MAYBE We are left with following problem, upon which TcT provides the certificate MAYBE. Strict Trs: { eq(X, Y) -> false() , eq(n__0(), n__0()) -> true() , eq(n__s(X), n__s(Y)) -> eq(activate(X), activate(Y)) , activate(X) -> X , activate(n__0()) -> 0() , activate(n__s(X)) -> s(X) , activate(n__inf(X)) -> inf(activate(X)) , activate(n__take(X1, X2)) -> take(activate(X1), activate(X2)) , activate(n__length(X)) -> length(activate(X)) , inf(X) -> cons(X, n__inf(n__s(X))) , inf(X) -> n__inf(X) , take(X1, X2) -> n__take(X1, X2) , take(0(), X) -> nil() , take(s(X), cons(Y, L)) -> cons(activate(Y), n__take(activate(X), activate(L))) , 0() -> n__0() , s(X) -> n__s(X) , length(X) -> n__length(X) , length(cons(X, L)) -> s(n__length(activate(L))) , length(nil()) -> 0() } Obligation: runtime complexity Answer: MAYBE None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'WithProblem (timeout of 60 seconds)' failed due to the following reason: Computation stopped due to timeout after 60.0 seconds. 2) 'Best' failed due to the following reason: None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'WithProblem (timeout of 30 seconds) (timeout of 60 seconds)' failed due to the following reason: Computation stopped due to timeout after 30.0 seconds. 2) 'Best' failed due to the following reason: None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'bsearch-popstar (timeout of 60 seconds)' failed due to the following reason: The processor is inapplicable, reason: Processor only applicable for innermost runtime complexity analysis 2) 'Polynomial Path Order (PS) (timeout of 60 seconds)' failed due to the following reason: The processor is inapplicable, reason: Processor only applicable for innermost runtime complexity analysis 3) 'Fastest (timeout of 5 seconds) (timeout of 60 seconds)' failed due to the following reason: None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'Bounds with minimal-enrichment and initial automaton 'match'' failed due to the following reason: match-boundness of the problem could not be verified. 2) 'Bounds with perSymbol-enrichment and initial automaton 'match'' failed due to the following reason: match-boundness of the problem could not be verified. 3) 'Innermost Weak Dependency Pairs (timeout of 60 seconds)' failed due to the following reason: We add the following weak dependency pairs: Strict DPs: { eq^#(X, Y) -> c_1() , eq^#(n__0(), n__0()) -> c_2() , eq^#(n__s(X), n__s(Y)) -> c_3(eq^#(activate(X), activate(Y))) , activate^#(X) -> c_4(X) , activate^#(n__0()) -> c_5(0^#()) , activate^#(n__s(X)) -> c_6(s^#(X)) , activate^#(n__inf(X)) -> c_7(inf^#(activate(X))) , activate^#(n__take(X1, X2)) -> c_8(take^#(activate(X1), activate(X2))) , activate^#(n__length(X)) -> c_9(length^#(activate(X))) , 0^#() -> c_15() , s^#(X) -> c_16(X) , inf^#(X) -> c_10(X, X) , inf^#(X) -> c_11(X) , take^#(X1, X2) -> c_12(X1, X2) , take^#(0(), X) -> c_13() , take^#(s(X), cons(Y, L)) -> c_14(activate^#(Y), activate^#(X), activate^#(L)) , length^#(X) -> c_17(X) , length^#(cons(X, L)) -> c_18(s^#(n__length(activate(L)))) , length^#(nil()) -> c_19(0^#()) } and mark the set of starting terms. We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { eq^#(X, Y) -> c_1() , eq^#(n__0(), n__0()) -> c_2() , eq^#(n__s(X), n__s(Y)) -> c_3(eq^#(activate(X), activate(Y))) , activate^#(X) -> c_4(X) , activate^#(n__0()) -> c_5(0^#()) , activate^#(n__s(X)) -> c_6(s^#(X)) , activate^#(n__inf(X)) -> c_7(inf^#(activate(X))) , activate^#(n__take(X1, X2)) -> c_8(take^#(activate(X1), activate(X2))) , activate^#(n__length(X)) -> c_9(length^#(activate(X))) , 0^#() -> c_15() , s^#(X) -> c_16(X) , inf^#(X) -> c_10(X, X) , inf^#(X) -> c_11(X) , take^#(X1, X2) -> c_12(X1, X2) , take^#(0(), X) -> c_13() , take^#(s(X), cons(Y, L)) -> c_14(activate^#(Y), activate^#(X), activate^#(L)) , length^#(X) -> c_17(X) , length^#(cons(X, L)) -> c_18(s^#(n__length(activate(L)))) , length^#(nil()) -> c_19(0^#()) } Strict Trs: { eq(X, Y) -> false() , eq(n__0(), n__0()) -> true() , eq(n__s(X), n__s(Y)) -> eq(activate(X), activate(Y)) , activate(X) -> X , activate(n__0()) -> 0() , activate(n__s(X)) -> s(X) , activate(n__inf(X)) -> inf(activate(X)) , activate(n__take(X1, X2)) -> take(activate(X1), activate(X2)) , activate(n__length(X)) -> length(activate(X)) , inf(X) -> cons(X, n__inf(n__s(X))) , inf(X) -> n__inf(X) , take(X1, X2) -> n__take(X1, X2) , take(0(), X) -> nil() , take(s(X), cons(Y, L)) -> cons(activate(Y), n__take(activate(X), activate(L))) , 0() -> n__0() , s(X) -> n__s(X) , length(X) -> n__length(X) , length(cons(X, L)) -> s(n__length(activate(L))) , length(nil()) -> 0() } Obligation: runtime complexity Answer: MAYBE We estimate the number of application of {1,2,10,15} by applications of Pre({1,2,10,15}) = {3,4,5,8,11,12,13,14,17,19}. Here rules are labeled as follows: DPs: { 1: eq^#(X, Y) -> c_1() , 2: eq^#(n__0(), n__0()) -> c_2() , 3: eq^#(n__s(X), n__s(Y)) -> c_3(eq^#(activate(X), activate(Y))) , 4: activate^#(X) -> c_4(X) , 5: activate^#(n__0()) -> c_5(0^#()) , 6: activate^#(n__s(X)) -> c_6(s^#(X)) , 7: activate^#(n__inf(X)) -> c_7(inf^#(activate(X))) , 8: activate^#(n__take(X1, X2)) -> c_8(take^#(activate(X1), activate(X2))) , 9: activate^#(n__length(X)) -> c_9(length^#(activate(X))) , 10: 0^#() -> c_15() , 11: s^#(X) -> c_16(X) , 12: inf^#(X) -> c_10(X, X) , 13: inf^#(X) -> c_11(X) , 14: take^#(X1, X2) -> c_12(X1, X2) , 15: take^#(0(), X) -> c_13() , 16: take^#(s(X), cons(Y, L)) -> c_14(activate^#(Y), activate^#(X), activate^#(L)) , 17: length^#(X) -> c_17(X) , 18: length^#(cons(X, L)) -> c_18(s^#(n__length(activate(L)))) , 19: length^#(nil()) -> c_19(0^#()) } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { eq^#(n__s(X), n__s(Y)) -> c_3(eq^#(activate(X), activate(Y))) , activate^#(X) -> c_4(X) , activate^#(n__0()) -> c_5(0^#()) , activate^#(n__s(X)) -> c_6(s^#(X)) , activate^#(n__inf(X)) -> c_7(inf^#(activate(X))) , activate^#(n__take(X1, X2)) -> c_8(take^#(activate(X1), activate(X2))) , activate^#(n__length(X)) -> c_9(length^#(activate(X))) , s^#(X) -> c_16(X) , inf^#(X) -> c_10(X, X) , inf^#(X) -> c_11(X) , take^#(X1, X2) -> c_12(X1, X2) , take^#(s(X), cons(Y, L)) -> c_14(activate^#(Y), activate^#(X), activate^#(L)) , length^#(X) -> c_17(X) , length^#(cons(X, L)) -> c_18(s^#(n__length(activate(L)))) , length^#(nil()) -> c_19(0^#()) } Strict Trs: { eq(X, Y) -> false() , eq(n__0(), n__0()) -> true() , eq(n__s(X), n__s(Y)) -> eq(activate(X), activate(Y)) , activate(X) -> X , activate(n__0()) -> 0() , activate(n__s(X)) -> s(X) , activate(n__inf(X)) -> inf(activate(X)) , activate(n__take(X1, X2)) -> take(activate(X1), activate(X2)) , activate(n__length(X)) -> length(activate(X)) , inf(X) -> cons(X, n__inf(n__s(X))) , inf(X) -> n__inf(X) , take(X1, X2) -> n__take(X1, X2) , take(0(), X) -> nil() , take(s(X), cons(Y, L)) -> cons(activate(Y), n__take(activate(X), activate(L))) , 0() -> n__0() , s(X) -> n__s(X) , length(X) -> n__length(X) , length(cons(X, L)) -> s(n__length(activate(L))) , length(nil()) -> 0() } Weak DPs: { eq^#(X, Y) -> c_1() , eq^#(n__0(), n__0()) -> c_2() , 0^#() -> c_15() , take^#(0(), X) -> c_13() } Obligation: runtime complexity Answer: MAYBE We estimate the number of application of {3,15} by applications of Pre({3,15}) = {2,7,8,9,10,11,12,13}. Here rules are labeled as follows: DPs: { 1: eq^#(n__s(X), n__s(Y)) -> c_3(eq^#(activate(X), activate(Y))) , 2: activate^#(X) -> c_4(X) , 3: activate^#(n__0()) -> c_5(0^#()) , 4: activate^#(n__s(X)) -> c_6(s^#(X)) , 5: activate^#(n__inf(X)) -> c_7(inf^#(activate(X))) , 6: activate^#(n__take(X1, X2)) -> c_8(take^#(activate(X1), activate(X2))) , 7: activate^#(n__length(X)) -> c_9(length^#(activate(X))) , 8: s^#(X) -> c_16(X) , 9: inf^#(X) -> c_10(X, X) , 10: inf^#(X) -> c_11(X) , 11: take^#(X1, X2) -> c_12(X1, X2) , 12: take^#(s(X), cons(Y, L)) -> c_14(activate^#(Y), activate^#(X), activate^#(L)) , 13: length^#(X) -> c_17(X) , 14: length^#(cons(X, L)) -> c_18(s^#(n__length(activate(L)))) , 15: length^#(nil()) -> c_19(0^#()) , 16: eq^#(X, Y) -> c_1() , 17: eq^#(n__0(), n__0()) -> c_2() , 18: 0^#() -> c_15() , 19: take^#(0(), X) -> c_13() } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { eq^#(n__s(X), n__s(Y)) -> c_3(eq^#(activate(X), activate(Y))) , activate^#(X) -> c_4(X) , activate^#(n__s(X)) -> c_6(s^#(X)) , activate^#(n__inf(X)) -> c_7(inf^#(activate(X))) , activate^#(n__take(X1, X2)) -> c_8(take^#(activate(X1), activate(X2))) , activate^#(n__length(X)) -> c_9(length^#(activate(X))) , s^#(X) -> c_16(X) , inf^#(X) -> c_10(X, X) , inf^#(X) -> c_11(X) , take^#(X1, X2) -> c_12(X1, X2) , take^#(s(X), cons(Y, L)) -> c_14(activate^#(Y), activate^#(X), activate^#(L)) , length^#(X) -> c_17(X) , length^#(cons(X, L)) -> c_18(s^#(n__length(activate(L)))) } Strict Trs: { eq(X, Y) -> false() , eq(n__0(), n__0()) -> true() , eq(n__s(X), n__s(Y)) -> eq(activate(X), activate(Y)) , activate(X) -> X , activate(n__0()) -> 0() , activate(n__s(X)) -> s(X) , activate(n__inf(X)) -> inf(activate(X)) , activate(n__take(X1, X2)) -> take(activate(X1), activate(X2)) , activate(n__length(X)) -> length(activate(X)) , inf(X) -> cons(X, n__inf(n__s(X))) , inf(X) -> n__inf(X) , take(X1, X2) -> n__take(X1, X2) , take(0(), X) -> nil() , take(s(X), cons(Y, L)) -> cons(activate(Y), n__take(activate(X), activate(L))) , 0() -> n__0() , s(X) -> n__s(X) , length(X) -> n__length(X) , length(cons(X, L)) -> s(n__length(activate(L))) , length(nil()) -> 0() } Weak DPs: { eq^#(X, Y) -> c_1() , eq^#(n__0(), n__0()) -> c_2() , activate^#(n__0()) -> c_5(0^#()) , 0^#() -> c_15() , take^#(0(), X) -> c_13() , length^#(nil()) -> c_19(0^#()) } Obligation: runtime complexity Answer: MAYBE Empty strict component of the problem is NOT empty. Arrrr..