MAYBE We are left with following problem, upon which TcT provides the certificate MAYBE. Strict Trs: { sel(s(X), cons(Y, Z)) -> sel(X, Z) , sel(0(), cons(X, Z)) -> X , first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z)) , first(0(), Z) -> nil() , from(X) -> cons(X, from(s(X))) , sel1(s(X), cons(Y, Z)) -> sel1(X, Z) , sel1(0(), cons(X, Z)) -> quote(X) , quote(sel(X, Z)) -> sel1(X, Z) , quote(s(X)) -> s1(quote(X)) , quote(0()) -> 01() , first1(s(X), cons(Y, Z)) -> cons1(quote(Y), first1(X, Z)) , first1(0(), Z) -> nil1() , quote1(cons(X, Z)) -> cons1(quote(X), quote1(Z)) , quote1(first(X, Z)) -> first1(X, Z) , quote1(nil()) -> nil1() , unquote(01()) -> 0() , unquote(s1(X)) -> s(unquote(X)) , unquote1(nil1()) -> nil() , unquote1(cons1(X, Z)) -> fcons(unquote(X), unquote1(Z)) , fcons(X, Z) -> cons(X, Z) } 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: { sel^#(s(X), cons(Y, Z)) -> c_1(sel^#(X, Z)) , sel^#(0(), cons(X, Z)) -> c_2(X) , first^#(s(X), cons(Y, Z)) -> c_3(Y, first^#(X, Z)) , first^#(0(), Z) -> c_4() , from^#(X) -> c_5(X, from^#(s(X))) , sel1^#(s(X), cons(Y, Z)) -> c_6(sel1^#(X, Z)) , sel1^#(0(), cons(X, Z)) -> c_7(quote^#(X)) , quote^#(sel(X, Z)) -> c_8(sel1^#(X, Z)) , quote^#(s(X)) -> c_9(quote^#(X)) , quote^#(0()) -> c_10() , first1^#(s(X), cons(Y, Z)) -> c_11(quote^#(Y), first1^#(X, Z)) , first1^#(0(), Z) -> c_12() , quote1^#(cons(X, Z)) -> c_13(quote^#(X), quote1^#(Z)) , quote1^#(first(X, Z)) -> c_14(first1^#(X, Z)) , quote1^#(nil()) -> c_15() , unquote^#(01()) -> c_16() , unquote^#(s1(X)) -> c_17(unquote^#(X)) , unquote1^#(nil1()) -> c_18() , unquote1^#(cons1(X, Z)) -> c_19(fcons^#(unquote(X), unquote1(Z))) , fcons^#(X, Z) -> c_20(X, Z) } and mark the set of starting terms. We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { sel^#(s(X), cons(Y, Z)) -> c_1(sel^#(X, Z)) , sel^#(0(), cons(X, Z)) -> c_2(X) , first^#(s(X), cons(Y, Z)) -> c_3(Y, first^#(X, Z)) , first^#(0(), Z) -> c_4() , from^#(X) -> c_5(X, from^#(s(X))) , sel1^#(s(X), cons(Y, Z)) -> c_6(sel1^#(X, Z)) , sel1^#(0(), cons(X, Z)) -> c_7(quote^#(X)) , quote^#(sel(X, Z)) -> c_8(sel1^#(X, Z)) , quote^#(s(X)) -> c_9(quote^#(X)) , quote^#(0()) -> c_10() , first1^#(s(X), cons(Y, Z)) -> c_11(quote^#(Y), first1^#(X, Z)) , first1^#(0(), Z) -> c_12() , quote1^#(cons(X, Z)) -> c_13(quote^#(X), quote1^#(Z)) , quote1^#(first(X, Z)) -> c_14(first1^#(X, Z)) , quote1^#(nil()) -> c_15() , unquote^#(01()) -> c_16() , unquote^#(s1(X)) -> c_17(unquote^#(X)) , unquote1^#(nil1()) -> c_18() , unquote1^#(cons1(X, Z)) -> c_19(fcons^#(unquote(X), unquote1(Z))) , fcons^#(X, Z) -> c_20(X, Z) } Strict Trs: { sel(s(X), cons(Y, Z)) -> sel(X, Z) , sel(0(), cons(X, Z)) -> X , first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z)) , first(0(), Z) -> nil() , from(X) -> cons(X, from(s(X))) , sel1(s(X), cons(Y, Z)) -> sel1(X, Z) , sel1(0(), cons(X, Z)) -> quote(X) , quote(sel(X, Z)) -> sel1(X, Z) , quote(s(X)) -> s1(quote(X)) , quote(0()) -> 01() , first1(s(X), cons(Y, Z)) -> cons1(quote(Y), first1(X, Z)) , first1(0(), Z) -> nil1() , quote1(cons(X, Z)) -> cons1(quote(X), quote1(Z)) , quote1(first(X, Z)) -> first1(X, Z) , quote1(nil()) -> nil1() , unquote(01()) -> 0() , unquote(s1(X)) -> s(unquote(X)) , unquote1(nil1()) -> nil() , unquote1(cons1(X, Z)) -> fcons(unquote(X), unquote1(Z)) , fcons(X, Z) -> cons(X, Z) } Obligation: runtime complexity Answer: MAYBE We estimate the number of application of {4,10,12,15,16,18} by applications of Pre({4,10,12,15,16,18}) = {2,3,5,7,9,11,13,14,17,20}. Here rules are labeled as follows: DPs: { 1: sel^#(s(X), cons(Y, Z)) -> c_1(sel^#(X, Z)) , 2: sel^#(0(), cons(X, Z)) -> c_2(X) , 3: first^#(s(X), cons(Y, Z)) -> c_3(Y, first^#(X, Z)) , 4: first^#(0(), Z) -> c_4() , 5: from^#(X) -> c_5(X, from^#(s(X))) , 6: sel1^#(s(X), cons(Y, Z)) -> c_6(sel1^#(X, Z)) , 7: sel1^#(0(), cons(X, Z)) -> c_7(quote^#(X)) , 8: quote^#(sel(X, Z)) -> c_8(sel1^#(X, Z)) , 9: quote^#(s(X)) -> c_9(quote^#(X)) , 10: quote^#(0()) -> c_10() , 11: first1^#(s(X), cons(Y, Z)) -> c_11(quote^#(Y), first1^#(X, Z)) , 12: first1^#(0(), Z) -> c_12() , 13: quote1^#(cons(X, Z)) -> c_13(quote^#(X), quote1^#(Z)) , 14: quote1^#(first(X, Z)) -> c_14(first1^#(X, Z)) , 15: quote1^#(nil()) -> c_15() , 16: unquote^#(01()) -> c_16() , 17: unquote^#(s1(X)) -> c_17(unquote^#(X)) , 18: unquote1^#(nil1()) -> c_18() , 19: unquote1^#(cons1(X, Z)) -> c_19(fcons^#(unquote(X), unquote1(Z))) , 20: fcons^#(X, Z) -> c_20(X, Z) } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { sel^#(s(X), cons(Y, Z)) -> c_1(sel^#(X, Z)) , sel^#(0(), cons(X, Z)) -> c_2(X) , first^#(s(X), cons(Y, Z)) -> c_3(Y, first^#(X, Z)) , from^#(X) -> c_5(X, from^#(s(X))) , sel1^#(s(X), cons(Y, Z)) -> c_6(sel1^#(X, Z)) , sel1^#(0(), cons(X, Z)) -> c_7(quote^#(X)) , quote^#(sel(X, Z)) -> c_8(sel1^#(X, Z)) , quote^#(s(X)) -> c_9(quote^#(X)) , first1^#(s(X), cons(Y, Z)) -> c_11(quote^#(Y), first1^#(X, Z)) , quote1^#(cons(X, Z)) -> c_13(quote^#(X), quote1^#(Z)) , quote1^#(first(X, Z)) -> c_14(first1^#(X, Z)) , unquote^#(s1(X)) -> c_17(unquote^#(X)) , unquote1^#(cons1(X, Z)) -> c_19(fcons^#(unquote(X), unquote1(Z))) , fcons^#(X, Z) -> c_20(X, Z) } Strict Trs: { sel(s(X), cons(Y, Z)) -> sel(X, Z) , sel(0(), cons(X, Z)) -> X , first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z)) , first(0(), Z) -> nil() , from(X) -> cons(X, from(s(X))) , sel1(s(X), cons(Y, Z)) -> sel1(X, Z) , sel1(0(), cons(X, Z)) -> quote(X) , quote(sel(X, Z)) -> sel1(X, Z) , quote(s(X)) -> s1(quote(X)) , quote(0()) -> 01() , first1(s(X), cons(Y, Z)) -> cons1(quote(Y), first1(X, Z)) , first1(0(), Z) -> nil1() , quote1(cons(X, Z)) -> cons1(quote(X), quote1(Z)) , quote1(first(X, Z)) -> first1(X, Z) , quote1(nil()) -> nil1() , unquote(01()) -> 0() , unquote(s1(X)) -> s(unquote(X)) , unquote1(nil1()) -> nil() , unquote1(cons1(X, Z)) -> fcons(unquote(X), unquote1(Z)) , fcons(X, Z) -> cons(X, Z) } Weak DPs: { first^#(0(), Z) -> c_4() , quote^#(0()) -> c_10() , first1^#(0(), Z) -> c_12() , quote1^#(nil()) -> c_15() , unquote^#(01()) -> c_16() , unquote1^#(nil1()) -> c_18() } Obligation: runtime complexity Answer: MAYBE Empty strict component of the problem is NOT empty. Arrrr..