MAYBE We are left with following problem, upon which TcT provides the certificate MAYBE. Strict Trs: { a__terms(N) -> cons(recip(a__sqr(mark(N))), terms(s(N))) , a__terms(X) -> terms(X) , a__sqr(X) -> sqr(X) , a__sqr(s(X)) -> s(add(sqr(X), dbl(X))) , a__sqr(0()) -> 0() , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(recip(X)) -> recip(mark(X)) , mark(terms(X)) -> a__terms(mark(X)) , mark(s(X)) -> s(X) , mark(0()) -> 0() , mark(add(X1, X2)) -> a__add(mark(X1), mark(X2)) , mark(sqr(X)) -> a__sqr(mark(X)) , mark(dbl(X)) -> a__dbl(mark(X)) , mark(nil()) -> nil() , mark(first(X1, X2)) -> a__first(mark(X1), mark(X2)) , a__dbl(X) -> dbl(X) , a__dbl(s(X)) -> s(s(dbl(X))) , a__dbl(0()) -> 0() , a__add(X1, X2) -> add(X1, X2) , a__add(s(X), Y) -> s(add(X, Y)) , a__add(0(), X) -> mark(X) , a__first(X1, X2) -> first(X1, X2) , a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z)) , a__first(0(), X) -> nil() } 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__terms^#(N) -> c_1(a__sqr^#(mark(N)), N) , a__terms^#(X) -> c_2(X) , a__sqr^#(X) -> c_3(X) , a__sqr^#(s(X)) -> c_4(X, X) , a__sqr^#(0()) -> c_5() , mark^#(cons(X1, X2)) -> c_6(mark^#(X1), X2) , mark^#(recip(X)) -> c_7(mark^#(X)) , mark^#(terms(X)) -> c_8(a__terms^#(mark(X))) , mark^#(s(X)) -> c_9(X) , mark^#(0()) -> c_10() , mark^#(add(X1, X2)) -> c_11(a__add^#(mark(X1), mark(X2))) , mark^#(sqr(X)) -> c_12(a__sqr^#(mark(X))) , mark^#(dbl(X)) -> c_13(a__dbl^#(mark(X))) , mark^#(nil()) -> c_14() , mark^#(first(X1, X2)) -> c_15(a__first^#(mark(X1), mark(X2))) , a__add^#(X1, X2) -> c_19(X1, X2) , a__add^#(s(X), Y) -> c_20(X, Y) , a__add^#(0(), X) -> c_21(mark^#(X)) , a__dbl^#(X) -> c_16(X) , a__dbl^#(s(X)) -> c_17(X) , a__dbl^#(0()) -> c_18() , a__first^#(X1, X2) -> c_22(X1, X2) , a__first^#(s(X), cons(Y, Z)) -> c_23(mark^#(Y), X, Z) , a__first^#(0(), X) -> c_24() } and mark the set of starting terms. We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { a__terms^#(N) -> c_1(a__sqr^#(mark(N)), N) , a__terms^#(X) -> c_2(X) , a__sqr^#(X) -> c_3(X) , a__sqr^#(s(X)) -> c_4(X, X) , a__sqr^#(0()) -> c_5() , mark^#(cons(X1, X2)) -> c_6(mark^#(X1), X2) , mark^#(recip(X)) -> c_7(mark^#(X)) , mark^#(terms(X)) -> c_8(a__terms^#(mark(X))) , mark^#(s(X)) -> c_9(X) , mark^#(0()) -> c_10() , mark^#(add(X1, X2)) -> c_11(a__add^#(mark(X1), mark(X2))) , mark^#(sqr(X)) -> c_12(a__sqr^#(mark(X))) , mark^#(dbl(X)) -> c_13(a__dbl^#(mark(X))) , mark^#(nil()) -> c_14() , mark^#(first(X1, X2)) -> c_15(a__first^#(mark(X1), mark(X2))) , a__add^#(X1, X2) -> c_19(X1, X2) , a__add^#(s(X), Y) -> c_20(X, Y) , a__add^#(0(), X) -> c_21(mark^#(X)) , a__dbl^#(X) -> c_16(X) , a__dbl^#(s(X)) -> c_17(X) , a__dbl^#(0()) -> c_18() , a__first^#(X1, X2) -> c_22(X1, X2) , a__first^#(s(X), cons(Y, Z)) -> c_23(mark^#(Y), X, Z) , a__first^#(0(), X) -> c_24() } Strict Trs: { a__terms(N) -> cons(recip(a__sqr(mark(N))), terms(s(N))) , a__terms(X) -> terms(X) , a__sqr(X) -> sqr(X) , a__sqr(s(X)) -> s(add(sqr(X), dbl(X))) , a__sqr(0()) -> 0() , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(recip(X)) -> recip(mark(X)) , mark(terms(X)) -> a__terms(mark(X)) , mark(s(X)) -> s(X) , mark(0()) -> 0() , mark(add(X1, X2)) -> a__add(mark(X1), mark(X2)) , mark(sqr(X)) -> a__sqr(mark(X)) , mark(dbl(X)) -> a__dbl(mark(X)) , mark(nil()) -> nil() , mark(first(X1, X2)) -> a__first(mark(X1), mark(X2)) , a__dbl(X) -> dbl(X) , a__dbl(s(X)) -> s(s(dbl(X))) , a__dbl(0()) -> 0() , a__add(X1, X2) -> add(X1, X2) , a__add(s(X), Y) -> s(add(X, Y)) , a__add(0(), X) -> mark(X) , a__first(X1, X2) -> first(X1, X2) , a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z)) , a__first(0(), X) -> nil() } Obligation: runtime complexity Answer: MAYBE We estimate the number of application of {5,10,14,21,24} by applications of Pre({5,10,14,21,24}) = {1,2,3,4,6,7,9,12,13,15,16,17,18,19,20,22,23}. Here rules are labeled as follows: DPs: { 1: a__terms^#(N) -> c_1(a__sqr^#(mark(N)), N) , 2: a__terms^#(X) -> c_2(X) , 3: a__sqr^#(X) -> c_3(X) , 4: a__sqr^#(s(X)) -> c_4(X, X) , 5: a__sqr^#(0()) -> c_5() , 6: mark^#(cons(X1, X2)) -> c_6(mark^#(X1), X2) , 7: mark^#(recip(X)) -> c_7(mark^#(X)) , 8: mark^#(terms(X)) -> c_8(a__terms^#(mark(X))) , 9: mark^#(s(X)) -> c_9(X) , 10: mark^#(0()) -> c_10() , 11: mark^#(add(X1, X2)) -> c_11(a__add^#(mark(X1), mark(X2))) , 12: mark^#(sqr(X)) -> c_12(a__sqr^#(mark(X))) , 13: mark^#(dbl(X)) -> c_13(a__dbl^#(mark(X))) , 14: mark^#(nil()) -> c_14() , 15: mark^#(first(X1, X2)) -> c_15(a__first^#(mark(X1), mark(X2))) , 16: a__add^#(X1, X2) -> c_19(X1, X2) , 17: a__add^#(s(X), Y) -> c_20(X, Y) , 18: a__add^#(0(), X) -> c_21(mark^#(X)) , 19: a__dbl^#(X) -> c_16(X) , 20: a__dbl^#(s(X)) -> c_17(X) , 21: a__dbl^#(0()) -> c_18() , 22: a__first^#(X1, X2) -> c_22(X1, X2) , 23: a__first^#(s(X), cons(Y, Z)) -> c_23(mark^#(Y), X, Z) , 24: a__first^#(0(), X) -> c_24() } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { a__terms^#(N) -> c_1(a__sqr^#(mark(N)), N) , a__terms^#(X) -> c_2(X) , a__sqr^#(X) -> c_3(X) , a__sqr^#(s(X)) -> c_4(X, X) , mark^#(cons(X1, X2)) -> c_6(mark^#(X1), X2) , mark^#(recip(X)) -> c_7(mark^#(X)) , mark^#(terms(X)) -> c_8(a__terms^#(mark(X))) , mark^#(s(X)) -> c_9(X) , mark^#(add(X1, X2)) -> c_11(a__add^#(mark(X1), mark(X2))) , mark^#(sqr(X)) -> c_12(a__sqr^#(mark(X))) , mark^#(dbl(X)) -> c_13(a__dbl^#(mark(X))) , mark^#(first(X1, X2)) -> c_15(a__first^#(mark(X1), mark(X2))) , a__add^#(X1, X2) -> c_19(X1, X2) , a__add^#(s(X), Y) -> c_20(X, Y) , a__add^#(0(), X) -> c_21(mark^#(X)) , a__dbl^#(X) -> c_16(X) , a__dbl^#(s(X)) -> c_17(X) , a__first^#(X1, X2) -> c_22(X1, X2) , a__first^#(s(X), cons(Y, Z)) -> c_23(mark^#(Y), X, Z) } Strict Trs: { a__terms(N) -> cons(recip(a__sqr(mark(N))), terms(s(N))) , a__terms(X) -> terms(X) , a__sqr(X) -> sqr(X) , a__sqr(s(X)) -> s(add(sqr(X), dbl(X))) , a__sqr(0()) -> 0() , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(recip(X)) -> recip(mark(X)) , mark(terms(X)) -> a__terms(mark(X)) , mark(s(X)) -> s(X) , mark(0()) -> 0() , mark(add(X1, X2)) -> a__add(mark(X1), mark(X2)) , mark(sqr(X)) -> a__sqr(mark(X)) , mark(dbl(X)) -> a__dbl(mark(X)) , mark(nil()) -> nil() , mark(first(X1, X2)) -> a__first(mark(X1), mark(X2)) , a__dbl(X) -> dbl(X) , a__dbl(s(X)) -> s(s(dbl(X))) , a__dbl(0()) -> 0() , a__add(X1, X2) -> add(X1, X2) , a__add(s(X), Y) -> s(add(X, Y)) , a__add(0(), X) -> mark(X) , a__first(X1, X2) -> first(X1, X2) , a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z)) , a__first(0(), X) -> nil() } Weak DPs: { a__sqr^#(0()) -> c_5() , mark^#(0()) -> c_10() , mark^#(nil()) -> c_14() , a__dbl^#(0()) -> c_18() , a__first^#(0(), X) -> c_24() } Obligation: runtime complexity Answer: MAYBE Empty strict component of the problem is NOT empty. Arrrr..