/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination of the given RelTRS could be proven: (0) RelTRS (1) RelTRS Reverse [SOUND, 0 ms] (2) RelTRS (3) RelTRSRRRProof [EQUIVALENT, 29 ms] (4) RelTRS (5) RIsEmptyProof [EQUIVALENT, 0 ms] (6) YES ---------------------------------------- (0) Obligation: Relative term rewrite system: The relative TRS consists of the following R rules: f(g(x)) -> x The relative TRS consists of the following S rules: a -> h(g(f(a))) ---------------------------------------- (1) RelTRS Reverse (SOUND) We have reversed the following relative TRS [REVERSE]: The set of rules R is f(g(x)) -> x The set of rules S is a -> h(g(f(a))) We have obtained the following relative TRS: The set of rules R is g(f(x)) -> x The set of rules S is a'(x) -> a'(f(g(h(x)))) ---------------------------------------- (2) Obligation: Relative term rewrite system: The relative TRS consists of the following R rules: g(f(x)) -> x The relative TRS consists of the following S rules: a'(x) -> a'(f(g(h(x)))) ---------------------------------------- (3) RelTRSRRRProof (EQUIVALENT) We used the following monotonic ordering for rule removal: Matrix interpretation [MATRO] to (N^2, +, *, >=, >) : <<< POL(g(x_1)) = [[0], [0]] + [[1, 1], [0, 2]] * x_1 >>> <<< POL(f(x_1)) = [[0], [1]] + [[1, 0], [0, 1]] * x_1 >>> <<< POL(a'(x_1)) = [[2], [2]] + [[1, 0], [0, 0]] * x_1 >>> <<< POL(h(x_1)) = [[0], [0]] + [[1, 0], [0, 0]] * x_1 >>> With this ordering the following rules can be removed [MATRO] because they are oriented strictly: Rules from R: g(f(x)) -> x Rules from S: none ---------------------------------------- (4) Obligation: Relative term rewrite system: R is empty. The relative TRS consists of the following S rules: a'(x) -> a'(f(g(h(x)))) ---------------------------------------- (5) RIsEmptyProof (EQUIVALENT) The TRS R is empty. Hence, termination is trivially proven. ---------------------------------------- (6) YES