YES(O(1),O(n^1)) We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { primes() -> sieve(from(s(s(0())))) , sieve(X) -> n__sieve(X) , sieve(cons(X, Y)) -> cons(X, n__filter(X, n__sieve(activate(Y)))) , from(X) -> cons(X, n__from(n__s(X))) , from(X) -> n__from(X) , s(X) -> n__s(X) , cons(X1, X2) -> n__cons(X1, X2) , head(cons(X, Y)) -> X , tail(cons(X, Y)) -> activate(Y) , activate(X) -> X , activate(n__from(X)) -> from(activate(X)) , activate(n__s(X)) -> s(activate(X)) , activate(n__filter(X1, X2)) -> filter(activate(X1), activate(X2)) , activate(n__cons(X1, X2)) -> cons(activate(X1), X2) , activate(n__sieve(X)) -> sieve(activate(X)) , if(true(), X, Y) -> activate(X) , if(false(), X, Y) -> activate(Y) , filter(X1, X2) -> n__filter(X1, X2) , filter(s(s(X)), cons(Y, Z)) -> if(divides(s(s(X)), Y), n__filter(n__s(n__s(X)), activate(Z)), n__cons(Y, n__filter(X, n__sieve(Y)))) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) Arguments of following rules are not normal-forms: { sieve(cons(X, Y)) -> cons(X, n__filter(X, n__sieve(activate(Y)))) , head(cons(X, Y)) -> X , tail(cons(X, Y)) -> activate(Y) , filter(s(s(X)), cons(Y, Z)) -> if(divides(s(s(X)), Y), n__filter(n__s(n__s(X)), activate(Z)), n__cons(Y, n__filter(X, n__sieve(Y)))) } All above mentioned rules can be savely removed. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { primes() -> sieve(from(s(s(0())))) , sieve(X) -> n__sieve(X) , from(X) -> cons(X, n__from(n__s(X))) , from(X) -> n__from(X) , s(X) -> n__s(X) , cons(X1, X2) -> n__cons(X1, X2) , activate(X) -> X , activate(n__from(X)) -> from(activate(X)) , activate(n__s(X)) -> s(activate(X)) , activate(n__filter(X1, X2)) -> filter(activate(X1), activate(X2)) , activate(n__cons(X1, X2)) -> cons(activate(X1), X2) , activate(n__sieve(X)) -> sieve(activate(X)) , if(true(), X, Y) -> activate(X) , if(false(), X, Y) -> activate(Y) , filter(X1, X2) -> n__filter(X1, X2) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We add the following innermost weak dependency pairs: Strict DPs: { primes^#() -> c_1(sieve^#(from(s(s(0()))))) , sieve^#(X) -> c_2() , from^#(X) -> c_3(cons^#(X, n__from(n__s(X)))) , from^#(X) -> c_4() , cons^#(X1, X2) -> c_6() , s^#(X) -> c_5() , activate^#(X) -> c_7() , activate^#(n__from(X)) -> c_8(from^#(activate(X))) , activate^#(n__s(X)) -> c_9(s^#(activate(X))) , activate^#(n__filter(X1, X2)) -> c_10(filter^#(activate(X1), activate(X2))) , activate^#(n__cons(X1, X2)) -> c_11(cons^#(activate(X1), X2)) , activate^#(n__sieve(X)) -> c_12(sieve^#(activate(X))) , filter^#(X1, X2) -> c_15() , if^#(true(), X, Y) -> c_13(activate^#(X)) , if^#(false(), X, Y) -> c_14(activate^#(Y)) } and mark the set of starting terms. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { primes^#() -> c_1(sieve^#(from(s(s(0()))))) , sieve^#(X) -> c_2() , from^#(X) -> c_3(cons^#(X, n__from(n__s(X)))) , from^#(X) -> c_4() , cons^#(X1, X2) -> c_6() , s^#(X) -> c_5() , activate^#(X) -> c_7() , activate^#(n__from(X)) -> c_8(from^#(activate(X))) , activate^#(n__s(X)) -> c_9(s^#(activate(X))) , activate^#(n__filter(X1, X2)) -> c_10(filter^#(activate(X1), activate(X2))) , activate^#(n__cons(X1, X2)) -> c_11(cons^#(activate(X1), X2)) , activate^#(n__sieve(X)) -> c_12(sieve^#(activate(X))) , filter^#(X1, X2) -> c_15() , if^#(true(), X, Y) -> c_13(activate^#(X)) , if^#(false(), X, Y) -> c_14(activate^#(Y)) } Strict Trs: { primes() -> sieve(from(s(s(0())))) , sieve(X) -> n__sieve(X) , from(X) -> cons(X, n__from(n__s(X))) , from(X) -> n__from(X) , s(X) -> n__s(X) , cons(X1, X2) -> n__cons(X1, X2) , activate(X) -> X , activate(n__from(X)) -> from(activate(X)) , activate(n__s(X)) -> s(activate(X)) , activate(n__filter(X1, X2)) -> filter(activate(X1), activate(X2)) , activate(n__cons(X1, X2)) -> cons(activate(X1), X2) , activate(n__sieve(X)) -> sieve(activate(X)) , if(true(), X, Y) -> activate(X) , if(false(), X, Y) -> activate(Y) , filter(X1, X2) -> n__filter(X1, X2) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We replace rewrite rules by usable rules: Strict Usable Rules: { sieve(X) -> n__sieve(X) , from(X) -> cons(X, n__from(n__s(X))) , from(X) -> n__from(X) , s(X) -> n__s(X) , cons(X1, X2) -> n__cons(X1, X2) , activate(X) -> X , activate(n__from(X)) -> from(activate(X)) , activate(n__s(X)) -> s(activate(X)) , activate(n__filter(X1, X2)) -> filter(activate(X1), activate(X2)) , activate(n__cons(X1, X2)) -> cons(activate(X1), X2) , activate(n__sieve(X)) -> sieve(activate(X)) , filter(X1, X2) -> n__filter(X1, X2) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { primes^#() -> c_1(sieve^#(from(s(s(0()))))) , sieve^#(X) -> c_2() , from^#(X) -> c_3(cons^#(X, n__from(n__s(X)))) , from^#(X) -> c_4() , cons^#(X1, X2) -> c_6() , s^#(X) -> c_5() , activate^#(X) -> c_7() , activate^#(n__from(X)) -> c_8(from^#(activate(X))) , activate^#(n__s(X)) -> c_9(s^#(activate(X))) , activate^#(n__filter(X1, X2)) -> c_10(filter^#(activate(X1), activate(X2))) , activate^#(n__cons(X1, X2)) -> c_11(cons^#(activate(X1), X2)) , activate^#(n__sieve(X)) -> c_12(sieve^#(activate(X))) , filter^#(X1, X2) -> c_15() , if^#(true(), X, Y) -> c_13(activate^#(X)) , if^#(false(), X, Y) -> c_14(activate^#(Y)) } Strict Trs: { sieve(X) -> n__sieve(X) , from(X) -> cons(X, n__from(n__s(X))) , from(X) -> n__from(X) , s(X) -> n__s(X) , cons(X1, X2) -> n__cons(X1, X2) , activate(X) -> X , activate(n__from(X)) -> from(activate(X)) , activate(n__s(X)) -> s(activate(X)) , activate(n__filter(X1, X2)) -> filter(activate(X1), activate(X2)) , activate(n__cons(X1, X2)) -> cons(activate(X1), X2) , activate(n__sieve(X)) -> sieve(activate(X)) , filter(X1, X2) -> n__filter(X1, X2) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) The weightgap principle applies (using the following constant growth matrix-interpretation) The following argument positions are usable: Uargs(sieve) = {1}, Uargs(from) = {1}, Uargs(s) = {1}, Uargs(cons) = {1}, Uargs(filter) = {1, 2}, Uargs(c_1) = {1}, Uargs(sieve^#) = {1}, Uargs(from^#) = {1}, Uargs(c_3) = {1}, Uargs(cons^#) = {1}, Uargs(s^#) = {1}, Uargs(c_8) = {1}, Uargs(c_9) = {1}, Uargs(c_10) = {1}, Uargs(filter^#) = {1, 2}, Uargs(c_11) = {1}, Uargs(c_12) = {1}, Uargs(c_13) = {1}, Uargs(c_14) = {1} TcT has computed the following constructor-restricted matrix interpretation. [sieve](x1) = [1 0] x1 + [2] [0 1] [2] [from](x1) = [1 0] x1 + [2] [0 1] [2] [s](x1) = [1 0] x1 + [1] [0 1] [1] [0] = [0] [0] [cons](x1, x2) = [1 0] x1 + [1] [0 1] [2] [n__from](x1) = [1 0] x1 + [0] [0 1] [2] [n__s](x1) = [1 0] x1 + [0] [0 1] [1] [activate](x1) = [1 2] x1 + [1] [0 2] [0] [true] = [1] [2] [false] = [1] [1] [filter](x1, x2) = [1 0] x1 + [1 0] x2 + [1] [0 1] [0 1] [2] [n__filter](x1, x2) = [1 0] x1 + [1 0] x2 + [0] [0 1] [0 1] [2] [n__cons](x1, x2) = [1 0] x1 + [0] [0 1] [2] [n__sieve](x1) = [1 0] x1 + [0] [0 1] [2] [primes^#] = [1] [1] [c_1](x1) = [1 0] x1 + [1] [0 1] [2] [sieve^#](x1) = [1 0] x1 + [1] [0 0] [2] [c_2] = [1] [1] [from^#](x1) = [1 0] x1 + [1] [0 0] [2] [c_3](x1) = [1 0] x1 + [1] [0 1] [2] [cons^#](x1, x2) = [1 0] x1 + [1] [0 0] [2] [c_4] = [1] [1] [s^#](x1) = [1 0] x1 + [1] [0 0] [2] [c_5] = [1] [1] [c_6] = [1] [1] [activate^#](x1) = [1 2] x1 + [0] [0 0] [0] [c_7] = [1] [1] [c_8](x1) = [1 0] x1 + [2] [0 1] [2] [c_9](x1) = [1 0] x1 + [2] [0 1] [2] [c_10](x1) = [1 0] x1 + [2] [0 1] [2] [filter^#](x1, x2) = [1 0] x1 + [1 0] x2 + [2] [0 0] [0 0] [2] [c_11](x1) = [1 0] x1 + [2] [0 1] [2] [c_12](x1) = [1 0] x1 + [2] [0 1] [2] [if^#](x1, x2, x3) = [1 1] x1 + [1 2] x2 + [1 2] x3 + [1] [1 1] [2 2] [1 1] [1] [c_13](x1) = [1 0] x1 + [2] [0 1] [2] [c_14](x1) = [1 0] x1 + [2] [0 1] [2] [c_15] = [2] [1] The following symbols are considered usable {sieve, from, s, cons, activate, filter, primes^#, sieve^#, from^#, cons^#, s^#, activate^#, filter^#, if^#} The order satisfies the following ordering constraints: [sieve(X)] = [1 0] X + [2] [0 1] [2] > [1 0] X + [0] [0 1] [2] = [n__sieve(X)] [from(X)] = [1 0] X + [2] [0 1] [2] > [1 0] X + [1] [0 1] [2] = [cons(X, n__from(n__s(X)))] [from(X)] = [1 0] X + [2] [0 1] [2] > [1 0] X + [0] [0 1] [2] = [n__from(X)] [s(X)] = [1 0] X + [1] [0 1] [1] > [1 0] X + [0] [0 1] [1] = [n__s(X)] [cons(X1, X2)] = [1 0] X1 + [1] [0 1] [2] > [1 0] X1 + [0] [0 1] [2] = [n__cons(X1, X2)] [activate(X)] = [1 2] X + [1] [0 2] [0] > [1 0] X + [0] [0 1] [0] = [X] [activate(n__from(X))] = [1 2] X + [5] [0 2] [4] > [1 2] X + [3] [0 2] [2] = [from(activate(X))] [activate(n__s(X))] = [1 2] X + [3] [0 2] [2] > [1 2] X + [2] [0 2] [1] = [s(activate(X))] [activate(n__filter(X1, X2))] = [1 2] X1 + [1 2] X2 + [5] [0 2] [0 2] [4] > [1 2] X1 + [1 2] X2 + [3] [0 2] [0 2] [2] = [filter(activate(X1), activate(X2))] [activate(n__cons(X1, X2))] = [1 2] X1 + [5] [0 2] [4] > [1 2] X1 + [2] [0 2] [2] = [cons(activate(X1), X2)] [activate(n__sieve(X))] = [1 2] X + [5] [0 2] [4] > [1 2] X + [3] [0 2] [2] = [sieve(activate(X))] [filter(X1, X2)] = [1 0] X1 + [1 0] X2 + [1] [0 1] [0 1] [2] > [1 0] X1 + [1 0] X2 + [0] [0 1] [0 1] [2] = [n__filter(X1, X2)] [primes^#()] = [1] [1] ? [6] [4] = [c_1(sieve^#(from(s(s(0())))))] [sieve^#(X)] = [1 0] X + [1] [0 0] [2] >= [1] [1] = [c_2()] [from^#(X)] = [1 0] X + [1] [0 0] [2] ? [1 0] X + [2] [0 0] [4] = [c_3(cons^#(X, n__from(n__s(X))))] [from^#(X)] = [1 0] X + [1] [0 0] [2] >= [1] [1] = [c_4()] [cons^#(X1, X2)] = [1 0] X1 + [1] [0 0] [2] >= [1] [1] = [c_6()] [s^#(X)] = [1 0] X + [1] [0 0] [2] >= [1] [1] = [c_5()] [activate^#(X)] = [1 2] X + [0] [0 0] [0] ? [1] [1] = [c_7()] [activate^#(n__from(X))] = [1 2] X + [4] [0 0] [0] ? [1 2] X + [4] [0 0] [4] = [c_8(from^#(activate(X)))] [activate^#(n__s(X))] = [1 2] X + [2] [0 0] [0] ? [1 2] X + [4] [0 0] [4] = [c_9(s^#(activate(X)))] [activate^#(n__filter(X1, X2))] = [1 2] X1 + [1 2] X2 + [4] [0 0] [0 0] [0] ? [1 2] X1 + [1 2] X2 + [6] [0 0] [0 0] [4] = [c_10(filter^#(activate(X1), activate(X2)))] [activate^#(n__cons(X1, X2))] = [1 2] X1 + [4] [0 0] [0] ? [1 2] X1 + [4] [0 0] [4] = [c_11(cons^#(activate(X1), X2))] [activate^#(n__sieve(X))] = [1 2] X + [4] [0 0] [0] ? [1 2] X + [4] [0 0] [4] = [c_12(sieve^#(activate(X)))] [filter^#(X1, X2)] = [1 0] X1 + [1 0] X2 + [2] [0 0] [0 0] [2] >= [2] [1] = [c_15()] [if^#(true(), X, Y)] = [1 2] X + [1 2] Y + [4] [2 2] [1 1] [4] > [1 2] X + [2] [0 0] [2] = [c_13(activate^#(X))] [if^#(false(), X, Y)] = [1 2] X + [1 2] Y + [3] [2 2] [1 1] [3] > [1 2] Y + [2] [0 0] [2] = [c_14(activate^#(Y))] Further, it can be verified that all rules not oriented are covered by the weightgap condition. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Strict DPs: { primes^#() -> c_1(sieve^#(from(s(s(0()))))) , sieve^#(X) -> c_2() , from^#(X) -> c_3(cons^#(X, n__from(n__s(X)))) , from^#(X) -> c_4() , cons^#(X1, X2) -> c_6() , s^#(X) -> c_5() , activate^#(X) -> c_7() , activate^#(n__from(X)) -> c_8(from^#(activate(X))) , activate^#(n__s(X)) -> c_9(s^#(activate(X))) , activate^#(n__filter(X1, X2)) -> c_10(filter^#(activate(X1), activate(X2))) , activate^#(n__cons(X1, X2)) -> c_11(cons^#(activate(X1), X2)) , activate^#(n__sieve(X)) -> c_12(sieve^#(activate(X))) , filter^#(X1, X2) -> c_15() } Weak DPs: { if^#(true(), X, Y) -> c_13(activate^#(X)) , if^#(false(), X, Y) -> c_14(activate^#(Y)) } Weak Trs: { sieve(X) -> n__sieve(X) , from(X) -> cons(X, n__from(n__s(X))) , from(X) -> n__from(X) , s(X) -> n__s(X) , cons(X1, X2) -> n__cons(X1, X2) , activate(X) -> X , activate(n__from(X)) -> from(activate(X)) , activate(n__s(X)) -> s(activate(X)) , activate(n__filter(X1, X2)) -> filter(activate(X1), activate(X2)) , activate(n__cons(X1, X2)) -> cons(activate(X1), X2) , activate(n__sieve(X)) -> sieve(activate(X)) , filter(X1, X2) -> n__filter(X1, X2) } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) We estimate the number of application of {2,4,5,6,13} by applications of Pre({2,4,5,6,13}) = {1,3,8,9,10,11,12}. Here rules are labeled as follows: DPs: { 1: primes^#() -> c_1(sieve^#(from(s(s(0()))))) , 2: sieve^#(X) -> c_2() , 3: from^#(X) -> c_3(cons^#(X, n__from(n__s(X)))) , 4: from^#(X) -> c_4() , 5: cons^#(X1, X2) -> c_6() , 6: s^#(X) -> c_5() , 7: activate^#(X) -> c_7() , 8: activate^#(n__from(X)) -> c_8(from^#(activate(X))) , 9: activate^#(n__s(X)) -> c_9(s^#(activate(X))) , 10: activate^#(n__filter(X1, X2)) -> c_10(filter^#(activate(X1), activate(X2))) , 11: activate^#(n__cons(X1, X2)) -> c_11(cons^#(activate(X1), X2)) , 12: activate^#(n__sieve(X)) -> c_12(sieve^#(activate(X))) , 13: filter^#(X1, X2) -> c_15() , 14: if^#(true(), X, Y) -> c_13(activate^#(X)) , 15: if^#(false(), X, Y) -> c_14(activate^#(Y)) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Strict DPs: { primes^#() -> c_1(sieve^#(from(s(s(0()))))) , from^#(X) -> c_3(cons^#(X, n__from(n__s(X)))) , activate^#(X) -> c_7() , activate^#(n__from(X)) -> c_8(from^#(activate(X))) , activate^#(n__s(X)) -> c_9(s^#(activate(X))) , activate^#(n__filter(X1, X2)) -> c_10(filter^#(activate(X1), activate(X2))) , activate^#(n__cons(X1, X2)) -> c_11(cons^#(activate(X1), X2)) , activate^#(n__sieve(X)) -> c_12(sieve^#(activate(X))) } Weak DPs: { sieve^#(X) -> c_2() , from^#(X) -> c_4() , cons^#(X1, X2) -> c_6() , s^#(X) -> c_5() , filter^#(X1, X2) -> c_15() , if^#(true(), X, Y) -> c_13(activate^#(X)) , if^#(false(), X, Y) -> c_14(activate^#(Y)) } Weak Trs: { sieve(X) -> n__sieve(X) , from(X) -> cons(X, n__from(n__s(X))) , from(X) -> n__from(X) , s(X) -> n__s(X) , cons(X1, X2) -> n__cons(X1, X2) , activate(X) -> X , activate(n__from(X)) -> from(activate(X)) , activate(n__s(X)) -> s(activate(X)) , activate(n__filter(X1, X2)) -> filter(activate(X1), activate(X2)) , activate(n__cons(X1, X2)) -> cons(activate(X1), X2) , activate(n__sieve(X)) -> sieve(activate(X)) , filter(X1, X2) -> n__filter(X1, X2) } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) We estimate the number of application of {1,2} by applications of Pre({1,2}) = {4}. Here rules are labeled as follows: DPs: { 1: primes^#() -> c_1(sieve^#(from(s(s(0()))))) , 2: from^#(X) -> c_3(cons^#(X, n__from(n__s(X)))) , 3: activate^#(X) -> c_7() , 4: activate^#(n__from(X)) -> c_8(from^#(activate(X))) , 5: activate^#(n__s(X)) -> c_9(s^#(activate(X))) , 6: activate^#(n__filter(X1, X2)) -> c_10(filter^#(activate(X1), activate(X2))) , 7: activate^#(n__cons(X1, X2)) -> c_11(cons^#(activate(X1), X2)) , 8: activate^#(n__sieve(X)) -> c_12(sieve^#(activate(X))) , 9: sieve^#(X) -> c_2() , 10: from^#(X) -> c_4() , 11: cons^#(X1, X2) -> c_6() , 12: s^#(X) -> c_5() , 13: filter^#(X1, X2) -> c_15() , 14: if^#(true(), X, Y) -> c_13(activate^#(X)) , 15: if^#(false(), X, Y) -> c_14(activate^#(Y)) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Strict DPs: { activate^#(X) -> c_7() , activate^#(n__from(X)) -> c_8(from^#(activate(X))) , activate^#(n__s(X)) -> c_9(s^#(activate(X))) , activate^#(n__filter(X1, X2)) -> c_10(filter^#(activate(X1), activate(X2))) , activate^#(n__cons(X1, X2)) -> c_11(cons^#(activate(X1), X2)) , activate^#(n__sieve(X)) -> c_12(sieve^#(activate(X))) } Weak DPs: { primes^#() -> c_1(sieve^#(from(s(s(0()))))) , sieve^#(X) -> c_2() , from^#(X) -> c_3(cons^#(X, n__from(n__s(X)))) , from^#(X) -> c_4() , cons^#(X1, X2) -> c_6() , s^#(X) -> c_5() , filter^#(X1, X2) -> c_15() , if^#(true(), X, Y) -> c_13(activate^#(X)) , if^#(false(), X, Y) -> c_14(activate^#(Y)) } Weak Trs: { sieve(X) -> n__sieve(X) , from(X) -> cons(X, n__from(n__s(X))) , from(X) -> n__from(X) , s(X) -> n__s(X) , cons(X1, X2) -> n__cons(X1, X2) , activate(X) -> X , activate(n__from(X)) -> from(activate(X)) , activate(n__s(X)) -> s(activate(X)) , activate(n__filter(X1, X2)) -> filter(activate(X1), activate(X2)) , activate(n__cons(X1, X2)) -> cons(activate(X1), X2) , activate(n__sieve(X)) -> sieve(activate(X)) , filter(X1, X2) -> n__filter(X1, X2) } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) The following weak DPs constitute a sub-graph of the DG that is closed under successors. The DPs are removed. { primes^#() -> c_1(sieve^#(from(s(s(0()))))) , sieve^#(X) -> c_2() , from^#(X) -> c_3(cons^#(X, n__from(n__s(X)))) , from^#(X) -> c_4() , cons^#(X1, X2) -> c_6() , s^#(X) -> c_5() , filter^#(X1, X2) -> c_15() } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Strict DPs: { activate^#(X) -> c_7() , activate^#(n__from(X)) -> c_8(from^#(activate(X))) , activate^#(n__s(X)) -> c_9(s^#(activate(X))) , activate^#(n__filter(X1, X2)) -> c_10(filter^#(activate(X1), activate(X2))) , activate^#(n__cons(X1, X2)) -> c_11(cons^#(activate(X1), X2)) , activate^#(n__sieve(X)) -> c_12(sieve^#(activate(X))) } Weak DPs: { if^#(true(), X, Y) -> c_13(activate^#(X)) , if^#(false(), X, Y) -> c_14(activate^#(Y)) } Weak Trs: { sieve(X) -> n__sieve(X) , from(X) -> cons(X, n__from(n__s(X))) , from(X) -> n__from(X) , s(X) -> n__s(X) , cons(X1, X2) -> n__cons(X1, X2) , activate(X) -> X , activate(n__from(X)) -> from(activate(X)) , activate(n__s(X)) -> s(activate(X)) , activate(n__filter(X1, X2)) -> filter(activate(X1), activate(X2)) , activate(n__cons(X1, X2)) -> cons(activate(X1), X2) , activate(n__sieve(X)) -> sieve(activate(X)) , filter(X1, X2) -> n__filter(X1, X2) } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) Due to missing edges in the dependency-graph, the right-hand sides of following rules could be simplified: { activate^#(n__from(X)) -> c_8(from^#(activate(X))) , activate^#(n__s(X)) -> c_9(s^#(activate(X))) , activate^#(n__filter(X1, X2)) -> c_10(filter^#(activate(X1), activate(X2))) , activate^#(n__cons(X1, X2)) -> c_11(cons^#(activate(X1), X2)) , activate^#(n__sieve(X)) -> c_12(sieve^#(activate(X))) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Strict DPs: { activate^#(X) -> c_1() , activate^#(n__from(X)) -> c_2() , activate^#(n__s(X)) -> c_3() , activate^#(n__filter(X1, X2)) -> c_4() , activate^#(n__cons(X1, X2)) -> c_5() , activate^#(n__sieve(X)) -> c_6() } Weak DPs: { if^#(true(), X, Y) -> c_7(activate^#(X)) , if^#(false(), X, Y) -> c_8(activate^#(Y)) } Weak Trs: { sieve(X) -> n__sieve(X) , from(X) -> cons(X, n__from(n__s(X))) , from(X) -> n__from(X) , s(X) -> n__s(X) , cons(X1, X2) -> n__cons(X1, X2) , activate(X) -> X , activate(n__from(X)) -> from(activate(X)) , activate(n__s(X)) -> s(activate(X)) , activate(n__filter(X1, X2)) -> filter(activate(X1), activate(X2)) , activate(n__cons(X1, X2)) -> cons(activate(X1), X2) , activate(n__sieve(X)) -> sieve(activate(X)) , filter(X1, X2) -> n__filter(X1, X2) } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) No rule is usable, rules are removed from the input problem. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Strict DPs: { activate^#(X) -> c_1() , activate^#(n__from(X)) -> c_2() , activate^#(n__s(X)) -> c_3() , activate^#(n__filter(X1, X2)) -> c_4() , activate^#(n__cons(X1, X2)) -> c_5() , activate^#(n__sieve(X)) -> c_6() } Weak DPs: { if^#(true(), X, Y) -> c_7(activate^#(X)) , if^#(false(), X, Y) -> c_8(activate^#(Y)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) Consider the dependency graph 1: activate^#(X) -> c_1() 2: activate^#(n__from(X)) -> c_2() 3: activate^#(n__s(X)) -> c_3() 4: activate^#(n__filter(X1, X2)) -> c_4() 5: activate^#(n__cons(X1, X2)) -> c_5() 6: activate^#(n__sieve(X)) -> c_6() 7: if^#(true(), X, Y) -> c_7(activate^#(X)) -->_1 activate^#(n__sieve(X)) -> c_6() :6 -->_1 activate^#(n__cons(X1, X2)) -> c_5() :5 -->_1 activate^#(n__filter(X1, X2)) -> c_4() :4 -->_1 activate^#(n__s(X)) -> c_3() :3 -->_1 activate^#(n__from(X)) -> c_2() :2 -->_1 activate^#(X) -> c_1() :1 8: if^#(false(), X, Y) -> c_8(activate^#(Y)) -->_1 activate^#(n__sieve(X)) -> c_6() :6 -->_1 activate^#(n__cons(X1, X2)) -> c_5() :5 -->_1 activate^#(n__filter(X1, X2)) -> c_4() :4 -->_1 activate^#(n__s(X)) -> c_3() :3 -->_1 activate^#(n__from(X)) -> c_2() :2 -->_1 activate^#(X) -> c_1() :1 Following roots of the dependency graph are removed, as the considered set of starting terms is closed under reduction with respect to these rules (modulo compound contexts). { if^#(true(), X, Y) -> c_7(activate^#(X)) , if^#(false(), X, Y) -> c_8(activate^#(Y)) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Strict DPs: { activate^#(X) -> c_1() , activate^#(n__from(X)) -> c_2() , activate^#(n__s(X)) -> c_3() , activate^#(n__filter(X1, X2)) -> c_4() , activate^#(n__cons(X1, X2)) -> c_5() , activate^#(n__sieve(X)) -> c_6() } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) Consider the dependency graph 1: activate^#(X) -> c_1() 2: activate^#(n__from(X)) -> c_2() 3: activate^#(n__s(X)) -> c_3() 4: activate^#(n__filter(X1, X2)) -> c_4() 5: activate^#(n__cons(X1, X2)) -> c_5() 6: activate^#(n__sieve(X)) -> c_6() Following roots of the dependency graph are removed, as the considered set of starting terms is closed under reduction with respect to these rules (modulo compound contexts). { activate^#(X) -> c_1() , activate^#(n__from(X)) -> c_2() , activate^#(n__s(X)) -> c_3() , activate^#(n__filter(X1, X2)) -> c_4() , activate^#(n__cons(X1, X2)) -> c_5() , activate^#(n__sieve(X)) -> c_6() } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Rules: Empty Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) Empty rules are trivially bounded Hurray, we answered YES(O(1),O(n^1))