/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The Derivational Complexity (full) of the given DCpxTrs could be proven to be BOUNDS(INF, INF).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), 239 ms] (4) CpxRelTRS (5) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms] (6) TRS for Loop Detection (7) DecreasingLoopProof [LOWER BOUND(ID), 0 ms] (8) BEST (9) proven lower bound (10) LowerBoundPropagationProof [FINISHED, 0 ms] (11) BOUNDS(n^1, INF) (12) TRS for Loop Detection (13) InfiniteLowerBoundProof [FINISHED, 64 ms] (14) BOUNDS(INF, INF) ---------------------------------------- (0) Obligation: The Derivational Complexity (full) of the given DCpxTrs could be proven to be BOUNDS(INF, INF). The TRS R consists of the following rules: zeros -> cons(0, n__zeros) and(tt, X) -> activate(X) length(nil) -> 0 length(cons(N, L)) -> s(length(activate(L))) take(0, IL) -> nil take(s(M), cons(N, IL)) -> cons(N, n__take(M, activate(IL))) zeros -> n__zeros take(X1, X2) -> n__take(X1, X2) activate(n__zeros) -> zeros activate(n__take(X1, X2)) -> take(activate(X1), activate(X2)) activate(X) -> X S is empty. Rewrite Strategy: FULL ---------------------------------------- (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(cons(x_1, x_2)) -> cons(encArg(x_1), encArg(x_2)) encArg(0) -> 0 encArg(n__zeros) -> n__zeros encArg(tt) -> tt encArg(nil) -> nil encArg(s(x_1)) -> s(encArg(x_1)) encArg(n__take(x_1, x_2)) -> n__take(encArg(x_1), encArg(x_2)) encArg(cons_zeros) -> zeros encArg(cons_and(x_1, x_2)) -> and(encArg(x_1), encArg(x_2)) encArg(cons_length(x_1)) -> length(encArg(x_1)) encArg(cons_take(x_1, x_2)) -> take(encArg(x_1), encArg(x_2)) encArg(cons_activate(x_1)) -> activate(encArg(x_1)) encode_zeros -> zeros encode_cons(x_1, x_2) -> cons(encArg(x_1), encArg(x_2)) encode_0 -> 0 encode_n__zeros -> n__zeros encode_and(x_1, x_2) -> and(encArg(x_1), encArg(x_2)) encode_tt -> tt encode_activate(x_1) -> activate(encArg(x_1)) encode_length(x_1) -> length(encArg(x_1)) encode_nil -> nil encode_s(x_1) -> s(encArg(x_1)) encode_take(x_1, x_2) -> take(encArg(x_1), encArg(x_2)) encode_n__take(x_1, x_2) -> n__take(encArg(x_1), encArg(x_2)) ---------------------------------------- (2) Obligation: The Runtime Complexity (full) of the given CpxRelTRS could be proven to be BOUNDS(INF, INF). The TRS R consists of the following rules: zeros -> cons(0, n__zeros) and(tt, X) -> activate(X) length(nil) -> 0 length(cons(N, L)) -> s(length(activate(L))) take(0, IL) -> nil take(s(M), cons(N, IL)) -> cons(N, n__take(M, activate(IL))) zeros -> n__zeros take(X1, X2) -> n__take(X1, X2) activate(n__zeros) -> zeros activate(n__take(X1, X2)) -> take(activate(X1), activate(X2)) activate(X) -> X The (relative) TRS S consists of the following rules: encArg(cons(x_1, x_2)) -> cons(encArg(x_1), encArg(x_2)) encArg(0) -> 0 encArg(n__zeros) -> n__zeros encArg(tt) -> tt encArg(nil) -> nil encArg(s(x_1)) -> s(encArg(x_1)) encArg(n__take(x_1, x_2)) -> n__take(encArg(x_1), encArg(x_2)) encArg(cons_zeros) -> zeros encArg(cons_and(x_1, x_2)) -> and(encArg(x_1), encArg(x_2)) encArg(cons_length(x_1)) -> length(encArg(x_1)) encArg(cons_take(x_1, x_2)) -> take(encArg(x_1), encArg(x_2)) encArg(cons_activate(x_1)) -> activate(encArg(x_1)) encode_zeros -> zeros encode_cons(x_1, x_2) -> cons(encArg(x_1), encArg(x_2)) encode_0 -> 0 encode_n__zeros -> n__zeros encode_and(x_1, x_2) -> and(encArg(x_1), encArg(x_2)) encode_tt -> tt encode_activate(x_1) -> activate(encArg(x_1)) encode_length(x_1) -> length(encArg(x_1)) encode_nil -> nil encode_s(x_1) -> s(encArg(x_1)) encode_take(x_1, x_2) -> take(encArg(x_1), encArg(x_2)) encode_n__take(x_1, x_2) -> n__take(encArg(x_1), encArg(x_2)) Rewrite Strategy: FULL ---------------------------------------- (3) SInnermostTerminationProof (BOTH CONCRETE BOUNDS(ID, ID)) proved innermost termination of relative rules ---------------------------------------- (4) Obligation: The Runtime Complexity (full) of the given CpxRelTRS could be proven to be BOUNDS(INF, INF). The TRS R consists of the following rules: zeros -> cons(0, n__zeros) and(tt, X) -> activate(X) length(nil) -> 0 length(cons(N, L)) -> s(length(activate(L))) take(0, IL) -> nil take(s(M), cons(N, IL)) -> cons(N, n__take(M, activate(IL))) zeros -> n__zeros take(X1, X2) -> n__take(X1, X2) activate(n__zeros) -> zeros activate(n__take(X1, X2)) -> take(activate(X1), activate(X2)) activate(X) -> X The (relative) TRS S consists of the following rules: encArg(cons(x_1, x_2)) -> cons(encArg(x_1), encArg(x_2)) encArg(0) -> 0 encArg(n__zeros) -> n__zeros encArg(tt) -> tt encArg(nil) -> nil encArg(s(x_1)) -> s(encArg(x_1)) encArg(n__take(x_1, x_2)) -> n__take(encArg(x_1), encArg(x_2)) encArg(cons_zeros) -> zeros encArg(cons_and(x_1, x_2)) -> and(encArg(x_1), encArg(x_2)) encArg(cons_length(x_1)) -> length(encArg(x_1)) encArg(cons_take(x_1, x_2)) -> take(encArg(x_1), encArg(x_2)) encArg(cons_activate(x_1)) -> activate(encArg(x_1)) encode_zeros -> zeros encode_cons(x_1, x_2) -> cons(encArg(x_1), encArg(x_2)) encode_0 -> 0 encode_n__zeros -> n__zeros encode_and(x_1, x_2) -> and(encArg(x_1), encArg(x_2)) encode_tt -> tt encode_activate(x_1) -> activate(encArg(x_1)) encode_length(x_1) -> length(encArg(x_1)) encode_nil -> nil encode_s(x_1) -> s(encArg(x_1)) encode_take(x_1, x_2) -> take(encArg(x_1), encArg(x_2)) encode_n__take(x_1, x_2) -> n__take(encArg(x_1), encArg(x_2)) Rewrite Strategy: FULL ---------------------------------------- (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 (full) of the given CpxRelTRS could be proven to be BOUNDS(INF, INF). The TRS R consists of the following rules: zeros -> cons(0, n__zeros) and(tt, X) -> activate(X) length(nil) -> 0 length(cons(N, L)) -> s(length(activate(L))) take(0, IL) -> nil take(s(M), cons(N, IL)) -> cons(N, n__take(M, activate(IL))) zeros -> n__zeros take(X1, X2) -> n__take(X1, X2) activate(n__zeros) -> zeros activate(n__take(X1, X2)) -> take(activate(X1), activate(X2)) activate(X) -> X The (relative) TRS S consists of the following rules: encArg(cons(x_1, x_2)) -> cons(encArg(x_1), encArg(x_2)) encArg(0) -> 0 encArg(n__zeros) -> n__zeros encArg(tt) -> tt encArg(nil) -> nil encArg(s(x_1)) -> s(encArg(x_1)) encArg(n__take(x_1, x_2)) -> n__take(encArg(x_1), encArg(x_2)) encArg(cons_zeros) -> zeros encArg(cons_and(x_1, x_2)) -> and(encArg(x_1), encArg(x_2)) encArg(cons_length(x_1)) -> length(encArg(x_1)) encArg(cons_take(x_1, x_2)) -> take(encArg(x_1), encArg(x_2)) encArg(cons_activate(x_1)) -> activate(encArg(x_1)) encode_zeros -> zeros encode_cons(x_1, x_2) -> cons(encArg(x_1), encArg(x_2)) encode_0 -> 0 encode_n__zeros -> n__zeros encode_and(x_1, x_2) -> and(encArg(x_1), encArg(x_2)) encode_tt -> tt encode_activate(x_1) -> activate(encArg(x_1)) encode_length(x_1) -> length(encArg(x_1)) encode_nil -> nil encode_s(x_1) -> s(encArg(x_1)) encode_take(x_1, x_2) -> take(encArg(x_1), encArg(x_2)) encode_n__take(x_1, x_2) -> n__take(encArg(x_1), encArg(x_2)) Rewrite Strategy: FULL ---------------------------------------- (7) DecreasingLoopProof (LOWER BOUND(ID)) The following loop(s) give(s) rise to the lower bound Omega(n^1): The rewrite sequence activate(n__take(X1, X2)) ->^+ take(activate(X1), activate(X2)) gives rise to a decreasing loop by considering the right hand sides subterm at position [0]. The pumping substitution is [X1 / n__take(X1, X2)]. The result substitution is [ ]. ---------------------------------------- (8) Complex Obligation (BEST) ---------------------------------------- (9) Obligation: Proved the lower bound n^1 for the following obligation: The Runtime Complexity (full) of the given CpxRelTRS could be proven to be BOUNDS(INF, INF). The TRS R consists of the following rules: zeros -> cons(0, n__zeros) and(tt, X) -> activate(X) length(nil) -> 0 length(cons(N, L)) -> s(length(activate(L))) take(0, IL) -> nil take(s(M), cons(N, IL)) -> cons(N, n__take(M, activate(IL))) zeros -> n__zeros take(X1, X2) -> n__take(X1, X2) activate(n__zeros) -> zeros activate(n__take(X1, X2)) -> take(activate(X1), activate(X2)) activate(X) -> X The (relative) TRS S consists of the following rules: encArg(cons(x_1, x_2)) -> cons(encArg(x_1), encArg(x_2)) encArg(0) -> 0 encArg(n__zeros) -> n__zeros encArg(tt) -> tt encArg(nil) -> nil encArg(s(x_1)) -> s(encArg(x_1)) encArg(n__take(x_1, x_2)) -> n__take(encArg(x_1), encArg(x_2)) encArg(cons_zeros) -> zeros encArg(cons_and(x_1, x_2)) -> and(encArg(x_1), encArg(x_2)) encArg(cons_length(x_1)) -> length(encArg(x_1)) encArg(cons_take(x_1, x_2)) -> take(encArg(x_1), encArg(x_2)) encArg(cons_activate(x_1)) -> activate(encArg(x_1)) encode_zeros -> zeros encode_cons(x_1, x_2) -> cons(encArg(x_1), encArg(x_2)) encode_0 -> 0 encode_n__zeros -> n__zeros encode_and(x_1, x_2) -> and(encArg(x_1), encArg(x_2)) encode_tt -> tt encode_activate(x_1) -> activate(encArg(x_1)) encode_length(x_1) -> length(encArg(x_1)) encode_nil -> nil encode_s(x_1) -> s(encArg(x_1)) encode_take(x_1, x_2) -> take(encArg(x_1), encArg(x_2)) encode_n__take(x_1, x_2) -> n__take(encArg(x_1), encArg(x_2)) Rewrite Strategy: FULL ---------------------------------------- (10) LowerBoundPropagationProof (FINISHED) Propagated lower bound. ---------------------------------------- (11) BOUNDS(n^1, INF) ---------------------------------------- (12) Obligation: Analyzing the following TRS for decreasing loops: The Runtime Complexity (full) of the given CpxRelTRS could be proven to be BOUNDS(INF, INF). The TRS R consists of the following rules: zeros -> cons(0, n__zeros) and(tt, X) -> activate(X) length(nil) -> 0 length(cons(N, L)) -> s(length(activate(L))) take(0, IL) -> nil take(s(M), cons(N, IL)) -> cons(N, n__take(M, activate(IL))) zeros -> n__zeros take(X1, X2) -> n__take(X1, X2) activate(n__zeros) -> zeros activate(n__take(X1, X2)) -> take(activate(X1), activate(X2)) activate(X) -> X The (relative) TRS S consists of the following rules: encArg(cons(x_1, x_2)) -> cons(encArg(x_1), encArg(x_2)) encArg(0) -> 0 encArg(n__zeros) -> n__zeros encArg(tt) -> tt encArg(nil) -> nil encArg(s(x_1)) -> s(encArg(x_1)) encArg(n__take(x_1, x_2)) -> n__take(encArg(x_1), encArg(x_2)) encArg(cons_zeros) -> zeros encArg(cons_and(x_1, x_2)) -> and(encArg(x_1), encArg(x_2)) encArg(cons_length(x_1)) -> length(encArg(x_1)) encArg(cons_take(x_1, x_2)) -> take(encArg(x_1), encArg(x_2)) encArg(cons_activate(x_1)) -> activate(encArg(x_1)) encode_zeros -> zeros encode_cons(x_1, x_2) -> cons(encArg(x_1), encArg(x_2)) encode_0 -> 0 encode_n__zeros -> n__zeros encode_and(x_1, x_2) -> and(encArg(x_1), encArg(x_2)) encode_tt -> tt encode_activate(x_1) -> activate(encArg(x_1)) encode_length(x_1) -> length(encArg(x_1)) encode_nil -> nil encode_s(x_1) -> s(encArg(x_1)) encode_take(x_1, x_2) -> take(encArg(x_1), encArg(x_2)) encode_n__take(x_1, x_2) -> n__take(encArg(x_1), encArg(x_2)) Rewrite Strategy: FULL ---------------------------------------- (13) InfiniteLowerBoundProof (FINISHED) The following loop proves infinite runtime complexity: The rewrite sequence length(cons(N, n__zeros)) ->^+ s(length(cons(0, n__zeros))) gives rise to a decreasing loop by considering the right hand sides subterm at position [0]. The pumping substitution is [ ]. The result substitution is [N / 0]. ---------------------------------------- (14) BOUNDS(INF, INF)