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(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(X) , activate(n__take(X1, X2)) -> take(X1, X2) , take(X1, X2) -> n__take(X1, X2) , take(s(N), cons(X, XS)) -> cons(X, n__take(N, activate(XS))) , take(0(), XS) -> nil() , sel(s(N), cons(X, XS)) -> sel(N, activate(XS)) , 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^#(X)) , activate^#(n__take(X1, X2)) -> c_7(take^#(X1, X2)) , take^#(X1, X2) -> c_8() , take^#(s(N), cons(X, XS)) -> c_9(activate^#(XS)) , take^#(0(), XS) -> c_10() , sel^#(s(N), cons(X, XS)) -> c_11(sel^#(N, activate(XS))) , 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^#(X)) , activate^#(n__take(X1, X2)) -> c_7(take^#(X1, X2)) , take^#(X1, X2) -> c_8() , take^#(s(N), cons(X, XS)) -> c_9(activate^#(XS)) , take^#(0(), XS) -> c_10() , sel^#(s(N), cons(X, XS)) -> c_11(sel^#(N, activate(XS))) , sel^#(0(), cons(X, XS)) -> c_12() } Strict Trs: { from(X) -> cons(X, n__from(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(X) , activate(n__take(X1, X2)) -> take(X1, X2) , take(X1, X2) -> n__take(X1, X2) , take(s(N), cons(X, XS)) -> cons(X, n__take(N, activate(XS))) , take(0(), XS) -> nil() , sel(s(N), cons(X, XS)) -> sel(N, activate(XS)) , 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(s(X))) , from(X) -> n__from(X) , activate(X) -> X , activate(n__from(X)) -> from(X) , activate(n__take(X1, X2)) -> take(X1, X2) , take(X1, X2) -> n__take(X1, X2) , take(s(N), cons(X, XS)) -> cons(X, n__take(N, activate(XS))) , take(0(), XS) -> nil() } 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^#(X)) , activate^#(n__take(X1, X2)) -> c_7(take^#(X1, X2)) , take^#(X1, X2) -> c_8() , take^#(s(N), cons(X, XS)) -> c_9(activate^#(XS)) , take^#(0(), XS) -> c_10() , sel^#(s(N), cons(X, XS)) -> c_11(sel^#(N, activate(XS))) , sel^#(0(), cons(X, XS)) -> c_12() } Strict Trs: { from(X) -> cons(X, n__from(s(X))) , from(X) -> n__from(X) , activate(X) -> X , activate(n__from(X)) -> from(X) , activate(n__take(X1, X2)) -> take(X1, X2) , take(X1, X2) -> n__take(X1, X2) , take(s(N), cons(X, XS)) -> cons(X, n__take(N, activate(XS))) , take(0(), XS) -> nil() } 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(cons) = {2}, Uargs(n__take) = {2}, Uargs(head^#) = {1}, Uargs(c_4) = {1}, Uargs(c_6) = {1}, Uargs(c_7) = {1}, Uargs(c_9) = {1}, Uargs(sel^#) = {2}, Uargs(c_11) = {1} TcT has computed the following constructor-restricted matrix interpretation. [from](x1) = [1] [0] [cons](x1, x2) = [1 0] x2 + [0] [0 0] [0] [n__from](x1) = [0] [0] [s](x1) = [1 0] x1 + [2] [0 0] [0] [activate](x1) = [1 0] x1 + [2] [0 1] [0] [take](x1, x2) = [1 0] x1 + [1 0] x2 + [1] [0 0] [0 0] [0] [0] = [0] [0] [nil] = [0] [0] [n__take](x1, x2) = [1 0] x1 + [1 0] x2 + [0] [0 0] [0 0] [0] [from^#](x1) = [0 0] x1 + [1] [1 1] [1] [c_1] = [0] [1] [c_2] = [1] [1] [head^#](x1) = [2 0] x1 + [2] [0 0] [2] [c_3] = [1] [1] [2nd^#](x1) = [2 2] x1 + [2] [1 2] [2] [c_4](x1) = [1 0] x1 + [1] [0 1] [2] [activate^#](x1) = [0 0] x1 + [2] [1 1] [2] [c_5] = [1] [1] [c_6](x1) = [1 0] x1 + [2] [0 1] [2] [c_7](x1) = [1 0] x1 + [1] [0 1] [1] [take^#](x1, x2) = [0 0] x1 + [0 0] x2 + [2] [1 1] [2 2] [2] [c_8] = [1] [1] [c_9](x1) = [1 0] x1 + [1] [0 1] [1] [c_10] = [1] [1] [sel^#](x1, x2) = [2 0] x2 + [0] [0 0] [0] [c_11](x1) = [1 0] x1 + [2] [0 1] [2] [c_12] = [1] [1] The following symbols are considered usable {from, activate, take, from^#, head^#, 2nd^#, activate^#, take^#, sel^#} The order satisfies the following ordering constraints: [from(X)] = [1] [0] > [0] [0] = [cons(X, n__from(s(X)))] [from(X)] = [1] [0] > [0] [0] = [n__from(X)] [activate(X)] = [1 0] X + [2] [0 1] [0] > [1 0] X + [0] [0 1] [0] = [X] [activate(n__from(X))] = [2] [0] > [1] [0] = [from(X)] [activate(n__take(X1, X2))] = [1 0] X1 + [1 0] X2 + [2] [0 0] [0 0] [0] > [1 0] X1 + [1 0] X2 + [1] [0 0] [0 0] [0] = [take(X1, X2)] [take(X1, X2)] = [1 0] X1 + [1 0] X2 + [1] [0 0] [0 0] [0] > [1 0] X1 + [1 0] X2 + [0] [0 0] [0 0] [0] = [n__take(X1, X2)] [take(s(N), cons(X, XS))] = [1 0] XS + [1 0] N + [3] [0 0] [0 0] [0] > [1 0] XS + [1 0] N + [2] [0 0] [0 0] [0] = [cons(X, n__take(N, activate(XS)))] [take(0(), XS)] = [1 0] XS + [1] [0 0] [0] > [0] [0] = [nil()] [from^#(X)] = [0 0] X + [1] [1 1] [1] > [0] [1] = [c_1()] [from^#(X)] = [0 0] X + [1] [1 1] [1] >= [1] [1] = [c_2()] [head^#(cons(X, XS))] = [2 0] XS + [2] [0 0] [2] > [1] [1] = [c_3()] [2nd^#(cons(X, XS))] = [2 0] XS + [2] [1 0] [2] ? [2 0] XS + [7] [0 0] [4] = [c_4(head^#(activate(XS)))] [activate^#(X)] = [0 0] X + [2] [1 1] [2] > [1] [1] = [c_5()] [activate^#(n__from(X))] = [2] [2] ? [0 0] X + [3] [1 1] [3] = [c_6(from^#(X))] [activate^#(n__take(X1, X2))] = [0 0] X1 + [0 0] X2 + [2] [1 0] [1 0] [2] ? [0 0] X1 + [0 0] X2 + [3] [1 1] [2 2] [3] = [c_7(take^#(X1, X2))] [take^#(X1, X2)] = [0 0] X1 + [0 0] X2 + [2] [1 1] [2 2] [2] > [1] [1] = [c_8()] [take^#(s(N), cons(X, XS))] = [0 0] XS + [0 0] N + [2] [2 0] [1 0] [4] ? [0 0] XS + [3] [1 1] [3] = [c_9(activate^#(XS))] [take^#(0(), XS)] = [0 0] XS + [2] [2 2] [2] > [1] [1] = [c_10()] [sel^#(s(N), cons(X, XS))] = [2 0] XS + [0] [0 0] [0] ? [2 0] XS + [6] [0 0] [2] = [c_11(sel^#(N, activate(XS)))] [sel^#(0(), cons(X, XS))] = [2 0] XS + [0] [0 0] [0] ? [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(n^1)). Strict DPs: { from^#(X) -> c_2() , 2nd^#(cons(X, XS)) -> c_4(head^#(activate(XS))) , activate^#(n__from(X)) -> c_6(from^#(X)) , activate^#(n__take(X1, X2)) -> c_7(take^#(X1, X2)) , take^#(s(N), cons(X, XS)) -> c_9(activate^#(XS)) , sel^#(s(N), cons(X, XS)) -> c_11(sel^#(N, activate(XS))) , sel^#(0(), cons(X, XS)) -> c_12() } Weak DPs: { from^#(X) -> c_1() , head^#(cons(X, XS)) -> c_3() , activate^#(X) -> c_5() , take^#(X1, X2) -> c_8() , take^#(0(), XS) -> c_10() } Weak Trs: { from(X) -> cons(X, n__from(s(X))) , from(X) -> n__from(X) , activate(X) -> X , activate(n__from(X)) -> from(X) , activate(n__take(X1, X2)) -> take(X1, X2) , take(X1, X2) -> n__take(X1, X2) , take(s(N), cons(X, XS)) -> cons(X, n__take(N, activate(XS))) , take(0(), XS) -> nil() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We estimate the number of application of {1,2,7} by applications of Pre({1,2,7}) = {3,6}. Here rules are labeled as follows: DPs: { 1: from^#(X) -> c_2() , 2: 2nd^#(cons(X, XS)) -> c_4(head^#(activate(XS))) , 3: activate^#(n__from(X)) -> c_6(from^#(X)) , 4: activate^#(n__take(X1, X2)) -> c_7(take^#(X1, X2)) , 5: take^#(s(N), cons(X, XS)) -> c_9(activate^#(XS)) , 6: sel^#(s(N), cons(X, XS)) -> c_11(sel^#(N, activate(XS))) , 7: sel^#(0(), cons(X, XS)) -> c_12() , 8: from^#(X) -> c_1() , 9: head^#(cons(X, XS)) -> c_3() , 10: activate^#(X) -> c_5() , 11: take^#(X1, X2) -> c_8() , 12: take^#(0(), XS) -> c_10() } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { activate^#(n__from(X)) -> c_6(from^#(X)) , activate^#(n__take(X1, X2)) -> c_7(take^#(X1, X2)) , take^#(s(N), cons(X, XS)) -> c_9(activate^#(XS)) , sel^#(s(N), cons(X, XS)) -> c_11(sel^#(N, activate(XS))) } 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() , take^#(X1, X2) -> c_8() , take^#(0(), XS) -> c_10() , sel^#(0(), cons(X, XS)) -> c_12() } Weak Trs: { from(X) -> cons(X, n__from(s(X))) , from(X) -> n__from(X) , activate(X) -> X , activate(n__from(X)) -> from(X) , activate(n__take(X1, X2)) -> take(X1, X2) , take(X1, X2) -> n__take(X1, X2) , take(s(N), cons(X, XS)) -> cons(X, n__take(N, activate(XS))) , take(0(), XS) -> nil() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We estimate the number of application of {1} by applications of Pre({1}) = {3}. Here rules are labeled as follows: DPs: { 1: activate^#(n__from(X)) -> c_6(from^#(X)) , 2: activate^#(n__take(X1, X2)) -> c_7(take^#(X1, X2)) , 3: take^#(s(N), cons(X, XS)) -> c_9(activate^#(XS)) , 4: sel^#(s(N), cons(X, XS)) -> c_11(sel^#(N, activate(XS))) , 5: from^#(X) -> c_1() , 6: from^#(X) -> c_2() , 7: head^#(cons(X, XS)) -> c_3() , 8: 2nd^#(cons(X, XS)) -> c_4(head^#(activate(XS))) , 9: activate^#(X) -> c_5() , 10: take^#(X1, X2) -> c_8() , 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(n^1)). Strict DPs: { activate^#(n__take(X1, X2)) -> c_7(take^#(X1, X2)) , take^#(s(N), cons(X, XS)) -> c_9(activate^#(XS)) , sel^#(s(N), cons(X, XS)) -> c_11(sel^#(N, activate(XS))) } 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^#(X)) , take^#(X1, X2) -> c_8() , take^#(0(), XS) -> c_10() , sel^#(0(), cons(X, XS)) -> c_12() } Weak Trs: { from(X) -> cons(X, n__from(s(X))) , from(X) -> n__from(X) , activate(X) -> X , activate(n__from(X)) -> from(X) , activate(n__take(X1, X2)) -> take(X1, X2) , take(X1, X2) -> n__take(X1, X2) , take(s(N), cons(X, XS)) -> cons(X, n__take(N, activate(XS))) , take(0(), XS) -> nil() } 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. { 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^#(X)) , take^#(X1, X2) -> c_8() , 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(n^1)). Strict DPs: { activate^#(n__take(X1, X2)) -> c_7(take^#(X1, X2)) , take^#(s(N), cons(X, XS)) -> c_9(activate^#(XS)) , sel^#(s(N), cons(X, XS)) -> c_11(sel^#(N, activate(XS))) } Weak Trs: { from(X) -> cons(X, n__from(s(X))) , from(X) -> n__from(X) , activate(X) -> X , activate(n__from(X)) -> from(X) , activate(n__take(X1, X2)) -> take(X1, X2) , take(X1, X2) -> n__take(X1, X2) , take(s(N), cons(X, XS)) -> cons(X, n__take(N, activate(XS))) , take(0(), XS) -> nil() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'Small Polynomial Path Order (PS,1-bounded)' to orient following rules strictly. DPs: { 1: activate^#(n__take(X1, X2)) -> c_7(take^#(X1, X2)) , 2: take^#(s(N), cons(X, XS)) -> c_9(activate^#(XS)) , 3: sel^#(s(N), cons(X, XS)) -> c_11(sel^#(N, activate(XS))) } Trs: { from(X) -> n__from(X) , take(0(), XS) -> nil() } Sub-proof: ---------- The input was oriented with the instance of 'Small Polynomial Path Order (PS,1-bounded)' as induced by the safe mapping safe(from) = {1}, safe(cons) = {1, 2}, safe(n__from) = {1}, safe(s) = {1}, safe(head) = {}, safe(2nd) = {}, safe(activate) = {}, safe(take) = {}, safe(0) = {}, safe(nil) = {}, safe(n__take) = {1, 2}, safe(sel) = {}, safe(from^#) = {}, safe(c_1) = {}, safe(c_2) = {}, safe(head^#) = {}, safe(c_3) = {}, safe(2nd^#) = {}, safe(c_4) = {}, safe(activate^#) = {}, safe(c_5) = {}, safe(c_6) = {}, safe(c_7) = {}, safe(take^#) = {1}, safe(c_8) = {}, safe(c_9) = {}, safe(c_10) = {}, safe(sel^#) = {2}, safe(c_11) = {}, safe(c_12) = {} and precedence sel^# > from, from ~ activate, take ~ activate^#, take ~ take^#, activate^# ~ take^# . Following symbols are considered recursive: {activate^#, take^#, sel^#} The recursion depth is 1. Further, following argument filtering is employed: pi(from) = [], pi(cons) = [2], pi(n__from) = [], pi(s) = [1], pi(head) = [], pi(2nd) = [], pi(activate) = [], pi(take) = [], pi(0) = [], pi(nil) = [], pi(n__take) = [2], pi(sel) = [], pi(from^#) = [], pi(c_1) = [], pi(c_2) = [], pi(head^#) = [], pi(c_3) = [], pi(2nd^#) = [], pi(c_4) = [], pi(activate^#) = [1], pi(c_5) = [], pi(c_6) = [], pi(c_7) = [1], pi(take^#) = [2], pi(c_8) = [], pi(c_9) = [1], pi(c_10) = [], pi(sel^#) = [1], pi(c_11) = [1], pi(c_12) = [] Usable defined function symbols are a subset of: {from^#, head^#, 2nd^#, activate^#, take^#, sel^#} For your convenience, here are the satisfied ordering constraints: pi(activate^#(n__take(X1, X2))) = activate^#(n__take(; X2);) > c_7(take^#(X2;);) = pi(c_7(take^#(X1, X2))) pi(take^#(s(N), cons(X, XS))) = take^#(cons(; XS);) > c_9(activate^#(XS;);) = pi(c_9(activate^#(XS))) pi(sel^#(s(N), cons(X, XS))) = sel^#(s(; N);) > c_11(sel^#(N;);) = pi(c_11(sel^#(N, activate(XS)))) The strictly oriented rules are moved into the weak component. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak DPs: { activate^#(n__take(X1, X2)) -> c_7(take^#(X1, X2)) , take^#(s(N), cons(X, XS)) -> c_9(activate^#(XS)) , sel^#(s(N), cons(X, XS)) -> c_11(sel^#(N, activate(XS))) } Weak Trs: { from(X) -> cons(X, n__from(s(X))) , from(X) -> n__from(X) , activate(X) -> X , activate(n__from(X)) -> from(X) , activate(n__take(X1, X2)) -> take(X1, X2) , take(X1, X2) -> n__take(X1, X2) , take(s(N), cons(X, XS)) -> cons(X, n__take(N, activate(XS))) , take(0(), XS) -> nil() } 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__take(X1, X2)) -> c_7(take^#(X1, X2)) , take^#(s(N), cons(X, XS)) -> c_9(activate^#(XS)) , sel^#(s(N), cons(X, XS)) -> c_11(sel^#(N, activate(XS))) } 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(s(X))) , from(X) -> n__from(X) , activate(X) -> X , activate(n__from(X)) -> from(X) , activate(n__take(X1, X2)) -> take(X1, X2) , take(X1, X2) -> n__take(X1, X2) , take(s(N), cons(X, XS)) -> cons(X, n__take(N, activate(XS))) , take(0(), XS) -> nil() } 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))