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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) MRRProof [EQUIVALENT, 25 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 51 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 37 ms] (10) QDP (11) DependencyGraphProof [EQUIVALENT, 0 ms] (12) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(x1) -> b(c(x1)) b(a(b(x1))) -> c(x1) c(c(x1)) -> a(a(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(x1) -> B(c(x1)) A(x1) -> C(x1) B(a(b(x1))) -> C(x1) C(c(x1)) -> A(a(b(x1))) C(c(x1)) -> A(b(x1)) C(c(x1)) -> B(x1) The TRS R consists of the following rules: a(x1) -> b(c(x1)) b(a(b(x1))) -> c(x1) c(c(x1)) -> a(a(b(x1))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) 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: C(c(x1)) -> A(b(x1)) C(c(x1)) -> B(x1) Used ordering: Polynomial interpretation [POLO]: POL(A(x_1)) = 2 + x_1 POL(B(x_1)) = 1 + x_1 POL(C(x_1)) = 2 + x_1 POL(a(x_1)) = 1 + x_1 POL(b(x_1)) = x_1 POL(c(x_1)) = 1 + x_1 ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: A(x1) -> B(c(x1)) A(x1) -> C(x1) B(a(b(x1))) -> C(x1) C(c(x1)) -> A(a(b(x1))) The TRS R consists of the following rules: a(x1) -> b(c(x1)) b(a(b(x1))) -> c(x1) c(c(x1)) -> a(a(b(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. A(x1) -> B(c(x1)) The remaining pairs can at least be oriented weakly. Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: <<< POL(A(x_1)) = [[1A]] + [[0A, 0A, 1A]] * x_1 >>> <<< POL(B(x_1)) = [[0A]] + [[-I, 0A, -I]] * x_1 >>> <<< POL(c(x_1)) = [[0A], [0A], [1A]] + [[0A, 1A, 1A], [-I, -I, 0A], [0A, 1A, 0A]] * x_1 >>> <<< POL(C(x_1)) = [[1A]] + [[0A, -I, -I]] * x_1 >>> <<< POL(a(x_1)) = [[1A], [0A], [0A]] + [[0A, 1A, 1A], [0A, 0A, 0A], [-I, 0A, 0A]] * x_1 >>> <<< POL(b(x_1)) = [[1A], [0A], [0A]] + [[0A, 1A, 0A], [-I, 0A, -I], [-I, 0A, -I]] * x_1 >>> The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: b(a(b(x1))) -> c(x1) c(c(x1)) -> a(a(b(x1))) a(x1) -> b(c(x1)) ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: A(x1) -> C(x1) B(a(b(x1))) -> C(x1) C(c(x1)) -> A(a(b(x1))) The TRS R consists of the following rules: a(x1) -> b(c(x1)) b(a(b(x1))) -> c(x1) c(c(x1)) -> a(a(b(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 1 SCC with 1 less node. ---------------------------------------- (8) Obligation: Q DP problem: The TRS P consists of the following rules: C(c(x1)) -> A(a(b(x1))) A(x1) -> C(x1) The TRS R consists of the following rules: a(x1) -> b(c(x1)) b(a(b(x1))) -> c(x1) c(c(x1)) -> a(a(b(x1))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (9) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. A(x1) -> C(x1) The remaining pairs can at least be oriented weakly. Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: <<< POL(C(x_1)) = [[0A]] + [[-I, -I, 0A]] * x_1 >>> <<< POL(c(x_1)) = [[0A], [0A], [1A]] + [[-I, -I, 0A], [0A, 0A, 1A], [1A, 0A, 0A]] * x_1 >>> <<< POL(A(x_1)) = [[1A]] + [[0A, 0A, 1A]] * x_1 >>> <<< POL(a(x_1)) = [[1A], [1A], [0A]] + [[0A, 0A, 0A], [1A, 0A, 1A], [0A, -I, 0A]] * x_1 >>> <<< POL(b(x_1)) = [[-I], [0A], [0A]] + [[0A, -I, -I], [1A, 0A, 0A], [0A, -I, -I]] * x_1 >>> The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: b(a(b(x1))) -> c(x1) c(c(x1)) -> a(a(b(x1))) a(x1) -> b(c(x1)) ---------------------------------------- (10) Obligation: Q DP problem: The TRS P consists of the following rules: C(c(x1)) -> A(a(b(x1))) The TRS R consists of the following rules: a(x1) -> b(c(x1)) b(a(b(x1))) -> c(x1) c(c(x1)) -> a(a(b(x1))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (11) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. ---------------------------------------- (12) TRUE