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: { from(X) -> cons(X, n__from(n__s(X))) , from(X) -> n__from(X) , head(cons(X, XS)) -> X , 2nd(cons(X, XS)) -> head(activate(XS)) , activate(X) -> X , activate(n__from(X)) -> from(activate(X)) , activate(n__s(X)) -> s(activate(X)) , activate(n__take(X1, X2)) -> take(activate(X1), activate(X2)) , take(X1, X2) -> n__take(X1, X2) , take(0(), XS) -> nil() , take(s(N), cons(X, XS)) -> cons(X, n__take(N, activate(XS))) , s(X) -> n__s(X) , sel(0(), cons(X, XS)) -> X , sel(s(N), cons(X, XS)) -> sel(N, activate(XS)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) Arguments of following rules are not normal-forms: { take(s(N), cons(X, XS)) -> cons(X, n__take(N, activate(XS))) , sel(s(N), cons(X, XS)) -> sel(N, 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: { from(X) -> cons(X, n__from(n__s(X))) , from(X) -> n__from(X) , head(cons(X, XS)) -> X , 2nd(cons(X, XS)) -> head(activate(XS)) , activate(X) -> X , activate(n__from(X)) -> from(activate(X)) , activate(n__s(X)) -> s(activate(X)) , activate(n__take(X1, X2)) -> take(activate(X1), activate(X2)) , take(X1, X2) -> n__take(X1, X2) , take(0(), XS) -> nil() , s(X) -> n__s(X) , sel(0(), cons(X, XS)) -> X } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We add the following innermost weak dependency pairs: Strict DPs: { from^#(X) -> c_1() , from^#(X) -> c_2() , head^#(cons(X, XS)) -> c_3() , 2nd^#(cons(X, XS)) -> c_4(head^#(activate(XS))) , activate^#(X) -> c_5() , activate^#(n__from(X)) -> c_6(from^#(activate(X))) , activate^#(n__s(X)) -> c_7(s^#(activate(X))) , activate^#(n__take(X1, X2)) -> c_8(take^#(activate(X1), activate(X2))) , s^#(X) -> c_11() , take^#(X1, X2) -> c_9() , take^#(0(), XS) -> c_10() , sel^#(0(), cons(X, XS)) -> c_12() } 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: { from^#(X) -> c_1() , from^#(X) -> c_2() , head^#(cons(X, XS)) -> c_3() , 2nd^#(cons(X, XS)) -> c_4(head^#(activate(XS))) , activate^#(X) -> c_5() , activate^#(n__from(X)) -> c_6(from^#(activate(X))) , activate^#(n__s(X)) -> c_7(s^#(activate(X))) , activate^#(n__take(X1, X2)) -> c_8(take^#(activate(X1), activate(X2))) , s^#(X) -> c_11() , take^#(X1, X2) -> c_9() , take^#(0(), XS) -> c_10() , sel^#(0(), cons(X, XS)) -> c_12() } Strict Trs: { from(X) -> cons(X, n__from(n__s(X))) , from(X) -> n__from(X) , head(cons(X, XS)) -> X , 2nd(cons(X, XS)) -> head(activate(XS)) , activate(X) -> X , activate(n__from(X)) -> from(activate(X)) , activate(n__s(X)) -> s(activate(X)) , activate(n__take(X1, X2)) -> take(activate(X1), activate(X2)) , take(X1, X2) -> n__take(X1, X2) , take(0(), XS) -> nil() , s(X) -> n__s(X) , sel(0(), cons(X, XS)) -> X } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We replace rewrite rules by usable rules: Strict Usable Rules: { from(X) -> cons(X, n__from(n__s(X))) , from(X) -> n__from(X) , activate(X) -> X , activate(n__from(X)) -> from(activate(X)) , activate(n__s(X)) -> s(activate(X)) , activate(n__take(X1, X2)) -> take(activate(X1), activate(X2)) , take(X1, X2) -> n__take(X1, X2) , take(0(), XS) -> nil() , 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: { from^#(X) -> c_1() , from^#(X) -> c_2() , head^#(cons(X, XS)) -> c_3() , 2nd^#(cons(X, XS)) -> c_4(head^#(activate(XS))) , activate^#(X) -> c_5() , activate^#(n__from(X)) -> c_6(from^#(activate(X))) , activate^#(n__s(X)) -> c_7(s^#(activate(X))) , activate^#(n__take(X1, X2)) -> c_8(take^#(activate(X1), activate(X2))) , s^#(X) -> c_11() , take^#(X1, X2) -> c_9() , take^#(0(), XS) -> c_10() , sel^#(0(), cons(X, XS)) -> c_12() } Strict Trs: { from(X) -> cons(X, n__from(n__s(X))) , from(X) -> n__from(X) , activate(X) -> X , activate(n__from(X)) -> from(activate(X)) , activate(n__s(X)) -> s(activate(X)) , activate(n__take(X1, X2)) -> take(activate(X1), activate(X2)) , take(X1, X2) -> n__take(X1, X2) , take(0(), XS) -> nil() , s(X) -> n__s(X) } 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(from) = {1}, Uargs(take) = {1, 2}, Uargs(s) = {1}, Uargs(from^#) = {1}, Uargs(head^#) = {1}, Uargs(c_4) = {1}, Uargs(c_6) = {1}, Uargs(c_7) = {1}, Uargs(s^#) = {1}, Uargs(c_8) = {1}, Uargs(take^#) = {1, 2} TcT has computed the following constructor-restricted matrix interpretation. [from](x1) = [1 0] x1 + [1] [0 1] [2] [cons](x1, x2) = [1 0] x2 + [0] [0 1] [0] [n__from](x1) = [1 0] x1 + [0] [0 1] [1] [n__s](x1) = [1 0] x1 + [0] [0 1] [1] [activate](x1) = [1 2] x1 + [1] [0 2] [0] [take](x1, x2) = [1 0] x1 + [1 0] x2 + [1] [0 1] [0 1] [2] [0] = [0] [0] [nil] = [0] [1] [s](x1) = [1 0] x1 + [1] [0 1] [1] [n__take](x1, x2) = [1 0] x1 + [1 0] x2 + [0] [0 1] [0 1] [2] [from^#](x1) = [1 0] x1 + [1] [0 0] [2] [c_1] = [1] [1] [c_2] = [0] [1] [head^#](x1) = [1 0] x1 + [1] [0 0] [2] [c_3] = [0] [1] [2nd^#](x1) = [2 2] x1 + [2] [1 1] [2] [c_4](x1) = [1 0] x1 + [2] [0 1] [2] [activate^#](x1) = [1 2] x1 + [0] [0 0] [0] [c_5] = [1] [1] [c_6](x1) = [1 0] x1 + [2] [0 1] [2] [c_7](x1) = [1 0] x1 + [2] [0 1] [2] [s^#](x1) = [1 0] x1 + [1] [0 0] [2] [c_8](x1) = [1 0] x1 + [2] [0 1] [2] [take^#](x1, x2) = [1 0] x1 + [1 0] x2 + [2] [0 0] [0 0] [2] [c_9] = [1] [1] [c_10] = [2] [1] [c_11] = [0] [1] [sel^#](x1, x2) = [1 1] x1 + [2 2] x2 + [2] [2 2] [1 2] [2] [c_12] = [1] [1] The following symbols are considered usable {from, activate, take, s, from^#, head^#, 2nd^#, activate^#, s^#, take^#, sel^#} The order satisfies the following ordering constraints: [from(X)] = [1 0] X + [1] [0 1] [2] > [1 0] X + [0] [0 1] [2] = [cons(X, n__from(n__s(X)))] [from(X)] = [1 0] X + [1] [0 1] [2] > [1 0] X + [0] [0 1] [1] = [n__from(X)] [activate(X)] = [1 2] X + [1] [0 2] [0] > [1 0] X + [0] [0 1] [0] = [X] [activate(n__from(X))] = [1 2] X + [3] [0 2] [2] > [1 2] X + [2] [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__take(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] = [take(activate(X1), activate(X2))] [take(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__take(X1, X2)] [take(0(), XS)] = [1 0] XS + [1] [0 1] [2] > [0] [1] = [nil()] [s(X)] = [1 0] X + [1] [0 1] [1] > [1 0] X + [0] [0 1] [1] = [n__s(X)] [from^#(X)] = [1 0] X + [1] [0 0] [2] >= [1] [1] = [c_1()] [from^#(X)] = [1 0] X + [1] [0 0] [2] > [0] [1] = [c_2()] [head^#(cons(X, XS))] = [1 0] XS + [1] [0 0] [2] > [0] [1] = [c_3()] [2nd^#(cons(X, XS))] = [2 2] XS + [2] [1 1] [2] ? [1 2] XS + [4] [0 0] [4] = [c_4(head^#(activate(XS)))] [activate^#(X)] = [1 2] X + [0] [0 0] [0] ? [1] [1] = [c_5()] [activate^#(n__from(X))] = [1 2] X + [2] [0 0] [0] ? [1 2] X + [4] [0 0] [4] = [c_6(from^#(activate(X)))] [activate^#(n__s(X))] = [1 2] X + [2] [0 0] [0] ? [1 2] X + [4] [0 0] [4] = [c_7(s^#(activate(X)))] [activate^#(n__take(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_8(take^#(activate(X1), activate(X2)))] [s^#(X)] = [1 0] X + [1] [0 0] [2] > [0] [1] = [c_11()] [take^#(X1, X2)] = [1 0] X1 + [1 0] X2 + [2] [0 0] [0 0] [2] > [1] [1] = [c_9()] [take^#(0(), XS)] = [1 0] XS + [2] [0 0] [2] >= [2] [1] = [c_10()] [sel^#(0(), cons(X, XS))] = [2 2] XS + [2] [1 2] [2] > [1] [1] = [c_12()] 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: { from^#(X) -> c_1() , 2nd^#(cons(X, XS)) -> c_4(head^#(activate(XS))) , activate^#(X) -> c_5() , activate^#(n__from(X)) -> c_6(from^#(activate(X))) , activate^#(n__s(X)) -> c_7(s^#(activate(X))) , activate^#(n__take(X1, X2)) -> c_8(take^#(activate(X1), activate(X2))) , take^#(0(), XS) -> c_10() } Weak DPs: { from^#(X) -> c_2() , head^#(cons(X, XS)) -> c_3() , s^#(X) -> c_11() , take^#(X1, X2) -> c_9() , sel^#(0(), cons(X, XS)) -> c_12() } Weak Trs: { from(X) -> cons(X, n__from(n__s(X))) , from(X) -> n__from(X) , activate(X) -> X , activate(n__from(X)) -> from(activate(X)) , activate(n__s(X)) -> s(activate(X)) , activate(n__take(X1, X2)) -> take(activate(X1), activate(X2)) , take(X1, X2) -> n__take(X1, X2) , take(0(), XS) -> nil() , s(X) -> n__s(X) } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) We estimate the number of application of {1,2,3,5,7} by applications of Pre({1,2,3,5,7}) = {4,6}. Here rules are labeled as follows: DPs: { 1: from^#(X) -> c_1() , 2: 2nd^#(cons(X, XS)) -> c_4(head^#(activate(XS))) , 3: activate^#(X) -> c_5() , 4: activate^#(n__from(X)) -> c_6(from^#(activate(X))) , 5: activate^#(n__s(X)) -> c_7(s^#(activate(X))) , 6: activate^#(n__take(X1, X2)) -> c_8(take^#(activate(X1), activate(X2))) , 7: take^#(0(), XS) -> c_10() , 8: from^#(X) -> c_2() , 9: head^#(cons(X, XS)) -> c_3() , 10: s^#(X) -> c_11() , 11: take^#(X1, X2) -> c_9() , 12: sel^#(0(), cons(X, XS)) -> c_12() } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Strict DPs: { activate^#(n__from(X)) -> c_6(from^#(activate(X))) , activate^#(n__take(X1, X2)) -> c_8(take^#(activate(X1), activate(X2))) } Weak DPs: { from^#(X) -> c_1() , from^#(X) -> c_2() , head^#(cons(X, XS)) -> c_3() , 2nd^#(cons(X, XS)) -> c_4(head^#(activate(XS))) , activate^#(X) -> c_5() , activate^#(n__s(X)) -> c_7(s^#(activate(X))) , s^#(X) -> c_11() , take^#(X1, X2) -> c_9() , take^#(0(), XS) -> c_10() , sel^#(0(), cons(X, XS)) -> c_12() } Weak Trs: { from(X) -> cons(X, n__from(n__s(X))) , from(X) -> n__from(X) , activate(X) -> X , activate(n__from(X)) -> from(activate(X)) , activate(n__s(X)) -> s(activate(X)) , activate(n__take(X1, X2)) -> take(activate(X1), activate(X2)) , take(X1, X2) -> n__take(X1, X2) , take(0(), XS) -> nil() , s(X) -> n__s(X) } 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}) = {}. Here rules are labeled as follows: DPs: { 1: activate^#(n__from(X)) -> c_6(from^#(activate(X))) , 2: activate^#(n__take(X1, X2)) -> c_8(take^#(activate(X1), activate(X2))) , 3: from^#(X) -> c_1() , 4: from^#(X) -> c_2() , 5: head^#(cons(X, XS)) -> c_3() , 6: 2nd^#(cons(X, XS)) -> c_4(head^#(activate(XS))) , 7: activate^#(X) -> c_5() , 8: activate^#(n__s(X)) -> c_7(s^#(activate(X))) , 9: s^#(X) -> c_11() , 10: take^#(X1, X2) -> c_9() , 11: take^#(0(), XS) -> c_10() , 12: sel^#(0(), cons(X, XS)) -> c_12() } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak DPs: { from^#(X) -> c_1() , from^#(X) -> c_2() , head^#(cons(X, XS)) -> c_3() , 2nd^#(cons(X, XS)) -> c_4(head^#(activate(XS))) , activate^#(X) -> c_5() , activate^#(n__from(X)) -> c_6(from^#(activate(X))) , activate^#(n__s(X)) -> c_7(s^#(activate(X))) , activate^#(n__take(X1, X2)) -> c_8(take^#(activate(X1), activate(X2))) , s^#(X) -> c_11() , take^#(X1, X2) -> c_9() , take^#(0(), XS) -> c_10() , sel^#(0(), cons(X, XS)) -> c_12() } Weak Trs: { from(X) -> cons(X, n__from(n__s(X))) , from(X) -> n__from(X) , activate(X) -> X , activate(n__from(X)) -> from(activate(X)) , activate(n__s(X)) -> s(activate(X)) , activate(n__take(X1, X2)) -> take(activate(X1), activate(X2)) , take(X1, X2) -> n__take(X1, X2) , take(0(), XS) -> nil() , 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. { from^#(X) -> c_1() , from^#(X) -> c_2() , head^#(cons(X, XS)) -> c_3() , 2nd^#(cons(X, XS)) -> c_4(head^#(activate(XS))) , activate^#(X) -> c_5() , activate^#(n__from(X)) -> c_6(from^#(activate(X))) , activate^#(n__s(X)) -> c_7(s^#(activate(X))) , activate^#(n__take(X1, X2)) -> c_8(take^#(activate(X1), activate(X2))) , s^#(X) -> c_11() , take^#(X1, X2) -> c_9() , take^#(0(), XS) -> c_10() , sel^#(0(), cons(X, XS)) -> c_12() } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak Trs: { from(X) -> cons(X, n__from(n__s(X))) , from(X) -> n__from(X) , activate(X) -> X , activate(n__from(X)) -> from(activate(X)) , activate(n__s(X)) -> s(activate(X)) , activate(n__take(X1, X2)) -> take(activate(X1), activate(X2)) , take(X1, X2) -> n__take(X1, X2) , take(0(), XS) -> nil() , 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))