MAYBE We are left with following problem, upon which TcT provides the certificate MAYBE. Strict Trs: { a__primes() -> a__sieve(a__from(s(s(0())))) , a__primes() -> primes() , a__sieve(X) -> sieve(X) , a__sieve(cons(X, Y)) -> cons(mark(X), filter(X, sieve(Y))) , a__from(X) -> cons(mark(X), from(s(X))) , a__from(X) -> from(X) , mark(s(X)) -> s(mark(X)) , mark(0()) -> 0() , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(from(X)) -> a__from(mark(X)) , mark(true()) -> true() , mark(false()) -> false() , mark(divides(X1, X2)) -> divides(mark(X1), mark(X2)) , mark(filter(X1, X2)) -> a__filter(mark(X1), mark(X2)) , mark(sieve(X)) -> a__sieve(mark(X)) , mark(primes()) -> a__primes() , mark(head(X)) -> a__head(mark(X)) , mark(tail(X)) -> a__tail(mark(X)) , mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3) , a__head(X) -> head(X) , a__head(cons(X, Y)) -> mark(X) , a__tail(X) -> tail(X) , a__tail(cons(X, Y)) -> mark(Y) , a__if(X1, X2, X3) -> if(X1, X2, X3) , a__if(true(), X, Y) -> mark(X) , a__if(false(), X, Y) -> mark(Y) , a__filter(X1, X2) -> filter(X1, X2) , a__filter(s(s(X)), cons(Y, Z)) -> a__if(divides(s(s(mark(X))), mark(Y)), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))) } 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) '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 2) '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 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__primes^#() -> c_1(a__sieve^#(a__from(s(s(0()))))) , a__primes^#() -> c_2() , a__sieve^#(X) -> c_3(X) , a__sieve^#(cons(X, Y)) -> c_4(mark^#(X), X, Y) , mark^#(s(X)) -> c_7(mark^#(X)) , mark^#(0()) -> c_8() , mark^#(cons(X1, X2)) -> c_9(mark^#(X1), X2) , mark^#(from(X)) -> c_10(a__from^#(mark(X))) , mark^#(true()) -> c_11() , mark^#(false()) -> c_12() , mark^#(divides(X1, X2)) -> c_13(mark^#(X1), mark^#(X2)) , mark^#(filter(X1, X2)) -> c_14(a__filter^#(mark(X1), mark(X2))) , mark^#(sieve(X)) -> c_15(a__sieve^#(mark(X))) , mark^#(primes()) -> c_16(a__primes^#()) , mark^#(head(X)) -> c_17(a__head^#(mark(X))) , mark^#(tail(X)) -> c_18(a__tail^#(mark(X))) , mark^#(if(X1, X2, X3)) -> c_19(a__if^#(mark(X1), X2, X3)) , a__from^#(X) -> c_5(mark^#(X), X) , a__from^#(X) -> c_6(X) , a__filter^#(X1, X2) -> c_27(X1, X2) , a__filter^#(s(s(X)), cons(Y, Z)) -> c_28(a__if^#(divides(s(s(mark(X))), mark(Y)), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))) , a__head^#(X) -> c_20(X) , a__head^#(cons(X, Y)) -> c_21(mark^#(X)) , a__tail^#(X) -> c_22(X) , a__tail^#(cons(X, Y)) -> c_23(mark^#(Y)) , a__if^#(X1, X2, X3) -> c_24(X1, X2, X3) , a__if^#(true(), X, Y) -> c_25(mark^#(X)) , a__if^#(false(), X, Y) -> c_26(mark^#(Y)) } and mark the set of starting terms. We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { a__primes^#() -> c_1(a__sieve^#(a__from(s(s(0()))))) , a__primes^#() -> c_2() , a__sieve^#(X) -> c_3(X) , a__sieve^#(cons(X, Y)) -> c_4(mark^#(X), X, Y) , mark^#(s(X)) -> c_7(mark^#(X)) , mark^#(0()) -> c_8() , mark^#(cons(X1, X2)) -> c_9(mark^#(X1), X2) , mark^#(from(X)) -> c_10(a__from^#(mark(X))) , mark^#(true()) -> c_11() , mark^#(false()) -> c_12() , mark^#(divides(X1, X2)) -> c_13(mark^#(X1), mark^#(X2)) , mark^#(filter(X1, X2)) -> c_14(a__filter^#(mark(X1), mark(X2))) , mark^#(sieve(X)) -> c_15(a__sieve^#(mark(X))) , mark^#(primes()) -> c_16(a__primes^#()) , mark^#(head(X)) -> c_17(a__head^#(mark(X))) , mark^#(tail(X)) -> c_18(a__tail^#(mark(X))) , mark^#(if(X1, X2, X3)) -> c_19(a__if^#(mark(X1), X2, X3)) , a__from^#(X) -> c_5(mark^#(X), X) , a__from^#(X) -> c_6(X) , a__filter^#(X1, X2) -> c_27(X1, X2) , a__filter^#(s(s(X)), cons(Y, Z)) -> c_28(a__if^#(divides(s(s(mark(X))), mark(Y)), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))) , a__head^#(X) -> c_20(X) , a__head^#(cons(X, Y)) -> c_21(mark^#(X)) , a__tail^#(X) -> c_22(X) , a__tail^#(cons(X, Y)) -> c_23(mark^#(Y)) , a__if^#(X1, X2, X3) -> c_24(X1, X2, X3) , a__if^#(true(), X, Y) -> c_25(mark^#(X)) , a__if^#(false(), X, Y) -> c_26(mark^#(Y)) } Strict Trs: { a__primes() -> a__sieve(a__from(s(s(0())))) , a__primes() -> primes() , a__sieve(X) -> sieve(X) , a__sieve(cons(X, Y)) -> cons(mark(X), filter(X, sieve(Y))) , a__from(X) -> cons(mark(X), from(s(X))) , a__from(X) -> from(X) , mark(s(X)) -> s(mark(X)) , mark(0()) -> 0() , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(from(X)) -> a__from(mark(X)) , mark(true()) -> true() , mark(false()) -> false() , mark(divides(X1, X2)) -> divides(mark(X1), mark(X2)) , mark(filter(X1, X2)) -> a__filter(mark(X1), mark(X2)) , mark(sieve(X)) -> a__sieve(mark(X)) , mark(primes()) -> a__primes() , mark(head(X)) -> a__head(mark(X)) , mark(tail(X)) -> a__tail(mark(X)) , mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3) , a__head(X) -> head(X) , a__head(cons(X, Y)) -> mark(X) , a__tail(X) -> tail(X) , a__tail(cons(X, Y)) -> mark(Y) , a__if(X1, X2, X3) -> if(X1, X2, X3) , a__if(true(), X, Y) -> mark(X) , a__if(false(), X, Y) -> mark(Y) , a__filter(X1, X2) -> filter(X1, X2) , a__filter(s(s(X)), cons(Y, Z)) -> a__if(divides(s(s(mark(X))), mark(Y)), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))) } Obligation: runtime complexity Answer: MAYBE We estimate the number of application of {2,6,9,10} by applications of Pre({2,6,9,10}) = {3,4,5,7,11,14,18,19,20,22,23,24,25,26,27,28}. Here rules are labeled as follows: DPs: { 1: a__primes^#() -> c_1(a__sieve^#(a__from(s(s(0()))))) , 2: a__primes^#() -> c_2() , 3: a__sieve^#(X) -> c_3(X) , 4: a__sieve^#(cons(X, Y)) -> c_4(mark^#(X), X, Y) , 5: mark^#(s(X)) -> c_7(mark^#(X)) , 6: mark^#(0()) -> c_8() , 7: mark^#(cons(X1, X2)) -> c_9(mark^#(X1), X2) , 8: mark^#(from(X)) -> c_10(a__from^#(mark(X))) , 9: mark^#(true()) -> c_11() , 10: mark^#(false()) -> c_12() , 11: mark^#(divides(X1, X2)) -> c_13(mark^#(X1), mark^#(X2)) , 12: mark^#(filter(X1, X2)) -> c_14(a__filter^#(mark(X1), mark(X2))) , 13: mark^#(sieve(X)) -> c_15(a__sieve^#(mark(X))) , 14: mark^#(primes()) -> c_16(a__primes^#()) , 15: mark^#(head(X)) -> c_17(a__head^#(mark(X))) , 16: mark^#(tail(X)) -> c_18(a__tail^#(mark(X))) , 17: mark^#(if(X1, X2, X3)) -> c_19(a__if^#(mark(X1), X2, X3)) , 18: a__from^#(X) -> c_5(mark^#(X), X) , 19: a__from^#(X) -> c_6(X) , 20: a__filter^#(X1, X2) -> c_27(X1, X2) , 21: a__filter^#(s(s(X)), cons(Y, Z)) -> c_28(a__if^#(divides(s(s(mark(X))), mark(Y)), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))) , 22: a__head^#(X) -> c_20(X) , 23: a__head^#(cons(X, Y)) -> c_21(mark^#(X)) , 24: a__tail^#(X) -> c_22(X) , 25: a__tail^#(cons(X, Y)) -> c_23(mark^#(Y)) , 26: a__if^#(X1, X2, X3) -> c_24(X1, X2, X3) , 27: a__if^#(true(), X, Y) -> c_25(mark^#(X)) , 28: a__if^#(false(), X, Y) -> c_26(mark^#(Y)) } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { a__primes^#() -> c_1(a__sieve^#(a__from(s(s(0()))))) , a__sieve^#(X) -> c_3(X) , a__sieve^#(cons(X, Y)) -> c_4(mark^#(X), X, Y) , mark^#(s(X)) -> c_7(mark^#(X)) , mark^#(cons(X1, X2)) -> c_9(mark^#(X1), X2) , mark^#(from(X)) -> c_10(a__from^#(mark(X))) , mark^#(divides(X1, X2)) -> c_13(mark^#(X1), mark^#(X2)) , mark^#(filter(X1, X2)) -> c_14(a__filter^#(mark(X1), mark(X2))) , mark^#(sieve(X)) -> c_15(a__sieve^#(mark(X))) , mark^#(primes()) -> c_16(a__primes^#()) , mark^#(head(X)) -> c_17(a__head^#(mark(X))) , mark^#(tail(X)) -> c_18(a__tail^#(mark(X))) , mark^#(if(X1, X2, X3)) -> c_19(a__if^#(mark(X1), X2, X3)) , a__from^#(X) -> c_5(mark^#(X), X) , a__from^#(X) -> c_6(X) , a__filter^#(X1, X2) -> c_27(X1, X2) , a__filter^#(s(s(X)), cons(Y, Z)) -> c_28(a__if^#(divides(s(s(mark(X))), mark(Y)), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))) , a__head^#(X) -> c_20(X) , a__head^#(cons(X, Y)) -> c_21(mark^#(X)) , a__tail^#(X) -> c_22(X) , a__tail^#(cons(X, Y)) -> c_23(mark^#(Y)) , a__if^#(X1, X2, X3) -> c_24(X1, X2, X3) , a__if^#(true(), X, Y) -> c_25(mark^#(X)) , a__if^#(false(), X, Y) -> c_26(mark^#(Y)) } Strict Trs: { a__primes() -> a__sieve(a__from(s(s(0())))) , a__primes() -> primes() , a__sieve(X) -> sieve(X) , a__sieve(cons(X, Y)) -> cons(mark(X), filter(X, sieve(Y))) , a__from(X) -> cons(mark(X), from(s(X))) , a__from(X) -> from(X) , mark(s(X)) -> s(mark(X)) , mark(0()) -> 0() , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(from(X)) -> a__from(mark(X)) , mark(true()) -> true() , mark(false()) -> false() , mark(divides(X1, X2)) -> divides(mark(X1), mark(X2)) , mark(filter(X1, X2)) -> a__filter(mark(X1), mark(X2)) , mark(sieve(X)) -> a__sieve(mark(X)) , mark(primes()) -> a__primes() , mark(head(X)) -> a__head(mark(X)) , mark(tail(X)) -> a__tail(mark(X)) , mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3) , a__head(X) -> head(X) , a__head(cons(X, Y)) -> mark(X) , a__tail(X) -> tail(X) , a__tail(cons(X, Y)) -> mark(Y) , a__if(X1, X2, X3) -> if(X1, X2, X3) , a__if(true(), X, Y) -> mark(X) , a__if(false(), X, Y) -> mark(Y) , a__filter(X1, X2) -> filter(X1, X2) , a__filter(s(s(X)), cons(Y, Z)) -> a__if(divides(s(s(mark(X))), mark(Y)), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))) } Weak DPs: { a__primes^#() -> c_2() , mark^#(0()) -> c_8() , mark^#(true()) -> c_11() , mark^#(false()) -> c_12() } Obligation: runtime complexity Answer: MAYBE Empty strict component of the problem is NOT empty. Arrrr..