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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms]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) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) UsableRulesProof [EQUIVALENT, 0 ms] (8) QDP (9) MNOCProof [EQUIVALENT, 4 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 108 ms] (12) QDP (13) QDPOrderProof [EQUIVALENT, 59 ms] (14) QDP (15) DependencyGraphProof [EQUIVALENT, 0 ms] (16) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(x1)) -> b(b(x1)) c(c(b(x1))) -> d(c(a(x1))) a(x1) -> d(c(c(x1))) c(d(x1)) -> b(c(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: a(a(x1)) -> b(b(x1)) b(c(c(x1))) -> a(c(d(x1))) a(x1) -> c(c(d(x1))) d(c(x1)) -> c(b(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: A(a(x1)) -> B(b(x1)) A(a(x1)) -> B(x1) B(c(c(x1))) -> A(c(d(x1))) B(c(c(x1))) -> D(x1) A(x1) -> D(x1) D(c(x1)) -> B(x1) The TRS R consists of the following rules: a(a(x1)) -> b(b(x1)) b(c(c(x1))) -> a(c(d(x1))) a(x1) -> c(c(d(x1))) d(c(x1)) -> c(b(x1)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: A(x1) -> D(x1) D(c(x1)) -> B(x1) B(c(c(x1))) -> A(c(d(x1))) B(c(c(x1))) -> D(x1) The TRS R consists of the following rules: a(a(x1)) -> b(b(x1)) b(c(c(x1))) -> a(c(d(x1))) a(x1) -> c(c(d(x1))) d(c(x1)) -> c(b(x1)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (7) 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. ---------------------------------------- (8) Obligation: Q DP problem: The TRS P consists of the following rules: A(x1) -> D(x1) D(c(x1)) -> B(x1) B(c(c(x1))) -> A(c(d(x1))) B(c(c(x1))) -> D(x1) The TRS R consists of the following rules: d(c(x1)) -> c(b(x1)) b(c(c(x1))) -> a(c(d(x1))) a(x1) -> c(c(d(x1))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (9) MNOCProof (EQUIVALENT) We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R. ---------------------------------------- (10) Obligation: Q DP problem: The TRS P consists of the following rules: A(x1) -> D(x1) D(c(x1)) -> B(x1) B(c(c(x1))) -> A(c(d(x1))) B(c(c(x1))) -> D(x1) The TRS R consists of the following rules: d(c(x1)) -> c(b(x1)) b(c(c(x1))) -> a(c(d(x1))) a(x1) -> c(c(d(x1))) The set Q consists of the following terms: d(c(x0)) b(c(c(x0))) a(x0) 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(c(c(x1))) -> D(x1) The remaining pairs can at least be oriented weakly. Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: <<< POL(A(x_1)) = [[0A]] + [[-I, 0A, 0A]] * x_1 >>> <<< POL(D(x_1)) = [[0A]] + [[-I, 0A, 0A]] * x_1 >>> <<< POL(c(x_1)) = [[1A], [0A], [0A]] + [[0A, 0A, 1A], [0A, 0A, -I], [-I, 0A, 0A]] * x_1 >>> <<< POL(B(x_1)) = [[0A]] + [[0A, 0A, 0A]] * x_1 >>> <<< POL(d(x_1)) = [[0A], [0A], [0A]] + [[0A, 0A, 1A], [-I, -I, 0A], [-I, -I, 0A]] * x_1 >>> <<< POL(b(x_1)) = [[0A], [0A], [0A]] + [[-I, 0A, 0A], [-I, 0A, 0A], [-I, 0A, 0A]] * x_1 >>> <<< POL(a(x_1)) = [[1A], [1A], [1A]] + [[0A, 0A, 1A], [0A, 0A, 1A], [0A, 0A, 1A]] * x_1 >>> The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: d(c(x1)) -> c(b(x1)) b(c(c(x1))) -> a(c(d(x1))) a(x1) -> c(c(d(x1))) ---------------------------------------- (12) Obligation: Q DP problem: The TRS P consists of the following rules: A(x1) -> D(x1) D(c(x1)) -> B(x1) B(c(c(x1))) -> A(c(d(x1))) The TRS R consists of the following rules: d(c(x1)) -> c(b(x1)) b(c(c(x1))) -> a(c(d(x1))) a(x1) -> c(c(d(x1))) The set Q consists of the following terms: d(c(x0)) b(c(c(x0))) a(x0) 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. A(x1) -> D(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(D(x_1)) = [[0A]] + [[-I, -I, 0A]] * x_1 >>> <<< POL(c(x_1)) = [[0A], [1A], [-I]] + [[0A, 0A, 0A], [0A, 0A, 1A], [0A, -I, 0A]] * x_1 >>> <<< POL(B(x_1)) = [[0A]] + [[0A, -I, 0A]] * x_1 >>> <<< POL(d(x_1)) = [[0A], [0A], [0A]] + [[-I, -I, 0A], [-I, 0A, 1A], [-I, -I, 0A]] * x_1 >>> <<< POL(b(x_1)) = [[-I], [-I], [-I]] + [[0A, -I, -I], [0A, -I, 0A], [0A, -I, 0A]] * x_1 >>> <<< POL(a(x_1)) = [[1A], [1A], [1A]] + [[0A, 0A, 1A], [-I, 0A, 1A], [-I, 0A, 1A]] * x_1 >>> The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: d(c(x1)) -> c(b(x1)) b(c(c(x1))) -> a(c(d(x1))) a(x1) -> c(c(d(x1))) ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: D(c(x1)) -> B(x1) B(c(c(x1))) -> A(c(d(x1))) The TRS R consists of the following rules: d(c(x1)) -> c(b(x1)) b(c(c(x1))) -> a(c(d(x1))) a(x1) -> c(c(d(x1))) The set Q consists of the following terms: d(c(x0)) b(c(c(x0))) a(x0) 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 2 less nodes. ---------------------------------------- (16) TRUE