YES(O(1),O(n^1)) 2.69/1.07 YES(O(1),O(n^1)) 2.69/1.07 2.69/1.07 We are left with following problem, upon which TcT provides the 2.69/1.07 certificate YES(O(1),O(n^1)). 2.69/1.07 2.69/1.07 Strict Trs: 2.69/1.07 { natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) 2.69/1.07 , natsFrom(X) -> n__natsFrom(X) 2.69/1.07 , fst(pair(XS, YS)) -> XS 2.69/1.07 , snd(pair(XS, YS)) -> YS 2.69/1.07 , splitAt(0(), XS) -> pair(nil(), XS) 2.69/1.07 , splitAt(s(N), cons(X, XS)) -> 2.69/1.07 u(splitAt(N, activate(XS)), N, X, activate(XS)) 2.69/1.07 , s(X) -> n__s(X) 2.69/1.07 , u(pair(YS, ZS), N, X, XS) -> pair(cons(activate(X), YS), ZS) 2.69/1.07 , activate(X) -> X 2.69/1.07 , activate(n__natsFrom(X)) -> natsFrom(activate(X)) 2.69/1.07 , activate(n__s(X)) -> s(activate(X)) 2.69/1.07 , head(cons(N, XS)) -> N 2.69/1.07 , tail(cons(N, XS)) -> activate(XS) 2.69/1.07 , sel(N, XS) -> head(afterNth(N, XS)) 2.69/1.07 , afterNth(N, XS) -> snd(splitAt(N, XS)) 2.69/1.07 , take(N, XS) -> fst(splitAt(N, XS)) } 2.69/1.07 Obligation: 2.69/1.07 innermost runtime complexity 2.69/1.07 Answer: 2.69/1.07 YES(O(1),O(n^1)) 2.69/1.07 2.69/1.07 Arguments of following rules are not normal-forms: 2.69/1.07 2.69/1.07 { splitAt(s(N), cons(X, XS)) -> 2.69/1.07 u(splitAt(N, activate(XS)), N, X, activate(XS)) } 2.69/1.07 2.69/1.07 All above mentioned rules can be savely removed. 2.69/1.07 2.69/1.07 We are left with following problem, upon which TcT provides the 2.69/1.07 certificate YES(O(1),O(n^1)). 2.69/1.07 2.69/1.07 Strict Trs: 2.69/1.07 { natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) 2.69/1.07 , natsFrom(X) -> n__natsFrom(X) 2.69/1.07 , fst(pair(XS, YS)) -> XS 2.69/1.07 , snd(pair(XS, YS)) -> YS 2.69/1.07 , splitAt(0(), XS) -> pair(nil(), XS) 2.69/1.07 , s(X) -> n__s(X) 2.69/1.07 , u(pair(YS, ZS), N, X, XS) -> pair(cons(activate(X), YS), ZS) 2.69/1.07 , activate(X) -> X 2.69/1.07 , activate(n__natsFrom(X)) -> natsFrom(activate(X)) 2.69/1.07 , activate(n__s(X)) -> s(activate(X)) 2.69/1.07 , head(cons(N, XS)) -> N 2.69/1.07 , tail(cons(N, XS)) -> activate(XS) 2.69/1.07 , sel(N, XS) -> head(afterNth(N, XS)) 2.69/1.07 , afterNth(N, XS) -> snd(splitAt(N, XS)) 2.69/1.07 , take(N, XS) -> fst(splitAt(N, XS)) } 2.69/1.07 Obligation: 2.69/1.07 innermost runtime complexity 2.69/1.07 Answer: 2.69/1.07 YES(O(1),O(n^1)) 2.69/1.07 2.69/1.07 We add the following weak dependency pairs: 2.69/1.07 2.69/1.07 Strict DPs: 2.69/1.07 { natsFrom^#(N) -> c_1() 2.69/1.07 , natsFrom^#(X) -> c_2() 2.69/1.07 , fst^#(pair(XS, YS)) -> c_3() 2.69/1.07 , snd^#(pair(XS, YS)) -> c_4() 2.69/1.07 , splitAt^#(0(), XS) -> c_5() 2.69/1.07 , s^#(X) -> c_6() 2.69/1.07 , u^#(pair(YS, ZS), N, X, XS) -> c_7(activate^#(X)) 2.69/1.07 , activate^#(X) -> c_8() 2.69/1.07 , activate^#(n__natsFrom(X)) -> c_9(natsFrom^#(activate(X))) 2.69/1.07 , activate^#(n__s(X)) -> c_10(s^#(activate(X))) 2.69/1.07 , head^#(cons(N, XS)) -> c_11() 2.69/1.07 , tail^#(cons(N, XS)) -> c_12(activate^#(XS)) 2.69/1.07 , sel^#(N, XS) -> c_13(head^#(afterNth(N, XS))) 2.69/1.07 , afterNth^#(N, XS) -> c_14(snd^#(splitAt(N, XS))) 2.69/1.07 , take^#(N, XS) -> c_15(fst^#(splitAt(N, XS))) } 2.69/1.07 2.69/1.07 and mark the set of starting terms. 2.69/1.07 2.69/1.07 We are left with following problem, upon which TcT provides the 2.69/1.07 certificate YES(O(1),O(n^1)). 2.69/1.07 2.69/1.07 Strict DPs: 2.69/1.07 { natsFrom^#(N) -> c_1() 2.69/1.07 , natsFrom^#(X) -> c_2() 2.69/1.07 , fst^#(pair(XS, YS)) -> c_3() 2.69/1.07 , snd^#(pair(XS, YS)) -> c_4() 2.69/1.07 , splitAt^#(0(), XS) -> c_5() 2.69/1.07 , s^#(X) -> c_6() 2.69/1.07 , u^#(pair(YS, ZS), N, X, XS) -> c_7(activate^#(X)) 2.69/1.07 , activate^#(X) -> c_8() 2.69/1.07 , activate^#(n__natsFrom(X)) -> c_9(natsFrom^#(activate(X))) 2.69/1.07 , activate^#(n__s(X)) -> c_10(s^#(activate(X))) 2.69/1.07 , head^#(cons(N, XS)) -> c_11() 2.69/1.07 , tail^#(cons(N, XS)) -> c_12(activate^#(XS)) 2.69/1.07 , sel^#(N, XS) -> c_13(head^#(afterNth(N, XS))) 2.69/1.07 , afterNth^#(N, XS) -> c_14(snd^#(splitAt(N, XS))) 2.69/1.07 , take^#(N, XS) -> c_15(fst^#(splitAt(N, XS))) } 2.69/1.07 Strict Trs: 2.69/1.07 { natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) 2.69/1.07 , natsFrom(X) -> n__natsFrom(X) 2.69/1.07 , fst(pair(XS, YS)) -> XS 2.69/1.07 , snd(pair(XS, YS)) -> YS 2.69/1.07 , splitAt(0(), XS) -> pair(nil(), XS) 2.69/1.07 , s(X) -> n__s(X) 2.69/1.07 , u(pair(YS, ZS), N, X, XS) -> pair(cons(activate(X), YS), ZS) 2.69/1.07 , activate(X) -> X 2.69/1.07 , activate(n__natsFrom(X)) -> natsFrom(activate(X)) 2.69/1.07 , activate(n__s(X)) -> s(activate(X)) 2.69/1.07 , head(cons(N, XS)) -> N 2.69/1.07 , tail(cons(N, XS)) -> activate(XS) 2.69/1.07 , sel(N, XS) -> head(afterNth(N, XS)) 2.69/1.07 , afterNth(N, XS) -> snd(splitAt(N, XS)) 2.69/1.07 , take(N, XS) -> fst(splitAt(N, XS)) } 2.69/1.07 Obligation: 2.69/1.07 innermost runtime complexity 2.69/1.07 Answer: 2.69/1.07 YES(O(1),O(n^1)) 2.69/1.07 2.69/1.07 We replace rewrite rules by usable rules: 2.69/1.07 2.69/1.07 Strict Usable Rules: 2.69/1.07 { natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) 2.69/1.07 , natsFrom(X) -> n__natsFrom(X) 2.69/1.07 , snd(pair(XS, YS)) -> YS 2.69/1.07 , splitAt(0(), XS) -> pair(nil(), XS) 2.69/1.07 , s(X) -> n__s(X) 2.69/1.07 , activate(X) -> X 2.69/1.07 , activate(n__natsFrom(X)) -> natsFrom(activate(X)) 2.69/1.07 , activate(n__s(X)) -> s(activate(X)) 2.69/1.07 , afterNth(N, XS) -> snd(splitAt(N, XS)) } 2.69/1.07 2.69/1.07 We are left with following problem, upon which TcT provides the 2.69/1.07 certificate YES(O(1),O(n^1)). 2.69/1.07 2.69/1.07 Strict DPs: 2.69/1.07 { natsFrom^#(N) -> c_1() 2.69/1.07 , natsFrom^#(X) -> c_2() 2.69/1.07 , fst^#(pair(XS, YS)) -> c_3() 2.69/1.07 , snd^#(pair(XS, YS)) -> c_4() 2.69/1.07 , splitAt^#(0(), XS) -> c_5() 2.69/1.07 , s^#(X) -> c_6() 2.69/1.07 , u^#(pair(YS, ZS), N, X, XS) -> c_7(activate^#(X)) 2.69/1.07 , activate^#(X) -> c_8() 2.69/1.07 , activate^#(n__natsFrom(X)) -> c_9(natsFrom^#(activate(X))) 2.69/1.07 , activate^#(n__s(X)) -> c_10(s^#(activate(X))) 2.69/1.07 , head^#(cons(N, XS)) -> c_11() 2.69/1.07 , tail^#(cons(N, XS)) -> c_12(activate^#(XS)) 2.69/1.07 , sel^#(N, XS) -> c_13(head^#(afterNth(N, XS))) 2.69/1.07 , afterNth^#(N, XS) -> c_14(snd^#(splitAt(N, XS))) 2.69/1.07 , take^#(N, XS) -> c_15(fst^#(splitAt(N, XS))) } 2.69/1.07 Strict Trs: 2.69/1.07 { natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) 2.69/1.07 , natsFrom(X) -> n__natsFrom(X) 2.69/1.07 , snd(pair(XS, YS)) -> YS 2.69/1.07 , splitAt(0(), XS) -> pair(nil(), XS) 2.69/1.07 , s(X) -> n__s(X) 2.69/1.07 , activate(X) -> X 2.69/1.07 , activate(n__natsFrom(X)) -> natsFrom(activate(X)) 2.69/1.07 , activate(n__s(X)) -> s(activate(X)) 2.69/1.07 , afterNth(N, XS) -> snd(splitAt(N, XS)) } 2.69/1.07 Obligation: 2.69/1.07 innermost runtime complexity 2.69/1.07 Answer: 2.69/1.07 YES(O(1),O(n^1)) 2.69/1.07 2.69/1.07 The weightgap principle applies (using the following constant 2.69/1.07 growth matrix-interpretation) 2.69/1.07 2.69/1.07 The following argument positions are usable: 2.69/1.07 Uargs(natsFrom) = {1}, Uargs(snd) = {1}, Uargs(s) = {1}, 2.69/1.07 Uargs(natsFrom^#) = {1}, Uargs(fst^#) = {1}, Uargs(snd^#) = {1}, 2.69/1.07 Uargs(s^#) = {1}, Uargs(c_7) = {1}, Uargs(c_9) = {1}, 2.69/1.07 Uargs(c_10) = {1}, Uargs(head^#) = {1}, Uargs(c_12) = {1}, 2.69/1.07 Uargs(c_13) = {1}, Uargs(c_14) = {1}, Uargs(c_15) = {1} 2.69/1.07 2.69/1.07 TcT has computed the following constructor-restricted matrix 2.69/1.07 interpretation. 2.69/1.07 2.69/1.07 [natsFrom](x1) = [1 0] x1 + [2] 2.69/1.07 [0 1] [2] 2.69/1.07 2.69/1.07 [cons](x1, x2) = [1 0] x2 + [0] 2.69/1.07 [0 1] [0] 2.69/1.07 2.69/1.07 [n__natsFrom](x1) = [1 0] x1 + [1] 2.69/1.07 [0 1] [1] 2.69/1.07 2.69/1.07 [n__s](x1) = [1 0] x1 + [0] 2.69/1.07 [0 1] [1] 2.69/1.07 2.69/1.07 [pair](x1, x2) = [1 0] x1 + [1 0] x2 + [0] 2.69/1.07 [0 0] [0 1] [1] 2.69/1.07 2.69/1.07 [snd](x1) = [1 1] x1 + [0] 2.69/1.07 [0 1] [0] 2.69/1.07 2.69/1.07 [splitAt](x1, x2) = [1 0] x1 + [2 0] x2 + [0] 2.69/1.07 [1 0] [0 1] [0] 2.69/1.07 2.69/1.07 [0] = [1] 2.69/1.07 [0] 2.69/1.07 2.69/1.07 [nil] = [0] 2.69/1.07 [2] 2.69/1.07 2.69/1.07 [s](x1) = [1 0] x1 + [1] 2.69/1.07 [0 1] [1] 2.69/1.07 2.69/1.07 [activate](x1) = [1 2] x1 + [2] 2.69/1.07 [2 1] [2] 2.69/1.07 2.69/1.07 [afterNth](x1, x2) = [2 0] x1 + [2 1] x2 + [1] 2.69/1.07 [2 0] [0 2] [0] 2.69/1.07 2.69/1.07 [natsFrom^#](x1) = [1 0] x1 + [2] 2.69/1.07 [0 0] [2] 2.69/1.07 2.69/1.07 [c_1] = [1] 2.69/1.07 [1] 2.69/1.07 2.69/1.07 [c_2] = [1] 2.69/1.07 [1] 2.69/1.07 2.69/1.07 [fst^#](x1) = [1 0] x1 + [2] 2.69/1.07 [0 0] [2] 2.69/1.07 2.69/1.07 [c_3] = [1] 2.69/1.07 [1] 2.69/1.07 2.69/1.07 [snd^#](x1) = [1 0] x1 + [2] 2.69/1.07 [0 0] [2] 2.69/1.07 2.69/1.07 [c_4] = [1] 2.69/1.07 [1] 2.69/1.07 2.69/1.07 [splitAt^#](x1, x2) = [2 1] x1 + [2] 2.69/1.07 [2 2] [2] 2.69/1.07 2.69/1.07 [c_5] = [1] 2.69/1.07 [1] 2.69/1.07 2.69/1.07 [s^#](x1) = [1 0] x1 + [2] 2.69/1.07 [0 0] [2] 2.69/1.07 2.69/1.07 [c_6] = [1] 2.69/1.07 [1] 2.69/1.07 2.69/1.07 [u^#](x1, x2, x3, x4) = [2 2] x1 + [1 2] x3 + [2] 2.69/1.07 [2 2] [1 2] [2] 2.69/1.07 2.69/1.07 [c_7](x1) = [1 0] x1 + [1] 2.69/1.07 [0 1] [1] 2.69/1.07 2.69/1.07 [activate^#](x1) = [1 2] x1 + [1] 2.69/1.07 [2 2] [2] 2.69/1.07 2.69/1.07 [c_8] = [0] 2.69/1.07 [1] 2.69/1.07 2.69/1.07 [c_9](x1) = [1 0] x1 + [2] 2.69/1.07 [0 1] [2] 2.69/1.07 2.69/1.07 [c_10](x1) = [1 0] x1 + [2] 2.69/1.07 [0 1] [2] 2.69/1.07 2.69/1.07 [head^#](x1) = [1 0] x1 + [1] 2.69/1.07 [0 0] [2] 2.69/1.07 2.69/1.07 [c_11] = [1] 2.69/1.07 [1] 2.69/1.07 2.69/1.07 [tail^#](x1) = [2 2] x1 + [2] 2.69/1.07 [1 1] [2] 2.69/1.07 2.69/1.07 [c_12](x1) = [1 0] x1 + [1] 2.69/1.07 [0 1] [1] 2.69/1.07 2.69/1.07 [sel^#](x1, x2) = [2 1] x1 + [2 2] x2 + [1] 2.69/1.07 [2 1] [1 2] [1] 2.69/1.07 2.69/1.07 [c_13](x1) = [1 0] x1 + [2] 2.69/1.07 [0 1] [2] 2.69/1.07 2.69/1.07 [afterNth^#](x1, x2) = [2 2] x1 + [2 2] x2 + [1] 2.69/1.07 [1 2] [2 1] [1] 2.69/1.07 2.69/1.07 [c_14](x1) = [1 0] x1 + [2] 2.69/1.07 [0 1] [2] 2.69/1.07 2.69/1.07 [take^#](x1, x2) = [2 1] x1 + [2 1] x2 + [2] 2.69/1.07 [1 1] [2 2] [1] 2.69/1.07 2.69/1.07 [c_15](x1) = [1 0] x1 + [2] 2.69/1.07 [0 1] [2] 2.69/1.07 2.69/1.07 The order satisfies the following ordering constraints: 2.69/1.07 2.69/1.07 [natsFrom(N)] = [1 0] N + [2] 2.69/1.07 [0 1] [2] 2.69/1.07 > [1 0] N + [1] 2.69/1.07 [0 1] [2] 2.69/1.07 = [cons(N, n__natsFrom(n__s(N)))] 2.69/1.07 2.69/1.07 [natsFrom(X)] = [1 0] X + [2] 2.69/1.07 [0 1] [2] 2.69/1.07 > [1 0] X + [1] 2.69/1.07 [0 1] [1] 2.69/1.07 = [n__natsFrom(X)] 2.69/1.07 2.69/1.07 [snd(pair(XS, YS))] = [1 0] XS + [1 1] YS + [1] 2.69/1.07 [0 0] [0 1] [1] 2.69/1.07 > [1 0] YS + [0] 2.69/1.07 [0 1] [0] 2.69/1.07 = [YS] 2.69/1.07 2.69/1.07 [splitAt(0(), XS)] = [2 0] XS + [1] 2.69/1.07 [0 1] [1] 2.69/1.07 > [1 0] XS + [0] 2.69/1.07 [0 1] [1] 2.69/1.07 = [pair(nil(), XS)] 2.69/1.07 2.69/1.07 [s(X)] = [1 0] X + [1] 2.69/1.07 [0 1] [1] 2.69/1.07 > [1 0] X + [0] 2.69/1.07 [0 1] [1] 2.69/1.07 = [n__s(X)] 2.69/1.07 2.69/1.07 [activate(X)] = [1 2] X + [2] 2.69/1.07 [2 1] [2] 2.69/1.07 > [1 0] X + [0] 2.69/1.07 [0 1] [0] 2.69/1.07 = [X] 2.69/1.07 2.69/1.07 [activate(n__natsFrom(X))] = [1 2] X + [5] 2.69/1.07 [2 1] [5] 2.69/1.07 > [1 2] X + [4] 2.69/1.07 [2 1] [4] 2.69/1.07 = [natsFrom(activate(X))] 2.69/1.07 2.69/1.07 [activate(n__s(X))] = [1 2] X + [4] 2.69/1.07 [2 1] [3] 2.69/1.07 > [1 2] X + [3] 2.69/1.07 [2 1] [3] 2.69/1.07 = [s(activate(X))] 2.69/1.07 2.69/1.07 [afterNth(N, XS)] = [2 0] N + [2 1] XS + [1] 2.69/1.07 [2 0] [0 2] [0] 2.69/1.07 > [2 0] N + [2 1] XS + [0] 2.69/1.07 [1 0] [0 1] [0] 2.69/1.07 = [snd(splitAt(N, XS))] 2.69/1.07 2.69/1.07 [natsFrom^#(N)] = [1 0] N + [2] 2.69/1.07 [0 0] [2] 2.69/1.07 > [1] 2.69/1.07 [1] 2.69/1.07 = [c_1()] 2.69/1.07 2.69/1.07 [natsFrom^#(X)] = [1 0] X + [2] 2.69/1.07 [0 0] [2] 2.69/1.07 > [1] 2.69/1.07 [1] 2.69/1.07 = [c_2()] 2.69/1.07 2.69/1.07 [fst^#(pair(XS, YS))] = [1 0] XS + [1 0] YS + [2] 2.69/1.07 [0 0] [0 0] [2] 2.69/1.07 > [1] 2.69/1.07 [1] 2.69/1.07 = [c_3()] 2.69/1.07 2.69/1.07 [snd^#(pair(XS, YS))] = [1 0] XS + [1 0] YS + [2] 2.69/1.07 [0 0] [0 0] [2] 2.69/1.07 > [1] 2.69/1.07 [1] 2.69/1.07 = [c_4()] 2.69/1.07 2.69/1.07 [splitAt^#(0(), XS)] = [4] 2.69/1.07 [4] 2.69/1.07 > [1] 2.69/1.07 [1] 2.69/1.07 = [c_5()] 2.69/1.07 2.69/1.07 [s^#(X)] = [1 0] X + [2] 2.69/1.07 [0 0] [2] 2.69/1.07 > [1] 2.69/1.07 [1] 2.69/1.07 = [c_6()] 2.69/1.08 2.69/1.08 [u^#(pair(YS, ZS), N, X, XS)] = [2 0] YS + [1 2] X + [2 2] ZS + [4] 2.69/1.08 [2 0] [1 2] [2 2] [4] 2.69/1.08 ? [1 2] X + [2] 2.69/1.08 [2 2] [3] 2.69/1.08 = [c_7(activate^#(X))] 2.69/1.08 2.69/1.08 [activate^#(X)] = [1 2] X + [1] 2.69/1.08 [2 2] [2] 2.69/1.08 > [0] 2.69/1.08 [1] 2.69/1.08 = [c_8()] 2.69/1.08 2.69/1.08 [activate^#(n__natsFrom(X))] = [1 2] X + [4] 2.69/1.08 [2 2] [6] 2.69/1.08 ? [1 2] X + [6] 2.69/1.08 [0 0] [4] 2.69/1.08 = [c_9(natsFrom^#(activate(X)))] 2.69/1.08 2.69/1.08 [activate^#(n__s(X))] = [1 2] X + [3] 2.69/1.08 [2 2] [4] 2.69/1.08 ? [1 2] X + [6] 2.69/1.08 [0 0] [4] 2.69/1.08 = [c_10(s^#(activate(X)))] 2.69/1.08 2.69/1.08 [head^#(cons(N, XS))] = [1 0] XS + [1] 2.69/1.08 [0 0] [2] 2.69/1.08 >= [1] 2.69/1.08 [1] 2.69/1.08 = [c_11()] 2.69/1.08 2.69/1.08 [tail^#(cons(N, XS))] = [2 2] XS + [2] 2.69/1.08 [1 1] [2] 2.69/1.08 ? [1 2] XS + [2] 2.69/1.08 [2 2] [3] 2.69/1.08 = [c_12(activate^#(XS))] 2.69/1.08 2.69/1.08 [sel^#(N, XS)] = [2 1] N + [2 2] XS + [1] 2.69/1.08 [2 1] [1 2] [1] 2.69/1.08 ? [2 0] N + [2 1] XS + [4] 2.69/1.08 [0 0] [0 0] [4] 2.69/1.08 = [c_13(head^#(afterNth(N, XS)))] 2.69/1.08 2.69/1.08 [afterNth^#(N, XS)] = [2 2] N + [2 2] XS + [1] 2.69/1.08 [1 2] [2 1] [1] 2.69/1.08 ? [1 0] N + [2 0] XS + [4] 2.69/1.08 [0 0] [0 0] [4] 2.69/1.08 = [c_14(snd^#(splitAt(N, XS)))] 2.69/1.08 2.69/1.08 [take^#(N, XS)] = [2 1] N + [2 1] XS + [2] 2.69/1.08 [1 1] [2 2] [1] 2.69/1.08 ? [1 0] N + [2 0] XS + [4] 2.69/1.08 [0 0] [0 0] [4] 2.69/1.08 = [c_15(fst^#(splitAt(N, XS)))] 2.69/1.08 2.69/1.08 2.69/1.08 Further, it can be verified that all rules not oriented are covered by the weightgap condition. 2.69/1.08 2.69/1.08 We are left with following problem, upon which TcT provides the 2.69/1.08 certificate YES(O(1),O(1)). 2.69/1.08 2.69/1.08 Strict DPs: 2.69/1.08 { u^#(pair(YS, ZS), N, X, XS) -> c_7(activate^#(X)) 2.69/1.08 , activate^#(n__natsFrom(X)) -> c_9(natsFrom^#(activate(X))) 2.69/1.08 , activate^#(n__s(X)) -> c_10(s^#(activate(X))) 2.69/1.08 , head^#(cons(N, XS)) -> c_11() 2.69/1.08 , tail^#(cons(N, XS)) -> c_12(activate^#(XS)) 2.69/1.08 , sel^#(N, XS) -> c_13(head^#(afterNth(N, XS))) 2.69/1.08 , afterNth^#(N, XS) -> c_14(snd^#(splitAt(N, XS))) 2.69/1.08 , take^#(N, XS) -> c_15(fst^#(splitAt(N, XS))) } 2.69/1.08 Weak DPs: 2.69/1.08 { natsFrom^#(N) -> c_1() 2.69/1.08 , natsFrom^#(X) -> c_2() 2.69/1.08 , fst^#(pair(XS, YS)) -> c_3() 2.69/1.08 , snd^#(pair(XS, YS)) -> c_4() 2.69/1.08 , splitAt^#(0(), XS) -> c_5() 2.69/1.08 , s^#(X) -> c_6() 2.69/1.08 , activate^#(X) -> c_8() } 2.69/1.08 Weak Trs: 2.69/1.08 { natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) 2.69/1.08 , natsFrom(X) -> n__natsFrom(X) 2.69/1.08 , snd(pair(XS, YS)) -> YS 2.69/1.08 , splitAt(0(), XS) -> pair(nil(), XS) 2.69/1.08 , s(X) -> n__s(X) 2.69/1.08 , activate(X) -> X 2.69/1.08 , activate(n__natsFrom(X)) -> natsFrom(activate(X)) 2.69/1.08 , activate(n__s(X)) -> s(activate(X)) 2.69/1.08 , afterNth(N, XS) -> snd(splitAt(N, XS)) } 2.69/1.08 Obligation: 2.69/1.08 innermost runtime complexity 2.69/1.08 Answer: 2.69/1.08 YES(O(1),O(1)) 2.69/1.08 2.69/1.08 We estimate the number of application of {2,3,4,7,8} by 2.69/1.08 applications of Pre({2,3,4,7,8}) = {1,5,6}. Here rules are labeled 2.69/1.08 as follows: 2.69/1.08 2.69/1.08 DPs: 2.69/1.08 { 1: u^#(pair(YS, ZS), N, X, XS) -> c_7(activate^#(X)) 2.69/1.08 , 2: activate^#(n__natsFrom(X)) -> c_9(natsFrom^#(activate(X))) 2.69/1.08 , 3: activate^#(n__s(X)) -> c_10(s^#(activate(X))) 2.69/1.08 , 4: head^#(cons(N, XS)) -> c_11() 2.69/1.08 , 5: tail^#(cons(N, XS)) -> c_12(activate^#(XS)) 2.69/1.08 , 6: sel^#(N, XS) -> c_13(head^#(afterNth(N, XS))) 2.69/1.08 , 7: afterNth^#(N, XS) -> c_14(snd^#(splitAt(N, XS))) 2.69/1.08 , 8: take^#(N, XS) -> c_15(fst^#(splitAt(N, XS))) 2.69/1.08 , 9: natsFrom^#(N) -> c_1() 2.69/1.08 , 10: natsFrom^#(X) -> c_2() 2.69/1.08 , 11: fst^#(pair(XS, YS)) -> c_3() 2.69/1.08 , 12: snd^#(pair(XS, YS)) -> c_4() 2.69/1.08 , 13: splitAt^#(0(), XS) -> c_5() 2.69/1.08 , 14: s^#(X) -> c_6() 2.69/1.08 , 15: activate^#(X) -> c_8() } 2.69/1.08 2.69/1.08 We are left with following problem, upon which TcT provides the 2.69/1.08 certificate YES(O(1),O(1)). 2.69/1.08 2.69/1.08 Strict DPs: 2.69/1.08 { u^#(pair(YS, ZS), N, X, XS) -> c_7(activate^#(X)) 2.69/1.08 , tail^#(cons(N, XS)) -> c_12(activate^#(XS)) 2.69/1.08 , sel^#(N, XS) -> c_13(head^#(afterNth(N, XS))) } 2.69/1.08 Weak DPs: 2.69/1.08 { natsFrom^#(N) -> c_1() 2.69/1.08 , natsFrom^#(X) -> c_2() 2.69/1.08 , fst^#(pair(XS, YS)) -> c_3() 2.69/1.08 , snd^#(pair(XS, YS)) -> c_4() 2.69/1.08 , splitAt^#(0(), XS) -> c_5() 2.69/1.08 , s^#(X) -> c_6() 2.69/1.08 , activate^#(X) -> c_8() 2.69/1.08 , activate^#(n__natsFrom(X)) -> c_9(natsFrom^#(activate(X))) 2.69/1.08 , activate^#(n__s(X)) -> c_10(s^#(activate(X))) 2.69/1.08 , head^#(cons(N, XS)) -> c_11() 2.69/1.08 , afterNth^#(N, XS) -> c_14(snd^#(splitAt(N, XS))) 2.69/1.08 , take^#(N, XS) -> c_15(fst^#(splitAt(N, XS))) } 2.69/1.08 Weak Trs: 2.69/1.08 { natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) 2.69/1.08 , natsFrom(X) -> n__natsFrom(X) 2.69/1.08 , snd(pair(XS, YS)) -> YS 2.69/1.08 , splitAt(0(), XS) -> pair(nil(), XS) 2.69/1.08 , s(X) -> n__s(X) 2.69/1.08 , activate(X) -> X 2.69/1.08 , activate(n__natsFrom(X)) -> natsFrom(activate(X)) 2.69/1.08 , activate(n__s(X)) -> s(activate(X)) 2.69/1.08 , afterNth(N, XS) -> snd(splitAt(N, XS)) } 2.69/1.08 Obligation: 2.69/1.08 innermost runtime complexity 2.69/1.08 Answer: 2.69/1.08 YES(O(1),O(1)) 2.69/1.08 2.69/1.08 We estimate the number of application of {1,2,3} by applications of 2.69/1.08 Pre({1,2,3}) = {}. Here rules are labeled as follows: 2.69/1.08 2.69/1.08 DPs: 2.69/1.08 { 1: u^#(pair(YS, ZS), N, X, XS) -> c_7(activate^#(X)) 2.69/1.08 , 2: tail^#(cons(N, XS)) -> c_12(activate^#(XS)) 2.69/1.08 , 3: sel^#(N, XS) -> c_13(head^#(afterNth(N, XS))) 2.69/1.08 , 4: natsFrom^#(N) -> c_1() 2.69/1.08 , 5: natsFrom^#(X) -> c_2() 2.69/1.08 , 6: fst^#(pair(XS, YS)) -> c_3() 2.69/1.08 , 7: snd^#(pair(XS, YS)) -> c_4() 2.69/1.08 , 8: splitAt^#(0(), XS) -> c_5() 2.69/1.08 , 9: s^#(X) -> c_6() 2.69/1.08 , 10: activate^#(X) -> c_8() 2.69/1.08 , 11: activate^#(n__natsFrom(X)) -> c_9(natsFrom^#(activate(X))) 2.69/1.08 , 12: activate^#(n__s(X)) -> c_10(s^#(activate(X))) 2.69/1.08 , 13: head^#(cons(N, XS)) -> c_11() 2.69/1.08 , 14: afterNth^#(N, XS) -> c_14(snd^#(splitAt(N, XS))) 2.69/1.08 , 15: take^#(N, XS) -> c_15(fst^#(splitAt(N, XS))) } 2.69/1.08 2.69/1.08 We are left with following problem, upon which TcT provides the 2.69/1.08 certificate YES(O(1),O(1)). 2.69/1.08 2.69/1.08 Weak DPs: 2.69/1.08 { natsFrom^#(N) -> c_1() 2.69/1.08 , natsFrom^#(X) -> c_2() 2.69/1.08 , fst^#(pair(XS, YS)) -> c_3() 2.69/1.08 , snd^#(pair(XS, YS)) -> c_4() 2.69/1.08 , splitAt^#(0(), XS) -> c_5() 2.69/1.08 , s^#(X) -> c_6() 2.69/1.08 , u^#(pair(YS, ZS), N, X, XS) -> c_7(activate^#(X)) 2.69/1.08 , activate^#(X) -> c_8() 2.69/1.08 , activate^#(n__natsFrom(X)) -> c_9(natsFrom^#(activate(X))) 2.69/1.08 , activate^#(n__s(X)) -> c_10(s^#(activate(X))) 2.69/1.08 , head^#(cons(N, XS)) -> c_11() 2.69/1.08 , tail^#(cons(N, XS)) -> c_12(activate^#(XS)) 2.69/1.08 , sel^#(N, XS) -> c_13(head^#(afterNth(N, XS))) 2.69/1.08 , afterNth^#(N, XS) -> c_14(snd^#(splitAt(N, XS))) 2.69/1.08 , take^#(N, XS) -> c_15(fst^#(splitAt(N, XS))) } 2.69/1.08 Weak Trs: 2.69/1.08 { natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) 2.69/1.08 , natsFrom(X) -> n__natsFrom(X) 2.69/1.08 , snd(pair(XS, YS)) -> YS 2.69/1.08 , splitAt(0(), XS) -> pair(nil(), XS) 2.69/1.08 , s(X) -> n__s(X) 2.69/1.08 , activate(X) -> X 2.69/1.08 , activate(n__natsFrom(X)) -> natsFrom(activate(X)) 2.69/1.08 , activate(n__s(X)) -> s(activate(X)) 2.69/1.08 , afterNth(N, XS) -> snd(splitAt(N, XS)) } 2.69/1.08 Obligation: 2.69/1.08 innermost runtime complexity 2.69/1.08 Answer: 2.69/1.08 YES(O(1),O(1)) 2.69/1.08 2.69/1.08 The following weak DPs constitute a sub-graph of the DG that is 2.69/1.08 closed under successors. The DPs are removed. 2.69/1.08 2.69/1.08 { natsFrom^#(N) -> c_1() 2.69/1.08 , natsFrom^#(X) -> c_2() 2.69/1.08 , fst^#(pair(XS, YS)) -> c_3() 2.69/1.08 , snd^#(pair(XS, YS)) -> c_4() 2.69/1.08 , splitAt^#(0(), XS) -> c_5() 2.69/1.08 , s^#(X) -> c_6() 2.69/1.08 , u^#(pair(YS, ZS), N, X, XS) -> c_7(activate^#(X)) 2.69/1.08 , activate^#(X) -> c_8() 2.69/1.08 , activate^#(n__natsFrom(X)) -> c_9(natsFrom^#(activate(X))) 2.69/1.08 , activate^#(n__s(X)) -> c_10(s^#(activate(X))) 2.69/1.08 , head^#(cons(N, XS)) -> c_11() 2.69/1.08 , tail^#(cons(N, XS)) -> c_12(activate^#(XS)) 2.69/1.08 , sel^#(N, XS) -> c_13(head^#(afterNth(N, XS))) 2.69/1.08 , afterNth^#(N, XS) -> c_14(snd^#(splitAt(N, XS))) 2.69/1.08 , take^#(N, XS) -> c_15(fst^#(splitAt(N, XS))) } 2.69/1.08 2.69/1.08 We are left with following problem, upon which TcT provides the 2.69/1.08 certificate YES(O(1),O(1)). 2.69/1.08 2.69/1.08 Weak Trs: 2.69/1.08 { natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) 2.69/1.08 , natsFrom(X) -> n__natsFrom(X) 2.69/1.08 , snd(pair(XS, YS)) -> YS 2.69/1.08 , splitAt(0(), XS) -> pair(nil(), XS) 2.69/1.08 , s(X) -> n__s(X) 2.69/1.08 , activate(X) -> X 2.69/1.08 , activate(n__natsFrom(X)) -> natsFrom(activate(X)) 2.69/1.08 , activate(n__s(X)) -> s(activate(X)) 2.69/1.08 , afterNth(N, XS) -> snd(splitAt(N, XS)) } 2.69/1.08 Obligation: 2.69/1.08 innermost runtime complexity 2.69/1.08 Answer: 2.69/1.08 YES(O(1),O(1)) 2.69/1.08 2.69/1.08 No rule is usable, rules are removed from the input problem. 2.69/1.08 2.69/1.08 We are left with following problem, upon which TcT provides the 2.69/1.08 certificate YES(O(1),O(1)). 2.69/1.08 2.69/1.08 Rules: Empty 2.69/1.08 Obligation: 2.69/1.08 innermost runtime complexity 2.69/1.08 Answer: 2.69/1.08 YES(O(1),O(1)) 2.69/1.08 2.69/1.08 Empty rules are trivially bounded 2.69/1.08 2.69/1.08 Hurray, we answered YES(O(1),O(n^1)) 2.69/1.08 EOF