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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 13 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) MRRProof [EQUIVALENT, 0 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 892 ms] (8) QDP (9) PisEmptyProof [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(x1) -> b(x1) a(b(x1)) -> c(a(c(a(x1)))) c(b(c(x1))) -> b(x1) 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: A(b(x1)) -> C(a(c(a(x1)))) A(b(x1)) -> A(c(a(x1))) A(b(x1)) -> C(a(x1)) A(b(x1)) -> A(x1) The TRS R consists of the following rules: a(x1) -> b(x1) a(b(x1)) -> c(a(c(a(x1)))) c(b(c(x1))) -> b(x1) 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 1 SCC with 2 less nodes. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: A(b(x1)) -> A(x1) A(b(x1)) -> A(c(a(x1))) The TRS R consists of the following rules: a(x1) -> b(x1) a(b(x1)) -> c(a(c(a(x1)))) c(b(c(x1))) -> b(x1) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) MRRProof (EQUIVALENT) By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. Strictly oriented dependency pairs: A(b(x1)) -> A(x1) Used ordering: Polynomial interpretation [POLO]: POL(A(x_1)) = 2*x_1 POL(a(x_1)) = 2 + x_1 POL(b(x_1)) = 2 + x_1 POL(c(x_1)) = x_1 ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: A(b(x1)) -> A(c(a(x1))) The TRS R consists of the following rules: a(x1) -> b(x1) a(b(x1)) -> c(a(c(a(x1)))) c(b(c(x1))) -> b(x1) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (7) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. A(b(x1)) -> A(c(a(x1))) The remaining pairs can at least be oriented weakly. Used ordering: Matrix interpretation [MATRO] with arctic integers [ARCTIC,STERNAGEL_THIEMANN_RTA14]: <<< POL(A(x_1)) = [[0A]] + [[-1A, 2A, -I]] * x_1 >>> <<< POL(b(x_1)) = [[0A], [-1A], [-1A]] + [[-I, -I, -1A], [-I, -I, -1A], [-1A, 2A, -I]] * x_1 >>> <<< POL(c(x_1)) = [[0A], [-I], [1A]] + [[-1A, -I, -I], [-I, -1A, -I], [-I, -I, 1A]] * x_1 >>> <<< POL(a(x_1)) = [[0A], [-1A], [-1A]] + [[-I, -I, -1A], [-I, -I, -1A], [-1A, 2A, -I]] * x_1 >>> The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: a(x1) -> b(x1) a(b(x1)) -> c(a(c(a(x1)))) c(b(c(x1))) -> b(x1) ---------------------------------------- (8) Obligation: Q DP problem: P is empty. The TRS R consists of the following rules: a(x1) -> b(x1) a(b(x1)) -> c(a(c(a(x1)))) c(b(c(x1))) -> b(x1) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (9) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (10) YES