/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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) QDP (5) TransformationProof [EQUIVALENT, 0 ms] (6) QDP (7) TransformationProof [EQUIVALENT, 0 ms] (8) QDP (9) TransformationProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 14 ms] (12) QDP (13) QDPOrderProof [EQUIVALENT, 14 ms] (14) QDP (15) DependencyGraphProof [EQUIVALENT, 0 ms] (16) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(c(a, z, x)) -> b(a, z) b(x, b(z, y)) -> f(b(f(f(z)), c(x, z, y))) b(y, z) -> z 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(c(a, z, x)) -> B(a, z) B(x, b(z, y)) -> F(b(f(f(z)), c(x, z, y))) B(x, b(z, y)) -> B(f(f(z)), c(x, z, y)) B(x, b(z, y)) -> F(f(z)) B(x, b(z, y)) -> F(z) The TRS R consists of the following rules: f(c(a, z, x)) -> b(a, z) b(x, b(z, y)) -> f(b(f(f(z)), c(x, z, y))) b(y, z) -> z 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 1 less node. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: B(x, b(z, y)) -> F(b(f(f(z)), c(x, z, y))) F(c(a, z, x)) -> B(a, z) B(x, b(z, y)) -> F(f(z)) B(x, b(z, y)) -> F(z) The TRS R consists of the following rules: f(c(a, z, x)) -> b(a, z) b(x, b(z, y)) -> f(b(f(f(z)), c(x, z, y))) b(y, z) -> z Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) TransformationProof (EQUIVALENT) By instantiating [LPAR04] the rule B(x, b(z, y)) -> F(b(f(f(z)), c(x, z, y))) we obtained the following new rules [LPAR04]: (B(a, b(x1, x2)) -> F(b(f(f(x1)), c(a, x1, x2))),B(a, b(x1, x2)) -> F(b(f(f(x1)), c(a, x1, x2)))) ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: F(c(a, z, x)) -> B(a, z) B(x, b(z, y)) -> F(f(z)) B(x, b(z, y)) -> F(z) B(a, b(x1, x2)) -> F(b(f(f(x1)), c(a, x1, x2))) The TRS R consists of the following rules: f(c(a, z, x)) -> b(a, z) b(x, b(z, y)) -> f(b(f(f(z)), c(x, z, y))) b(y, z) -> z Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (7) TransformationProof (EQUIVALENT) By instantiating [LPAR04] the rule B(x, b(z, y)) -> F(f(z)) we obtained the following new rules [LPAR04]: (B(a, b(x1, x2)) -> F(f(x1)),B(a, b(x1, x2)) -> F(f(x1))) ---------------------------------------- (8) Obligation: Q DP problem: The TRS P consists of the following rules: F(c(a, z, x)) -> B(a, z) B(x, b(z, y)) -> F(z) B(a, b(x1, x2)) -> F(b(f(f(x1)), c(a, x1, x2))) B(a, b(x1, x2)) -> F(f(x1)) The TRS R consists of the following rules: f(c(a, z, x)) -> b(a, z) b(x, b(z, y)) -> f(b(f(f(z)), c(x, z, y))) b(y, z) -> z Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (9) TransformationProof (EQUIVALENT) By instantiating [LPAR04] the rule B(x, b(z, y)) -> F(z) we obtained the following new rules [LPAR04]: (B(a, b(x1, x2)) -> F(x1),B(a, b(x1, x2)) -> F(x1)) ---------------------------------------- (10) Obligation: Q DP problem: The TRS P consists of the following rules: F(c(a, z, x)) -> B(a, z) B(a, b(x1, x2)) -> F(b(f(f(x1)), c(a, x1, x2))) B(a, b(x1, x2)) -> F(f(x1)) B(a, b(x1, x2)) -> F(x1) The TRS R consists of the following rules: f(c(a, z, x)) -> b(a, z) b(x, b(z, y)) -> f(b(f(f(z)), c(x, z, y))) b(y, z) -> z 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. B(a, b(x1, x2)) -> F(f(x1)) B(a, b(x1, x2)) -> F(x1) The remaining pairs can at least be oriented weakly. Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: <<< POL(F(x_1)) = [[0A]] + [[0A]] * x_1 >>> <<< POL(c(x_1, x_2, x_3)) = [[-I]] + [[-I]] * x_1 + [[1A]] * x_2 + [[1A]] * x_3 >>> <<< POL(a) = [[0A]] >>> <<< POL(B(x_1, x_2)) = [[0A]] + [[-I]] * x_1 + [[1A]] * x_2 >>> <<< POL(b(x_1, x_2)) = [[1A]] + [[1A]] * x_1 + [[1A]] * x_2 >>> <<< POL(f(x_1)) = [[1A]] + [[0A]] * x_1 >>> The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: b(x, b(z, y)) -> f(b(f(f(z)), c(x, z, y))) f(c(a, z, x)) -> b(a, z) b(y, z) -> z ---------------------------------------- (12) Obligation: Q DP problem: The TRS P consists of the following rules: F(c(a, z, x)) -> B(a, z) B(a, b(x1, x2)) -> F(b(f(f(x1)), c(a, x1, x2))) The TRS R consists of the following rules: f(c(a, z, x)) -> b(a, z) b(x, b(z, y)) -> f(b(f(f(z)), c(x, z, y))) b(y, z) -> z Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (13) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. F(c(a, z, x)) -> B(a, z) The remaining pairs can at least be oriented weakly. Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: <<< POL(F(x_1)) = [[-I]] + [[0A]] * x_1 >>> <<< POL(c(x_1, x_2, x_3)) = [[-I]] + [[1A]] * x_1 + [[1A]] * x_2 + [[-I]] * x_3 >>> <<< POL(a) = [[0A]] >>> <<< POL(B(x_1, x_2)) = [[-I]] + [[-I]] * x_1 + [[0A]] * x_2 >>> <<< POL(b(x_1, x_2)) = [[1A]] + [[1A]] * x_1 + [[0A]] * x_2 >>> <<< POL(f(x_1)) = [[-I]] + [[0A]] * x_1 >>> The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: b(x, b(z, y)) -> f(b(f(f(z)), c(x, z, y))) f(c(a, z, x)) -> b(a, z) b(y, z) -> z ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: B(a, b(x1, x2)) -> F(b(f(f(x1)), c(a, x1, x2))) The TRS R consists of the following rules: f(c(a, z, x)) -> b(a, z) b(x, b(z, y)) -> f(b(f(f(z)), c(x, z, y))) b(y, z) -> z Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (15) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. ---------------------------------------- (16) TRUE