MAYBE We are left with following problem, upon which TcT provides the certificate MAYBE. Strict Trs: { a__dbl(X) -> dbl(X) , a__dbl(0()) -> 0() , a__dbl(s(X)) -> s(s(dbl(X))) , a__dbls(X) -> dbls(X) , a__dbls(nil()) -> nil() , a__dbls(cons(X, Y)) -> cons(dbl(X), dbls(Y)) , a__sel(X1, X2) -> sel(X1, X2) , a__sel(0(), cons(X, Y)) -> mark(X) , a__sel(s(X), cons(Y, Z)) -> a__sel(mark(X), mark(Z)) , mark(0()) -> 0() , mark(s(X)) -> s(X) , mark(dbl(X)) -> a__dbl(mark(X)) , mark(nil()) -> nil() , mark(cons(X1, X2)) -> cons(X1, X2) , mark(dbls(X)) -> a__dbls(mark(X)) , mark(sel(X1, X2)) -> a__sel(mark(X1), mark(X2)) , mark(indx(X1, X2)) -> a__indx(mark(X1), X2) , mark(from(X)) -> a__from(X) , a__indx(X1, X2) -> indx(X1, X2) , a__indx(nil(), X) -> nil() , a__indx(cons(X, Y), Z) -> cons(sel(X, Z), indx(Y, Z)) , a__from(X) -> cons(X, from(s(X))) , a__from(X) -> from(X) } 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 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) 'Innermost Weak Dependency Pairs (timeout of 60 seconds)' failed due to the following reason: We add the following weak dependency pairs: Strict DPs: { a__dbl^#(X) -> c_1(X) , a__dbl^#(0()) -> c_2() , a__dbl^#(s(X)) -> c_3(X) , a__dbls^#(X) -> c_4(X) , a__dbls^#(nil()) -> c_5() , a__dbls^#(cons(X, Y)) -> c_6(X, Y) , a__sel^#(X1, X2) -> c_7(X1, X2) , a__sel^#(0(), cons(X, Y)) -> c_8(mark^#(X)) , a__sel^#(s(X), cons(Y, Z)) -> c_9(a__sel^#(mark(X), mark(Z))) , mark^#(0()) -> c_10() , mark^#(s(X)) -> c_11(X) , mark^#(dbl(X)) -> c_12(a__dbl^#(mark(X))) , mark^#(nil()) -> c_13() , mark^#(cons(X1, X2)) -> c_14(X1, X2) , mark^#(dbls(X)) -> c_15(a__dbls^#(mark(X))) , mark^#(sel(X1, X2)) -> c_16(a__sel^#(mark(X1), mark(X2))) , mark^#(indx(X1, X2)) -> c_17(a__indx^#(mark(X1), X2)) , mark^#(from(X)) -> c_18(a__from^#(X)) , a__indx^#(X1, X2) -> c_19(X1, X2) , a__indx^#(nil(), X) -> c_20() , a__indx^#(cons(X, Y), Z) -> c_21(X, Z, Y, Z) , a__from^#(X) -> c_22(X, X) , a__from^#(X) -> c_23(X) } and mark the set of starting terms. We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { a__dbl^#(X) -> c_1(X) , a__dbl^#(0()) -> c_2() , a__dbl^#(s(X)) -> c_3(X) , a__dbls^#(X) -> c_4(X) , a__dbls^#(nil()) -> c_5() , a__dbls^#(cons(X, Y)) -> c_6(X, Y) , a__sel^#(X1, X2) -> c_7(X1, X2) , a__sel^#(0(), cons(X, Y)) -> c_8(mark^#(X)) , a__sel^#(s(X), cons(Y, Z)) -> c_9(a__sel^#(mark(X), mark(Z))) , mark^#(0()) -> c_10() , mark^#(s(X)) -> c_11(X) , mark^#(dbl(X)) -> c_12(a__dbl^#(mark(X))) , mark^#(nil()) -> c_13() , mark^#(cons(X1, X2)) -> c_14(X1, X2) , mark^#(dbls(X)) -> c_15(a__dbls^#(mark(X))) , mark^#(sel(X1, X2)) -> c_16(a__sel^#(mark(X1), mark(X2))) , mark^#(indx(X1, X2)) -> c_17(a__indx^#(mark(X1), X2)) , mark^#(from(X)) -> c_18(a__from^#(X)) , a__indx^#(X1, X2) -> c_19(X1, X2) , a__indx^#(nil(), X) -> c_20() , a__indx^#(cons(X, Y), Z) -> c_21(X, Z, Y, Z) , a__from^#(X) -> c_22(X, X) , a__from^#(X) -> c_23(X) } Strict Trs: { a__dbl(X) -> dbl(X) , a__dbl(0()) -> 0() , a__dbl(s(X)) -> s(s(dbl(X))) , a__dbls(X) -> dbls(X) , a__dbls(nil()) -> nil() , a__dbls(cons(X, Y)) -> cons(dbl(X), dbls(Y)) , a__sel(X1, X2) -> sel(X1, X2) , a__sel(0(), cons(X, Y)) -> mark(X) , a__sel(s(X), cons(Y, Z)) -> a__sel(mark(X), mark(Z)) , mark(0()) -> 0() , mark(s(X)) -> s(X) , mark(dbl(X)) -> a__dbl(mark(X)) , mark(nil()) -> nil() , mark(cons(X1, X2)) -> cons(X1, X2) , mark(dbls(X)) -> a__dbls(mark(X)) , mark(sel(X1, X2)) -> a__sel(mark(X1), mark(X2)) , mark(indx(X1, X2)) -> a__indx(mark(X1), X2) , mark(from(X)) -> a__from(X) , a__indx(X1, X2) -> indx(X1, X2) , a__indx(nil(), X) -> nil() , a__indx(cons(X, Y), Z) -> cons(sel(X, Z), indx(Y, Z)) , a__from(X) -> cons(X, from(s(X))) , a__from(X) -> from(X) } Obligation: runtime complexity Answer: MAYBE We estimate the number of application of {2,5,10,13,20} by applications of Pre({2,5,10,13,20}) = {1,3,4,6,7,8,11,12,14,15,17,19,21,22,23}. Here rules are labeled as follows: DPs: { 1: a__dbl^#(X) -> c_1(X) , 2: a__dbl^#(0()) -> c_2() , 3: a__dbl^#(s(X)) -> c_3(X) , 4: a__dbls^#(X) -> c_4(X) , 5: a__dbls^#(nil()) -> c_5() , 6: a__dbls^#(cons(X, Y)) -> c_6(X, Y) , 7: a__sel^#(X1, X2) -> c_7(X1, X2) , 8: a__sel^#(0(), cons(X, Y)) -> c_8(mark^#(X)) , 9: a__sel^#(s(X), cons(Y, Z)) -> c_9(a__sel^#(mark(X), mark(Z))) , 10: mark^#(0()) -> c_10() , 11: mark^#(s(X)) -> c_11(X) , 12: mark^#(dbl(X)) -> c_12(a__dbl^#(mark(X))) , 13: mark^#(nil()) -> c_13() , 14: mark^#(cons(X1, X2)) -> c_14(X1, X2) , 15: mark^#(dbls(X)) -> c_15(a__dbls^#(mark(X))) , 16: mark^#(sel(X1, X2)) -> c_16(a__sel^#(mark(X1), mark(X2))) , 17: mark^#(indx(X1, X2)) -> c_17(a__indx^#(mark(X1), X2)) , 18: mark^#(from(X)) -> c_18(a__from^#(X)) , 19: a__indx^#(X1, X2) -> c_19(X1, X2) , 20: a__indx^#(nil(), X) -> c_20() , 21: a__indx^#(cons(X, Y), Z) -> c_21(X, Z, Y, Z) , 22: a__from^#(X) -> c_22(X, X) , 23: a__from^#(X) -> c_23(X) } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { a__dbl^#(X) -> c_1(X) , a__dbl^#(s(X)) -> c_3(X) , a__dbls^#(X) -> c_4(X) , a__dbls^#(cons(X, Y)) -> c_6(X, Y) , a__sel^#(X1, X2) -> c_7(X1, X2) , a__sel^#(0(), cons(X, Y)) -> c_8(mark^#(X)) , a__sel^#(s(X), cons(Y, Z)) -> c_9(a__sel^#(mark(X), mark(Z))) , mark^#(s(X)) -> c_11(X) , mark^#(dbl(X)) -> c_12(a__dbl^#(mark(X))) , mark^#(cons(X1, X2)) -> c_14(X1, X2) , mark^#(dbls(X)) -> c_15(a__dbls^#(mark(X))) , mark^#(sel(X1, X2)) -> c_16(a__sel^#(mark(X1), mark(X2))) , mark^#(indx(X1, X2)) -> c_17(a__indx^#(mark(X1), X2)) , mark^#(from(X)) -> c_18(a__from^#(X)) , a__indx^#(X1, X2) -> c_19(X1, X2) , a__indx^#(cons(X, Y), Z) -> c_21(X, Z, Y, Z) , a__from^#(X) -> c_22(X, X) , a__from^#(X) -> c_23(X) } Strict Trs: { a__dbl(X) -> dbl(X) , a__dbl(0()) -> 0() , a__dbl(s(X)) -> s(s(dbl(X))) , a__dbls(X) -> dbls(X) , a__dbls(nil()) -> nil() , a__dbls(cons(X, Y)) -> cons(dbl(X), dbls(Y)) , a__sel(X1, X2) -> sel(X1, X2) , a__sel(0(), cons(X, Y)) -> mark(X) , a__sel(s(X), cons(Y, Z)) -> a__sel(mark(X), mark(Z)) , mark(0()) -> 0() , mark(s(X)) -> s(X) , mark(dbl(X)) -> a__dbl(mark(X)) , mark(nil()) -> nil() , mark(cons(X1, X2)) -> cons(X1, X2) , mark(dbls(X)) -> a__dbls(mark(X)) , mark(sel(X1, X2)) -> a__sel(mark(X1), mark(X2)) , mark(indx(X1, X2)) -> a__indx(mark(X1), X2) , mark(from(X)) -> a__from(X) , a__indx(X1, X2) -> indx(X1, X2) , a__indx(nil(), X) -> nil() , a__indx(cons(X, Y), Z) -> cons(sel(X, Z), indx(Y, Z)) , a__from(X) -> cons(X, from(s(X))) , a__from(X) -> from(X) } Weak DPs: { a__dbl^#(0()) -> c_2() , a__dbls^#(nil()) -> c_5() , mark^#(0()) -> c_10() , mark^#(nil()) -> c_13() , a__indx^#(nil(), X) -> c_20() } Obligation: runtime complexity Answer: MAYBE Empty strict component of the problem is NOT empty. Arrrr..