YES proof of /export/starexec/sandbox/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms]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) QDPOrderProof [EQUIVALENT, 76 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 57 ms] (10) QDP (11) PisEmptyProof [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(x1) -> x1 a(x1) -> b(x1) b(a(c(x1))) -> c(c(a(a(x1)))) c(x1) -> b(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(x1) -> x1 a(x1) -> b(x1) c(a(b(x1))) -> a(a(c(c(x1)))) c(x1) -> 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: C(a(b(x1))) -> A(a(c(c(x1)))) C(a(b(x1))) -> A(c(c(x1))) C(a(b(x1))) -> C(c(x1)) C(a(b(x1))) -> C(x1) The TRS R consists of the following rules: a(x1) -> x1 a(x1) -> b(x1) c(a(b(x1))) -> a(a(c(c(x1)))) c(x1) -> 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: C(a(b(x1))) -> C(x1) C(a(b(x1))) -> C(c(x1)) The TRS R consists of the following rules: a(x1) -> x1 a(x1) -> b(x1) c(a(b(x1))) -> a(a(c(c(x1)))) 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. C(a(b(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)) = [[-I]] + [[0A, 0A, -I]] * x_1 >>> <<< POL(a(x_1)) = [[0A], [-I], [1A]] + [[0A, 0A, 0A], [1A, 0A, -I], [0A, 0A, 0A]] * x_1 >>> <<< POL(b(x_1)) = [[-I], [-I], [0A]] + [[0A, 0A, 0A], [-I, -I, -I], [0A, 0A, 0A]] * x_1 >>> <<< POL(c(x_1)) = [[0A], [0A], [0A]] + [[0A, 0A, 0A], [0A, 0A, 0A], [0A, 0A, 0A]] * x_1 >>> The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: c(a(b(x1))) -> a(a(c(c(x1)))) c(x1) -> b(x1) a(x1) -> x1 a(x1) -> b(x1) ---------------------------------------- (8) Obligation: Q DP problem: The TRS P consists of the following rules: C(a(b(x1))) -> C(c(x1)) The TRS R consists of the following rules: a(x1) -> x1 a(x1) -> b(x1) c(a(b(x1))) -> a(a(c(c(x1)))) c(x1) -> 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. C(a(b(x1))) -> C(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)) = [[-I]] + [[0A, -I, -I]] * x_1 >>> <<< POL(a(x_1)) = [[0A], [0A], [0A]] + [[0A, 0A, -I], [1A, 0A, 0A], [0A, 0A, 0A]] * x_1 >>> <<< POL(b(x_1)) = [[-I], [0A], [0A]] + [[0A, -I, -I], [1A, -I, 0A], [-I, -I, -I]] * x_1 >>> <<< POL(c(x_1)) = [[-I], [0A], [0A]] + [[0A, -I, -I], [1A, -I, 0A], [0A, -I, -I]] * x_1 >>> The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: c(a(b(x1))) -> a(a(c(c(x1)))) c(x1) -> b(x1) a(x1) -> x1 a(x1) -> b(x1) ---------------------------------------- (10) Obligation: Q DP problem: P is empty. The TRS R consists of the following rules: a(x1) -> x1 a(x1) -> b(x1) c(a(b(x1))) -> a(a(c(c(x1)))) c(x1) -> b(x1) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (11) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (12) YES