/export/starexec/sandbox2/solver/bin/starexec_run_rcdcRelativeAlsoLower /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- WORST_CASE(NON_POLY, ?) proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The Derivational Complexity (innermost) of the given DCpxTrs could be proven to be BOUNDS(INF, INF). (0) DCpxTrs (1) DerivationalComplexityToRuntimeComplexityProof [BOTH BOUNDS(ID, ID), 0 ms] (2) CpxRelTRS (3) SInnermostTerminationProof [BOTH CONCRETE BOUNDS(ID, ID), 176 ms] (4) CpxRelTRS (5) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms] (6) TRS for Loop Detection (7) InfiniteLowerBoundProof [FINISHED, 0 ms] (8) BOUNDS(INF, INF) ---------------------------------------- (0) Obligation: The Derivational Complexity (innermost) of the given DCpxTrs could be proven to be BOUNDS(INF, INF). The TRS R consists of the following rules: ap(ap(g, x), y) -> y ap(f, x) -> ap(f, app(g, x)) S is empty. Rewrite Strategy: INNERMOST ---------------------------------------- (1) DerivationalComplexityToRuntimeComplexityProof (BOTH BOUNDS(ID, ID)) The following rules have been added to S to convert the given derivational complexity problem to a runtime complexity problem: encArg(g) -> g encArg(f) -> f encArg(app(x_1, x_2)) -> app(encArg(x_1), encArg(x_2)) encArg(cons_ap(x_1, x_2)) -> ap(encArg(x_1), encArg(x_2)) encode_ap(x_1, x_2) -> ap(encArg(x_1), encArg(x_2)) encode_g -> g encode_f -> f encode_app(x_1, x_2) -> app(encArg(x_1), encArg(x_2)) ---------------------------------------- (2) Obligation: The Runtime Complexity (innermost) of the given CpxRelTRS could be proven to be BOUNDS(INF, INF). The TRS R consists of the following rules: ap(ap(g, x), y) -> y ap(f, x) -> ap(f, app(g, x)) The (relative) TRS S consists of the following rules: encArg(g) -> g encArg(f) -> f encArg(app(x_1, x_2)) -> app(encArg(x_1), encArg(x_2)) encArg(cons_ap(x_1, x_2)) -> ap(encArg(x_1), encArg(x_2)) encode_ap(x_1, x_2) -> ap(encArg(x_1), encArg(x_2)) encode_g -> g encode_f -> f encode_app(x_1, x_2) -> app(encArg(x_1), encArg(x_2)) Rewrite Strategy: INNERMOST ---------------------------------------- (3) SInnermostTerminationProof (BOTH CONCRETE BOUNDS(ID, ID)) proved innermost termination of relative rules ---------------------------------------- (4) Obligation: The Runtime Complexity (innermost) of the given CpxRelTRS could be proven to be BOUNDS(INF, INF). The TRS R consists of the following rules: ap(ap(g, x), y) -> y ap(f, x) -> ap(f, app(g, x)) The (relative) TRS S consists of the following rules: encArg(g) -> g encArg(f) -> f encArg(app(x_1, x_2)) -> app(encArg(x_1), encArg(x_2)) encArg(cons_ap(x_1, x_2)) -> ap(encArg(x_1), encArg(x_2)) encode_ap(x_1, x_2) -> ap(encArg(x_1), encArg(x_2)) encode_g -> g encode_f -> f encode_app(x_1, x_2) -> app(encArg(x_1), encArg(x_2)) Rewrite Strategy: INNERMOST ---------------------------------------- (5) RelTrsToDecreasingLoopProblemProof (LOWER BOUND(ID)) Transformed a relative TRS into a decreasing-loop problem. ---------------------------------------- (6) Obligation: Analyzing the following TRS for decreasing loops: The Runtime Complexity (innermost) of the given CpxRelTRS could be proven to be BOUNDS(INF, INF). The TRS R consists of the following rules: ap(ap(g, x), y) -> y ap(f, x) -> ap(f, app(g, x)) The (relative) TRS S consists of the following rules: encArg(g) -> g encArg(f) -> f encArg(app(x_1, x_2)) -> app(encArg(x_1), encArg(x_2)) encArg(cons_ap(x_1, x_2)) -> ap(encArg(x_1), encArg(x_2)) encode_ap(x_1, x_2) -> ap(encArg(x_1), encArg(x_2)) encode_g -> g encode_f -> f encode_app(x_1, x_2) -> app(encArg(x_1), encArg(x_2)) Rewrite Strategy: INNERMOST ---------------------------------------- (7) InfiniteLowerBoundProof (FINISHED) The following loop proves infinite runtime complexity: The rewrite sequence ap(f, x) ->^+ ap(f, app(g, x)) gives rise to a decreasing loop by considering the right hand sides subterm at position []. The pumping substitution is [ ]. The result substitution is [x / app(g, x)]. ---------------------------------------- (8) BOUNDS(INF, INF)