YES(O(1),O(n^1)) 0.00/0.86 YES(O(1),O(n^1)) 0.00/0.86 0.00/0.86 We are left with following problem, upon which TcT provides the 0.00/0.86 certificate YES(O(1),O(n^1)). 0.00/0.86 0.00/0.86 Strict Trs: 0.00/0.86 { terms(N) -> cons(recip(sqr(N)), n__terms(n__s(N))) 0.00/0.86 , terms(X) -> n__terms(X) 0.00/0.86 , sqr(0()) -> 0() 0.00/0.86 , sqr(s(X)) -> s(add(sqr(X), dbl(X))) 0.00/0.86 , s(X) -> n__s(X) 0.00/0.86 , add(0(), X) -> X 0.00/0.86 , add(s(X), Y) -> s(add(X, Y)) 0.00/0.86 , dbl(0()) -> 0() 0.00/0.86 , dbl(s(X)) -> s(s(dbl(X))) 0.00/0.86 , first(X1, X2) -> n__first(X1, X2) 0.00/0.86 , first(0(), X) -> nil() 0.00/0.86 , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z))) 0.00/0.86 , activate(X) -> X 0.00/0.86 , activate(n__terms(X)) -> terms(activate(X)) 0.00/0.86 , activate(n__s(X)) -> s(activate(X)) 0.00/0.86 , activate(n__first(X1, X2)) -> first(activate(X1), activate(X2)) 0.00/0.86 , half(0()) -> 0() 0.00/0.86 , half(s(0())) -> 0() 0.00/0.86 , half(s(s(X))) -> s(half(X)) 0.00/0.86 , half(dbl(X)) -> X } 0.00/0.86 Obligation: 0.00/0.86 innermost runtime complexity 0.00/0.86 Answer: 0.00/0.86 YES(O(1),O(n^1)) 0.00/0.86 0.00/0.86 Arguments of following rules are not normal-forms: 0.00/0.86 0.00/0.86 { sqr(s(X)) -> s(add(sqr(X), dbl(X))) 0.00/0.86 , add(s(X), Y) -> s(add(X, Y)) 0.00/0.86 , dbl(s(X)) -> s(s(dbl(X))) 0.00/0.86 , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z))) 0.00/0.86 , half(s(0())) -> 0() 0.00/0.86 , half(s(s(X))) -> s(half(X)) } 0.00/0.86 0.00/0.86 All above mentioned rules can be savely removed. 0.00/0.86 0.00/0.86 We are left with following problem, upon which TcT provides the 0.00/0.86 certificate YES(O(1),O(n^1)). 0.00/0.86 0.00/0.86 Strict Trs: 0.00/0.86 { terms(N) -> cons(recip(sqr(N)), n__terms(n__s(N))) 0.00/0.86 , terms(X) -> n__terms(X) 0.00/0.86 , sqr(0()) -> 0() 0.00/0.86 , s(X) -> n__s(X) 0.00/0.86 , add(0(), X) -> X 0.00/0.86 , dbl(0()) -> 0() 0.00/0.86 , first(X1, X2) -> n__first(X1, X2) 0.00/0.86 , first(0(), X) -> nil() 0.00/0.86 , activate(X) -> X 0.00/0.86 , activate(n__terms(X)) -> terms(activate(X)) 0.00/0.86 , activate(n__s(X)) -> s(activate(X)) 0.00/0.86 , activate(n__first(X1, X2)) -> first(activate(X1), activate(X2)) 0.00/0.86 , half(0()) -> 0() 0.00/0.86 , half(dbl(X)) -> X } 0.00/0.86 Obligation: 0.00/0.86 innermost runtime complexity 0.00/0.86 Answer: 0.00/0.86 YES(O(1),O(n^1)) 0.00/0.86 0.00/0.86 The weightgap principle applies (using the following nonconstant 0.00/0.86 growth matrix-interpretation) 0.00/0.86 0.00/0.86 The following argument positions are usable: 0.00/0.86 Uargs(terms) = {1}, Uargs(cons) = {1}, Uargs(recip) = {1}, 0.00/0.86 Uargs(s) = {1}, Uargs(first) = {1, 2} 0.00/0.86 0.00/0.86 TcT has computed the following matrix interpretation satisfying 0.00/0.86 not(EDA) and not(IDA(1)). 0.00/0.86 0.00/0.86 [terms](x1) = [1] x1 + [5] 0.00/0.86 0.00/0.86 [cons](x1, x2) = [1] x1 + [1] x2 + [0] 0.00/0.86 0.00/0.86 [recip](x1) = [1] x1 + [0] 0.00/0.86 0.00/0.86 [sqr](x1) = [1] 0.00/0.86 0.00/0.86 [n__terms](x1) = [1] x1 + [7] 0.00/0.86 0.00/0.86 [n__s](x1) = [1] x1 + [7] 0.00/0.86 0.00/0.86 [0] = [7] 0.00/0.86 0.00/0.86 [s](x1) = [1] x1 + [7] 0.00/0.86 0.00/0.86 [add](x1, x2) = [1] x1 + [1] x2 + [7] 0.00/0.86 0.00/0.86 [dbl](x1) = [1] x1 + [7] 0.00/0.86 0.00/0.86 [first](x1, x2) = [1] x1 + [1] x2 + [3] 0.00/0.86 0.00/0.86 [nil] = [1] 0.00/0.86 0.00/0.86 [n__first](x1, x2) = [1] x1 + [1] x2 + [6] 0.00/0.86 0.00/0.86 [activate](x1) = [1] x1 + [6] 0.00/0.86 0.00/0.86 [half](x1) = [1] x1 + [7] 0.00/0.86 0.00/0.86 The order satisfies the following ordering constraints: 0.00/0.86 0.00/0.86 [terms(N)] = [1] N + [5] 0.00/0.86 ? [1] N + [15] 0.00/0.86 = [cons(recip(sqr(N)), n__terms(n__s(N)))] 0.00/0.86 0.00/0.86 [terms(X)] = [1] X + [5] 0.00/0.86 ? [1] X + [7] 0.00/0.86 = [n__terms(X)] 0.00/0.86 0.00/0.86 [sqr(0())] = [1] 0.00/0.86 ? [7] 0.00/0.86 = [0()] 0.00/0.86 0.00/0.86 [s(X)] = [1] X + [7] 0.00/0.86 >= [1] X + [7] 0.00/0.86 = [n__s(X)] 0.00/0.86 0.00/0.86 [add(0(), X)] = [1] X + [14] 0.00/0.86 > [1] X + [0] 0.00/0.86 = [X] 0.00/0.86 0.00/0.86 [dbl(0())] = [14] 0.00/0.86 > [7] 0.00/0.86 = [0()] 0.00/0.86 0.00/0.86 [first(X1, X2)] = [1] X1 + [1] X2 + [3] 0.00/0.86 ? [1] X1 + [1] X2 + [6] 0.00/0.86 = [n__first(X1, X2)] 0.00/0.86 0.00/0.86 [first(0(), X)] = [1] X + [10] 0.00/0.86 > [1] 0.00/0.86 = [nil()] 0.00/0.86 0.00/0.86 [activate(X)] = [1] X + [6] 0.00/0.86 > [1] X + [0] 0.00/0.86 = [X] 0.00/0.86 0.00/0.86 [activate(n__terms(X))] = [1] X + [13] 0.00/0.86 > [1] X + [11] 0.00/0.86 = [terms(activate(X))] 0.00/0.86 0.00/0.86 [activate(n__s(X))] = [1] X + [13] 0.00/0.86 >= [1] X + [13] 0.00/0.86 = [s(activate(X))] 0.00/0.86 0.00/0.86 [activate(n__first(X1, X2))] = [1] X1 + [1] X2 + [12] 0.00/0.86 ? [1] X1 + [1] X2 + [15] 0.00/0.86 = [first(activate(X1), activate(X2))] 0.00/0.86 0.00/0.86 [half(0())] = [14] 0.00/0.86 > [7] 0.00/0.86 = [0()] 0.00/0.86 0.00/0.86 [half(dbl(X))] = [1] X + [14] 0.00/0.86 > [1] X + [0] 0.00/0.86 = [X] 0.00/0.86 0.00/0.86 0.00/0.86 Further, it can be verified that all rules not oriented are covered by the weightgap condition. 0.00/0.86 0.00/0.86 We are left with following problem, upon which TcT provides the 0.00/0.86 certificate YES(O(1),O(n^1)). 0.00/0.86 0.00/0.86 Strict Trs: 0.00/0.86 { terms(N) -> cons(recip(sqr(N)), n__terms(n__s(N))) 0.00/0.86 , terms(X) -> n__terms(X) 0.00/0.86 , sqr(0()) -> 0() 0.00/0.86 , s(X) -> n__s(X) 0.00/0.86 , first(X1, X2) -> n__first(X1, X2) 0.00/0.86 , activate(n__s(X)) -> s(activate(X)) 0.00/0.86 , activate(n__first(X1, X2)) -> first(activate(X1), activate(X2)) } 0.00/0.86 Weak Trs: 0.00/0.86 { add(0(), X) -> X 0.00/0.86 , dbl(0()) -> 0() 0.00/0.86 , first(0(), X) -> nil() 0.00/0.86 , activate(X) -> X 0.00/0.86 , activate(n__terms(X)) -> terms(activate(X)) 0.00/0.86 , half(0()) -> 0() 0.00/0.86 , half(dbl(X)) -> X } 0.00/0.86 Obligation: 0.00/0.86 innermost runtime complexity 0.00/0.86 Answer: 0.00/0.86 YES(O(1),O(n^1)) 0.00/0.86 0.00/0.86 The weightgap principle applies (using the following nonconstant 0.00/0.86 growth matrix-interpretation) 0.00/0.86 0.00/0.86 The following argument positions are usable: 0.00/0.86 Uargs(terms) = {1}, Uargs(cons) = {1}, Uargs(recip) = {1}, 0.00/0.86 Uargs(s) = {1}, Uargs(first) = {1, 2} 0.00/0.86 0.00/0.86 TcT has computed the following matrix interpretation satisfying 0.00/0.86 not(EDA) and not(IDA(1)). 0.00/0.86 0.00/0.86 [terms](x1) = [1] x1 + [1] 0.00/0.86 0.00/0.86 [cons](x1, x2) = [1] x1 + [0] 0.00/0.86 0.00/0.86 [recip](x1) = [1] x1 + [0] 0.00/0.86 0.00/0.86 [sqr](x1) = [1] x1 + [0] 0.00/0.86 0.00/0.86 [n__terms](x1) = [1] x1 + [4] 0.00/0.86 0.00/0.86 [n__s](x1) = [1] x1 + [0] 0.00/0.86 0.00/0.86 [0] = [5] 0.00/0.86 0.00/0.86 [s](x1) = [1] x1 + [0] 0.00/0.86 0.00/0.86 [add](x1, x2) = [1] x1 + [1] x2 + [1] 0.00/0.86 0.00/0.86 [dbl](x1) = [1] x1 + [7] 0.00/0.86 0.00/0.86 [first](x1, x2) = [1] x1 + [1] x2 + [0] 0.00/0.86 0.00/0.86 [nil] = [5] 0.00/0.86 0.00/0.86 [n__first](x1, x2) = [1] x1 + [1] x2 + [1] 0.00/0.86 0.00/0.86 [activate](x1) = [1] x1 + [0] 0.00/0.86 0.00/0.86 [half](x1) = [1] x1 + [7] 0.00/0.86 0.00/0.86 The order satisfies the following ordering constraints: 0.00/0.86 0.00/0.86 [terms(N)] = [1] N + [1] 0.00/0.86 > [1] N + [0] 0.00/0.86 = [cons(recip(sqr(N)), n__terms(n__s(N)))] 0.00/0.86 0.00/0.86 [terms(X)] = [1] X + [1] 0.00/0.86 ? [1] X + [4] 0.00/0.86 = [n__terms(X)] 0.00/0.86 0.00/0.86 [sqr(0())] = [5] 0.00/0.86 >= [5] 0.00/0.86 = [0()] 0.00/0.86 0.00/0.86 [s(X)] = [1] X + [0] 0.00/0.86 >= [1] X + [0] 0.00/0.86 = [n__s(X)] 0.00/0.86 0.00/0.86 [add(0(), X)] = [1] X + [6] 0.00/0.86 > [1] X + [0] 0.00/0.86 = [X] 0.00/0.86 0.00/0.86 [dbl(0())] = [12] 0.00/0.86 > [5] 0.00/0.86 = [0()] 0.00/0.86 0.00/0.86 [first(X1, X2)] = [1] X1 + [1] X2 + [0] 0.00/0.86 ? [1] X1 + [1] X2 + [1] 0.00/0.86 = [n__first(X1, X2)] 0.00/0.86 0.00/0.86 [first(0(), X)] = [1] X + [5] 0.00/0.86 >= [5] 0.00/0.86 = [nil()] 0.00/0.86 0.00/0.86 [activate(X)] = [1] X + [0] 0.00/0.86 >= [1] X + [0] 0.00/0.86 = [X] 0.00/0.86 0.00/0.86 [activate(n__terms(X))] = [1] X + [4] 0.00/0.86 > [1] X + [1] 0.00/0.86 = [terms(activate(X))] 0.00/0.86 0.00/0.86 [activate(n__s(X))] = [1] X + [0] 0.00/0.86 >= [1] X + [0] 0.00/0.86 = [s(activate(X))] 0.00/0.86 0.00/0.86 [activate(n__first(X1, X2))] = [1] X1 + [1] X2 + [1] 0.00/0.86 > [1] X1 + [1] X2 + [0] 0.00/0.86 = [first(activate(X1), activate(X2))] 0.00/0.86 0.00/0.86 [half(0())] = [12] 0.00/0.86 > [5] 0.00/0.86 = [0()] 0.00/0.86 0.00/0.86 [half(dbl(X))] = [1] X + [14] 0.00/0.86 > [1] X + [0] 0.00/0.86 = [X] 0.00/0.86 0.00/0.86 0.00/0.86 Further, it can be verified that all rules not oriented are covered by the weightgap condition. 0.00/0.86 0.00/0.86 We are left with following problem, upon which TcT provides the 0.00/0.86 certificate YES(O(1),O(n^1)). 0.00/0.86 0.00/0.86 Strict Trs: 0.00/0.86 { terms(X) -> n__terms(X) 0.00/0.86 , sqr(0()) -> 0() 0.00/0.86 , s(X) -> n__s(X) 0.00/0.86 , first(X1, X2) -> n__first(X1, X2) 0.00/0.86 , activate(n__s(X)) -> s(activate(X)) } 0.00/0.86 Weak Trs: 0.00/0.86 { terms(N) -> cons(recip(sqr(N)), n__terms(n__s(N))) 0.00/0.86 , add(0(), X) -> X 0.00/0.86 , dbl(0()) -> 0() 0.00/0.86 , first(0(), X) -> nil() 0.00/0.86 , activate(X) -> X 0.00/0.86 , activate(n__terms(X)) -> terms(activate(X)) 0.00/0.86 , activate(n__first(X1, X2)) -> first(activate(X1), activate(X2)) 0.00/0.86 , half(0()) -> 0() 0.00/0.86 , half(dbl(X)) -> X } 0.00/0.86 Obligation: 0.00/0.86 innermost runtime complexity 0.00/0.86 Answer: 0.00/0.86 YES(O(1),O(n^1)) 0.00/0.86 0.00/0.86 The weightgap principle applies (using the following nonconstant 0.00/0.86 growth matrix-interpretation) 0.00/0.86 0.00/0.86 The following argument positions are usable: 0.00/0.86 Uargs(terms) = {1}, Uargs(cons) = {1}, Uargs(recip) = {1}, 0.00/0.86 Uargs(s) = {1}, Uargs(first) = {1, 2} 0.00/0.86 0.00/0.86 TcT has computed the following matrix interpretation satisfying 0.00/0.86 not(EDA) and not(IDA(1)). 0.00/0.86 0.00/0.86 [terms](x1) = [1] x1 + [1] 0.00/0.86 0.00/0.86 [cons](x1, x2) = [1] x1 + [0] 0.00/0.86 0.00/0.86 [recip](x1) = [1] x1 + [0] 0.00/0.86 0.00/0.86 [sqr](x1) = [1] x1 + [1] 0.00/0.86 0.00/0.86 [n__terms](x1) = [1] x1 + [1] 0.00/0.86 0.00/0.86 [n__s](x1) = [1] x1 + [0] 0.00/0.86 0.00/0.86 [0] = [7] 0.00/0.86 0.00/0.86 [s](x1) = [1] x1 + [0] 0.00/0.86 0.00/0.86 [add](x1, x2) = [1] x1 + [1] x2 + [7] 0.00/0.86 0.00/0.86 [dbl](x1) = [1] x1 + [7] 0.00/0.86 0.00/0.86 [first](x1, x2) = [1] x1 + [1] x2 + [0] 0.00/0.86 0.00/0.86 [nil] = [7] 0.00/0.86 0.00/0.86 [n__first](x1, x2) = [1] x1 + [1] x2 + [0] 0.00/0.86 0.00/0.86 [activate](x1) = [1] x1 + [0] 0.00/0.86 0.00/0.86 [half](x1) = [1] x1 + [7] 0.00/0.86 0.00/0.86 The order satisfies the following ordering constraints: 0.00/0.86 0.00/0.86 [terms(N)] = [1] N + [1] 0.00/0.86 >= [1] N + [1] 0.00/0.86 = [cons(recip(sqr(N)), n__terms(n__s(N)))] 0.00/0.86 0.00/0.86 [terms(X)] = [1] X + [1] 0.00/0.86 >= [1] X + [1] 0.00/0.86 = [n__terms(X)] 0.00/0.86 0.00/0.86 [sqr(0())] = [8] 0.00/0.86 > [7] 0.00/0.86 = [0()] 0.00/0.86 0.00/0.86 [s(X)] = [1] X + [0] 0.00/0.86 >= [1] X + [0] 0.00/0.86 = [n__s(X)] 0.00/0.86 0.00/0.86 [add(0(), X)] = [1] X + [14] 0.00/0.86 > [1] X + [0] 0.00/0.86 = [X] 0.00/0.86 0.00/0.86 [dbl(0())] = [14] 0.00/0.86 > [7] 0.00/0.86 = [0()] 0.00/0.86 0.00/0.86 [first(X1, X2)] = [1] X1 + [1] X2 + [0] 0.00/0.86 >= [1] X1 + [1] X2 + [0] 0.00/0.86 = [n__first(X1, X2)] 0.00/0.86 0.00/0.86 [first(0(), X)] = [1] X + [7] 0.00/0.86 >= [7] 0.00/0.86 = [nil()] 0.00/0.86 0.00/0.86 [activate(X)] = [1] X + [0] 0.00/0.86 >= [1] X + [0] 0.00/0.86 = [X] 0.00/0.86 0.00/0.86 [activate(n__terms(X))] = [1] X + [1] 0.00/0.86 >= [1] X + [1] 0.00/0.86 = [terms(activate(X))] 0.00/0.86 0.00/0.86 [activate(n__s(X))] = [1] X + [0] 0.00/0.86 >= [1] X + [0] 0.00/0.86 = [s(activate(X))] 0.00/0.86 0.00/0.86 [activate(n__first(X1, X2))] = [1] X1 + [1] X2 + [0] 0.00/0.86 >= [1] X1 + [1] X2 + [0] 0.00/0.86 = [first(activate(X1), activate(X2))] 0.00/0.86 0.00/0.86 [half(0())] = [14] 0.00/0.86 > [7] 0.00/0.86 = [0()] 0.00/0.86 0.00/0.86 [half(dbl(X))] = [1] X + [14] 0.00/0.86 > [1] X + [0] 0.00/0.86 = [X] 0.00/0.86 0.00/0.86 0.00/0.86 Further, it can be verified that all rules not oriented are covered by the weightgap condition. 0.00/0.86 0.00/0.86 We are left with following problem, upon which TcT provides the 0.00/0.86 certificate YES(O(1),O(n^1)). 0.00/0.86 0.00/0.86 Strict Trs: 0.00/0.86 { terms(X) -> n__terms(X) 0.00/0.86 , s(X) -> n__s(X) 0.00/0.86 , first(X1, X2) -> n__first(X1, X2) 0.00/0.86 , activate(n__s(X)) -> s(activate(X)) } 0.00/0.86 Weak Trs: 0.00/0.86 { terms(N) -> cons(recip(sqr(N)), n__terms(n__s(N))) 0.00/0.86 , sqr(0()) -> 0() 0.00/0.86 , add(0(), X) -> X 0.00/0.86 , dbl(0()) -> 0() 0.00/0.86 , first(0(), X) -> nil() 0.00/0.86 , activate(X) -> X 0.00/0.86 , activate(n__terms(X)) -> terms(activate(X)) 0.00/0.86 , activate(n__first(X1, X2)) -> first(activate(X1), activate(X2)) 0.00/0.86 , half(0()) -> 0() 0.00/0.86 , half(dbl(X)) -> X } 0.00/0.86 Obligation: 0.00/0.86 innermost runtime complexity 0.00/0.86 Answer: 0.00/0.86 YES(O(1),O(n^1)) 0.00/0.86 0.00/0.86 The weightgap principle applies (using the following nonconstant 0.00/0.86 growth matrix-interpretation) 0.00/0.86 0.00/0.86 The following argument positions are usable: 0.00/0.86 Uargs(terms) = {1}, Uargs(cons) = {1}, Uargs(recip) = {1}, 0.00/0.86 Uargs(s) = {1}, Uargs(first) = {1, 2} 0.00/0.86 0.00/0.86 TcT has computed the following matrix interpretation satisfying 0.00/0.86 not(EDA) and not(IDA(1)). 0.00/0.86 0.00/0.86 [terms](x1) = [1] x1 + [0] 0.00/0.86 0.00/0.86 [cons](x1, x2) = [1] x1 + [0] 0.00/0.86 0.00/0.86 [recip](x1) = [1] x1 + [0] 0.00/0.86 0.00/0.86 [sqr](x1) = [1] x1 + [0] 0.00/0.86 0.00/0.86 [n__terms](x1) = [1] x1 + [0] 0.00/0.86 0.00/0.86 [n__s](x1) = [1] x1 + [0] 0.00/0.86 0.00/0.86 [0] = [7] 0.00/0.86 0.00/0.86 [s](x1) = [1] x1 + [4] 0.00/0.86 0.00/0.86 [add](x1, x2) = [1] x1 + [1] x2 + [7] 0.00/0.86 0.00/0.86 [dbl](x1) = [1] x1 + [7] 0.00/0.86 0.00/0.86 [first](x1, x2) = [1] x1 + [1] x2 + [0] 0.00/0.86 0.00/0.86 [nil] = [7] 0.00/0.86 0.00/0.86 [n__first](x1, x2) = [1] x1 + [1] x2 + [4] 0.00/0.86 0.00/0.86 [activate](x1) = [1] x1 + [0] 0.00/0.86 0.00/0.86 [half](x1) = [1] x1 + [7] 0.00/0.86 0.00/0.86 The order satisfies the following ordering constraints: 0.00/0.86 0.00/0.86 [terms(N)] = [1] N + [0] 0.00/0.86 >= [1] N + [0] 0.00/0.86 = [cons(recip(sqr(N)), n__terms(n__s(N)))] 0.00/0.86 0.00/0.86 [terms(X)] = [1] X + [0] 0.00/0.86 >= [1] X + [0] 0.00/0.86 = [n__terms(X)] 0.00/0.86 0.00/0.86 [sqr(0())] = [7] 0.00/0.86 >= [7] 0.00/0.86 = [0()] 0.00/0.86 0.00/0.86 [s(X)] = [1] X + [4] 0.00/0.86 > [1] X + [0] 0.00/0.86 = [n__s(X)] 0.00/0.86 0.00/0.86 [add(0(), X)] = [1] X + [14] 0.00/0.86 > [1] X + [0] 0.00/0.86 = [X] 0.00/0.86 0.00/0.86 [dbl(0())] = [14] 0.00/0.86 > [7] 0.00/0.86 = [0()] 0.00/0.86 0.00/0.86 [first(X1, X2)] = [1] X1 + [1] X2 + [0] 0.00/0.86 ? [1] X1 + [1] X2 + [4] 0.00/0.86 = [n__first(X1, X2)] 0.00/0.86 0.00/0.86 [first(0(), X)] = [1] X + [7] 0.00/0.86 >= [7] 0.00/0.86 = [nil()] 0.00/0.86 0.00/0.86 [activate(X)] = [1] X + [0] 0.00/0.86 >= [1] X + [0] 0.00/0.86 = [X] 0.00/0.86 0.00/0.86 [activate(n__terms(X))] = [1] X + [0] 0.00/0.86 >= [1] X + [0] 0.00/0.86 = [terms(activate(X))] 0.00/0.86 0.00/0.86 [activate(n__s(X))] = [1] X + [0] 0.00/0.86 ? [1] X + [4] 0.00/0.86 = [s(activate(X))] 0.00/0.86 0.00/0.86 [activate(n__first(X1, X2))] = [1] X1 + [1] X2 + [4] 0.00/0.86 > [1] X1 + [1] X2 + [0] 0.00/0.86 = [first(activate(X1), activate(X2))] 0.00/0.86 0.00/0.86 [half(0())] = [14] 0.00/0.86 > [7] 0.00/0.86 = [0()] 0.00/0.86 0.00/0.86 [half(dbl(X))] = [1] X + [14] 0.00/0.86 > [1] X + [0] 0.00/0.86 = [X] 0.00/0.86 0.00/0.86 0.00/0.86 Further, it can be verified that all rules not oriented are covered by the weightgap condition. 0.00/0.86 0.00/0.86 We are left with following problem, upon which TcT provides the 0.00/0.86 certificate YES(O(1),O(n^1)). 0.00/0.86 0.00/0.86 Strict Trs: 0.00/0.86 { terms(X) -> n__terms(X) 0.00/0.86 , first(X1, X2) -> n__first(X1, X2) 0.00/0.86 , activate(n__s(X)) -> s(activate(X)) } 0.00/0.86 Weak Trs: 0.00/0.86 { terms(N) -> cons(recip(sqr(N)), n__terms(n__s(N))) 0.00/0.86 , sqr(0()) -> 0() 0.00/0.86 , s(X) -> n__s(X) 0.00/0.86 , add(0(), X) -> X 0.00/0.86 , dbl(0()) -> 0() 0.00/0.86 , first(0(), X) -> nil() 0.00/0.86 , activate(X) -> X 0.00/0.86 , activate(n__terms(X)) -> terms(activate(X)) 0.00/0.86 , activate(n__first(X1, X2)) -> first(activate(X1), activate(X2)) 0.00/0.86 , half(0()) -> 0() 0.00/0.86 , half(dbl(X)) -> X } 0.00/0.86 Obligation: 0.00/0.86 innermost runtime complexity 0.00/0.86 Answer: 0.00/0.86 YES(O(1),O(n^1)) 0.00/0.86 0.00/0.86 We use the processor 'matrix interpretation of dimension 1' to 0.00/0.86 orient following rules strictly. 0.00/0.86 0.00/0.86 Trs: { terms(X) -> n__terms(X) } 0.00/0.86 0.00/0.86 The induced complexity on above rules (modulo remaining rules) is 0.00/0.86 YES(?,O(n^1)) . These rules are moved into the corresponding weak 0.00/0.86 component(s). 0.00/0.86 0.00/0.86 Sub-proof: 0.00/0.86 ---------- 0.00/0.86 The following argument positions are usable: 0.00/0.86 Uargs(terms) = {1}, Uargs(cons) = {1}, Uargs(recip) = {1}, 0.00/0.86 Uargs(s) = {1}, Uargs(first) = {1, 2} 0.00/0.86 0.00/0.86 TcT has computed the following constructor-based matrix 0.00/0.86 interpretation satisfying not(EDA). 0.00/0.86 0.00/0.86 [terms](x1) = [1] x1 + [3] 0.00/0.86 0.00/0.86 [cons](x1, x2) = [1] x1 + [1] 0.00/0.86 0.00/0.86 [recip](x1) = [1] x1 + [1] 0.00/0.86 0.00/0.86 [sqr](x1) = [0] 0.00/0.86 0.00/0.86 [n__terms](x1) = [1] x1 + [2] 0.00/0.86 0.00/0.86 [n__s](x1) = [1] x1 + [0] 0.00/0.86 0.00/0.86 [0] = [0] 0.00/0.86 0.00/0.86 [s](x1) = [1] x1 + [0] 0.00/0.86 0.00/0.86 [add](x1, x2) = [7] x1 + [7] x2 + [7] 0.00/0.86 0.00/0.86 [dbl](x1) = [2] x1 + [0] 0.00/0.86 0.00/0.86 [first](x1, x2) = [1] x1 + [1] x2 + [3] 0.00/0.86 0.00/0.86 [nil] = [3] 0.00/0.86 0.00/0.86 [n__first](x1, x2) = [1] x1 + [1] x2 + [3] 0.00/0.86 0.00/0.86 [activate](x1) = [3] x1 + [0] 0.00/0.86 0.00/0.86 [half](x1) = [4] x1 + [1] 0.00/0.86 0.00/0.86 The order satisfies the following ordering constraints: 0.00/0.86 0.00/0.86 [terms(N)] = [1] N + [3] 0.00/0.86 > [2] 0.00/0.86 = [cons(recip(sqr(N)), n__terms(n__s(N)))] 0.00/0.86 0.00/0.86 [terms(X)] = [1] X + [3] 0.00/0.86 > [1] X + [2] 0.00/0.86 = [n__terms(X)] 0.00/0.86 0.00/0.86 [sqr(0())] = [0] 0.00/0.86 >= [0] 0.00/0.86 = [0()] 0.00/0.86 0.00/0.86 [s(X)] = [1] X + [0] 0.00/0.86 >= [1] X + [0] 0.00/0.86 = [n__s(X)] 0.00/0.86 0.00/0.86 [add(0(), X)] = [7] X + [7] 0.00/0.86 > [1] X + [0] 0.00/0.86 = [X] 0.00/0.86 0.00/0.86 [dbl(0())] = [0] 0.00/0.86 >= [0] 0.00/0.86 = [0()] 0.00/0.86 0.00/0.86 [first(X1, X2)] = [1] X1 + [1] X2 + [3] 0.00/0.86 >= [1] X1 + [1] X2 + [3] 0.00/0.86 = [n__first(X1, X2)] 0.00/0.86 0.00/0.86 [first(0(), X)] = [1] X + [3] 0.00/0.86 >= [3] 0.00/0.86 = [nil()] 0.00/0.86 0.00/0.86 [activate(X)] = [3] X + [0] 0.00/0.86 >= [1] X + [0] 0.00/0.86 = [X] 0.00/0.86 0.00/0.86 [activate(n__terms(X))] = [3] X + [6] 0.00/0.86 > [3] X + [3] 0.00/0.86 = [terms(activate(X))] 0.00/0.86 0.00/0.86 [activate(n__s(X))] = [3] X + [0] 0.00/0.86 >= [3] X + [0] 0.00/0.86 = [s(activate(X))] 0.00/0.86 0.00/0.86 [activate(n__first(X1, X2))] = [3] X1 + [3] X2 + [9] 0.00/0.86 > [3] X1 + [3] X2 + [3] 0.00/0.86 = [first(activate(X1), activate(X2))] 0.00/0.86 0.00/0.86 [half(0())] = [1] 0.00/0.86 > [0] 0.00/0.86 = [0()] 0.00/0.86 0.00/0.86 [half(dbl(X))] = [8] X + [1] 0.00/0.86 > [1] X + [0] 0.00/0.86 = [X] 0.00/0.86 0.00/0.86 0.00/0.86 We return to the main proof. 0.00/0.86 0.00/0.86 We are left with following problem, upon which TcT provides the 0.00/0.86 certificate YES(O(1),O(n^1)). 0.00/0.86 0.00/0.86 Strict Trs: 0.00/0.86 { first(X1, X2) -> n__first(X1, X2) 0.00/0.86 , activate(n__s(X)) -> s(activate(X)) } 0.00/0.86 Weak Trs: 0.00/0.86 { terms(N) -> cons(recip(sqr(N)), n__terms(n__s(N))) 0.00/0.86 , terms(X) -> n__terms(X) 0.00/0.86 , sqr(0()) -> 0() 0.00/0.86 , s(X) -> n__s(X) 0.00/0.86 , add(0(), X) -> X 0.00/0.86 , dbl(0()) -> 0() 0.00/0.86 , first(0(), X) -> nil() 0.00/0.86 , activate(X) -> X 0.00/0.86 , activate(n__terms(X)) -> terms(activate(X)) 0.00/0.86 , activate(n__first(X1, X2)) -> first(activate(X1), activate(X2)) 0.00/0.86 , half(0()) -> 0() 0.00/0.86 , half(dbl(X)) -> X } 0.00/0.86 Obligation: 0.00/0.86 innermost runtime complexity 0.00/0.86 Answer: 0.00/0.86 YES(O(1),O(n^1)) 0.00/0.86 0.00/0.86 We use the processor 'matrix interpretation of dimension 1' to 0.00/0.86 orient following rules strictly. 0.00/0.86 0.00/0.86 Trs: 0.00/0.86 { first(X1, X2) -> n__first(X1, X2) 0.00/0.86 , activate(n__s(X)) -> s(activate(X)) } 0.00/0.86 0.00/0.86 The induced complexity on above rules (modulo remaining rules) is 0.00/0.86 YES(?,O(n^1)) . These rules are moved into the corresponding weak 0.00/0.86 component(s). 0.00/0.86 0.00/0.86 Sub-proof: 0.00/0.86 ---------- 0.00/0.86 The following argument positions are usable: 0.00/0.86 Uargs(terms) = {1}, Uargs(cons) = {1}, Uargs(recip) = {1}, 0.00/0.86 Uargs(s) = {1}, Uargs(first) = {1, 2} 0.00/0.86 0.00/0.86 TcT has computed the following constructor-based matrix 0.00/0.86 interpretation satisfying not(EDA). 0.00/0.86 0.00/0.86 [terms](x1) = [1] x1 + [4] 0.00/0.86 0.00/0.86 [cons](x1, x2) = [1] x1 + [0] 0.00/0.86 0.00/0.86 [recip](x1) = [1] x1 + [0] 0.00/0.86 0.00/0.86 [sqr](x1) = [4] 0.00/0.86 0.00/0.86 [n__terms](x1) = [1] x1 + [2] 0.00/0.86 0.00/0.86 [n__s](x1) = [1] x1 + [2] 0.00/0.86 0.00/0.86 [0] = [0] 0.00/0.86 0.00/0.86 [s](x1) = [1] x1 + [2] 0.00/0.86 0.00/0.86 [add](x1, x2) = [7] x1 + [7] x2 + [7] 0.00/0.86 0.00/0.86 [dbl](x1) = [1] x1 + [0] 0.00/0.86 0.00/0.86 [first](x1, x2) = [1] x1 + [1] x2 + [4] 0.00/0.86 0.00/0.86 [nil] = [3] 0.00/0.86 0.00/0.86 [n__first](x1, x2) = [1] x1 + [1] x2 + [2] 0.00/0.86 0.00/0.86 [activate](x1) = [4] x1 + [0] 0.00/0.86 0.00/0.86 [half](x1) = [2] x1 + [5] 0.00/0.86 0.00/0.86 The order satisfies the following ordering constraints: 0.00/0.86 0.00/0.86 [terms(N)] = [1] N + [4] 0.00/0.86 >= [4] 0.00/0.86 = [cons(recip(sqr(N)), n__terms(n__s(N)))] 0.00/0.86 0.00/0.86 [terms(X)] = [1] X + [4] 0.00/0.86 > [1] X + [2] 0.00/0.86 = [n__terms(X)] 0.00/0.86 0.00/0.86 [sqr(0())] = [4] 0.00/0.86 > [0] 0.00/0.86 = [0()] 0.00/0.86 0.00/0.86 [s(X)] = [1] X + [2] 0.00/0.86 >= [1] X + [2] 0.00/0.86 = [n__s(X)] 0.00/0.86 0.00/0.86 [add(0(), X)] = [7] X + [7] 0.00/0.86 > [1] X + [0] 0.00/0.86 = [X] 0.00/0.86 0.00/0.86 [dbl(0())] = [0] 0.00/0.86 >= [0] 0.00/0.86 = [0()] 0.00/0.86 0.00/0.86 [first(X1, X2)] = [1] X1 + [1] X2 + [4] 0.00/0.86 > [1] X1 + [1] X2 + [2] 0.00/0.86 = [n__first(X1, X2)] 0.00/0.86 0.00/0.86 [first(0(), X)] = [1] X + [4] 0.00/0.86 > [3] 0.00/0.86 = [nil()] 0.00/0.86 0.00/0.86 [activate(X)] = [4] X + [0] 0.00/0.86 >= [1] X + [0] 0.00/0.86 = [X] 0.00/0.86 0.00/0.86 [activate(n__terms(X))] = [4] X + [8] 0.00/0.86 > [4] X + [4] 0.00/0.86 = [terms(activate(X))] 0.00/0.86 0.00/0.86 [activate(n__s(X))] = [4] X + [8] 0.00/0.86 > [4] X + [2] 0.00/0.86 = [s(activate(X))] 0.00/0.86 0.00/0.86 [activate(n__first(X1, X2))] = [4] X1 + [4] X2 + [8] 0.00/0.86 > [4] X1 + [4] X2 + [4] 0.00/0.86 = [first(activate(X1), activate(X2))] 0.00/0.86 0.00/0.86 [half(0())] = [5] 0.00/0.86 > [0] 0.00/0.86 = [0()] 0.00/0.86 0.00/0.86 [half(dbl(X))] = [2] X + [5] 0.00/0.86 > [1] X + [0] 0.00/0.86 = [X] 0.00/0.86 0.00/0.86 0.00/0.86 We return to the main proof. 0.00/0.86 0.00/0.86 We are left with following problem, upon which TcT provides the 0.00/0.86 certificate YES(O(1),O(1)). 0.00/0.86 0.00/0.86 Weak Trs: 0.00/0.86 { terms(N) -> cons(recip(sqr(N)), n__terms(n__s(N))) 0.00/0.86 , terms(X) -> n__terms(X) 0.00/0.86 , sqr(0()) -> 0() 0.00/0.86 , s(X) -> n__s(X) 0.00/0.86 , add(0(), X) -> X 0.00/0.86 , dbl(0()) -> 0() 0.00/0.86 , first(X1, X2) -> n__first(X1, X2) 0.00/0.86 , first(0(), X) -> nil() 0.00/0.86 , activate(X) -> X 0.00/0.86 , activate(n__terms(X)) -> terms(activate(X)) 0.00/0.86 , activate(n__s(X)) -> s(activate(X)) 0.00/0.86 , activate(n__first(X1, X2)) -> first(activate(X1), activate(X2)) 0.00/0.86 , half(0()) -> 0() 0.00/0.86 , half(dbl(X)) -> X } 0.00/0.86 Obligation: 0.00/0.86 innermost runtime complexity 0.00/0.86 Answer: 0.00/0.86 YES(O(1),O(1)) 0.00/0.86 0.00/0.86 Empty rules are trivially bounded 0.00/0.86 0.00/0.86 Hurray, we answered YES(O(1),O(n^1)) 0.00/0.87 EOF