(a, h(x)) -> F(g(x), h(x)) F(a, h(x)) -> G(x) H(g(x)) -> H(a) G(h(x)) -> G(x) The TRS R consists of the following rules: f(a, h(x)) -> f(g(x), h(x)) h(g(x)) -> h(a) g(h(x)) -> g(x) h(h(x)) -> x Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 2 less nodes. ---------------------------------------- (4) Complex Obligation (AND) ---------------------------------------- (5) Obligation: Q DP problem: The TRS P consists of the following rules: G(h(x)) -> G(x) The TRS R consists of the following rules: f(a, h(x)) -> f(g(x), h(x)) h(g(x)) -> h(a) g(h(x)) -> g(x) h(h(x)) -> x Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (6) UsableRulesProof (EQUIVALENT) We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (7) Obligation: Q DP problem: The TRS P consists of the following rules: G(h(x)) -> G(x) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (8) QDPSizeChangeProof (EQUIVALENT) By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs: *G(h(x)) -> G(x) The graph contains the following edges 1 > 1 ---------------------------------------- (9) YES ---------------------------------------- (10) Obligation: Q DP problem: The TRS P consists of the following rules: F(a, h(x)) -> F(g(x), h(x)) The TRS R consists of the following rules: f(a, h(x)) -> f(g(x), h(x)) h(g(x)) -> h(a) g(h(x)) -> g(x) h(h(x)) -> x Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (11) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. F(a, h(x)) -> F(g(x), h(x)) The remaining pairs can at least be oriented weakly. Used ordering: Combined order from the following AFS and order. F(x1, x2) = F(x1) a = a h(x1) = h(x1) g(x1) = g Recursive path order with status [RPO]. Quasi-Precedence: a > [h_1, g] > F_1 Status: F_1: multiset status a: multiset status h_1: multiset status g: multiset status The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: g(h(x)) -> g(x) ---------------------------------------- (12) Obligation: Q DP problem: P is empty. The TRS R consists of the following rules: f(a, h(x)) -> f(g(x), h(x)) h(g(x)) -> h(a) g(h(x)) -> g(x) h(h(x)) -> x Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (13) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (14) YES