YES(O(1),O(n^1)) 0.00/0.80 YES(O(1),O(n^1)) 0.00/0.80 0.00/0.80 We are left with following problem, upon which TcT provides the 0.00/0.80 certificate YES(O(1),O(n^1)). 0.00/0.80 0.00/0.80 Strict Trs: 0.00/0.80 { terms(N) -> cons(recip(sqr(N)), n__terms(n__s(N))) 0.00/0.80 , terms(X) -> n__terms(X) 0.00/0.80 , sqr(0()) -> 0() 0.00/0.80 , sqr(s(X)) -> s(add(sqr(X), dbl(X))) 0.00/0.80 , s(X) -> n__s(X) 0.00/0.80 , add(0(), X) -> X 0.00/0.80 , add(s(X), Y) -> s(add(X, Y)) 0.00/0.80 , dbl(0()) -> 0() 0.00/0.80 , dbl(s(X)) -> s(s(dbl(X))) 0.00/0.80 , first(X1, X2) -> n__first(X1, X2) 0.00/0.80 , first(0(), X) -> nil() 0.00/0.80 , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z))) 0.00/0.80 , activate(X) -> X 0.00/0.80 , activate(n__terms(X)) -> terms(activate(X)) 0.00/0.80 , activate(n__s(X)) -> s(activate(X)) 0.00/0.80 , activate(n__first(X1, X2)) -> first(activate(X1), activate(X2)) } 0.00/0.80 Obligation: 0.00/0.80 innermost runtime complexity 0.00/0.80 Answer: 0.00/0.80 YES(O(1),O(n^1)) 0.00/0.80 0.00/0.80 Arguments of following rules are not normal-forms: 0.00/0.80 0.00/0.80 { sqr(s(X)) -> s(add(sqr(X), dbl(X))) 0.00/0.80 , add(s(X), Y) -> s(add(X, Y)) 0.00/0.80 , dbl(s(X)) -> s(s(dbl(X))) 0.00/0.80 , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z))) } 0.00/0.80 0.00/0.80 All above mentioned rules can be savely removed. 0.00/0.80 0.00/0.80 We are left with following problem, upon which TcT provides the 0.00/0.80 certificate YES(O(1),O(n^1)). 0.00/0.80 0.00/0.80 Strict Trs: 0.00/0.80 { terms(N) -> cons(recip(sqr(N)), n__terms(n__s(N))) 0.00/0.80 , terms(X) -> n__terms(X) 0.00/0.80 , sqr(0()) -> 0() 0.00/0.80 , s(X) -> n__s(X) 0.00/0.80 , add(0(), X) -> X 0.00/0.80 , dbl(0()) -> 0() 0.00/0.80 , first(X1, X2) -> n__first(X1, X2) 0.00/0.80 , first(0(), X) -> nil() 0.00/0.80 , activate(X) -> X 0.00/0.80 , activate(n__terms(X)) -> terms(activate(X)) 0.00/0.80 , activate(n__s(X)) -> s(activate(X)) 0.00/0.80 , activate(n__first(X1, X2)) -> first(activate(X1), activate(X2)) } 0.00/0.80 Obligation: 0.00/0.80 innermost runtime complexity 0.00/0.80 Answer: 0.00/0.80 YES(O(1),O(n^1)) 0.00/0.80 0.00/0.80 We add the following weak dependency pairs: 0.00/0.80 0.00/0.80 Strict DPs: 0.00/0.80 { terms^#(N) -> c_1(sqr^#(N)) 0.00/0.80 , terms^#(X) -> c_2() 0.00/0.80 , sqr^#(0()) -> c_3() 0.00/0.80 , s^#(X) -> c_4() 0.00/0.80 , add^#(0(), X) -> c_5() 0.00/0.80 , dbl^#(0()) -> c_6() 0.00/0.80 , first^#(X1, X2) -> c_7() 0.00/0.80 , first^#(0(), X) -> c_8() 0.00/0.80 , activate^#(X) -> c_9() 0.00/0.80 , activate^#(n__terms(X)) -> c_10(terms^#(activate(X))) 0.00/0.80 , activate^#(n__s(X)) -> c_11(s^#(activate(X))) 0.00/0.80 , activate^#(n__first(X1, X2)) -> 0.00/0.80 c_12(first^#(activate(X1), activate(X2))) } 0.00/0.80 0.00/0.80 and mark the set of starting terms. 0.00/0.80 0.00/0.80 We are left with following problem, upon which TcT provides the 0.00/0.80 certificate YES(O(1),O(n^1)). 0.00/0.80 0.00/0.80 Strict DPs: 0.00/0.80 { terms^#(N) -> c_1(sqr^#(N)) 0.00/0.80 , terms^#(X) -> c_2() 0.00/0.80 , sqr^#(0()) -> c_3() 0.00/0.80 , s^#(X) -> c_4() 0.00/0.80 , add^#(0(), X) -> c_5() 0.00/0.80 , dbl^#(0()) -> c_6() 0.00/0.80 , first^#(X1, X2) -> c_7() 0.00/0.80 , first^#(0(), X) -> c_8() 0.00/0.80 , activate^#(X) -> c_9() 0.00/0.80 , activate^#(n__terms(X)) -> c_10(terms^#(activate(X))) 0.00/0.80 , activate^#(n__s(X)) -> c_11(s^#(activate(X))) 0.00/0.80 , activate^#(n__first(X1, X2)) -> 0.00/0.80 c_12(first^#(activate(X1), activate(X2))) } 0.00/0.80 Strict Trs: 0.00/0.80 { terms(N) -> cons(recip(sqr(N)), n__terms(n__s(N))) 0.00/0.80 , terms(X) -> n__terms(X) 0.00/0.80 , sqr(0()) -> 0() 0.00/0.80 , s(X) -> n__s(X) 0.00/0.80 , add(0(), X) -> X 0.00/0.80 , dbl(0()) -> 0() 0.00/0.80 , first(X1, X2) -> n__first(X1, X2) 0.00/0.80 , first(0(), X) -> nil() 0.00/0.80 , activate(X) -> X 0.00/0.80 , activate(n__terms(X)) -> terms(activate(X)) 0.00/0.80 , activate(n__s(X)) -> s(activate(X)) 0.00/0.80 , activate(n__first(X1, X2)) -> first(activate(X1), activate(X2)) } 0.00/0.80 Obligation: 0.00/0.80 innermost runtime complexity 0.00/0.80 Answer: 0.00/0.80 YES(O(1),O(n^1)) 0.00/0.80 0.00/0.80 We replace rewrite rules by usable rules: 0.00/0.80 0.00/0.80 Strict Usable Rules: 0.00/0.80 { terms(N) -> cons(recip(sqr(N)), n__terms(n__s(N))) 0.00/0.80 , terms(X) -> n__terms(X) 0.00/0.80 , sqr(0()) -> 0() 0.00/0.80 , s(X) -> n__s(X) 0.00/0.80 , first(X1, X2) -> n__first(X1, X2) 0.00/0.80 , first(0(), X) -> nil() 0.00/0.80 , activate(X) -> X 0.00/0.80 , activate(n__terms(X)) -> terms(activate(X)) 0.00/0.80 , activate(n__s(X)) -> s(activate(X)) 0.00/0.80 , activate(n__first(X1, X2)) -> first(activate(X1), activate(X2)) } 0.00/0.80 0.00/0.80 We are left with following problem, upon which TcT provides the 0.00/0.80 certificate YES(O(1),O(n^1)). 0.00/0.80 0.00/0.80 Strict DPs: 0.00/0.80 { terms^#(N) -> c_1(sqr^#(N)) 0.00/0.80 , terms^#(X) -> c_2() 0.00/0.80 , sqr^#(0()) -> c_3() 0.00/0.80 , s^#(X) -> c_4() 0.00/0.80 , add^#(0(), X) -> c_5() 0.00/0.80 , dbl^#(0()) -> c_6() 0.00/0.80 , first^#(X1, X2) -> c_7() 0.00/0.80 , first^#(0(), X) -> c_8() 0.00/0.80 , activate^#(X) -> c_9() 0.00/0.80 , activate^#(n__terms(X)) -> c_10(terms^#(activate(X))) 0.00/0.80 , activate^#(n__s(X)) -> c_11(s^#(activate(X))) 0.00/0.80 , activate^#(n__first(X1, X2)) -> 0.00/0.80 c_12(first^#(activate(X1), activate(X2))) } 0.00/0.80 Strict Trs: 0.00/0.80 { terms(N) -> cons(recip(sqr(N)), n__terms(n__s(N))) 0.00/0.80 , terms(X) -> n__terms(X) 0.00/0.80 , sqr(0()) -> 0() 0.00/0.80 , s(X) -> n__s(X) 0.00/0.80 , first(X1, X2) -> n__first(X1, X2) 0.00/0.80 , first(0(), X) -> nil() 0.00/0.80 , activate(X) -> X 0.00/0.80 , activate(n__terms(X)) -> terms(activate(X)) 0.00/0.80 , activate(n__s(X)) -> s(activate(X)) 0.00/0.80 , activate(n__first(X1, X2)) -> first(activate(X1), activate(X2)) } 0.00/0.80 Obligation: 0.00/0.80 innermost runtime complexity 0.00/0.80 Answer: 0.00/0.80 YES(O(1),O(n^1)) 0.00/0.80 0.00/0.80 The weightgap principle applies (using the following constant 0.00/0.80 growth matrix-interpretation) 0.00/0.80 0.00/0.80 The following argument positions are usable: 0.00/0.80 Uargs(terms) = {1}, Uargs(cons) = {1}, Uargs(recip) = {1}, 0.00/0.80 Uargs(s) = {1}, Uargs(first) = {1, 2}, Uargs(terms^#) = {1}, 0.00/0.80 Uargs(c_1) = {1}, Uargs(s^#) = {1}, Uargs(first^#) = {1, 2}, 0.00/0.80 Uargs(c_10) = {1}, Uargs(c_11) = {1}, Uargs(c_12) = {1} 0.00/0.80 0.00/0.80 TcT has computed the following constructor-restricted matrix 0.00/0.80 interpretation. 0.00/0.80 0.00/0.80 [terms](x1) = [1 0] x1 + [2] 0.00/0.80 [0 1] [2] 0.00/0.80 0.00/0.80 [cons](x1, x2) = [1 0] x1 + [0] 0.00/0.80 [0 0] [1] 0.00/0.80 0.00/0.80 [recip](x1) = [1 0] x1 + [0] 0.00/0.80 [0 0] [2] 0.00/0.80 0.00/0.80 [sqr](x1) = [1] 0.00/0.80 [0] 0.00/0.80 0.00/0.80 [n__terms](x1) = [1 0] x1 + [0] 0.00/0.80 [0 1] [2] 0.00/0.80 0.00/0.80 [n__s](x1) = [1 0] x1 + [0] 0.00/0.80 [0 1] [2] 0.00/0.80 0.00/0.80 [0] = [0] 0.00/0.80 [0] 0.00/0.80 0.00/0.80 [s](x1) = [1 0] x1 + [1] 0.00/0.80 [0 1] [2] 0.00/0.80 0.00/0.80 [first](x1, x2) = [1 0] x1 + [1 0] x2 + [2] 0.00/0.80 [0 1] [0 1] [2] 0.00/0.80 0.00/0.80 [nil] = [1] 0.00/0.80 [1] 0.00/0.80 0.00/0.80 [n__first](x1, x2) = [1 0] x1 + [1 0] x2 + [0] 0.00/0.80 [0 1] [0 1] [2] 0.00/0.80 0.00/0.80 [activate](x1) = [1 2] x1 + [1] 0.00/0.80 [0 2] [0] 0.00/0.80 0.00/0.80 [terms^#](x1) = [1 0] x1 + [1] 0.00/0.80 [0 0] [2] 0.00/0.80 0.00/0.80 [c_1](x1) = [1 0] x1 + [1] 0.00/0.80 [0 1] [1] 0.00/0.80 0.00/0.80 [sqr^#](x1) = [1 0] x1 + [2] 0.00/0.80 [1 1] [2] 0.00/0.80 0.00/0.80 [c_2] = [0] 0.00/0.80 [1] 0.00/0.80 0.00/0.80 [c_3] = [1] 0.00/0.80 [1] 0.00/0.80 0.00/0.80 [s^#](x1) = [1 0] x1 + [1] 0.00/0.80 [0 0] [2] 0.00/0.80 0.00/0.80 [c_4] = [0] 0.00/0.80 [1] 0.00/0.80 0.00/0.80 [add^#](x1, x2) = [2 2] x1 + [2] 0.00/0.80 [1 2] [2] 0.00/0.80 0.00/0.80 [c_5] = [1] 0.00/0.80 [1] 0.00/0.80 0.00/0.80 [dbl^#](x1) = [1 2] x1 + [2] 0.00/0.80 [2 2] [2] 0.00/0.80 0.00/0.80 [c_6] = [1] 0.00/0.80 [1] 0.00/0.80 0.00/0.80 [first^#](x1, x2) = [1 0] x1 + [1 0] x2 + [2] 0.00/0.80 [0 0] [0 0] [2] 0.00/0.80 0.00/0.80 [c_7] = [1] 0.00/0.80 [1] 0.00/0.80 0.00/0.80 [c_8] = [1] 0.00/0.80 [1] 0.00/0.80 0.00/0.80 [activate^#](x1) = [1 2] x1 + [1] 0.00/0.80 [0 0] [0] 0.00/0.80 0.00/0.80 [c_9] = [0] 0.00/0.80 [1] 0.00/0.80 0.00/0.80 [c_10](x1) = [1 0] x1 + [2] 0.00/0.80 [0 1] [2] 0.00/0.80 0.00/0.80 [c_11](x1) = [1 0] x1 + [2] 0.00/0.80 [0 1] [2] 0.00/0.80 0.00/0.80 [c_12](x1) = [1 0] x1 + [2] 0.00/0.80 [0 1] [2] 0.00/0.80 0.00/0.80 The order satisfies the following ordering constraints: 0.00/0.80 0.00/0.80 [terms(N)] = [1 0] N + [2] 0.00/0.80 [0 1] [2] 0.00/0.80 > [1] 0.00/0.80 [1] 0.00/0.80 = [cons(recip(sqr(N)), n__terms(n__s(N)))] 0.00/0.80 0.00/0.80 [terms(X)] = [1 0] X + [2] 0.00/0.80 [0 1] [2] 0.00/0.80 > [1 0] X + [0] 0.00/0.80 [0 1] [2] 0.00/0.80 = [n__terms(X)] 0.00/0.80 0.00/0.80 [sqr(0())] = [1] 0.00/0.80 [0] 0.00/0.80 > [0] 0.00/0.80 [0] 0.00/0.80 = [0()] 0.00/0.80 0.00/0.80 [s(X)] = [1 0] X + [1] 0.00/0.80 [0 1] [2] 0.00/0.80 > [1 0] X + [0] 0.00/0.80 [0 1] [2] 0.00/0.80 = [n__s(X)] 0.00/0.80 0.00/0.80 [first(X1, X2)] = [1 0] X1 + [1 0] X2 + [2] 0.00/0.80 [0 1] [0 1] [2] 0.00/0.80 > [1 0] X1 + [1 0] X2 + [0] 0.00/0.80 [0 1] [0 1] [2] 0.00/0.80 = [n__first(X1, X2)] 0.00/0.80 0.00/0.80 [first(0(), X)] = [1 0] X + [2] 0.00/0.80 [0 1] [2] 0.00/0.80 > [1] 0.00/0.80 [1] 0.00/0.80 = [nil()] 0.00/0.80 0.00/0.80 [activate(X)] = [1 2] X + [1] 0.00/0.80 [0 2] [0] 0.00/0.80 > [1 0] X + [0] 0.00/0.80 [0 1] [0] 0.00/0.80 = [X] 0.00/0.80 0.00/0.80 [activate(n__terms(X))] = [1 2] X + [5] 0.00/0.80 [0 2] [4] 0.00/0.80 > [1 2] X + [3] 0.00/0.80 [0 2] [2] 0.00/0.80 = [terms(activate(X))] 0.00/0.80 0.00/0.80 [activate(n__s(X))] = [1 2] X + [5] 0.00/0.80 [0 2] [4] 0.00/0.80 > [1 2] X + [2] 0.00/0.80 [0 2] [2] 0.00/0.80 = [s(activate(X))] 0.00/0.80 0.00/0.80 [activate(n__first(X1, X2))] = [1 2] X1 + [1 2] X2 + [5] 0.00/0.80 [0 2] [0 2] [4] 0.00/0.80 > [1 2] X1 + [1 2] X2 + [4] 0.00/0.80 [0 2] [0 2] [2] 0.00/0.80 = [first(activate(X1), activate(X2))] 0.00/0.80 0.00/0.80 [terms^#(N)] = [1 0] N + [1] 0.00/0.80 [0 0] [2] 0.00/0.80 ? [1 0] N + [3] 0.00/0.80 [1 1] [3] 0.00/0.80 = [c_1(sqr^#(N))] 0.00/0.80 0.00/0.80 [terms^#(X)] = [1 0] X + [1] 0.00/0.80 [0 0] [2] 0.00/0.80 > [0] 0.00/0.80 [1] 0.00/0.80 = [c_2()] 0.00/0.80 0.00/0.80 [sqr^#(0())] = [2] 0.00/0.80 [2] 0.00/0.80 > [1] 0.00/0.80 [1] 0.00/0.80 = [c_3()] 0.00/0.80 0.00/0.80 [s^#(X)] = [1 0] X + [1] 0.00/0.80 [0 0] [2] 0.00/0.80 > [0] 0.00/0.80 [1] 0.00/0.80 = [c_4()] 0.00/0.81 0.00/0.81 [add^#(0(), X)] = [2] 0.00/0.81 [2] 0.00/0.81 > [1] 0.00/0.81 [1] 0.00/0.81 = [c_5()] 0.00/0.81 0.00/0.81 [dbl^#(0())] = [2] 0.00/0.81 [2] 0.00/0.81 > [1] 0.00/0.81 [1] 0.00/0.81 = [c_6()] 0.00/0.81 0.00/0.81 [first^#(X1, X2)] = [1 0] X1 + [1 0] X2 + [2] 0.00/0.81 [0 0] [0 0] [2] 0.00/0.81 > [1] 0.00/0.81 [1] 0.00/0.81 = [c_7()] 0.00/0.81 0.00/0.81 [first^#(0(), X)] = [1 0] X + [2] 0.00/0.81 [0 0] [2] 0.00/0.81 > [1] 0.00/0.81 [1] 0.00/0.81 = [c_8()] 0.00/0.81 0.00/0.81 [activate^#(X)] = [1 2] X + [1] 0.00/0.81 [0 0] [0] 0.00/0.81 ? [0] 0.00/0.81 [1] 0.00/0.81 = [c_9()] 0.00/0.81 0.00/0.81 [activate^#(n__terms(X))] = [1 2] X + [5] 0.00/0.81 [0 0] [0] 0.00/0.81 ? [1 2] X + [4] 0.00/0.81 [0 0] [4] 0.00/0.81 = [c_10(terms^#(activate(X)))] 0.00/0.81 0.00/0.81 [activate^#(n__s(X))] = [1 2] X + [5] 0.00/0.81 [0 0] [0] 0.00/0.81 ? [1 2] X + [4] 0.00/0.81 [0 0] [4] 0.00/0.81 = [c_11(s^#(activate(X)))] 0.00/0.81 0.00/0.81 [activate^#(n__first(X1, X2))] = [1 2] X1 + [1 2] X2 + [5] 0.00/0.81 [0 0] [0 0] [0] 0.00/0.81 ? [1 2] X1 + [1 2] X2 + [6] 0.00/0.81 [0 0] [0 0] [4] 0.00/0.81 = [c_12(first^#(activate(X1), activate(X2)))] 0.00/0.81 0.00/0.81 0.00/0.81 Further, it can be verified that all rules not oriented are covered by the weightgap condition. 0.00/0.81 0.00/0.81 We are left with following problem, upon which TcT provides the 0.00/0.81 certificate YES(O(1),O(1)). 0.00/0.81 0.00/0.81 Strict DPs: 0.00/0.81 { terms^#(N) -> c_1(sqr^#(N)) 0.00/0.81 , activate^#(X) -> c_9() 0.00/0.81 , activate^#(n__terms(X)) -> c_10(terms^#(activate(X))) 0.00/0.81 , activate^#(n__s(X)) -> c_11(s^#(activate(X))) 0.00/0.81 , activate^#(n__first(X1, X2)) -> 0.00/0.81 c_12(first^#(activate(X1), activate(X2))) } 0.00/0.81 Weak DPs: 0.00/0.81 { terms^#(X) -> c_2() 0.00/0.81 , sqr^#(0()) -> c_3() 0.00/0.81 , s^#(X) -> c_4() 0.00/0.81 , add^#(0(), X) -> c_5() 0.00/0.81 , dbl^#(0()) -> c_6() 0.00/0.81 , first^#(X1, X2) -> c_7() 0.00/0.81 , first^#(0(), X) -> c_8() } 0.00/0.81 Weak Trs: 0.00/0.81 { terms(N) -> cons(recip(sqr(N)), n__terms(n__s(N))) 0.00/0.81 , terms(X) -> n__terms(X) 0.00/0.81 , sqr(0()) -> 0() 0.00/0.81 , s(X) -> n__s(X) 0.00/0.81 , first(X1, X2) -> n__first(X1, X2) 0.00/0.81 , first(0(), X) -> nil() 0.00/0.81 , activate(X) -> X 0.00/0.81 , activate(n__terms(X)) -> terms(activate(X)) 0.00/0.81 , activate(n__s(X)) -> s(activate(X)) 0.00/0.81 , activate(n__first(X1, X2)) -> first(activate(X1), activate(X2)) } 0.00/0.81 Obligation: 0.00/0.81 innermost runtime complexity 0.00/0.81 Answer: 0.00/0.81 YES(O(1),O(1)) 0.00/0.81 0.00/0.81 We estimate the number of application of {1,2,4,5} by applications 0.00/0.81 of Pre({1,2,4,5}) = {3}. Here rules are labeled as follows: 0.00/0.81 0.00/0.81 DPs: 0.00/0.81 { 1: terms^#(N) -> c_1(sqr^#(N)) 0.00/0.81 , 2: activate^#(X) -> c_9() 0.00/0.81 , 3: activate^#(n__terms(X)) -> c_10(terms^#(activate(X))) 0.00/0.81 , 4: activate^#(n__s(X)) -> c_11(s^#(activate(X))) 0.00/0.81 , 5: activate^#(n__first(X1, X2)) -> 0.00/0.81 c_12(first^#(activate(X1), activate(X2))) 0.00/0.81 , 6: terms^#(X) -> c_2() 0.00/0.81 , 7: sqr^#(0()) -> c_3() 0.00/0.81 , 8: s^#(X) -> c_4() 0.00/0.81 , 9: add^#(0(), X) -> c_5() 0.00/0.81 , 10: dbl^#(0()) -> c_6() 0.00/0.81 , 11: first^#(X1, X2) -> c_7() 0.00/0.81 , 12: first^#(0(), X) -> c_8() } 0.00/0.81 0.00/0.81 We are left with following problem, upon which TcT provides the 0.00/0.81 certificate YES(O(1),O(1)). 0.00/0.81 0.00/0.81 Strict DPs: 0.00/0.81 { activate^#(n__terms(X)) -> c_10(terms^#(activate(X))) } 0.00/0.81 Weak DPs: 0.00/0.81 { terms^#(N) -> c_1(sqr^#(N)) 0.00/0.81 , terms^#(X) -> c_2() 0.00/0.81 , sqr^#(0()) -> c_3() 0.00/0.81 , s^#(X) -> c_4() 0.00/0.81 , add^#(0(), X) -> c_5() 0.00/0.81 , dbl^#(0()) -> c_6() 0.00/0.81 , first^#(X1, X2) -> c_7() 0.00/0.81 , first^#(0(), X) -> c_8() 0.00/0.81 , activate^#(X) -> c_9() 0.00/0.81 , activate^#(n__s(X)) -> c_11(s^#(activate(X))) 0.00/0.81 , activate^#(n__first(X1, X2)) -> 0.00/0.81 c_12(first^#(activate(X1), activate(X2))) } 0.00/0.81 Weak Trs: 0.00/0.81 { terms(N) -> cons(recip(sqr(N)), n__terms(n__s(N))) 0.00/0.81 , terms(X) -> n__terms(X) 0.00/0.81 , sqr(0()) -> 0() 0.00/0.81 , s(X) -> n__s(X) 0.00/0.81 , first(X1, X2) -> n__first(X1, X2) 0.00/0.81 , first(0(), X) -> nil() 0.00/0.81 , activate(X) -> X 0.00/0.81 , activate(n__terms(X)) -> terms(activate(X)) 0.00/0.81 , activate(n__s(X)) -> s(activate(X)) 0.00/0.81 , activate(n__first(X1, X2)) -> first(activate(X1), activate(X2)) } 0.00/0.81 Obligation: 0.00/0.81 innermost runtime complexity 0.00/0.81 Answer: 0.00/0.81 YES(O(1),O(1)) 0.00/0.81 0.00/0.81 We estimate the number of application of {1} by applications of 0.00/0.81 Pre({1}) = {}. Here rules are labeled as follows: 0.00/0.81 0.00/0.81 DPs: 0.00/0.81 { 1: activate^#(n__terms(X)) -> c_10(terms^#(activate(X))) 0.00/0.81 , 2: terms^#(N) -> c_1(sqr^#(N)) 0.00/0.81 , 3: terms^#(X) -> c_2() 0.00/0.81 , 4: sqr^#(0()) -> c_3() 0.00/0.81 , 5: s^#(X) -> c_4() 0.00/0.81 , 6: add^#(0(), X) -> c_5() 0.00/0.81 , 7: dbl^#(0()) -> c_6() 0.00/0.81 , 8: first^#(X1, X2) -> c_7() 0.00/0.81 , 9: first^#(0(), X) -> c_8() 0.00/0.81 , 10: activate^#(X) -> c_9() 0.00/0.81 , 11: activate^#(n__s(X)) -> c_11(s^#(activate(X))) 0.00/0.81 , 12: activate^#(n__first(X1, X2)) -> 0.00/0.81 c_12(first^#(activate(X1), activate(X2))) } 0.00/0.81 0.00/0.81 We are left with following problem, upon which TcT provides the 0.00/0.81 certificate YES(O(1),O(1)). 0.00/0.81 0.00/0.81 Weak DPs: 0.00/0.81 { terms^#(N) -> c_1(sqr^#(N)) 0.00/0.81 , terms^#(X) -> c_2() 0.00/0.81 , sqr^#(0()) -> c_3() 0.00/0.81 , s^#(X) -> c_4() 0.00/0.81 , add^#(0(), X) -> c_5() 0.00/0.81 , dbl^#(0()) -> c_6() 0.00/0.81 , first^#(X1, X2) -> c_7() 0.00/0.81 , first^#(0(), X) -> c_8() 0.00/0.81 , activate^#(X) -> c_9() 0.00/0.81 , activate^#(n__terms(X)) -> c_10(terms^#(activate(X))) 0.00/0.81 , activate^#(n__s(X)) -> c_11(s^#(activate(X))) 0.00/0.81 , activate^#(n__first(X1, X2)) -> 0.00/0.81 c_12(first^#(activate(X1), activate(X2))) } 0.00/0.81 Weak Trs: 0.00/0.81 { terms(N) -> cons(recip(sqr(N)), n__terms(n__s(N))) 0.00/0.81 , terms(X) -> n__terms(X) 0.00/0.81 , sqr(0()) -> 0() 0.00/0.81 , s(X) -> n__s(X) 0.00/0.81 , first(X1, X2) -> n__first(X1, X2) 0.00/0.81 , first(0(), X) -> nil() 0.00/0.81 , activate(X) -> X 0.00/0.81 , activate(n__terms(X)) -> terms(activate(X)) 0.00/0.81 , activate(n__s(X)) -> s(activate(X)) 0.00/0.81 , activate(n__first(X1, X2)) -> first(activate(X1), activate(X2)) } 0.00/0.81 Obligation: 0.00/0.81 innermost runtime complexity 0.00/0.81 Answer: 0.00/0.81 YES(O(1),O(1)) 0.00/0.81 0.00/0.81 The following weak DPs constitute a sub-graph of the DG that is 0.00/0.81 closed under successors. The DPs are removed. 0.00/0.81 0.00/0.81 { terms^#(N) -> c_1(sqr^#(N)) 0.00/0.81 , terms^#(X) -> c_2() 0.00/0.81 , sqr^#(0()) -> c_3() 0.00/0.81 , s^#(X) -> c_4() 0.00/0.81 , add^#(0(), X) -> c_5() 0.00/0.81 , dbl^#(0()) -> c_6() 0.00/0.81 , first^#(X1, X2) -> c_7() 0.00/0.81 , first^#(0(), X) -> c_8() 0.00/0.81 , activate^#(X) -> c_9() 0.00/0.81 , activate^#(n__terms(X)) -> c_10(terms^#(activate(X))) 0.00/0.81 , activate^#(n__s(X)) -> c_11(s^#(activate(X))) 0.00/0.81 , activate^#(n__first(X1, X2)) -> 0.00/0.81 c_12(first^#(activate(X1), activate(X2))) } 0.00/0.81 0.00/0.81 We are left with following problem, upon which TcT provides the 0.00/0.81 certificate YES(O(1),O(1)). 0.00/0.81 0.00/0.81 Weak Trs: 0.00/0.81 { terms(N) -> cons(recip(sqr(N)), n__terms(n__s(N))) 0.00/0.81 , terms(X) -> n__terms(X) 0.00/0.81 , sqr(0()) -> 0() 0.00/0.81 , s(X) -> n__s(X) 0.00/0.81 , first(X1, X2) -> n__first(X1, X2) 0.00/0.81 , first(0(), X) -> nil() 0.00/0.81 , activate(X) -> X 0.00/0.81 , activate(n__terms(X)) -> terms(activate(X)) 0.00/0.81 , activate(n__s(X)) -> s(activate(X)) 0.00/0.81 , activate(n__first(X1, X2)) -> first(activate(X1), activate(X2)) } 0.00/0.81 Obligation: 0.00/0.81 innermost runtime complexity 0.00/0.81 Answer: 0.00/0.81 YES(O(1),O(1)) 0.00/0.81 0.00/0.81 No rule is usable, rules are removed from the input problem. 0.00/0.81 0.00/0.81 We are left with following problem, upon which TcT provides the 0.00/0.81 certificate YES(O(1),O(1)). 0.00/0.81 0.00/0.81 Rules: Empty 0.00/0.81 Obligation: 0.00/0.81 innermost runtime complexity 0.00/0.81 Answer: 0.00/0.81 YES(O(1),O(1)) 0.00/0.81 0.00/0.81 Empty rules are trivially bounded 0.00/0.81 0.00/0.81 Hurray, we answered YES(O(1),O(n^1)) 0.00/0.81 EOF