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: { natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) , natsFrom(X) -> n__natsFrom(X) , fst(pair(XS, YS)) -> XS , snd(pair(XS, YS)) -> YS , splitAt(0(), XS) -> pair(nil(), XS) , splitAt(s(N), cons(X, XS)) -> u(splitAt(N, activate(XS)), N, X, activate(XS)) , s(X) -> n__s(X) , u(pair(YS, ZS), N, X, XS) -> pair(cons(activate(X), YS), ZS) , activate(X) -> X , activate(n__natsFrom(X)) -> natsFrom(activate(X)) , activate(n__s(X)) -> s(activate(X)) , head(cons(N, XS)) -> N , tail(cons(N, XS)) -> activate(XS) , sel(N, XS) -> head(afterNth(N, XS)) , afterNth(N, XS) -> snd(splitAt(N, XS)) , take(N, XS) -> fst(splitAt(N, XS)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) Arguments of following rules are not normal-forms: { splitAt(s(N), cons(X, XS)) -> u(splitAt(N, activate(XS)), N, X, activate(XS)) } 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: { natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) , natsFrom(X) -> n__natsFrom(X) , fst(pair(XS, YS)) -> XS , snd(pair(XS, YS)) -> YS , splitAt(0(), XS) -> pair(nil(), XS) , s(X) -> n__s(X) , u(pair(YS, ZS), N, X, XS) -> pair(cons(activate(X), YS), ZS) , activate(X) -> X , activate(n__natsFrom(X)) -> natsFrom(activate(X)) , activate(n__s(X)) -> s(activate(X)) , head(cons(N, XS)) -> N , tail(cons(N, XS)) -> activate(XS) , sel(N, XS) -> head(afterNth(N, XS)) , afterNth(N, XS) -> snd(splitAt(N, XS)) , take(N, XS) -> fst(splitAt(N, XS)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We add the following innermost weak dependency pairs: Strict DPs: { natsFrom^#(N) -> c_1() , natsFrom^#(X) -> c_2() , fst^#(pair(XS, YS)) -> c_3() , snd^#(pair(XS, YS)) -> c_4() , splitAt^#(0(), XS) -> c_5() , s^#(X) -> c_6() , u^#(pair(YS, ZS), N, X, XS) -> c_7(activate^#(X)) , activate^#(X) -> c_8() , activate^#(n__natsFrom(X)) -> c_9(natsFrom^#(activate(X))) , activate^#(n__s(X)) -> c_10(s^#(activate(X))) , head^#(cons(N, XS)) -> c_11() , tail^#(cons(N, XS)) -> c_12(activate^#(XS)) , sel^#(N, XS) -> c_13(head^#(afterNth(N, XS))) , afterNth^#(N, XS) -> c_14(snd^#(splitAt(N, XS))) , take^#(N, XS) -> c_15(fst^#(splitAt(N, XS))) } 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: { natsFrom^#(N) -> c_1() , natsFrom^#(X) -> c_2() , fst^#(pair(XS, YS)) -> c_3() , snd^#(pair(XS, YS)) -> c_4() , splitAt^#(0(), XS) -> c_5() , s^#(X) -> c_6() , u^#(pair(YS, ZS), N, X, XS) -> c_7(activate^#(X)) , activate^#(X) -> c_8() , activate^#(n__natsFrom(X)) -> c_9(natsFrom^#(activate(X))) , activate^#(n__s(X)) -> c_10(s^#(activate(X))) , head^#(cons(N, XS)) -> c_11() , tail^#(cons(N, XS)) -> c_12(activate^#(XS)) , sel^#(N, XS) -> c_13(head^#(afterNth(N, XS))) , afterNth^#(N, XS) -> c_14(snd^#(splitAt(N, XS))) , take^#(N, XS) -> c_15(fst^#(splitAt(N, XS))) } Strict Trs: { natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) , natsFrom(X) -> n__natsFrom(X) , fst(pair(XS, YS)) -> XS , snd(pair(XS, YS)) -> YS , splitAt(0(), XS) -> pair(nil(), XS) , s(X) -> n__s(X) , u(pair(YS, ZS), N, X, XS) -> pair(cons(activate(X), YS), ZS) , activate(X) -> X , activate(n__natsFrom(X)) -> natsFrom(activate(X)) , activate(n__s(X)) -> s(activate(X)) , head(cons(N, XS)) -> N , tail(cons(N, XS)) -> activate(XS) , sel(N, XS) -> head(afterNth(N, XS)) , afterNth(N, XS) -> snd(splitAt(N, XS)) , take(N, XS) -> fst(splitAt(N, XS)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We replace rewrite rules by usable rules: Strict Usable Rules: { natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) , natsFrom(X) -> n__natsFrom(X) , snd(pair(XS, YS)) -> YS , splitAt(0(), XS) -> pair(nil(), XS) , s(X) -> n__s(X) , activate(X) -> X , activate(n__natsFrom(X)) -> natsFrom(activate(X)) , activate(n__s(X)) -> s(activate(X)) , afterNth(N, XS) -> snd(splitAt(N, XS)) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { natsFrom^#(N) -> c_1() , natsFrom^#(X) -> c_2() , fst^#(pair(XS, YS)) -> c_3() , snd^#(pair(XS, YS)) -> c_4() , splitAt^#(0(), XS) -> c_5() , s^#(X) -> c_6() , u^#(pair(YS, ZS), N, X, XS) -> c_7(activate^#(X)) , activate^#(X) -> c_8() , activate^#(n__natsFrom(X)) -> c_9(natsFrom^#(activate(X))) , activate^#(n__s(X)) -> c_10(s^#(activate(X))) , head^#(cons(N, XS)) -> c_11() , tail^#(cons(N, XS)) -> c_12(activate^#(XS)) , sel^#(N, XS) -> c_13(head^#(afterNth(N, XS))) , afterNth^#(N, XS) -> c_14(snd^#(splitAt(N, XS))) , take^#(N, XS) -> c_15(fst^#(splitAt(N, XS))) } Strict Trs: { natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) , natsFrom(X) -> n__natsFrom(X) , snd(pair(XS, YS)) -> YS , splitAt(0(), XS) -> pair(nil(), XS) , s(X) -> n__s(X) , activate(X) -> X , activate(n__natsFrom(X)) -> natsFrom(activate(X)) , activate(n__s(X)) -> s(activate(X)) , afterNth(N, XS) -> snd(splitAt(N, XS)) } 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(natsFrom) = {1}, Uargs(snd) = {1}, Uargs(s) = {1}, Uargs(natsFrom^#) = {1}, Uargs(fst^#) = {1}, Uargs(snd^#) = {1}, Uargs(s^#) = {1}, Uargs(c_7) = {1}, Uargs(c_9) = {1}, Uargs(c_10) = {1}, Uargs(head^#) = {1}, Uargs(c_12) = {1}, Uargs(c_13) = {1}, Uargs(c_14) = {1}, Uargs(c_15) = {1} TcT has computed the following constructor-restricted matrix interpretation. [natsFrom](x1) = [1 0] x1 + [2] [0 1] [2] [cons](x1, x2) = [1 0] x2 + [0] [0 1] [0] [n__natsFrom](x1) = [1 0] x1 + [1] [0 1] [1] [n__s](x1) = [1 0] x1 + [0] [0 1] [1] [pair](x1, x2) = [1 0] x1 + [1 0] x2 + [0] [0 0] [0 1] [1] [snd](x1) = [1 1] x1 + [0] [0 1] [0] [splitAt](x1, x2) = [1 0] x1 + [2 0] x2 + [0] [1 0] [0 1] [0] [0] = [1] [0] [nil] = [0] [2] [s](x1) = [1 0] x1 + [1] [0 1] [1] [activate](x1) = [1 2] x1 + [2] [2 1] [2] [afterNth](x1, x2) = [2 0] x1 + [2 1] x2 + [1] [2 0] [0 2] [0] [natsFrom^#](x1) = [1 0] x1 + [2] [0 0] [2] [c_1] = [1] [1] [c_2] = [1] [1] [fst^#](x1) = [1 0] x1 + [2] [0 0] [2] [c_3] = [1] [1] [snd^#](x1) = [1 0] x1 + [2] [0 0] [2] [c_4] = [1] [1] [splitAt^#](x1, x2) = [2 1] x1 + [2] [2 2] [2] [c_5] = [1] [1] [s^#](x1) = [1 0] x1 + [2] [0 0] [2] [c_6] = [1] [1] [u^#](x1, x2, x3, x4) = [2 2] x1 + [1 2] x3 + [2] [2 2] [1 2] [2] [c_7](x1) = [1 0] x1 + [1] [0 1] [1] [activate^#](x1) = [1 2] x1 + [1] [2 2] [2] [c_8] = [0] [1] [c_9](x1) = [1 0] x1 + [2] [0 1] [2] [c_10](x1) = [1 0] x1 + [2] [0 1] [2] [head^#](x1) = [1 0] x1 + [1] [0 0] [2] [c_11] = [1] [1] [tail^#](x1) = [2 2] x1 + [2] [1 1] [2] [c_12](x1) = [1 0] x1 + [1] [0 1] [1] [sel^#](x1, x2) = [2 1] x1 + [2 2] x2 + [1] [2 1] [1 2] [1] [c_13](x1) = [1 0] x1 + [2] [0 1] [2] [afterNth^#](x1, x2) = [2 2] x1 + [2 2] x2 + [1] [1 2] [2 1] [1] [c_14](x1) = [1 0] x1 + [2] [0 1] [2] [take^#](x1, x2) = [2 1] x1 + [2 1] x2 + [2] [1 1] [2 2] [1] [c_15](x1) = [1 0] x1 + [2] [0 1] [2] The following symbols are considered usable {natsFrom, snd, splitAt, s, activate, afterNth, natsFrom^#, fst^#, snd^#, splitAt^#, s^#, u^#, activate^#, head^#, tail^#, sel^#, afterNth^#, take^#} The order satisfies the following ordering constraints: [natsFrom(N)] = [1 0] N + [2] [0 1] [2] > [1 0] N + [1] [0 1] [2] = [cons(N, n__natsFrom(n__s(N)))] [natsFrom(X)] = [1 0] X + [2] [0 1] [2] > [1 0] X + [1] [0 1] [1] = [n__natsFrom(X)] [snd(pair(XS, YS))] = [1 0] XS + [1 1] YS + [1] [0 0] [0 1] [1] > [1 0] YS + [0] [0 1] [0] = [YS] [splitAt(0(), XS)] = [2 0] XS + [1] [0 1] [1] > [1 0] XS + [0] [0 1] [1] = [pair(nil(), XS)] [s(X)] = [1 0] X + [1] [0 1] [1] > [1 0] X + [0] [0 1] [1] = [n__s(X)] [activate(X)] = [1 2] X + [2] [2 1] [2] > [1 0] X + [0] [0 1] [0] = [X] [activate(n__natsFrom(X))] = [1 2] X + [5] [2 1] [5] > [1 2] X + [4] [2 1] [4] = [natsFrom(activate(X))] [activate(n__s(X))] = [1 2] X + [4] [2 1] [3] > [1 2] X + [3] [2 1] [3] = [s(activate(X))] [afterNth(N, XS)] = [2 0] N + [2 1] XS + [1] [2 0] [0 2] [0] > [2 0] N + [2 1] XS + [0] [1 0] [0 1] [0] = [snd(splitAt(N, XS))] [natsFrom^#(N)] = [1 0] N + [2] [0 0] [2] > [1] [1] = [c_1()] [natsFrom^#(X)] = [1 0] X + [2] [0 0] [2] > [1] [1] = [c_2()] [fst^#(pair(XS, YS))] = [1 0] XS + [1 0] YS + [2] [0 0] [0 0] [2] > [1] [1] = [c_3()] [snd^#(pair(XS, YS))] = [1 0] XS + [1 0] YS + [2] [0 0] [0 0] [2] > [1] [1] = [c_4()] [splitAt^#(0(), XS)] = [4] [4] > [1] [1] = [c_5()] [s^#(X)] = [1 0] X + [2] [0 0] [2] > [1] [1] = [c_6()] [u^#(pair(YS, ZS), N, X, XS)] = [2 0] YS + [1 2] X + [2 2] ZS + [4] [2 0] [1 2] [2 2] [4] ? [1 2] X + [2] [2 2] [3] = [c_7(activate^#(X))] [activate^#(X)] = [1 2] X + [1] [2 2] [2] > [0] [1] = [c_8()] [activate^#(n__natsFrom(X))] = [1 2] X + [4] [2 2] [6] ? [1 2] X + [6] [0 0] [4] = [c_9(natsFrom^#(activate(X)))] [activate^#(n__s(X))] = [1 2] X + [3] [2 2] [4] ? [1 2] X + [6] [0 0] [4] = [c_10(s^#(activate(X)))] [head^#(cons(N, XS))] = [1 0] XS + [1] [0 0] [2] >= [1] [1] = [c_11()] [tail^#(cons(N, XS))] = [2 2] XS + [2] [1 1] [2] ? [1 2] XS + [2] [2 2] [3] = [c_12(activate^#(XS))] [sel^#(N, XS)] = [2 1] N + [2 2] XS + [1] [2 1] [1 2] [1] ? [2 0] N + [2 1] XS + [4] [0 0] [0 0] [4] = [c_13(head^#(afterNth(N, XS)))] [afterNth^#(N, XS)] = [2 2] N + [2 2] XS + [1] [1 2] [2 1] [1] ? [1 0] N + [2 0] XS + [4] [0 0] [0 0] [4] = [c_14(snd^#(splitAt(N, XS)))] [take^#(N, XS)] = [2 1] N + [2 1] XS + [2] [1 1] [2 2] [1] ? [1 0] N + [2 0] XS + [4] [0 0] [0 0] [4] = [c_15(fst^#(splitAt(N, XS)))] 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: { u^#(pair(YS, ZS), N, X, XS) -> c_7(activate^#(X)) , activate^#(n__natsFrom(X)) -> c_9(natsFrom^#(activate(X))) , activate^#(n__s(X)) -> c_10(s^#(activate(X))) , head^#(cons(N, XS)) -> c_11() , tail^#(cons(N, XS)) -> c_12(activate^#(XS)) , sel^#(N, XS) -> c_13(head^#(afterNth(N, XS))) , afterNth^#(N, XS) -> c_14(snd^#(splitAt(N, XS))) , take^#(N, XS) -> c_15(fst^#(splitAt(N, XS))) } Weak DPs: { natsFrom^#(N) -> c_1() , natsFrom^#(X) -> c_2() , fst^#(pair(XS, YS)) -> c_3() , snd^#(pair(XS, YS)) -> c_4() , splitAt^#(0(), XS) -> c_5() , s^#(X) -> c_6() , activate^#(X) -> c_8() } Weak Trs: { natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) , natsFrom(X) -> n__natsFrom(X) , snd(pair(XS, YS)) -> YS , splitAt(0(), XS) -> pair(nil(), XS) , s(X) -> n__s(X) , activate(X) -> X , activate(n__natsFrom(X)) -> natsFrom(activate(X)) , activate(n__s(X)) -> s(activate(X)) , afterNth(N, XS) -> snd(splitAt(N, XS)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) We estimate the number of application of {2,3,4,7,8} by applications of Pre({2,3,4,7,8}) = {1,5,6}. Here rules are labeled as follows: DPs: { 1: u^#(pair(YS, ZS), N, X, XS) -> c_7(activate^#(X)) , 2: activate^#(n__natsFrom(X)) -> c_9(natsFrom^#(activate(X))) , 3: activate^#(n__s(X)) -> c_10(s^#(activate(X))) , 4: head^#(cons(N, XS)) -> c_11() , 5: tail^#(cons(N, XS)) -> c_12(activate^#(XS)) , 6: sel^#(N, XS) -> c_13(head^#(afterNth(N, XS))) , 7: afterNth^#(N, XS) -> c_14(snd^#(splitAt(N, XS))) , 8: take^#(N, XS) -> c_15(fst^#(splitAt(N, XS))) , 9: natsFrom^#(N) -> c_1() , 10: natsFrom^#(X) -> c_2() , 11: fst^#(pair(XS, YS)) -> c_3() , 12: snd^#(pair(XS, YS)) -> c_4() , 13: splitAt^#(0(), XS) -> c_5() , 14: s^#(X) -> c_6() , 15: activate^#(X) -> c_8() } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Strict DPs: { u^#(pair(YS, ZS), N, X, XS) -> c_7(activate^#(X)) , tail^#(cons(N, XS)) -> c_12(activate^#(XS)) , sel^#(N, XS) -> c_13(head^#(afterNth(N, XS))) } Weak DPs: { natsFrom^#(N) -> c_1() , natsFrom^#(X) -> c_2() , fst^#(pair(XS, YS)) -> c_3() , snd^#(pair(XS, YS)) -> c_4() , splitAt^#(0(), XS) -> c_5() , s^#(X) -> c_6() , activate^#(X) -> c_8() , activate^#(n__natsFrom(X)) -> c_9(natsFrom^#(activate(X))) , activate^#(n__s(X)) -> c_10(s^#(activate(X))) , head^#(cons(N, XS)) -> c_11() , afterNth^#(N, XS) -> c_14(snd^#(splitAt(N, XS))) , take^#(N, XS) -> c_15(fst^#(splitAt(N, XS))) } Weak Trs: { natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) , natsFrom(X) -> n__natsFrom(X) , snd(pair(XS, YS)) -> YS , splitAt(0(), XS) -> pair(nil(), XS) , s(X) -> n__s(X) , activate(X) -> X , activate(n__natsFrom(X)) -> natsFrom(activate(X)) , activate(n__s(X)) -> s(activate(X)) , afterNth(N, XS) -> snd(splitAt(N, XS)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) We estimate the number of application of {1,2,3} by applications of Pre({1,2,3}) = {}. Here rules are labeled as follows: DPs: { 1: u^#(pair(YS, ZS), N, X, XS) -> c_7(activate^#(X)) , 2: tail^#(cons(N, XS)) -> c_12(activate^#(XS)) , 3: sel^#(N, XS) -> c_13(head^#(afterNth(N, XS))) , 4: natsFrom^#(N) -> c_1() , 5: natsFrom^#(X) -> c_2() , 6: fst^#(pair(XS, YS)) -> c_3() , 7: snd^#(pair(XS, YS)) -> c_4() , 8: splitAt^#(0(), XS) -> c_5() , 9: s^#(X) -> c_6() , 10: activate^#(X) -> c_8() , 11: activate^#(n__natsFrom(X)) -> c_9(natsFrom^#(activate(X))) , 12: activate^#(n__s(X)) -> c_10(s^#(activate(X))) , 13: head^#(cons(N, XS)) -> c_11() , 14: afterNth^#(N, XS) -> c_14(snd^#(splitAt(N, XS))) , 15: take^#(N, XS) -> c_15(fst^#(splitAt(N, XS))) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak DPs: { natsFrom^#(N) -> c_1() , natsFrom^#(X) -> c_2() , fst^#(pair(XS, YS)) -> c_3() , snd^#(pair(XS, YS)) -> c_4() , splitAt^#(0(), XS) -> c_5() , s^#(X) -> c_6() , u^#(pair(YS, ZS), N, X, XS) -> c_7(activate^#(X)) , activate^#(X) -> c_8() , activate^#(n__natsFrom(X)) -> c_9(natsFrom^#(activate(X))) , activate^#(n__s(X)) -> c_10(s^#(activate(X))) , head^#(cons(N, XS)) -> c_11() , tail^#(cons(N, XS)) -> c_12(activate^#(XS)) , sel^#(N, XS) -> c_13(head^#(afterNth(N, XS))) , afterNth^#(N, XS) -> c_14(snd^#(splitAt(N, XS))) , take^#(N, XS) -> c_15(fst^#(splitAt(N, XS))) } Weak Trs: { natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) , natsFrom(X) -> n__natsFrom(X) , snd(pair(XS, YS)) -> YS , splitAt(0(), XS) -> pair(nil(), XS) , s(X) -> n__s(X) , activate(X) -> X , activate(n__natsFrom(X)) -> natsFrom(activate(X)) , activate(n__s(X)) -> s(activate(X)) , afterNth(N, XS) -> snd(splitAt(N, XS)) } 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. { natsFrom^#(N) -> c_1() , natsFrom^#(X) -> c_2() , fst^#(pair(XS, YS)) -> c_3() , snd^#(pair(XS, YS)) -> c_4() , splitAt^#(0(), XS) -> c_5() , s^#(X) -> c_6() , u^#(pair(YS, ZS), N, X, XS) -> c_7(activate^#(X)) , activate^#(X) -> c_8() , activate^#(n__natsFrom(X)) -> c_9(natsFrom^#(activate(X))) , activate^#(n__s(X)) -> c_10(s^#(activate(X))) , head^#(cons(N, XS)) -> c_11() , tail^#(cons(N, XS)) -> c_12(activate^#(XS)) , sel^#(N, XS) -> c_13(head^#(afterNth(N, XS))) , afterNth^#(N, XS) -> c_14(snd^#(splitAt(N, XS))) , take^#(N, XS) -> c_15(fst^#(splitAt(N, XS))) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak Trs: { natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) , natsFrom(X) -> n__natsFrom(X) , snd(pair(XS, YS)) -> YS , splitAt(0(), XS) -> pair(nil(), XS) , s(X) -> n__s(X) , activate(X) -> X , activate(n__natsFrom(X)) -> natsFrom(activate(X)) , activate(n__s(X)) -> s(activate(X)) , afterNth(N, XS) -> snd(splitAt(N, XS)) } 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)). 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))