/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination of the given RelTRS could be proven: (0) RelTRS (1) RelTRS Reverse [EQUIVALENT, 0 ms] (2) RelTRS (3) FlatCCProof [EQUIVALENT, 0 ms] (4) RelTRS (5) RootLabelingProof [EQUIVALENT, 0 ms] (6) RelTRS (7) RelTRSRRRProof [EQUIVALENT, 248 ms] (8) RelTRS (9) RelTRSRRRProof [EQUIVALENT, 0 ms] (10) RelTRS (11) SIsEmptyProof [EQUIVALENT, 0 ms] (12) QTRS (13) QTRSRoofMatchBoundsTAProof [EQUIVALENT, 0 ms] (14) YES ---------------------------------------- (0) Obligation: Relative term rewrite system: The relative TRS consists of the following R rules: b(a(b(x1))) -> b(a(a(a(b(x1))))) The relative TRS consists of the following S rules: a(a(a(x1))) -> b(b(b(b(x1)))) ---------------------------------------- (1) RelTRS Reverse (EQUIVALENT) We have reversed the following relative TRS [REVERSE]: The set of rules R is b(a(b(x1))) -> b(a(a(a(b(x1))))) The set of rules S is a(a(a(x1))) -> b(b(b(b(x1)))) We have obtained the following relative TRS: The set of rules R is b(a(b(x1))) -> b(a(a(a(b(x1))))) The set of rules S is a(a(a(x1))) -> b(b(b(b(x1)))) ---------------------------------------- (2) Obligation: Relative term rewrite system: The relative TRS consists of the following R rules: b(a(b(x1))) -> b(a(a(a(b(x1))))) The relative TRS consists of the following S rules: a(a(a(x1))) -> b(b(b(b(x1)))) ---------------------------------------- (3) FlatCCProof (EQUIVALENT) We used flat context closure [ROOTLAB] ---------------------------------------- (4) Obligation: Relative term rewrite system: The relative TRS consists of the following R rules: b(a(b(x1))) -> b(a(a(a(b(x1))))) The relative TRS consists of the following S rules: b(a(a(a(x1)))) -> b(b(b(b(b(x1))))) a(a(a(a(x1)))) -> a(b(b(b(b(x1))))) ---------------------------------------- (5) RootLabelingProof (EQUIVALENT) We used plain root labeling [ROOTLAB] with the following heuristic: LabelAll: All function symbols get labeled ---------------------------------------- (6) Obligation: Relative term rewrite system: The relative TRS consists of the following R rules: b_{a_1}(a_{b_1}(b_{b_1}(x1))) -> b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))) b_{a_1}(a_{b_1}(b_{a_1}(x1))) -> b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))) The relative TRS consists of the following S rules: b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(x1)))) -> b_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(x1))))) b_{a_1}(a_{a_1}(a_{a_1}(a_{a_1}(x1)))) -> b_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))) a_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(x1)))) -> a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(x1))))) a_{a_1}(a_{a_1}(a_{a_1}(a_{a_1}(x1)))) -> a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))) ---------------------------------------- (7) RelTRSRRRProof (EQUIVALENT) We used the following monotonic ordering for rule removal: Matrix interpretation [MATRO] to (N^3, +, *, >=, >) : <<< POL(b_{a_1}(x_1)) = [[0], [0], [0]] + [[1, 0, 0], [0, 0, 0], [0, 0, 0]] * x_1 >>> <<< POL(a_{b_1}(x_1)) = [[0], [0], [0]] + [[1, 0, 0], [0, 0, 0], [0, 0, 0]] * x_1 >>> <<< POL(b_{b_1}(x_1)) = [[0], [0], [0]] + [[1, 0, 0], [0, 0, 0], [0, 0, 0]] * x_1 >>> <<< POL(a_{a_1}(x_1)) = [[0], [1], [0]] + [[1, 0, 1], [0, 1, 0], [0, 1, 0]] * x_1 >>> With this ordering the following rules can be removed [MATRO] because they are oriented strictly: Rules from R: none Rules from S: b_{a_1}(a_{a_1}(a_{a_1}(a_{a_1}(x1)))) -> b_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))) a_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(x1)))) -> a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(x1))))) a_{a_1}(a_{a_1}(a_{a_1}(a_{a_1}(x1)))) -> a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))) ---------------------------------------- (8) Obligation: Relative term rewrite system: The relative TRS consists of the following R rules: b_{a_1}(a_{b_1}(b_{b_1}(x1))) -> b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))) b_{a_1}(a_{b_1}(b_{a_1}(x1))) -> b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))) The relative TRS consists of the following S rules: b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(x1)))) -> b_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(x1))))) ---------------------------------------- (9) RelTRSRRRProof (EQUIVALENT) We used the following monotonic ordering for rule removal: Polynomial interpretation [POLO]: POL(a_{a_1}(x_1)) = x_1 POL(a_{b_1}(x_1)) = x_1 POL(b_{a_1}(x_1)) = 1 + x_1 POL(b_{b_1}(x_1)) = x_1 With this ordering the following rules can be removed [MATRO] because they are oriented strictly: Rules from R: none Rules from S: b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(x1)))) -> b_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(x1))))) ---------------------------------------- (10) Obligation: Relative term rewrite system: The relative TRS consists of the following R rules: b_{a_1}(a_{b_1}(b_{b_1}(x1))) -> b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))) b_{a_1}(a_{b_1}(b_{a_1}(x1))) -> b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))) S is empty. ---------------------------------------- (11) SIsEmptyProof (EQUIVALENT) The TRS S is empty. Hence, termination of R/S is equivalent to termination of R. ---------------------------------------- (12) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b_{a_1}(a_{b_1}(b_{b_1}(x1))) -> b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1))))) b_{a_1}(a_{b_1}(b_{a_1}(x1))) -> b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1))))) Q is empty. ---------------------------------------- (13) QTRSRoofMatchBoundsTAProof (EQUIVALENT) The TRS R could be shown to be Match-Bounded [TAB_LEFTLINEAR,TAB_NONLEFTLINEAR] by 1. Therefore it terminates. The compatible tree automaton used to show the Match-Boundedness is represented by: final states : [0, 1, 2, 3] transitions: b_{a_1}0(0) -> 0 b_{a_1}0(1) -> 0 b_{a_1}0(2) -> 0 b_{a_1}0(3) -> 0 a_{b_1}0(0) -> 1 a_{b_1}0(1) -> 1 a_{b_1}0(2) -> 1 a_{b_1}0(3) -> 1 b_{b_1}0(0) -> 2 b_{b_1}0(1) -> 2 b_{b_1}0(2) -> 2 b_{b_1}0(3) -> 2 a_{a_1}0(0) -> 3 a_{a_1}0(1) -> 3 a_{a_1}0(2) -> 3 a_{a_1}0(3) -> 3 b_{b_1}1(0) -> 7 a_{b_1}1(7) -> 6 a_{a_1}1(6) -> 5 a_{a_1}1(5) -> 4 b_{a_1}1(4) -> 0 b_{b_1}1(1) -> 7 b_{b_1}1(2) -> 7 b_{b_1}1(3) -> 7 b_{a_1}1(0) -> 7 b_{a_1}1(1) -> 7 b_{a_1}1(2) -> 7 b_{a_1}1(3) -> 7 4 -> 0 ---------------------------------------- (14) YES