/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 58 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 9 ms] (4) QTRS (5) RisEmptyProof [EQUIVALENT, 0 ms] (6) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(0) -> s(0) f(s(0)) -> s(s(0)) f(s(0)) -> *(s(s(0)), f(0)) f(+(x, s(0))) -> +(s(s(0)), f(x)) f(+(x, y)) -> *(f(x), f(y)) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(*(x_1, x_2)) = x_1 + x_2 POL(+(x_1, x_2)) = 2 + 2*x_1 + x_2 POL(0) = 0 POL(f(x_1)) = 1 + 2*x_1 POL(s(x_1)) = 2*x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: f(0) -> s(0) f(s(0)) -> s(s(0)) f(+(x, s(0))) -> +(s(s(0)), f(x)) f(+(x, y)) -> *(f(x), f(y)) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(s(0)) -> *(s(s(0)), f(0)) Q is empty. ---------------------------------------- (3) QTRSRRRProof (EQUIVALENT) Used ordering: Quasi precedence: f_1 > [0, *_2] > s_1 Status: f_1: multiset status s_1: multiset status 0: multiset status *_2: multiset status With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: f(s(0)) -> *(s(s(0)), f(0)) ---------------------------------------- (4) Obligation: Q restricted rewrite system: R is empty. Q is empty. ---------------------------------------- (5) RisEmptyProof (EQUIVALENT) The TRS R is empty. Hence, termination is trivially proven. ---------------------------------------- (6) YES