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: { U11(tt(), N, X, XS) -> U12(splitAt(activate(N), activate(XS)), activate(X)) , U12(pair(YS, ZS), X) -> pair(cons(activate(X), YS), ZS) , splitAt(0(), XS) -> pair(nil(), XS) , splitAt(s(N), cons(X, XS)) -> U11(tt(), N, X, activate(XS)) , activate(X) -> X , activate(n__natsFrom(X)) -> natsFrom(activate(X)) , activate(n__s(X)) -> s(activate(X)) , afterNth(N, XS) -> snd(splitAt(N, XS)) , snd(pair(X, Y)) -> Y , and(tt(), X) -> activate(X) , fst(pair(X, Y)) -> X , head(cons(N, XS)) -> N , natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) , natsFrom(X) -> n__natsFrom(X) , sel(N, XS) -> head(afterNth(N, XS)) , s(X) -> n__s(X) , tail(cons(N, XS)) -> activate(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)) -> U11(tt(), 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: { U11(tt(), N, X, XS) -> U12(splitAt(activate(N), activate(XS)), activate(X)) , U12(pair(YS, ZS), X) -> pair(cons(activate(X), YS), ZS) , splitAt(0(), XS) -> pair(nil(), XS) , activate(X) -> X , activate(n__natsFrom(X)) -> natsFrom(activate(X)) , activate(n__s(X)) -> s(activate(X)) , afterNth(N, XS) -> snd(splitAt(N, XS)) , snd(pair(X, Y)) -> Y , and(tt(), X) -> activate(X) , fst(pair(X, Y)) -> X , head(cons(N, XS)) -> N , natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) , natsFrom(X) -> n__natsFrom(X) , sel(N, XS) -> head(afterNth(N, XS)) , s(X) -> n__s(X) , tail(cons(N, XS)) -> activate(XS) , take(N, XS) -> fst(splitAt(N, XS)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We add the following dependency tuples: Strict DPs: { U11^#(tt(), N, X, XS) -> c_1(U12^#(splitAt(activate(N), activate(XS)), activate(X)), splitAt^#(activate(N), activate(XS)), activate^#(N), activate^#(XS), activate^#(X)) , U12^#(pair(YS, ZS), X) -> c_2(activate^#(X)) , splitAt^#(0(), XS) -> c_3() , activate^#(X) -> c_4() , activate^#(n__natsFrom(X)) -> c_5(natsFrom^#(activate(X)), activate^#(X)) , activate^#(n__s(X)) -> c_6(s^#(activate(X)), activate^#(X)) , natsFrom^#(N) -> c_12() , natsFrom^#(X) -> c_13() , s^#(X) -> c_15() , afterNth^#(N, XS) -> c_7(snd^#(splitAt(N, XS)), splitAt^#(N, XS)) , snd^#(pair(X, Y)) -> c_8() , and^#(tt(), X) -> c_9(activate^#(X)) , fst^#(pair(X, Y)) -> c_10() , head^#(cons(N, XS)) -> c_11() , sel^#(N, XS) -> c_14(head^#(afterNth(N, XS)), afterNth^#(N, XS)) , tail^#(cons(N, XS)) -> c_16(activate^#(XS)) , take^#(N, XS) -> c_17(fst^#(splitAt(N, XS)), 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: { U11^#(tt(), N, X, XS) -> c_1(U12^#(splitAt(activate(N), activate(XS)), activate(X)), splitAt^#(activate(N), activate(XS)), activate^#(N), activate^#(XS), activate^#(X)) , U12^#(pair(YS, ZS), X) -> c_2(activate^#(X)) , splitAt^#(0(), XS) -> c_3() , activate^#(X) -> c_4() , activate^#(n__natsFrom(X)) -> c_5(natsFrom^#(activate(X)), activate^#(X)) , activate^#(n__s(X)) -> c_6(s^#(activate(X)), activate^#(X)) , natsFrom^#(N) -> c_12() , natsFrom^#(X) -> c_13() , s^#(X) -> c_15() , afterNth^#(N, XS) -> c_7(snd^#(splitAt(N, XS)), splitAt^#(N, XS)) , snd^#(pair(X, Y)) -> c_8() , and^#(tt(), X) -> c_9(activate^#(X)) , fst^#(pair(X, Y)) -> c_10() , head^#(cons(N, XS)) -> c_11() , sel^#(N, XS) -> c_14(head^#(afterNth(N, XS)), afterNth^#(N, XS)) , tail^#(cons(N, XS)) -> c_16(activate^#(XS)) , take^#(N, XS) -> c_17(fst^#(splitAt(N, XS)), splitAt^#(N, XS)) } Weak Trs: { U11(tt(), N, X, XS) -> U12(splitAt(activate(N), activate(XS)), activate(X)) , U12(pair(YS, ZS), X) -> pair(cons(activate(X), YS), ZS) , splitAt(0(), XS) -> pair(nil(), XS) , activate(X) -> X , activate(n__natsFrom(X)) -> natsFrom(activate(X)) , activate(n__s(X)) -> s(activate(X)) , afterNth(N, XS) -> snd(splitAt(N, XS)) , snd(pair(X, Y)) -> Y , and(tt(), X) -> activate(X) , fst(pair(X, Y)) -> X , head(cons(N, XS)) -> N , natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) , natsFrom(X) -> n__natsFrom(X) , sel(N, XS) -> head(afterNth(N, XS)) , s(X) -> n__s(X) , tail(cons(N, XS)) -> activate(XS) , take(N, XS) -> fst(splitAt(N, XS)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We estimate the number of application of {3,4,7,8,9,11,13,14} by applications of Pre({3,4,7,8,9,11,13,14}) = {1,2,5,6,10,12,15,16,17}. Here rules are labeled as follows: DPs: { 1: U11^#(tt(), N, X, XS) -> c_1(U12^#(splitAt(activate(N), activate(XS)), activate(X)), splitAt^#(activate(N), activate(XS)), activate^#(N), activate^#(XS), activate^#(X)) , 2: U12^#(pair(YS, ZS), X) -> c_2(activate^#(X)) , 3: splitAt^#(0(), XS) -> c_3() , 4: activate^#(X) -> c_4() , 5: activate^#(n__natsFrom(X)) -> c_5(natsFrom^#(activate(X)), activate^#(X)) , 6: activate^#(n__s(X)) -> c_6(s^#(activate(X)), activate^#(X)) , 7: natsFrom^#(N) -> c_12() , 8: natsFrom^#(X) -> c_13() , 9: s^#(X) -> c_15() , 10: afterNth^#(N, XS) -> c_7(snd^#(splitAt(N, XS)), splitAt^#(N, XS)) , 11: snd^#(pair(X, Y)) -> c_8() , 12: and^#(tt(), X) -> c_9(activate^#(X)) , 13: fst^#(pair(X, Y)) -> c_10() , 14: head^#(cons(N, XS)) -> c_11() , 15: sel^#(N, XS) -> c_14(head^#(afterNth(N, XS)), afterNth^#(N, XS)) , 16: tail^#(cons(N, XS)) -> c_16(activate^#(XS)) , 17: take^#(N, XS) -> c_17(fst^#(splitAt(N, XS)), splitAt^#(N, XS)) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { U11^#(tt(), N, X, XS) -> c_1(U12^#(splitAt(activate(N), activate(XS)), activate(X)), splitAt^#(activate(N), activate(XS)), activate^#(N), activate^#(XS), activate^#(X)) , U12^#(pair(YS, ZS), X) -> c_2(activate^#(X)) , activate^#(n__natsFrom(X)) -> c_5(natsFrom^#(activate(X)), activate^#(X)) , activate^#(n__s(X)) -> c_6(s^#(activate(X)), activate^#(X)) , afterNth^#(N, XS) -> c_7(snd^#(splitAt(N, XS)), splitAt^#(N, XS)) , and^#(tt(), X) -> c_9(activate^#(X)) , sel^#(N, XS) -> c_14(head^#(afterNth(N, XS)), afterNth^#(N, XS)) , tail^#(cons(N, XS)) -> c_16(activate^#(XS)) , take^#(N, XS) -> c_17(fst^#(splitAt(N, XS)), splitAt^#(N, XS)) } Weak DPs: { splitAt^#(0(), XS) -> c_3() , activate^#(X) -> c_4() , natsFrom^#(N) -> c_12() , natsFrom^#(X) -> c_13() , s^#(X) -> c_15() , snd^#(pair(X, Y)) -> c_8() , fst^#(pair(X, Y)) -> c_10() , head^#(cons(N, XS)) -> c_11() } Weak Trs: { U11(tt(), N, X, XS) -> U12(splitAt(activate(N), activate(XS)), activate(X)) , U12(pair(YS, ZS), X) -> pair(cons(activate(X), YS), ZS) , splitAt(0(), XS) -> pair(nil(), XS) , activate(X) -> X , activate(n__natsFrom(X)) -> natsFrom(activate(X)) , activate(n__s(X)) -> s(activate(X)) , afterNth(N, XS) -> snd(splitAt(N, XS)) , snd(pair(X, Y)) -> Y , and(tt(), X) -> activate(X) , fst(pair(X, Y)) -> X , head(cons(N, XS)) -> N , natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) , natsFrom(X) -> n__natsFrom(X) , sel(N, XS) -> head(afterNth(N, XS)) , s(X) -> n__s(X) , tail(cons(N, XS)) -> activate(XS) , take(N, XS) -> fst(splitAt(N, XS)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We estimate the number of application of {5,9} by applications of Pre({5,9}) = {7}. Here rules are labeled as follows: DPs: { 1: U11^#(tt(), N, X, XS) -> c_1(U12^#(splitAt(activate(N), activate(XS)), activate(X)), splitAt^#(activate(N), activate(XS)), activate^#(N), activate^#(XS), activate^#(X)) , 2: U12^#(pair(YS, ZS), X) -> c_2(activate^#(X)) , 3: activate^#(n__natsFrom(X)) -> c_5(natsFrom^#(activate(X)), activate^#(X)) , 4: activate^#(n__s(X)) -> c_6(s^#(activate(X)), activate^#(X)) , 5: afterNth^#(N, XS) -> c_7(snd^#(splitAt(N, XS)), splitAt^#(N, XS)) , 6: and^#(tt(), X) -> c_9(activate^#(X)) , 7: sel^#(N, XS) -> c_14(head^#(afterNth(N, XS)), afterNth^#(N, XS)) , 8: tail^#(cons(N, XS)) -> c_16(activate^#(XS)) , 9: take^#(N, XS) -> c_17(fst^#(splitAt(N, XS)), splitAt^#(N, XS)) , 10: splitAt^#(0(), XS) -> c_3() , 11: activate^#(X) -> c_4() , 12: natsFrom^#(N) -> c_12() , 13: natsFrom^#(X) -> c_13() , 14: s^#(X) -> c_15() , 15: snd^#(pair(X, Y)) -> c_8() , 16: fst^#(pair(X, Y)) -> c_10() , 17: head^#(cons(N, XS)) -> c_11() } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { U11^#(tt(), N, X, XS) -> c_1(U12^#(splitAt(activate(N), activate(XS)), activate(X)), splitAt^#(activate(N), activate(XS)), activate^#(N), activate^#(XS), activate^#(X)) , U12^#(pair(YS, ZS), X) -> c_2(activate^#(X)) , activate^#(n__natsFrom(X)) -> c_5(natsFrom^#(activate(X)), activate^#(X)) , activate^#(n__s(X)) -> c_6(s^#(activate(X)), activate^#(X)) , and^#(tt(), X) -> c_9(activate^#(X)) , sel^#(N, XS) -> c_14(head^#(afterNth(N, XS)), afterNth^#(N, XS)) , tail^#(cons(N, XS)) -> c_16(activate^#(XS)) } Weak DPs: { splitAt^#(0(), XS) -> c_3() , activate^#(X) -> c_4() , natsFrom^#(N) -> c_12() , natsFrom^#(X) -> c_13() , s^#(X) -> c_15() , afterNth^#(N, XS) -> c_7(snd^#(splitAt(N, XS)), splitAt^#(N, XS)) , snd^#(pair(X, Y)) -> c_8() , fst^#(pair(X, Y)) -> c_10() , head^#(cons(N, XS)) -> c_11() , take^#(N, XS) -> c_17(fst^#(splitAt(N, XS)), splitAt^#(N, XS)) } Weak Trs: { U11(tt(), N, X, XS) -> U12(splitAt(activate(N), activate(XS)), activate(X)) , U12(pair(YS, ZS), X) -> pair(cons(activate(X), YS), ZS) , splitAt(0(), XS) -> pair(nil(), XS) , activate(X) -> X , activate(n__natsFrom(X)) -> natsFrom(activate(X)) , activate(n__s(X)) -> s(activate(X)) , afterNth(N, XS) -> snd(splitAt(N, XS)) , snd(pair(X, Y)) -> Y , and(tt(), X) -> activate(X) , fst(pair(X, Y)) -> X , head(cons(N, XS)) -> N , natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) , natsFrom(X) -> n__natsFrom(X) , sel(N, XS) -> head(afterNth(N, XS)) , s(X) -> n__s(X) , tail(cons(N, XS)) -> activate(XS) , take(N, XS) -> fst(splitAt(N, XS)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We estimate the number of application of {6} by applications of Pre({6}) = {}. Here rules are labeled as follows: DPs: { 1: U11^#(tt(), N, X, XS) -> c_1(U12^#(splitAt(activate(N), activate(XS)), activate(X)), splitAt^#(activate(N), activate(XS)), activate^#(N), activate^#(XS), activate^#(X)) , 2: U12^#(pair(YS, ZS), X) -> c_2(activate^#(X)) , 3: activate^#(n__natsFrom(X)) -> c_5(natsFrom^#(activate(X)), activate^#(X)) , 4: activate^#(n__s(X)) -> c_6(s^#(activate(X)), activate^#(X)) , 5: and^#(tt(), X) -> c_9(activate^#(X)) , 6: sel^#(N, XS) -> c_14(head^#(afterNth(N, XS)), afterNth^#(N, XS)) , 7: tail^#(cons(N, XS)) -> c_16(activate^#(XS)) , 8: splitAt^#(0(), XS) -> c_3() , 9: activate^#(X) -> c_4() , 10: natsFrom^#(N) -> c_12() , 11: natsFrom^#(X) -> c_13() , 12: s^#(X) -> c_15() , 13: afterNth^#(N, XS) -> c_7(snd^#(splitAt(N, XS)), splitAt^#(N, XS)) , 14: snd^#(pair(X, Y)) -> c_8() , 15: fst^#(pair(X, Y)) -> c_10() , 16: head^#(cons(N, XS)) -> c_11() , 17: take^#(N, XS) -> c_17(fst^#(splitAt(N, XS)), splitAt^#(N, XS)) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { U11^#(tt(), N, X, XS) -> c_1(U12^#(splitAt(activate(N), activate(XS)), activate(X)), splitAt^#(activate(N), activate(XS)), activate^#(N), activate^#(XS), activate^#(X)) , U12^#(pair(YS, ZS), X) -> c_2(activate^#(X)) , activate^#(n__natsFrom(X)) -> c_5(natsFrom^#(activate(X)), activate^#(X)) , activate^#(n__s(X)) -> c_6(s^#(activate(X)), activate^#(X)) , and^#(tt(), X) -> c_9(activate^#(X)) , tail^#(cons(N, XS)) -> c_16(activate^#(XS)) } Weak DPs: { splitAt^#(0(), XS) -> c_3() , activate^#(X) -> c_4() , natsFrom^#(N) -> c_12() , natsFrom^#(X) -> c_13() , s^#(X) -> c_15() , afterNth^#(N, XS) -> c_7(snd^#(splitAt(N, XS)), splitAt^#(N, XS)) , snd^#(pair(X, Y)) -> c_8() , fst^#(pair(X, Y)) -> c_10() , head^#(cons(N, XS)) -> c_11() , sel^#(N, XS) -> c_14(head^#(afterNth(N, XS)), afterNth^#(N, XS)) , take^#(N, XS) -> c_17(fst^#(splitAt(N, XS)), splitAt^#(N, XS)) } Weak Trs: { U11(tt(), N, X, XS) -> U12(splitAt(activate(N), activate(XS)), activate(X)) , U12(pair(YS, ZS), X) -> pair(cons(activate(X), YS), ZS) , splitAt(0(), XS) -> pair(nil(), XS) , activate(X) -> X , activate(n__natsFrom(X)) -> natsFrom(activate(X)) , activate(n__s(X)) -> s(activate(X)) , afterNth(N, XS) -> snd(splitAt(N, XS)) , snd(pair(X, Y)) -> Y , and(tt(), X) -> activate(X) , fst(pair(X, Y)) -> X , head(cons(N, XS)) -> N , natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) , natsFrom(X) -> n__natsFrom(X) , sel(N, XS) -> head(afterNth(N, XS)) , s(X) -> n__s(X) , tail(cons(N, XS)) -> activate(XS) , take(N, XS) -> fst(splitAt(N, XS)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) The following weak DPs constitute a sub-graph of the DG that is closed under successors. The DPs are removed. { splitAt^#(0(), XS) -> c_3() , activate^#(X) -> c_4() , natsFrom^#(N) -> c_12() , natsFrom^#(X) -> c_13() , s^#(X) -> c_15() , afterNth^#(N, XS) -> c_7(snd^#(splitAt(N, XS)), splitAt^#(N, XS)) , snd^#(pair(X, Y)) -> c_8() , fst^#(pair(X, Y)) -> c_10() , head^#(cons(N, XS)) -> c_11() , sel^#(N, XS) -> c_14(head^#(afterNth(N, XS)), afterNth^#(N, XS)) , take^#(N, XS) -> c_17(fst^#(splitAt(N, XS)), splitAt^#(N, XS)) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { U11^#(tt(), N, X, XS) -> c_1(U12^#(splitAt(activate(N), activate(XS)), activate(X)), splitAt^#(activate(N), activate(XS)), activate^#(N), activate^#(XS), activate^#(X)) , U12^#(pair(YS, ZS), X) -> c_2(activate^#(X)) , activate^#(n__natsFrom(X)) -> c_5(natsFrom^#(activate(X)), activate^#(X)) , activate^#(n__s(X)) -> c_6(s^#(activate(X)), activate^#(X)) , and^#(tt(), X) -> c_9(activate^#(X)) , tail^#(cons(N, XS)) -> c_16(activate^#(XS)) } Weak Trs: { U11(tt(), N, X, XS) -> U12(splitAt(activate(N), activate(XS)), activate(X)) , U12(pair(YS, ZS), X) -> pair(cons(activate(X), YS), ZS) , splitAt(0(), XS) -> pair(nil(), XS) , activate(X) -> X , activate(n__natsFrom(X)) -> natsFrom(activate(X)) , activate(n__s(X)) -> s(activate(X)) , afterNth(N, XS) -> snd(splitAt(N, XS)) , snd(pair(X, Y)) -> Y , and(tt(), X) -> activate(X) , fst(pair(X, Y)) -> X , head(cons(N, XS)) -> N , natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) , natsFrom(X) -> n__natsFrom(X) , sel(N, XS) -> head(afterNth(N, XS)) , s(X) -> n__s(X) , tail(cons(N, XS)) -> activate(XS) , take(N, XS) -> fst(splitAt(N, XS)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) Due to missing edges in the dependency-graph, the right-hand sides of following rules could be simplified: { U11^#(tt(), N, X, XS) -> c_1(U12^#(splitAt(activate(N), activate(XS)), activate(X)), splitAt^#(activate(N), activate(XS)), activate^#(N), activate^#(XS), activate^#(X)) , activate^#(n__natsFrom(X)) -> c_5(natsFrom^#(activate(X)), activate^#(X)) , activate^#(n__s(X)) -> c_6(s^#(activate(X)), activate^#(X)) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { U11^#(tt(), N, X, XS) -> c_1(U12^#(splitAt(activate(N), activate(XS)), activate(X)), activate^#(N), activate^#(XS), activate^#(X)) , U12^#(pair(YS, ZS), X) -> c_2(activate^#(X)) , activate^#(n__natsFrom(X)) -> c_3(activate^#(X)) , activate^#(n__s(X)) -> c_4(activate^#(X)) , and^#(tt(), X) -> c_5(activate^#(X)) , tail^#(cons(N, XS)) -> c_6(activate^#(XS)) } Weak Trs: { U11(tt(), N, X, XS) -> U12(splitAt(activate(N), activate(XS)), activate(X)) , U12(pair(YS, ZS), X) -> pair(cons(activate(X), YS), ZS) , splitAt(0(), XS) -> pair(nil(), XS) , activate(X) -> X , activate(n__natsFrom(X)) -> natsFrom(activate(X)) , activate(n__s(X)) -> s(activate(X)) , afterNth(N, XS) -> snd(splitAt(N, XS)) , snd(pair(X, Y)) -> Y , and(tt(), X) -> activate(X) , fst(pair(X, Y)) -> X , head(cons(N, XS)) -> N , natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) , natsFrom(X) -> n__natsFrom(X) , sel(N, XS) -> head(afterNth(N, XS)) , s(X) -> n__s(X) , tail(cons(N, XS)) -> activate(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: Weak Usable Rules: { splitAt(0(), XS) -> pair(nil(), XS) , activate(X) -> X , activate(n__natsFrom(X)) -> natsFrom(activate(X)) , activate(n__s(X)) -> s(activate(X)) , natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) , natsFrom(X) -> n__natsFrom(X) , s(X) -> n__s(X) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { U11^#(tt(), N, X, XS) -> c_1(U12^#(splitAt(activate(N), activate(XS)), activate(X)), activate^#(N), activate^#(XS), activate^#(X)) , U12^#(pair(YS, ZS), X) -> c_2(activate^#(X)) , activate^#(n__natsFrom(X)) -> c_3(activate^#(X)) , activate^#(n__s(X)) -> c_4(activate^#(X)) , and^#(tt(), X) -> c_5(activate^#(X)) , tail^#(cons(N, XS)) -> c_6(activate^#(XS)) } Weak Trs: { splitAt(0(), XS) -> pair(nil(), XS) , activate(X) -> X , activate(n__natsFrom(X)) -> natsFrom(activate(X)) , activate(n__s(X)) -> s(activate(X)) , natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) , natsFrom(X) -> n__natsFrom(X) , s(X) -> n__s(X) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) Consider the dependency graph 1: U11^#(tt(), N, X, XS) -> c_1(U12^#(splitAt(activate(N), activate(XS)), activate(X)), activate^#(N), activate^#(XS), activate^#(X)) -->_4 activate^#(n__s(X)) -> c_4(activate^#(X)) :4 -->_3 activate^#(n__s(X)) -> c_4(activate^#(X)) :4 -->_2 activate^#(n__s(X)) -> c_4(activate^#(X)) :4 -->_4 activate^#(n__natsFrom(X)) -> c_3(activate^#(X)) :3 -->_3 activate^#(n__natsFrom(X)) -> c_3(activate^#(X)) :3 -->_2 activate^#(n__natsFrom(X)) -> c_3(activate^#(X)) :3 -->_1 U12^#(pair(YS, ZS), X) -> c_2(activate^#(X)) :2 2: U12^#(pair(YS, ZS), X) -> c_2(activate^#(X)) -->_1 activate^#(n__s(X)) -> c_4(activate^#(X)) :4 -->_1 activate^#(n__natsFrom(X)) -> c_3(activate^#(X)) :3 3: activate^#(n__natsFrom(X)) -> c_3(activate^#(X)) -->_1 activate^#(n__s(X)) -> c_4(activate^#(X)) :4 -->_1 activate^#(n__natsFrom(X)) -> c_3(activate^#(X)) :3 4: activate^#(n__s(X)) -> c_4(activate^#(X)) -->_1 activate^#(n__s(X)) -> c_4(activate^#(X)) :4 -->_1 activate^#(n__natsFrom(X)) -> c_3(activate^#(X)) :3 5: and^#(tt(), X) -> c_5(activate^#(X)) -->_1 activate^#(n__s(X)) -> c_4(activate^#(X)) :4 -->_1 activate^#(n__natsFrom(X)) -> c_3(activate^#(X)) :3 6: tail^#(cons(N, XS)) -> c_6(activate^#(XS)) -->_1 activate^#(n__s(X)) -> c_4(activate^#(X)) :4 -->_1 activate^#(n__natsFrom(X)) -> c_3(activate^#(X)) :3 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). { and^#(tt(), X) -> c_5(activate^#(X)) , tail^#(cons(N, XS)) -> c_6(activate^#(XS)) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { U11^#(tt(), N, X, XS) -> c_1(U12^#(splitAt(activate(N), activate(XS)), activate(X)), activate^#(N), activate^#(XS), activate^#(X)) , U12^#(pair(YS, ZS), X) -> c_2(activate^#(X)) , activate^#(n__natsFrom(X)) -> c_3(activate^#(X)) , activate^#(n__s(X)) -> c_4(activate^#(X)) } Weak Trs: { splitAt(0(), XS) -> pair(nil(), XS) , activate(X) -> X , activate(n__natsFrom(X)) -> natsFrom(activate(X)) , activate(n__s(X)) -> s(activate(X)) , natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) , natsFrom(X) -> n__natsFrom(X) , s(X) -> n__s(X) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We analyse the complexity of following sub-problems (R) and (S). Problem (S) is obtained from the input problem by shifting strict rules from (R) into the weak component: Problem (R): ------------ Strict DPs: { U11^#(tt(), N, X, XS) -> c_1(U12^#(splitAt(activate(N), activate(XS)), activate(X)), activate^#(N), activate^#(XS), activate^#(X)) , activate^#(n__natsFrom(X)) -> c_3(activate^#(X)) , activate^#(n__s(X)) -> c_4(activate^#(X)) } Weak DPs: { U12^#(pair(YS, ZS), X) -> c_2(activate^#(X)) } Weak Trs: { splitAt(0(), XS) -> pair(nil(), XS) , activate(X) -> X , activate(n__natsFrom(X)) -> natsFrom(activate(X)) , activate(n__s(X)) -> s(activate(X)) , natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) , natsFrom(X) -> n__natsFrom(X) , s(X) -> n__s(X) } StartTerms: basic terms Defined Symbols: U11^# U12^# splitAt^# activate^# natsFrom^# s^# afterNth^# snd^# and^# fst^# head^# sel^# tail^# take^# Constructors: tt pair cons n__natsFrom n__s 0 nil Strategy: innermost Problem (S): ------------ Strict DPs: { U12^#(pair(YS, ZS), X) -> c_2(activate^#(X)) } Weak DPs: { U11^#(tt(), N, X, XS) -> c_1(U12^#(splitAt(activate(N), activate(XS)), activate(X)), activate^#(N), activate^#(XS), activate^#(X)) , activate^#(n__natsFrom(X)) -> c_3(activate^#(X)) , activate^#(n__s(X)) -> c_4(activate^#(X)) } Weak Trs: { splitAt(0(), XS) -> pair(nil(), XS) , activate(X) -> X , activate(n__natsFrom(X)) -> natsFrom(activate(X)) , activate(n__s(X)) -> s(activate(X)) , natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) , natsFrom(X) -> n__natsFrom(X) , s(X) -> n__s(X) } StartTerms: basic terms Defined Symbols: U11^# U12^# splitAt^# activate^# natsFrom^# s^# afterNth^# snd^# and^# fst^# head^# sel^# tail^# take^# Constructors: tt pair cons n__natsFrom n__s 0 nil Strategy: innermost Overall, the transformation results in the following sub-problem(s): Generated new problems: ----------------------- R) Strict DPs: { U11^#(tt(), N, X, XS) -> c_1(U12^#(splitAt(activate(N), activate(XS)), activate(X)), activate^#(N), activate^#(XS), activate^#(X)) , activate^#(n__natsFrom(X)) -> c_3(activate^#(X)) , activate^#(n__s(X)) -> c_4(activate^#(X)) } Weak DPs: { U12^#(pair(YS, ZS), X) -> c_2(activate^#(X)) } Weak Trs: { splitAt(0(), XS) -> pair(nil(), XS) , activate(X) -> X , activate(n__natsFrom(X)) -> natsFrom(activate(X)) , activate(n__s(X)) -> s(activate(X)) , natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) , natsFrom(X) -> n__natsFrom(X) , s(X) -> n__s(X) } StartTerms: basic terms Defined Symbols: U11^# U12^# splitAt^# activate^# natsFrom^# s^# afterNth^# snd^# and^# fst^# head^# sel^# tail^# take^# Constructors: tt pair cons n__natsFrom n__s 0 nil Strategy: innermost This problem was proven YES(O(1),O(n^1)). S) Strict DPs: { U12^#(pair(YS, ZS), X) -> c_2(activate^#(X)) } Weak DPs: { U11^#(tt(), N, X, XS) -> c_1(U12^#(splitAt(activate(N), activate(XS)), activate(X)), activate^#(N), activate^#(XS), activate^#(X)) , activate^#(n__natsFrom(X)) -> c_3(activate^#(X)) , activate^#(n__s(X)) -> c_4(activate^#(X)) } Weak Trs: { splitAt(0(), XS) -> pair(nil(), XS) , activate(X) -> X , activate(n__natsFrom(X)) -> natsFrom(activate(X)) , activate(n__s(X)) -> s(activate(X)) , natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) , natsFrom(X) -> n__natsFrom(X) , s(X) -> n__s(X) } StartTerms: basic terms Defined Symbols: U11^# U12^# splitAt^# activate^# natsFrom^# s^# afterNth^# snd^# and^# fst^# head^# sel^# tail^# take^# Constructors: tt pair cons n__natsFrom n__s 0 nil Strategy: innermost This problem was proven YES(O(1),O(1)). Proofs for generated problems: ------------------------------ R) We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { U11^#(tt(), N, X, XS) -> c_1(U12^#(splitAt(activate(N), activate(XS)), activate(X)), activate^#(N), activate^#(XS), activate^#(X)) , activate^#(n__natsFrom(X)) -> c_3(activate^#(X)) , activate^#(n__s(X)) -> c_4(activate^#(X)) } Weak DPs: { U12^#(pair(YS, ZS), X) -> c_2(activate^#(X)) } Weak Trs: { splitAt(0(), XS) -> pair(nil(), XS) , activate(X) -> X , activate(n__natsFrom(X)) -> natsFrom(activate(X)) , activate(n__s(X)) -> s(activate(X)) , natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) , natsFrom(X) -> n__natsFrom(X) , s(X) -> n__s(X) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. DPs: { 1: U11^#(tt(), N, X, XS) -> c_1(U12^#(splitAt(activate(N), activate(XS)), activate(X)), activate^#(N), activate^#(XS), activate^#(X)) , 3: activate^#(n__s(X)) -> c_4(activate^#(X)) } Trs: { activate(n__s(X)) -> s(activate(X)) } Sub-proof: ---------- The following argument positions are usable: Uargs(c_1) = {1, 2, 3, 4}, Uargs(c_2) = {1}, Uargs(c_3) = {1}, Uargs(c_4) = {1} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [U11](x1, x2, x3, x4) = [7] x1 + [7] x2 + [7] x3 + [7] x4 + [0] [tt] = [0] [U12](x1, x2) = [7] x1 + [7] x2 + [0] [splitAt](x1, x2) = [0] [activate](x1) = [2] x1 + [0] [pair](x1, x2) = [1] x1 + [0] [cons](x1, x2) = [1] x1 + [0] [afterNth](x1, x2) = [7] x1 + [7] x2 + [0] [snd](x1) = [7] x1 + [0] [and](x1, x2) = [7] x1 + [7] x2 + [0] [fst](x1) = [7] x1 + [0] [head](x1) = [7] x1 + [0] [natsFrom](x1) = [1] x1 + [0] [n__natsFrom](x1) = [1] x1 + [0] [n__s](x1) = [1] x1 + [4] [sel](x1, x2) = [7] x1 + [7] x2 + [0] [0] = [7] [nil] = [0] [s](x1) = [1] x1 + [4] [tail](x1) = [7] x1 + [0] [take](x1, x2) = [7] x1 + [7] x2 + [0] [U11^#](x1, x2, x3, x4) = [7] x1 + [7] x2 + [7] x3 + [7] x4 + [7] [c_1](x1, x2, x3, x4, x5) = [7] x1 + [7] x2 + [7] x3 + [7] x4 + [7] x5 + [0] [U12^#](x1, x2) = [2] x2 + [0] [splitAt^#](x1, x2) = [7] x1 + [7] x2 + [0] [activate^#](x1) = [2] x1 + [0] [c_2](x1) = [7] x1 + [0] [c_3] = [0] [c_4] = [0] [c_5](x1, x2) = [7] x1 + [7] x2 + [0] [natsFrom^#](x1) = [7] x1 + [0] [c_6](x1, x2) = [7] x1 + [7] x2 + [0] [s^#](x1) = [7] x1 + [0] [afterNth^#](x1, x2) = [7] x1 + [7] x2 + [0] [c_7](x1, x2) = [7] x1 + [7] x2 + [0] [snd^#](x1) = [7] x1 + [0] [c_8] = [0] [and^#](x1, x2) = [7] x1 + [7] x2 + [0] [c_9](x1) = [7] x1 + [0] [fst^#](x1) = [7] x1 + [0] [c_10] = [0] [head^#](x1) = [7] x1 + [0] [c_11] = [0] [c_12] = [0] [c_13] = [0] [sel^#](x1, x2) = [7] x1 + [7] x2 + [0] [c_14](x1, x2) = [7] x1 + [7] x2 + [0] [c_15] = [0] [tail^#](x1) = [7] x1 + [0] [c_16](x1) = [7] x1 + [0] [take^#](x1, x2) = [7] x1 + [7] x2 + [0] [c_17](x1, x2) = [7] x1 + [7] x2 + [0] [c] = [0] [c_1](x1, x2, x3, x4) = [1] x1 + [1] x2 + [1] x3 + [1] x4 + [3] [c_2](x1) = [1] x1 + [0] [c_3](x1) = [1] x1 + [0] [c_4](x1) = [1] x1 + [5] [c_5](x1) = [7] x1 + [0] [c_6](x1) = [7] x1 + [0] The following symbols are considered usable {splitAt, activate, natsFrom, s, U11^#, U12^#, activate^#} The order satisfies the following ordering constraints: [splitAt(0(), XS)] = [0] >= [0] = [pair(nil(), XS)] [activate(X)] = [2] X + [0] >= [1] X + [0] = [X] [activate(n__natsFrom(X))] = [2] X + [0] >= [2] X + [0] = [natsFrom(activate(X))] [activate(n__s(X))] = [2] X + [8] > [2] X + [4] = [s(activate(X))] [natsFrom(N)] = [1] N + [0] >= [1] N + [0] = [cons(N, n__natsFrom(n__s(N)))] [natsFrom(X)] = [1] X + [0] >= [1] X + [0] = [n__natsFrom(X)] [s(X)] = [1] X + [4] >= [1] X + [4] = [n__s(X)] [U11^#(tt(), N, X, XS)] = [7] N + [7] X + [7] XS + [7] > [2] N + [6] X + [2] XS + [3] = [c_1(U12^#(splitAt(activate(N), activate(XS)), activate(X)), activate^#(N), activate^#(XS), activate^#(X))] [U12^#(pair(YS, ZS), X)] = [2] X + [0] >= [2] X + [0] = [c_2(activate^#(X))] [activate^#(n__natsFrom(X))] = [2] X + [0] >= [2] X + [0] = [c_3(activate^#(X))] [activate^#(n__s(X))] = [2] X + [8] > [2] X + [5] = [c_4(activate^#(X))] We return to the main proof. Consider the set of all dependency pairs : { 1: U11^#(tt(), N, X, XS) -> c_1(U12^#(splitAt(activate(N), activate(XS)), activate(X)), activate^#(N), activate^#(XS), activate^#(X)) , 2: activate^#(n__natsFrom(X)) -> c_3(activate^#(X)) , 3: activate^#(n__s(X)) -> c_4(activate^#(X)) , 4: U12^#(pair(YS, ZS), X) -> c_2(activate^#(X)) } Processor 'matrix interpretation of dimension 1' induces the complexity certificate YES(?,O(n^1)) on application of dependency pairs {1,3}. These cover all (indirect) predecessors of dependency pairs {1,3,4}, their number of application is equally bounded. The dependency pairs are shifted into the weak component. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { activate^#(n__natsFrom(X)) -> c_3(activate^#(X)) } Weak DPs: { U11^#(tt(), N, X, XS) -> c_1(U12^#(splitAt(activate(N), activate(XS)), activate(X)), activate^#(N), activate^#(XS), activate^#(X)) , U12^#(pair(YS, ZS), X) -> c_2(activate^#(X)) , activate^#(n__s(X)) -> c_4(activate^#(X)) } Weak Trs: { splitAt(0(), XS) -> pair(nil(), XS) , activate(X) -> X , activate(n__natsFrom(X)) -> natsFrom(activate(X)) , activate(n__s(X)) -> s(activate(X)) , natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) , natsFrom(X) -> n__natsFrom(X) , s(X) -> n__s(X) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. DPs: { 1: activate^#(n__natsFrom(X)) -> c_3(activate^#(X)) , 2: U11^#(tt(), N, X, XS) -> c_1(U12^#(splitAt(activate(N), activate(XS)), activate(X)), activate^#(N), activate^#(XS), activate^#(X)) , 4: activate^#(n__s(X)) -> c_4(activate^#(X)) } Trs: { activate(n__natsFrom(X)) -> natsFrom(activate(X)) , activate(n__s(X)) -> s(activate(X)) , natsFrom(X) -> n__natsFrom(X) , s(X) -> n__s(X) } Sub-proof: ---------- The following argument positions are usable: Uargs(c_1) = {1, 2, 3, 4}, Uargs(c_2) = {1}, Uargs(c_3) = {1}, Uargs(c_4) = {1} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [U11](x1, x2, x3, x4) = [7] x1 + [7] x2 + [7] x3 + [7] x4 + [0] [tt] = [7] [U12](x1, x2) = [7] x1 + [7] x2 + [0] [splitAt](x1, x2) = [0] [activate](x1) = [2] x1 + [0] [pair](x1, x2) = [1] x1 + [0] [cons](x1, x2) = [1] x1 + [7] [afterNth](x1, x2) = [7] x1 + [7] x2 + [0] [snd](x1) = [7] x1 + [0] [and](x1, x2) = [7] x1 + [7] x2 + [0] [fst](x1) = [7] x1 + [0] [head](x1) = [7] x1 + [0] [natsFrom](x1) = [1] x1 + [7] [n__natsFrom](x1) = [1] x1 + [4] [n__s](x1) = [1] x1 + [4] [sel](x1, x2) = [7] x1 + [7] x2 + [0] [0] = [7] [nil] = [0] [s](x1) = [1] x1 + [7] [tail](x1) = [7] x1 + [0] [take](x1, x2) = [7] x1 + [7] x2 + [0] [U11^#](x1, x2, x3, x4) = [7] x2 + [7] x3 + [7] x4 + [7] [c_1](x1, x2, x3, x4, x5) = [7] x1 + [7] x2 + [7] x3 + [7] x4 + [7] x5 + [0] [U12^#](x1, x2) = [2] x2 + [0] [splitAt^#](x1, x2) = [7] x1 + [7] x2 + [0] [activate^#](x1) = [2] x1 + [0] [c_2](x1) = [7] x1 + [0] [c_3] = [0] [c_4] = [0] [c_5](x1, x2) = [7] x1 + [7] x2 + [0] [natsFrom^#](x1) = [7] x1 + [0] [c_6](x1, x2) = [7] x1 + [7] x2 + [0] [s^#](x1) = [7] x1 + [0] [afterNth^#](x1, x2) = [7] x1 + [7] x2 + [0] [c_7](x1, x2) = [7] x1 + [7] x2 + [0] [snd^#](x1) = [7] x1 + [0] [c_8] = [0] [and^#](x1, x2) = [7] x1 + [7] x2 + [0] [c_9](x1) = [7] x1 + [0] [fst^#](x1) = [7] x1 + [0] [c_10] = [0] [head^#](x1) = [7] x1 + [0] [c_11] = [0] [c_12] = [0] [c_13] = [0] [sel^#](x1, x2) = [7] x1 + [7] x2 + [0] [c_14](x1, x2) = [7] x1 + [7] x2 + [0] [c_15] = [0] [tail^#](x1) = [7] x1 + [0] [c_16](x1) = [7] x1 + [0] [take^#](x1, x2) = [7] x1 + [7] x2 + [0] [c_17](x1, x2) = [7] x1 + [7] x2 + [0] [c] = [0] [c_1](x1, x2, x3, x4) = [1] x1 + [2] x2 + [1] x3 + [1] x4 + [3] [c_2](x1) = [1] x1 + [0] [c_3](x1) = [1] x1 + [5] [c_4](x1) = [1] x1 + [5] [c_5](x1) = [7] x1 + [0] [c_6](x1) = [7] x1 + [0] The following symbols are considered usable {splitAt, activate, natsFrom, s, U11^#, U12^#, activate^#} The order satisfies the following ordering constraints: [splitAt(0(), XS)] = [0] >= [0] = [pair(nil(), XS)] [activate(X)] = [2] X + [0] >= [1] X + [0] = [X] [activate(n__natsFrom(X))] = [2] X + [8] > [2] X + [7] = [natsFrom(activate(X))] [activate(n__s(X))] = [2] X + [8] > [2] X + [7] = [s(activate(X))] [natsFrom(N)] = [1] N + [7] >= [1] N + [7] = [cons(N, n__natsFrom(n__s(N)))] [natsFrom(X)] = [1] X + [7] > [1] X + [4] = [n__natsFrom(X)] [s(X)] = [1] X + [7] > [1] X + [4] = [n__s(X)] [U11^#(tt(), N, X, XS)] = [7] N + [7] X + [7] XS + [7] > [4] N + [6] X + [2] XS + [3] = [c_1(U12^#(splitAt(activate(N), activate(XS)), activate(X)), activate^#(N), activate^#(XS), activate^#(X))] [U12^#(pair(YS, ZS), X)] = [2] X + [0] >= [2] X + [0] = [c_2(activate^#(X))] [activate^#(n__natsFrom(X))] = [2] X + [8] > [2] X + [5] = [c_3(activate^#(X))] [activate^#(n__s(X))] = [2] X + [8] > [2] X + [5] = [c_4(activate^#(X))] We return to the main proof. Consider the set of all dependency pairs : { 1: activate^#(n__natsFrom(X)) -> c_3(activate^#(X)) , 2: U11^#(tt(), N, X, XS) -> c_1(U12^#(splitAt(activate(N), activate(XS)), activate(X)), activate^#(N), activate^#(XS), activate^#(X)) , 3: U12^#(pair(YS, ZS), X) -> c_2(activate^#(X)) , 4: activate^#(n__s(X)) -> c_4(activate^#(X)) } Processor 'matrix interpretation of dimension 1' induces the complexity certificate YES(?,O(n^1)) on application of dependency pairs {1,2,4}. These cover all (indirect) predecessors of dependency pairs {1,2,3,4}, their number of application is equally bounded. The dependency pairs are shifted into the weak component. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak DPs: { U11^#(tt(), N, X, XS) -> c_1(U12^#(splitAt(activate(N), activate(XS)), activate(X)), activate^#(N), activate^#(XS), activate^#(X)) , U12^#(pair(YS, ZS), X) -> c_2(activate^#(X)) , activate^#(n__natsFrom(X)) -> c_3(activate^#(X)) , activate^#(n__s(X)) -> c_4(activate^#(X)) } Weak Trs: { splitAt(0(), XS) -> pair(nil(), XS) , activate(X) -> X , activate(n__natsFrom(X)) -> natsFrom(activate(X)) , activate(n__s(X)) -> s(activate(X)) , natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) , natsFrom(X) -> n__natsFrom(X) , s(X) -> n__s(X) } 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. { U11^#(tt(), N, X, XS) -> c_1(U12^#(splitAt(activate(N), activate(XS)), activate(X)), activate^#(N), activate^#(XS), activate^#(X)) , U12^#(pair(YS, ZS), X) -> c_2(activate^#(X)) , activate^#(n__natsFrom(X)) -> c_3(activate^#(X)) , activate^#(n__s(X)) -> c_4(activate^#(X)) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak Trs: { splitAt(0(), XS) -> pair(nil(), XS) , activate(X) -> X , activate(n__natsFrom(X)) -> natsFrom(activate(X)) , activate(n__s(X)) -> s(activate(X)) , natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) , natsFrom(X) -> n__natsFrom(X) , s(X) -> n__s(X) } 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 S) We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Strict DPs: { U12^#(pair(YS, ZS), X) -> c_2(activate^#(X)) } Weak DPs: { U11^#(tt(), N, X, XS) -> c_1(U12^#(splitAt(activate(N), activate(XS)), activate(X)), activate^#(N), activate^#(XS), activate^#(X)) , activate^#(n__natsFrom(X)) -> c_3(activate^#(X)) , activate^#(n__s(X)) -> c_4(activate^#(X)) } Weak Trs: { splitAt(0(), XS) -> pair(nil(), XS) , activate(X) -> X , activate(n__natsFrom(X)) -> natsFrom(activate(X)) , activate(n__s(X)) -> s(activate(X)) , natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) , natsFrom(X) -> n__natsFrom(X) , s(X) -> n__s(X) } 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. { activate^#(n__natsFrom(X)) -> c_3(activate^#(X)) , activate^#(n__s(X)) -> c_4(activate^#(X)) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Strict DPs: { U12^#(pair(YS, ZS), X) -> c_2(activate^#(X)) } Weak DPs: { U11^#(tt(), N, X, XS) -> c_1(U12^#(splitAt(activate(N), activate(XS)), activate(X)), activate^#(N), activate^#(XS), activate^#(X)) } Weak Trs: { splitAt(0(), XS) -> pair(nil(), XS) , activate(X) -> X , activate(n__natsFrom(X)) -> natsFrom(activate(X)) , activate(n__s(X)) -> s(activate(X)) , natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) , natsFrom(X) -> n__natsFrom(X) , s(X) -> n__s(X) } 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: { U11^#(tt(), N, X, XS) -> c_1(U12^#(splitAt(activate(N), activate(XS)), activate(X)), activate^#(N), activate^#(XS), activate^#(X)) , U12^#(pair(YS, ZS), X) -> c_2(activate^#(X)) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Strict DPs: { U12^#(pair(YS, ZS), X) -> c_1() } Weak DPs: { U11^#(tt(), N, X, XS) -> c_2(U12^#(splitAt(activate(N), activate(XS)), activate(X))) } Weak Trs: { splitAt(0(), XS) -> pair(nil(), XS) , activate(X) -> X , activate(n__natsFrom(X)) -> natsFrom(activate(X)) , activate(n__s(X)) -> s(activate(X)) , natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) , natsFrom(X) -> n__natsFrom(X) , s(X) -> n__s(X) } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) The dependency graph contains no loops, we remove all dependency pairs. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak Trs: { splitAt(0(), XS) -> pair(nil(), XS) , activate(X) -> X , activate(n__natsFrom(X)) -> natsFrom(activate(X)) , activate(n__s(X)) -> s(activate(X)) , natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) , natsFrom(X) -> n__natsFrom(X) , s(X) -> n__s(X) } 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))