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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 43 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 2 ms] (8) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(p(x1)) -> p(a(A(x1))) a(A(x1)) -> A(a(x1)) p(A(A(x1))) -> a(p(x1)) Q is empty. ---------------------------------------- (1) QTRS Reverse (EQUIVALENT) We applied the QTRS Reverse Processor [REVERSE]. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: p(a(x1)) -> A(a(p(x1))) A(a(x1)) -> a(A(x1)) A(A(p(x1))) -> p(a(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: P(a(x1)) -> A^1(a(p(x1))) P(a(x1)) -> P(x1) A^1(a(x1)) -> A^1(x1) A^1(A(p(x1))) -> P(a(x1)) The TRS R consists of the following rules: p(a(x1)) -> A(a(p(x1))) A(a(x1)) -> a(A(x1)) A(A(p(x1))) -> p(a(x1)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. P(a(x1)) -> P(x1) A^1(a(x1)) -> A^1(x1) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial interpretation [POLO]: POL(A(x_1)) = 4 + x_1 POL(A^1(x_1)) = 2*x_1 POL(P(x_1)) = 4*x_1 POL(a(x_1)) = 4 + x_1 POL(p(x_1)) = 4 + 2*x_1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: p(a(x1)) -> A(a(p(x1))) A(a(x1)) -> a(A(x1)) A(A(p(x1))) -> p(a(x1)) ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: P(a(x1)) -> A^1(a(p(x1))) A^1(A(p(x1))) -> P(a(x1)) The TRS R consists of the following rules: p(a(x1)) -> A(a(p(x1))) A(a(x1)) -> a(A(x1)) A(A(p(x1))) -> p(a(x1)) Q is empty. 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 2 less nodes. ---------------------------------------- (8) TRUE