YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination of the given RelTRS could be proven: (0) RelTRS (1) RelTRS Reverse [EQUIVALENT, 0 ms] (2) RelTRS (3) RootLabelingProof [EQUIVALENT, 0 ms] (4) RelTRS (5) RelTRSRRRProof [EQUIVALENT, 107 ms] (6) RelTRS (7) RIsEmptyProof [EQUIVALENT, 0 ms] (8) YES ---------------------------------------- (0) Obligation: Relative term rewrite system: The relative TRS consists of the following R rules: a(b(a(x1))) -> a(b(b(a(x1)))) The relative TRS consists of the following S rules: b(x1) -> b(b(x1)) ---------------------------------------- (1) RelTRS Reverse (EQUIVALENT) We have reversed the following relative TRS [REVERSE]: The set of rules R is a(b(a(x1))) -> a(b(b(a(x1)))) The set of rules S is b(x1) -> b(b(x1)) We have obtained the following relative TRS: The set of rules R is a(b(a(x1))) -> a(b(b(a(x1)))) The set of rules S is b(x1) -> b(b(x1)) ---------------------------------------- (2) Obligation: Relative term rewrite system: The relative TRS consists of the following R rules: a(b(a(x1))) -> a(b(b(a(x1)))) The relative TRS consists of the following S rules: b(x1) -> b(b(x1)) ---------------------------------------- (3) RootLabelingProof (EQUIVALENT) We used plain root labeling [ROOTLAB] with the following heuristic: LabelAll: All function symbols get labeled ---------------------------------------- (4) Obligation: Relative term rewrite system: The relative TRS consists of the following R rules: a_{b_1}(b_{a_1}(a_{a_1}(x1))) -> a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(x1)))) a_{b_1}(b_{a_1}(a_{b_1}(x1))) -> a_{b_1}(b_{b_1}(b_{a_1}(a_{b_1}(x1)))) The relative TRS consists of the following S rules: b_{a_1}(x1) -> b_{b_1}(b_{a_1}(x1)) b_{b_1}(x1) -> b_{b_1}(b_{b_1}(x1)) ---------------------------------------- (5) RelTRSRRRProof (EQUIVALENT) We used the following monotonic ordering for rule removal: Matrix interpretation [MATRO] to (N^2, +, *, >=, >) : <<< POL(a_{b_1}(x_1)) = [[0], [1]] + [[1, 2], [0, 2]] * x_1 >>> <<< POL(b_{a_1}(x_1)) = [[0], [2]] + [[1, 0], [2, 2]] * x_1 >>> <<< POL(a_{a_1}(x_1)) = [[0], [1]] + [[2, 0], [2, 0]] * x_1 >>> <<< POL(b_{b_1}(x_1)) = [[0], [2]] + [[1, 0], [2, 0]] * x_1 >>> With this ordering the following rules can be removed [MATRO] because they are oriented strictly: Rules from R: a_{b_1}(b_{a_1}(a_{a_1}(x1))) -> a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(x1)))) a_{b_1}(b_{a_1}(a_{b_1}(x1))) -> a_{b_1}(b_{b_1}(b_{a_1}(a_{b_1}(x1)))) Rules from S: none ---------------------------------------- (6) Obligation: Relative term rewrite system: R is empty. The relative TRS consists of the following S rules: b_{a_1}(x1) -> b_{b_1}(b_{a_1}(x1)) b_{b_1}(x1) -> b_{b_1}(b_{b_1}(x1)) ---------------------------------------- (7) RIsEmptyProof (EQUIVALENT) The TRS R is empty. Hence, termination is trivially proven. ---------------------------------------- (8) YES