MAYBE We are left with following problem, upon which TcT provides the certificate MAYBE. Strict Trs: { a__zeros() -> cons(0(), zeros()) , a__zeros() -> zeros() , a__U11(X1, X2) -> U11(X1, X2) , a__U11(tt(), L) -> s(a__length(mark(L))) , a__length(X) -> length(X) , a__length(cons(N, L)) -> a__U11(a__and(a__isNatList(L), isNat(N)), L) , a__length(nil()) -> 0() , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(0()) -> 0() , mark(zeros()) -> a__zeros() , mark(tt()) -> tt() , mark(s(X)) -> s(mark(X)) , mark(length(X)) -> a__length(mark(X)) , mark(isNatIList(X)) -> a__isNatIList(X) , mark(nil()) -> nil() , mark(isNatList(X)) -> a__isNatList(X) , mark(isNat(X)) -> a__isNat(X) , mark(U11(X1, X2)) -> a__U11(mark(X1), X2) , mark(and(X1, X2)) -> a__and(mark(X1), X2) , a__and(X1, X2) -> and(X1, X2) , a__and(tt(), X) -> mark(X) , a__isNat(X) -> isNat(X) , a__isNat(0()) -> tt() , a__isNat(s(V1)) -> a__isNat(V1) , a__isNat(length(V1)) -> a__isNatList(V1) , a__isNatList(X) -> isNatList(X) , a__isNatList(cons(V1, V2)) -> a__and(a__isNat(V1), isNatList(V2)) , a__isNatList(nil()) -> tt() , a__isNatIList(V) -> a__isNatList(V) , a__isNatIList(X) -> isNatIList(X) , a__isNatIList(cons(V1, V2)) -> a__and(a__isNat(V1), isNatIList(V2)) , a__isNatIList(zeros()) -> tt() } 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) '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 perSymbol-enrichment and initial automaton 'match'' failed due to the following reason: match-boundness of the problem could not be verified. 2) 'Bounds with minimal-enrichment and initial automaton 'match'' failed due to the following reason: match-boundness of the problem could not be verified. 3) '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) 'Innermost Weak Dependency Pairs (timeout of 60 seconds)' failed due to the following reason: We add the following weak dependency pairs: Strict DPs: { a__zeros^#() -> c_1() , a__zeros^#() -> c_2() , a__U11^#(X1, X2) -> c_3(X1, X2) , a__U11^#(tt(), L) -> c_4(a__length^#(mark(L))) , a__length^#(X) -> c_5(X) , a__length^#(cons(N, L)) -> c_6(a__U11^#(a__and(a__isNatList(L), isNat(N)), L)) , a__length^#(nil()) -> c_7() , mark^#(cons(X1, X2)) -> c_8(mark^#(X1), X2) , mark^#(0()) -> c_9() , mark^#(zeros()) -> c_10(a__zeros^#()) , mark^#(tt()) -> c_11() , mark^#(s(X)) -> c_12(mark^#(X)) , mark^#(length(X)) -> c_13(a__length^#(mark(X))) , mark^#(isNatIList(X)) -> c_14(a__isNatIList^#(X)) , mark^#(nil()) -> c_15() , mark^#(isNatList(X)) -> c_16(a__isNatList^#(X)) , mark^#(isNat(X)) -> c_17(a__isNat^#(X)) , mark^#(U11(X1, X2)) -> c_18(a__U11^#(mark(X1), X2)) , mark^#(and(X1, X2)) -> c_19(a__and^#(mark(X1), X2)) , a__isNatIList^#(V) -> c_29(a__isNatList^#(V)) , a__isNatIList^#(X) -> c_30(X) , a__isNatIList^#(cons(V1, V2)) -> c_31(a__and^#(a__isNat(V1), isNatIList(V2))) , a__isNatIList^#(zeros()) -> c_32() , a__isNatList^#(X) -> c_26(X) , a__isNatList^#(cons(V1, V2)) -> c_27(a__and^#(a__isNat(V1), isNatList(V2))) , a__isNatList^#(nil()) -> c_28() , a__isNat^#(X) -> c_22(X) , a__isNat^#(0()) -> c_23() , a__isNat^#(s(V1)) -> c_24(a__isNat^#(V1)) , a__isNat^#(length(V1)) -> c_25(a__isNatList^#(V1)) , a__and^#(X1, X2) -> c_20(X1, X2) , a__and^#(tt(), X) -> c_21(mark^#(X)) } and mark the set of starting terms. We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { a__zeros^#() -> c_1() , a__zeros^#() -> c_2() , a__U11^#(X1, X2) -> c_3(X1, X2) , a__U11^#(tt(), L) -> c_4(a__length^#(mark(L))) , a__length^#(X) -> c_5(X) , a__length^#(cons(N, L)) -> c_6(a__U11^#(a__and(a__isNatList(L), isNat(N)), L)) , a__length^#(nil()) -> c_7() , mark^#(cons(X1, X2)) -> c_8(mark^#(X1), X2) , mark^#(0()) -> c_9() , mark^#(zeros()) -> c_10(a__zeros^#()) , mark^#(tt()) -> c_11() , mark^#(s(X)) -> c_12(mark^#(X)) , mark^#(length(X)) -> c_13(a__length^#(mark(X))) , mark^#(isNatIList(X)) -> c_14(a__isNatIList^#(X)) , mark^#(nil()) -> c_15() , mark^#(isNatList(X)) -> c_16(a__isNatList^#(X)) , mark^#(isNat(X)) -> c_17(a__isNat^#(X)) , mark^#(U11(X1, X2)) -> c_18(a__U11^#(mark(X1), X2)) , mark^#(and(X1, X2)) -> c_19(a__and^#(mark(X1), X2)) , a__isNatIList^#(V) -> c_29(a__isNatList^#(V)) , a__isNatIList^#(X) -> c_30(X) , a__isNatIList^#(cons(V1, V2)) -> c_31(a__and^#(a__isNat(V1), isNatIList(V2))) , a__isNatIList^#(zeros()) -> c_32() , a__isNatList^#(X) -> c_26(X) , a__isNatList^#(cons(V1, V2)) -> c_27(a__and^#(a__isNat(V1), isNatList(V2))) , a__isNatList^#(nil()) -> c_28() , a__isNat^#(X) -> c_22(X) , a__isNat^#(0()) -> c_23() , a__isNat^#(s(V1)) -> c_24(a__isNat^#(V1)) , a__isNat^#(length(V1)) -> c_25(a__isNatList^#(V1)) , a__and^#(X1, X2) -> c_20(X1, X2) , a__and^#(tt(), X) -> c_21(mark^#(X)) } Strict Trs: { a__zeros() -> cons(0(), zeros()) , a__zeros() -> zeros() , a__U11(X1, X2) -> U11(X1, X2) , a__U11(tt(), L) -> s(a__length(mark(L))) , a__length(X) -> length(X) , a__length(cons(N, L)) -> a__U11(a__and(a__isNatList(L), isNat(N)), L) , a__length(nil()) -> 0() , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(0()) -> 0() , mark(zeros()) -> a__zeros() , mark(tt()) -> tt() , mark(s(X)) -> s(mark(X)) , mark(length(X)) -> a__length(mark(X)) , mark(isNatIList(X)) -> a__isNatIList(X) , mark(nil()) -> nil() , mark(isNatList(X)) -> a__isNatList(X) , mark(isNat(X)) -> a__isNat(X) , mark(U11(X1, X2)) -> a__U11(mark(X1), X2) , mark(and(X1, X2)) -> a__and(mark(X1), X2) , a__and(X1, X2) -> and(X1, X2) , a__and(tt(), X) -> mark(X) , a__isNat(X) -> isNat(X) , a__isNat(0()) -> tt() , a__isNat(s(V1)) -> a__isNat(V1) , a__isNat(length(V1)) -> a__isNatList(V1) , a__isNatList(X) -> isNatList(X) , a__isNatList(cons(V1, V2)) -> a__and(a__isNat(V1), isNatList(V2)) , a__isNatList(nil()) -> tt() , a__isNatIList(V) -> a__isNatList(V) , a__isNatIList(X) -> isNatIList(X) , a__isNatIList(cons(V1, V2)) -> a__and(a__isNat(V1), isNatIList(V2)) , a__isNatIList(zeros()) -> tt() } Obligation: runtime complexity Answer: MAYBE We estimate the number of application of {1,2,7,9,11,15,23,26,28} by applications of Pre({1,2,7,9,11,15,23,26,28}) = {3,4,5,8,10,12,13,14,16,17,20,21,24,27,29,30,31,32}. Here rules are labeled as follows: DPs: { 1: a__zeros^#() -> c_1() , 2: a__zeros^#() -> c_2() , 3: a__U11^#(X1, X2) -> c_3(X1, X2) , 4: a__U11^#(tt(), L) -> c_4(a__length^#(mark(L))) , 5: a__length^#(X) -> c_5(X) , 6: a__length^#(cons(N, L)) -> c_6(a__U11^#(a__and(a__isNatList(L), isNat(N)), L)) , 7: a__length^#(nil()) -> c_7() , 8: mark^#(cons(X1, X2)) -> c_8(mark^#(X1), X2) , 9: mark^#(0()) -> c_9() , 10: mark^#(zeros()) -> c_10(a__zeros^#()) , 11: mark^#(tt()) -> c_11() , 12: mark^#(s(X)) -> c_12(mark^#(X)) , 13: mark^#(length(X)) -> c_13(a__length^#(mark(X))) , 14: mark^#(isNatIList(X)) -> c_14(a__isNatIList^#(X)) , 15: mark^#(nil()) -> c_15() , 16: mark^#(isNatList(X)) -> c_16(a__isNatList^#(X)) , 17: mark^#(isNat(X)) -> c_17(a__isNat^#(X)) , 18: mark^#(U11(X1, X2)) -> c_18(a__U11^#(mark(X1), X2)) , 19: mark^#(and(X1, X2)) -> c_19(a__and^#(mark(X1), X2)) , 20: a__isNatIList^#(V) -> c_29(a__isNatList^#(V)) , 21: a__isNatIList^#(X) -> c_30(X) , 22: a__isNatIList^#(cons(V1, V2)) -> c_31(a__and^#(a__isNat(V1), isNatIList(V2))) , 23: a__isNatIList^#(zeros()) -> c_32() , 24: a__isNatList^#(X) -> c_26(X) , 25: a__isNatList^#(cons(V1, V2)) -> c_27(a__and^#(a__isNat(V1), isNatList(V2))) , 26: a__isNatList^#(nil()) -> c_28() , 27: a__isNat^#(X) -> c_22(X) , 28: a__isNat^#(0()) -> c_23() , 29: a__isNat^#(s(V1)) -> c_24(a__isNat^#(V1)) , 30: a__isNat^#(length(V1)) -> c_25(a__isNatList^#(V1)) , 31: a__and^#(X1, X2) -> c_20(X1, X2) , 32: a__and^#(tt(), X) -> c_21(mark^#(X)) } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { a__U11^#(X1, X2) -> c_3(X1, X2) , a__U11^#(tt(), L) -> c_4(a__length^#(mark(L))) , a__length^#(X) -> c_5(X) , a__length^#(cons(N, L)) -> c_6(a__U11^#(a__and(a__isNatList(L), isNat(N)), L)) , mark^#(cons(X1, X2)) -> c_8(mark^#(X1), X2) , mark^#(zeros()) -> c_10(a__zeros^#()) , mark^#(s(X)) -> c_12(mark^#(X)) , mark^#(length(X)) -> c_13(a__length^#(mark(X))) , mark^#(isNatIList(X)) -> c_14(a__isNatIList^#(X)) , mark^#(isNatList(X)) -> c_16(a__isNatList^#(X)) , mark^#(isNat(X)) -> c_17(a__isNat^#(X)) , mark^#(U11(X1, X2)) -> c_18(a__U11^#(mark(X1), X2)) , mark^#(and(X1, X2)) -> c_19(a__and^#(mark(X1), X2)) , a__isNatIList^#(V) -> c_29(a__isNatList^#(V)) , a__isNatIList^#(X) -> c_30(X) , a__isNatIList^#(cons(V1, V2)) -> c_31(a__and^#(a__isNat(V1), isNatIList(V2))) , a__isNatList^#(X) -> c_26(X) , a__isNatList^#(cons(V1, V2)) -> c_27(a__and^#(a__isNat(V1), isNatList(V2))) , a__isNat^#(X) -> c_22(X) , a__isNat^#(s(V1)) -> c_24(a__isNat^#(V1)) , a__isNat^#(length(V1)) -> c_25(a__isNatList^#(V1)) , a__and^#(X1, X2) -> c_20(X1, X2) , a__and^#(tt(), X) -> c_21(mark^#(X)) } Strict Trs: { a__zeros() -> cons(0(), zeros()) , a__zeros() -> zeros() , a__U11(X1, X2) -> U11(X1, X2) , a__U11(tt(), L) -> s(a__length(mark(L))) , a__length(X) -> length(X) , a__length(cons(N, L)) -> a__U11(a__and(a__isNatList(L), isNat(N)), L) , a__length(nil()) -> 0() , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(0()) -> 0() , mark(zeros()) -> a__zeros() , mark(tt()) -> tt() , mark(s(X)) -> s(mark(X)) , mark(length(X)) -> a__length(mark(X)) , mark(isNatIList(X)) -> a__isNatIList(X) , mark(nil()) -> nil() , mark(isNatList(X)) -> a__isNatList(X) , mark(isNat(X)) -> a__isNat(X) , mark(U11(X1, X2)) -> a__U11(mark(X1), X2) , mark(and(X1, X2)) -> a__and(mark(X1), X2) , a__and(X1, X2) -> and(X1, X2) , a__and(tt(), X) -> mark(X) , a__isNat(X) -> isNat(X) , a__isNat(0()) -> tt() , a__isNat(s(V1)) -> a__isNat(V1) , a__isNat(length(V1)) -> a__isNatList(V1) , a__isNatList(X) -> isNatList(X) , a__isNatList(cons(V1, V2)) -> a__and(a__isNat(V1), isNatList(V2)) , a__isNatList(nil()) -> tt() , a__isNatIList(V) -> a__isNatList(V) , a__isNatIList(X) -> isNatIList(X) , a__isNatIList(cons(V1, V2)) -> a__and(a__isNat(V1), isNatIList(V2)) , a__isNatIList(zeros()) -> tt() } Weak DPs: { a__zeros^#() -> c_1() , a__zeros^#() -> c_2() , a__length^#(nil()) -> c_7() , mark^#(0()) -> c_9() , mark^#(tt()) -> c_11() , mark^#(nil()) -> c_15() , a__isNatIList^#(zeros()) -> c_32() , a__isNatList^#(nil()) -> c_28() , a__isNat^#(0()) -> c_23() } Obligation: runtime complexity Answer: MAYBE We estimate the number of application of {6} by applications of Pre({6}) = {1,3,5,7,15,17,19,22,23}. Here rules are labeled as follows: DPs: { 1: a__U11^#(X1, X2) -> c_3(X1, X2) , 2: a__U11^#(tt(), L) -> c_4(a__length^#(mark(L))) , 3: a__length^#(X) -> c_5(X) , 4: a__length^#(cons(N, L)) -> c_6(a__U11^#(a__and(a__isNatList(L), isNat(N)), L)) , 5: mark^#(cons(X1, X2)) -> c_8(mark^#(X1), X2) , 6: mark^#(zeros()) -> c_10(a__zeros^#()) , 7: mark^#(s(X)) -> c_12(mark^#(X)) , 8: mark^#(length(X)) -> c_13(a__length^#(mark(X))) , 9: mark^#(isNatIList(X)) -> c_14(a__isNatIList^#(X)) , 10: mark^#(isNatList(X)) -> c_16(a__isNatList^#(X)) , 11: mark^#(isNat(X)) -> c_17(a__isNat^#(X)) , 12: mark^#(U11(X1, X2)) -> c_18(a__U11^#(mark(X1), X2)) , 13: mark^#(and(X1, X2)) -> c_19(a__and^#(mark(X1), X2)) , 14: a__isNatIList^#(V) -> c_29(a__isNatList^#(V)) , 15: a__isNatIList^#(X) -> c_30(X) , 16: a__isNatIList^#(cons(V1, V2)) -> c_31(a__and^#(a__isNat(V1), isNatIList(V2))) , 17: a__isNatList^#(X) -> c_26(X) , 18: a__isNatList^#(cons(V1, V2)) -> c_27(a__and^#(a__isNat(V1), isNatList(V2))) , 19: a__isNat^#(X) -> c_22(X) , 20: a__isNat^#(s(V1)) -> c_24(a__isNat^#(V1)) , 21: a__isNat^#(length(V1)) -> c_25(a__isNatList^#(V1)) , 22: a__and^#(X1, X2) -> c_20(X1, X2) , 23: a__and^#(tt(), X) -> c_21(mark^#(X)) , 24: a__zeros^#() -> c_1() , 25: a__zeros^#() -> c_2() , 26: a__length^#(nil()) -> c_7() , 27: mark^#(0()) -> c_9() , 28: mark^#(tt()) -> c_11() , 29: mark^#(nil()) -> c_15() , 30: a__isNatIList^#(zeros()) -> c_32() , 31: a__isNatList^#(nil()) -> c_28() , 32: a__isNat^#(0()) -> c_23() } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { a__U11^#(X1, X2) -> c_3(X1, X2) , a__U11^#(tt(), L) -> c_4(a__length^#(mark(L))) , a__length^#(X) -> c_5(X) , a__length^#(cons(N, L)) -> c_6(a__U11^#(a__and(a__isNatList(L), isNat(N)), L)) , mark^#(cons(X1, X2)) -> c_8(mark^#(X1), X2) , mark^#(s(X)) -> c_12(mark^#(X)) , mark^#(length(X)) -> c_13(a__length^#(mark(X))) , mark^#(isNatIList(X)) -> c_14(a__isNatIList^#(X)) , mark^#(isNatList(X)) -> c_16(a__isNatList^#(X)) , mark^#(isNat(X)) -> c_17(a__isNat^#(X)) , mark^#(U11(X1, X2)) -> c_18(a__U11^#(mark(X1), X2)) , mark^#(and(X1, X2)) -> c_19(a__and^#(mark(X1), X2)) , a__isNatIList^#(V) -> c_29(a__isNatList^#(V)) , a__isNatIList^#(X) -> c_30(X) , a__isNatIList^#(cons(V1, V2)) -> c_31(a__and^#(a__isNat(V1), isNatIList(V2))) , a__isNatList^#(X) -> c_26(X) , a__isNatList^#(cons(V1, V2)) -> c_27(a__and^#(a__isNat(V1), isNatList(V2))) , a__isNat^#(X) -> c_22(X) , a__isNat^#(s(V1)) -> c_24(a__isNat^#(V1)) , a__isNat^#(length(V1)) -> c_25(a__isNatList^#(V1)) , a__and^#(X1, X2) -> c_20(X1, X2) , a__and^#(tt(), X) -> c_21(mark^#(X)) } Strict Trs: { a__zeros() -> cons(0(), zeros()) , a__zeros() -> zeros() , a__U11(X1, X2) -> U11(X1, X2) , a__U11(tt(), L) -> s(a__length(mark(L))) , a__length(X) -> length(X) , a__length(cons(N, L)) -> a__U11(a__and(a__isNatList(L), isNat(N)), L) , a__length(nil()) -> 0() , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(0()) -> 0() , mark(zeros()) -> a__zeros() , mark(tt()) -> tt() , mark(s(X)) -> s(mark(X)) , mark(length(X)) -> a__length(mark(X)) , mark(isNatIList(X)) -> a__isNatIList(X) , mark(nil()) -> nil() , mark(isNatList(X)) -> a__isNatList(X) , mark(isNat(X)) -> a__isNat(X) , mark(U11(X1, X2)) -> a__U11(mark(X1), X2) , mark(and(X1, X2)) -> a__and(mark(X1), X2) , a__and(X1, X2) -> and(X1, X2) , a__and(tt(), X) -> mark(X) , a__isNat(X) -> isNat(X) , a__isNat(0()) -> tt() , a__isNat(s(V1)) -> a__isNat(V1) , a__isNat(length(V1)) -> a__isNatList(V1) , a__isNatList(X) -> isNatList(X) , a__isNatList(cons(V1, V2)) -> a__and(a__isNat(V1), isNatList(V2)) , a__isNatList(nil()) -> tt() , a__isNatIList(V) -> a__isNatList(V) , a__isNatIList(X) -> isNatIList(X) , a__isNatIList(cons(V1, V2)) -> a__and(a__isNat(V1), isNatIList(V2)) , a__isNatIList(zeros()) -> tt() } Weak DPs: { a__zeros^#() -> c_1() , a__zeros^#() -> c_2() , a__length^#(nil()) -> c_7() , mark^#(0()) -> c_9() , mark^#(zeros()) -> c_10(a__zeros^#()) , mark^#(tt()) -> c_11() , mark^#(nil()) -> c_15() , a__isNatIList^#(zeros()) -> c_32() , a__isNatList^#(nil()) -> c_28() , a__isNat^#(0()) -> c_23() } Obligation: runtime complexity Answer: MAYBE Empty strict component of the problem is NOT empty. Arrrr..