YES(O(1),O(n^1)) 0.00/0.95 YES(O(1),O(n^1)) 0.00/0.95 0.00/0.95 We are left with following problem, upon which TcT provides the 0.00/0.95 certificate YES(O(1),O(n^1)). 0.00/0.95 0.00/0.95 Strict Trs: 0.00/0.95 { fst(X1, X2) -> n__fst(X1, X2) 0.00/0.95 , fst(0(), Z) -> nil() 0.00/0.95 , fst(s(X), cons(Y, Z)) -> 0.00/0.95 cons(Y, n__fst(activate(X), activate(Z))) 0.00/0.95 , activate(X) -> X 0.00/0.95 , activate(n__fst(X1, X2)) -> fst(X1, X2) 0.00/0.95 , activate(n__from(X)) -> from(X) 0.00/0.95 , activate(n__add(X1, X2)) -> add(X1, X2) 0.00/0.95 , activate(n__len(X)) -> len(X) 0.00/0.95 , from(X) -> cons(X, n__from(s(X))) 0.00/0.95 , from(X) -> n__from(X) 0.00/0.95 , add(X1, X2) -> n__add(X1, X2) 0.00/0.95 , add(0(), X) -> X 0.00/0.95 , add(s(X), Y) -> s(n__add(activate(X), Y)) 0.00/0.95 , len(X) -> n__len(X) 0.00/0.95 , len(nil()) -> 0() 0.00/0.95 , len(cons(X, Z)) -> s(n__len(activate(Z))) } 0.00/0.95 Obligation: 0.00/0.95 innermost runtime complexity 0.00/0.95 Answer: 0.00/0.95 YES(O(1),O(n^1)) 0.00/0.95 0.00/0.95 The weightgap principle applies (using the following nonconstant 0.00/0.95 growth matrix-interpretation) 0.00/0.95 0.00/0.95 The following argument positions are usable: 0.00/0.95 Uargs(s) = {1}, Uargs(cons) = {2}, Uargs(n__fst) = {1, 2}, 0.00/0.95 Uargs(n__add) = {1}, Uargs(n__len) = {1} 0.00/0.95 0.00/0.95 TcT has computed the following matrix interpretation satisfying 0.00/0.95 not(EDA) and not(IDA(1)). 0.00/0.95 0.00/0.95 [fst](x1, x2) = [1] x1 + [1] x2 + [0] 0.00/0.95 0.00/0.95 [0] = [0] 0.00/0.95 0.00/0.95 [nil] = [1] 0.00/0.95 0.00/0.95 [s](x1) = [1] x1 + [0] 0.00/0.95 0.00/0.95 [cons](x1, x2) = [1] x2 + [0] 0.00/0.95 0.00/0.95 [n__fst](x1, x2) = [1] x1 + [1] x2 + [0] 0.00/0.95 0.00/0.95 [activate](x1) = [1] x1 + [0] 0.00/0.95 0.00/0.95 [from](x1) = [1] x1 + [5] 0.00/0.95 0.00/0.95 [n__from](x1) = [1] x1 + [0] 0.00/0.95 0.00/0.95 [add](x1, x2) = [1] x1 + [1] x2 + [0] 0.00/0.95 0.00/0.95 [n__add](x1, x2) = [1] x1 + [1] x2 + [0] 0.00/0.95 0.00/0.95 [len](x1) = [1] x1 + [0] 0.00/0.95 0.00/0.95 [n__len](x1) = [1] x1 + [0] 0.00/0.95 0.00/0.95 The order satisfies the following ordering constraints: 0.00/0.95 0.00/0.95 [fst(X1, X2)] = [1] X1 + [1] X2 + [0] 0.00/0.95 >= [1] X1 + [1] X2 + [0] 0.00/0.95 = [n__fst(X1, X2)] 0.00/0.95 0.00/0.95 [fst(0(), Z)] = [1] Z + [0] 0.00/0.95 ? [1] 0.00/0.95 = [nil()] 0.00/0.95 0.00/0.95 [fst(s(X), cons(Y, Z))] = [1] Z + [1] X + [0] 0.00/0.95 >= [1] Z + [1] X + [0] 0.00/0.95 = [cons(Y, n__fst(activate(X), activate(Z)))] 0.00/0.95 0.00/0.95 [activate(X)] = [1] X + [0] 0.00/0.95 >= [1] X + [0] 0.00/0.95 = [X] 0.00/0.95 0.00/0.95 [activate(n__fst(X1, X2))] = [1] X1 + [1] X2 + [0] 0.00/0.95 >= [1] X1 + [1] X2 + [0] 0.00/0.95 = [fst(X1, X2)] 0.00/0.95 0.00/0.95 [activate(n__from(X))] = [1] X + [0] 0.00/0.95 ? [1] X + [5] 0.00/0.95 = [from(X)] 0.00/0.95 0.00/0.95 [activate(n__add(X1, X2))] = [1] X1 + [1] X2 + [0] 0.00/0.95 >= [1] X1 + [1] X2 + [0] 0.00/0.95 = [add(X1, X2)] 0.00/0.95 0.00/0.95 [activate(n__len(X))] = [1] X + [0] 0.00/0.95 >= [1] X + [0] 0.00/0.95 = [len(X)] 0.00/0.95 0.00/0.95 [from(X)] = [1] X + [5] 0.00/0.95 > [1] X + [0] 0.00/0.95 = [cons(X, n__from(s(X)))] 0.00/0.95 0.00/0.95 [from(X)] = [1] X + [5] 0.00/0.95 > [1] X + [0] 0.00/0.95 = [n__from(X)] 0.00/0.95 0.00/0.95 [add(X1, X2)] = [1] X1 + [1] X2 + [0] 0.00/0.95 >= [1] X1 + [1] X2 + [0] 0.00/0.95 = [n__add(X1, X2)] 0.00/0.95 0.00/0.95 [add(0(), X)] = [1] X + [0] 0.00/0.95 >= [1] X + [0] 0.00/0.95 = [X] 0.00/0.95 0.00/0.95 [add(s(X), Y)] = [1] X + [1] Y + [0] 0.00/0.95 >= [1] X + [1] Y + [0] 0.00/0.95 = [s(n__add(activate(X), Y))] 0.00/0.95 0.00/0.95 [len(X)] = [1] X + [0] 0.00/0.95 >= [1] X + [0] 0.00/0.95 = [n__len(X)] 0.00/0.95 0.00/0.95 [len(nil())] = [1] 0.00/0.95 > [0] 0.00/0.95 = [0()] 0.00/0.95 0.00/0.95 [len(cons(X, Z))] = [1] Z + [0] 0.00/0.95 >= [1] Z + [0] 0.00/0.95 = [s(n__len(activate(Z)))] 0.00/0.95 0.00/0.95 0.00/0.95 Further, it can be verified that all rules not oriented are covered by the weightgap condition. 0.00/0.95 0.00/0.95 We are left with following problem, upon which TcT provides the 0.00/0.95 certificate YES(O(1),O(n^1)). 0.00/0.95 0.00/0.95 Strict Trs: 0.00/0.95 { fst(X1, X2) -> n__fst(X1, X2) 0.00/0.95 , fst(0(), Z) -> nil() 0.00/0.95 , fst(s(X), cons(Y, Z)) -> 0.00/0.95 cons(Y, n__fst(activate(X), activate(Z))) 0.00/0.95 , activate(X) -> X 0.00/0.95 , activate(n__fst(X1, X2)) -> fst(X1, X2) 0.00/0.95 , activate(n__from(X)) -> from(X) 0.00/0.95 , activate(n__add(X1, X2)) -> add(X1, X2) 0.00/0.95 , activate(n__len(X)) -> len(X) 0.00/0.95 , add(X1, X2) -> n__add(X1, X2) 0.00/0.95 , add(0(), X) -> X 0.00/0.95 , add(s(X), Y) -> s(n__add(activate(X), Y)) 0.00/0.95 , len(X) -> n__len(X) 0.00/0.95 , len(cons(X, Z)) -> s(n__len(activate(Z))) } 0.00/0.95 Weak Trs: 0.00/0.95 { from(X) -> cons(X, n__from(s(X))) 0.00/0.95 , from(X) -> n__from(X) 0.00/0.95 , len(nil()) -> 0() } 0.00/0.95 Obligation: 0.00/0.95 innermost runtime complexity 0.00/0.95 Answer: 0.00/0.95 YES(O(1),O(n^1)) 0.00/0.95 0.00/0.95 The weightgap principle applies (using the following nonconstant 0.00/0.95 growth matrix-interpretation) 0.00/0.95 0.00/0.95 The following argument positions are usable: 0.00/0.96 Uargs(s) = {1}, Uargs(cons) = {2}, Uargs(n__fst) = {1, 2}, 0.00/0.96 Uargs(n__add) = {1}, Uargs(n__len) = {1} 0.00/0.96 0.00/0.96 TcT has computed the following matrix interpretation satisfying 0.00/0.96 not(EDA) and not(IDA(1)). 0.00/0.96 0.00/0.96 [fst](x1, x2) = [1] x1 + [1] x2 + [0] 0.00/0.96 0.00/0.96 [0] = [0] 0.00/0.96 0.00/0.96 [nil] = [0] 0.00/0.96 0.00/0.96 [s](x1) = [1] x1 + [0] 0.00/0.96 0.00/0.96 [cons](x1, x2) = [1] x2 + [0] 0.00/0.96 0.00/0.96 [n__fst](x1, x2) = [1] x1 + [1] x2 + [4] 0.00/0.96 0.00/0.96 [activate](x1) = [1] x1 + [0] 0.00/0.96 0.00/0.96 [from](x1) = [1] x1 + [4] 0.00/0.96 0.00/0.96 [n__from](x1) = [1] x1 + [0] 0.00/0.96 0.00/0.96 [add](x1, x2) = [1] x1 + [1] x2 + [0] 0.00/0.96 0.00/0.96 [n__add](x1, x2) = [1] x1 + [1] x2 + [0] 0.00/0.96 0.00/0.96 [len](x1) = [1] x1 + [0] 0.00/0.96 0.00/0.96 [n__len](x1) = [1] x1 + [0] 0.00/0.96 0.00/0.96 The order satisfies the following ordering constraints: 0.00/0.96 0.00/0.96 [fst(X1, X2)] = [1] X1 + [1] X2 + [0] 0.00/0.96 ? [1] X1 + [1] X2 + [4] 0.00/0.96 = [n__fst(X1, X2)] 0.00/0.96 0.00/0.96 [fst(0(), Z)] = [1] Z + [0] 0.00/0.96 >= [0] 0.00/0.96 = [nil()] 0.00/0.96 0.00/0.96 [fst(s(X), cons(Y, Z))] = [1] Z + [1] X + [0] 0.00/0.96 ? [1] Z + [1] X + [4] 0.00/0.96 = [cons(Y, n__fst(activate(X), activate(Z)))] 0.00/0.96 0.00/0.96 [activate(X)] = [1] X + [0] 0.00/0.96 >= [1] X + [0] 0.00/0.96 = [X] 0.00/0.96 0.00/0.96 [activate(n__fst(X1, X2))] = [1] X1 + [1] X2 + [4] 0.00/0.96 > [1] X1 + [1] X2 + [0] 0.00/0.96 = [fst(X1, X2)] 0.00/0.96 0.00/0.96 [activate(n__from(X))] = [1] X + [0] 0.00/0.96 ? [1] X + [4] 0.00/0.96 = [from(X)] 0.00/0.96 0.00/0.96 [activate(n__add(X1, X2))] = [1] X1 + [1] X2 + [0] 0.00/0.96 >= [1] X1 + [1] X2 + [0] 0.00/0.96 = [add(X1, X2)] 0.00/0.96 0.00/0.96 [activate(n__len(X))] = [1] X + [0] 0.00/0.96 >= [1] X + [0] 0.00/0.96 = [len(X)] 0.00/0.96 0.00/0.96 [from(X)] = [1] X + [4] 0.00/0.96 > [1] X + [0] 0.00/0.96 = [cons(X, n__from(s(X)))] 0.00/0.96 0.00/0.96 [from(X)] = [1] X + [4] 0.00/0.96 > [1] X + [0] 0.00/0.96 = [n__from(X)] 0.00/0.96 0.00/0.96 [add(X1, X2)] = [1] X1 + [1] X2 + [0] 0.00/0.96 >= [1] X1 + [1] X2 + [0] 0.00/0.96 = [n__add(X1, X2)] 0.00/0.96 0.00/0.96 [add(0(), X)] = [1] X + [0] 0.00/0.96 >= [1] X + [0] 0.00/0.96 = [X] 0.00/0.96 0.00/0.96 [add(s(X), Y)] = [1] X + [1] Y + [0] 0.00/0.96 >= [1] X + [1] Y + [0] 0.00/0.96 = [s(n__add(activate(X), Y))] 0.00/0.96 0.00/0.96 [len(X)] = [1] X + [0] 0.00/0.96 >= [1] X + [0] 0.00/0.96 = [n__len(X)] 0.00/0.96 0.00/0.96 [len(nil())] = [0] 0.00/0.96 >= [0] 0.00/0.96 = [0()] 0.00/0.96 0.00/0.96 [len(cons(X, Z))] = [1] Z + [0] 0.00/0.96 >= [1] Z + [0] 0.00/0.96 = [s(n__len(activate(Z)))] 0.00/0.96 0.00/0.96 0.00/0.96 Further, it can be verified that all rules not oriented are covered by the weightgap condition. 0.00/0.96 0.00/0.96 We are left with following problem, upon which TcT provides the 0.00/0.96 certificate YES(O(1),O(n^1)). 0.00/0.96 0.00/0.96 Strict Trs: 0.00/0.96 { fst(X1, X2) -> n__fst(X1, X2) 0.00/0.96 , fst(0(), Z) -> nil() 0.00/0.96 , fst(s(X), cons(Y, Z)) -> 0.00/0.96 cons(Y, n__fst(activate(X), activate(Z))) 0.00/0.96 , activate(X) -> X 0.00/0.96 , activate(n__from(X)) -> from(X) 0.00/0.96 , activate(n__add(X1, X2)) -> add(X1, X2) 0.00/0.96 , activate(n__len(X)) -> len(X) 0.00/0.96 , add(X1, X2) -> n__add(X1, X2) 0.00/0.96 , add(0(), X) -> X 0.00/0.96 , add(s(X), Y) -> s(n__add(activate(X), Y)) 0.00/0.96 , len(X) -> n__len(X) 0.00/0.96 , len(cons(X, Z)) -> s(n__len(activate(Z))) } 0.00/0.96 Weak Trs: 0.00/0.96 { activate(n__fst(X1, X2)) -> fst(X1, X2) 0.00/0.96 , from(X) -> cons(X, n__from(s(X))) 0.00/0.96 , from(X) -> n__from(X) 0.00/0.96 , len(nil()) -> 0() } 0.00/0.96 Obligation: 0.00/0.96 innermost runtime complexity 0.00/0.96 Answer: 0.00/0.96 YES(O(1),O(n^1)) 0.00/0.96 0.00/0.96 The weightgap principle applies (using the following nonconstant 0.00/0.96 growth matrix-interpretation) 0.00/0.96 0.00/0.96 The following argument positions are usable: 0.00/0.96 Uargs(s) = {1}, Uargs(cons) = {2}, Uargs(n__fst) = {1, 2}, 0.00/0.96 Uargs(n__add) = {1}, Uargs(n__len) = {1} 0.00/0.96 0.00/0.96 TcT has computed the following matrix interpretation satisfying 0.00/0.96 not(EDA) and not(IDA(1)). 0.00/0.96 0.00/0.96 [fst](x1, x2) = [1] x1 + [1] x2 + [0] 0.00/0.96 0.00/0.96 [0] = [0] 0.00/0.96 0.00/0.96 [nil] = [0] 0.00/0.96 0.00/0.96 [s](x1) = [1] x1 + [0] 0.00/0.96 0.00/0.96 [cons](x1, x2) = [1] x2 + [0] 0.00/0.96 0.00/0.96 [n__fst](x1, x2) = [1] x1 + [1] x2 + [4] 0.00/0.96 0.00/0.96 [activate](x1) = [1] x1 + [0] 0.00/0.96 0.00/0.96 [from](x1) = [1] x1 + [7] 0.00/0.96 0.00/0.96 [n__from](x1) = [1] x1 + [0] 0.00/0.96 0.00/0.96 [add](x1, x2) = [1] x1 + [1] x2 + [1] 0.00/0.96 0.00/0.96 [n__add](x1, x2) = [1] x1 + [1] x2 + [0] 0.00/0.96 0.00/0.96 [len](x1) = [1] x1 + [0] 0.00/0.96 0.00/0.96 [n__len](x1) = [1] x1 + [0] 0.00/0.96 0.00/0.96 The order satisfies the following ordering constraints: 0.00/0.96 0.00/0.96 [fst(X1, X2)] = [1] X1 + [1] X2 + [0] 0.00/0.96 ? [1] X1 + [1] X2 + [4] 0.00/0.96 = [n__fst(X1, X2)] 0.00/0.96 0.00/0.96 [fst(0(), Z)] = [1] Z + [0] 0.00/0.96 >= [0] 0.00/0.96 = [nil()] 0.00/0.96 0.00/0.96 [fst(s(X), cons(Y, Z))] = [1] Z + [1] X + [0] 0.00/0.96 ? [1] Z + [1] X + [4] 0.00/0.96 = [cons(Y, n__fst(activate(X), activate(Z)))] 0.00/0.96 0.00/0.96 [activate(X)] = [1] X + [0] 0.00/0.96 >= [1] X + [0] 0.00/0.96 = [X] 0.00/0.96 0.00/0.96 [activate(n__fst(X1, X2))] = [1] X1 + [1] X2 + [4] 0.00/0.96 > [1] X1 + [1] X2 + [0] 0.00/0.96 = [fst(X1, X2)] 0.00/0.96 0.00/0.96 [activate(n__from(X))] = [1] X + [0] 0.00/0.96 ? [1] X + [7] 0.00/0.96 = [from(X)] 0.00/0.96 0.00/0.96 [activate(n__add(X1, X2))] = [1] X1 + [1] X2 + [0] 0.00/0.96 ? [1] X1 + [1] X2 + [1] 0.00/0.96 = [add(X1, X2)] 0.00/0.96 0.00/0.96 [activate(n__len(X))] = [1] X + [0] 0.00/0.96 >= [1] X + [0] 0.00/0.96 = [len(X)] 0.00/0.96 0.00/0.96 [from(X)] = [1] X + [7] 0.00/0.96 > [1] X + [0] 0.00/0.96 = [cons(X, n__from(s(X)))] 0.00/0.96 0.00/0.96 [from(X)] = [1] X + [7] 0.00/0.96 > [1] X + [0] 0.00/0.96 = [n__from(X)] 0.00/0.96 0.00/0.96 [add(X1, X2)] = [1] X1 + [1] X2 + [1] 0.00/0.96 > [1] X1 + [1] X2 + [0] 0.00/0.96 = [n__add(X1, X2)] 0.00/0.96 0.00/0.96 [add(0(), X)] = [1] X + [1] 0.00/0.96 > [1] X + [0] 0.00/0.96 = [X] 0.00/0.96 0.00/0.96 [add(s(X), Y)] = [1] X + [1] Y + [1] 0.00/0.96 > [1] X + [1] Y + [0] 0.00/0.96 = [s(n__add(activate(X), Y))] 0.00/0.96 0.00/0.96 [len(X)] = [1] X + [0] 0.00/0.96 >= [1] X + [0] 0.00/0.96 = [n__len(X)] 0.00/0.96 0.00/0.96 [len(nil())] = [0] 0.00/0.96 >= [0] 0.00/0.96 = [0()] 0.00/0.96 0.00/0.96 [len(cons(X, Z))] = [1] Z + [0] 0.00/0.96 >= [1] Z + [0] 0.00/0.96 = [s(n__len(activate(Z)))] 0.00/0.96 0.00/0.96 0.00/0.96 Further, it can be verified that all rules not oriented are covered by the weightgap condition. 0.00/0.96 0.00/0.96 We are left with following problem, upon which TcT provides the 0.00/0.96 certificate YES(O(1),O(n^1)). 0.00/0.96 0.00/0.96 Strict Trs: 0.00/0.96 { fst(X1, X2) -> n__fst(X1, X2) 0.00/0.96 , fst(0(), Z) -> nil() 0.00/0.96 , fst(s(X), cons(Y, Z)) -> 0.00/0.96 cons(Y, n__fst(activate(X), activate(Z))) 0.00/0.96 , activate(X) -> X 0.00/0.96 , activate(n__from(X)) -> from(X) 0.00/0.96 , activate(n__add(X1, X2)) -> add(X1, X2) 0.00/0.96 , activate(n__len(X)) -> len(X) 0.00/0.96 , len(X) -> n__len(X) 0.00/0.96 , len(cons(X, Z)) -> s(n__len(activate(Z))) } 0.00/0.96 Weak Trs: 0.00/0.96 { activate(n__fst(X1, X2)) -> fst(X1, X2) 0.00/0.96 , from(X) -> cons(X, n__from(s(X))) 0.00/0.96 , from(X) -> n__from(X) 0.00/0.96 , add(X1, X2) -> n__add(X1, X2) 0.00/0.96 , add(0(), X) -> X 0.00/0.96 , add(s(X), Y) -> s(n__add(activate(X), Y)) 0.00/0.96 , len(nil()) -> 0() } 0.00/0.96 Obligation: 0.00/0.96 innermost runtime complexity 0.00/0.96 Answer: 0.00/0.96 YES(O(1),O(n^1)) 0.00/0.96 0.00/0.96 The weightgap principle applies (using the following nonconstant 0.00/0.96 growth matrix-interpretation) 0.00/0.96 0.00/0.96 The following argument positions are usable: 0.00/0.96 Uargs(s) = {1}, Uargs(cons) = {2}, Uargs(n__fst) = {1, 2}, 0.00/0.96 Uargs(n__add) = {1}, Uargs(n__len) = {1} 0.00/0.96 0.00/0.96 TcT has computed the following matrix interpretation satisfying 0.00/0.96 not(EDA) and not(IDA(1)). 0.00/0.96 0.00/0.96 [fst](x1, x2) = [1] x1 + [1] x2 + [0] 0.00/0.96 0.00/0.96 [0] = [0] 0.00/0.96 0.00/0.96 [nil] = [5] 0.00/0.96 0.00/0.96 [s](x1) = [1] x1 + [0] 0.00/0.96 0.00/0.96 [cons](x1, x2) = [1] x2 + [0] 0.00/0.96 0.00/0.96 [n__fst](x1, x2) = [1] x1 + [1] x2 + [7] 0.00/0.96 0.00/0.96 [activate](x1) = [1] x1 + [1] 0.00/0.96 0.00/0.96 [from](x1) = [1] x1 + [4] 0.00/0.96 0.00/0.96 [n__from](x1) = [1] x1 + [0] 0.00/0.96 0.00/0.96 [add](x1, x2) = [1] x1 + [1] x2 + [4] 0.00/0.96 0.00/0.96 [n__add](x1, x2) = [1] x1 + [1] x2 + [0] 0.00/0.96 0.00/0.96 [len](x1) = [1] x1 + [0] 0.00/0.96 0.00/0.96 [n__len](x1) = [1] x1 + [0] 0.00/0.96 0.00/0.96 The order satisfies the following ordering constraints: 0.00/0.96 0.00/0.96 [fst(X1, X2)] = [1] X1 + [1] X2 + [0] 0.00/0.96 ? [1] X1 + [1] X2 + [7] 0.00/0.96 = [n__fst(X1, X2)] 0.00/0.96 0.00/0.96 [fst(0(), Z)] = [1] Z + [0] 0.00/0.96 ? [5] 0.00/0.96 = [nil()] 0.00/0.96 0.00/0.96 [fst(s(X), cons(Y, Z))] = [1] Z + [1] X + [0] 0.00/0.96 ? [1] Z + [1] X + [9] 0.00/0.96 = [cons(Y, n__fst(activate(X), activate(Z)))] 0.00/0.96 0.00/0.96 [activate(X)] = [1] X + [1] 0.00/0.96 > [1] X + [0] 0.00/0.96 = [X] 0.00/0.96 0.00/0.96 [activate(n__fst(X1, X2))] = [1] X1 + [1] X2 + [8] 0.00/0.96 > [1] X1 + [1] X2 + [0] 0.00/0.96 = [fst(X1, X2)] 0.00/0.96 0.00/0.96 [activate(n__from(X))] = [1] X + [1] 0.00/0.96 ? [1] X + [4] 0.00/0.96 = [from(X)] 0.00/0.96 0.00/0.96 [activate(n__add(X1, X2))] = [1] X1 + [1] X2 + [1] 0.00/0.96 ? [1] X1 + [1] X2 + [4] 0.00/0.96 = [add(X1, X2)] 0.00/0.96 0.00/0.96 [activate(n__len(X))] = [1] X + [1] 0.00/0.96 > [1] X + [0] 0.00/0.96 = [len(X)] 0.00/0.96 0.00/0.96 [from(X)] = [1] X + [4] 0.00/0.96 > [1] X + [0] 0.00/0.96 = [cons(X, n__from(s(X)))] 0.00/0.96 0.00/0.96 [from(X)] = [1] X + [4] 0.00/0.96 > [1] X + [0] 0.00/0.96 = [n__from(X)] 0.00/0.96 0.00/0.96 [add(X1, X2)] = [1] X1 + [1] X2 + [4] 0.00/0.96 > [1] X1 + [1] X2 + [0] 0.00/0.96 = [n__add(X1, X2)] 0.00/0.96 0.00/0.96 [add(0(), X)] = [1] X + [4] 0.00/0.96 > [1] X + [0] 0.00/0.96 = [X] 0.00/0.96 0.00/0.96 [add(s(X), Y)] = [1] X + [1] Y + [4] 0.00/0.96 > [1] X + [1] Y + [1] 0.00/0.96 = [s(n__add(activate(X), Y))] 0.00/0.96 0.00/0.96 [len(X)] = [1] X + [0] 0.00/0.96 >= [1] X + [0] 0.00/0.96 = [n__len(X)] 0.00/0.96 0.00/0.96 [len(nil())] = [5] 0.00/0.96 > [0] 0.00/0.96 = [0()] 0.00/0.96 0.00/0.96 [len(cons(X, Z))] = [1] Z + [0] 0.00/0.96 ? [1] Z + [1] 0.00/0.96 = [s(n__len(activate(Z)))] 0.00/0.96 0.00/0.96 0.00/0.96 Further, it can be verified that all rules not oriented are covered by the weightgap condition. 0.00/0.96 0.00/0.96 We are left with following problem, upon which TcT provides the 0.00/0.96 certificate YES(O(1),O(n^1)). 0.00/0.96 0.00/0.96 Strict Trs: 0.00/0.96 { fst(X1, X2) -> n__fst(X1, X2) 0.00/0.96 , fst(0(), Z) -> nil() 0.00/0.96 , fst(s(X), cons(Y, Z)) -> 0.00/0.96 cons(Y, n__fst(activate(X), activate(Z))) 0.00/0.96 , activate(n__from(X)) -> from(X) 0.00/0.96 , activate(n__add(X1, X2)) -> add(X1, X2) 0.00/0.96 , len(X) -> n__len(X) 0.00/0.96 , len(cons(X, Z)) -> s(n__len(activate(Z))) } 0.00/0.96 Weak Trs: 0.00/0.96 { activate(X) -> X 0.00/0.96 , activate(n__fst(X1, X2)) -> fst(X1, X2) 0.00/0.96 , activate(n__len(X)) -> len(X) 0.00/0.96 , from(X) -> cons(X, n__from(s(X))) 0.00/0.96 , from(X) -> n__from(X) 0.00/0.96 , add(X1, X2) -> n__add(X1, X2) 0.00/0.96 , add(0(), X) -> X 0.00/0.96 , add(s(X), Y) -> s(n__add(activate(X), Y)) 0.00/0.96 , len(nil()) -> 0() } 0.00/0.96 Obligation: 0.00/0.96 innermost runtime complexity 0.00/0.96 Answer: 0.00/0.96 YES(O(1),O(n^1)) 0.00/0.96 0.00/0.96 The weightgap principle applies (using the following nonconstant 0.00/0.96 growth matrix-interpretation) 0.00/0.96 0.00/0.96 The following argument positions are usable: 0.00/0.96 Uargs(s) = {1}, Uargs(cons) = {2}, Uargs(n__fst) = {1, 2}, 0.00/0.96 Uargs(n__add) = {1}, Uargs(n__len) = {1} 0.00/0.96 0.00/0.96 TcT has computed the following matrix interpretation satisfying 0.00/0.96 not(EDA) and not(IDA(1)). 0.00/0.96 0.00/0.96 [fst](x1, x2) = [1] x1 + [1] x2 + [1] 0.00/0.96 0.00/0.96 [0] = [0] 0.00/0.96 0.00/0.96 [nil] = [0] 0.00/0.96 0.00/0.96 [s](x1) = [1] x1 + [0] 0.00/0.96 0.00/0.96 [cons](x1, x2) = [1] x2 + [0] 0.00/0.96 0.00/0.96 [n__fst](x1, x2) = [1] x1 + [1] x2 + [4] 0.00/0.96 0.00/0.96 [activate](x1) = [1] x1 + [0] 0.00/0.96 0.00/0.96 [from](x1) = [1] x1 + [4] 0.00/0.96 0.00/0.96 [n__from](x1) = [1] x1 + [0] 0.00/0.96 0.00/0.96 [add](x1, x2) = [1] x1 + [1] x2 + [4] 0.00/0.96 0.00/0.96 [n__add](x1, x2) = [1] x1 + [1] x2 + [0] 0.00/0.96 0.00/0.96 [len](x1) = [1] x1 + [0] 0.00/0.96 0.00/0.96 [n__len](x1) = [1] x1 + [0] 0.00/0.96 0.00/0.96 The order satisfies the following ordering constraints: 0.00/0.96 0.00/0.96 [fst(X1, X2)] = [1] X1 + [1] X2 + [1] 0.00/0.96 ? [1] X1 + [1] X2 + [4] 0.00/0.96 = [n__fst(X1, X2)] 0.00/0.96 0.00/0.96 [fst(0(), Z)] = [1] Z + [1] 0.00/0.96 > [0] 0.00/0.96 = [nil()] 0.00/0.96 0.00/0.96 [fst(s(X), cons(Y, Z))] = [1] Z + [1] X + [1] 0.00/0.96 ? [1] Z + [1] X + [4] 0.00/0.96 = [cons(Y, n__fst(activate(X), activate(Z)))] 0.00/0.96 0.00/0.96 [activate(X)] = [1] X + [0] 0.00/0.96 >= [1] X + [0] 0.00/0.96 = [X] 0.00/0.96 0.00/0.96 [activate(n__fst(X1, X2))] = [1] X1 + [1] X2 + [4] 0.00/0.96 > [1] X1 + [1] X2 + [1] 0.00/0.96 = [fst(X1, X2)] 0.00/0.96 0.00/0.96 [activate(n__from(X))] = [1] X + [0] 0.00/0.96 ? [1] X + [4] 0.00/0.96 = [from(X)] 0.00/0.96 0.00/0.96 [activate(n__add(X1, X2))] = [1] X1 + [1] X2 + [0] 0.00/0.96 ? [1] X1 + [1] X2 + [4] 0.00/0.96 = [add(X1, X2)] 0.00/0.96 0.00/0.96 [activate(n__len(X))] = [1] X + [0] 0.00/0.96 >= [1] X + [0] 0.00/0.96 = [len(X)] 0.00/0.96 0.00/0.96 [from(X)] = [1] X + [4] 0.00/0.96 > [1] X + [0] 0.00/0.96 = [cons(X, n__from(s(X)))] 0.00/0.96 0.00/0.96 [from(X)] = [1] X + [4] 0.00/0.96 > [1] X + [0] 0.00/0.96 = [n__from(X)] 0.00/0.96 0.00/0.96 [add(X1, X2)] = [1] X1 + [1] X2 + [4] 0.00/0.96 > [1] X1 + [1] X2 + [0] 0.00/0.96 = [n__add(X1, X2)] 0.00/0.96 0.00/0.96 [add(0(), X)] = [1] X + [4] 0.00/0.96 > [1] X + [0] 0.00/0.96 = [X] 0.00/0.96 0.00/0.96 [add(s(X), Y)] = [1] X + [1] Y + [4] 0.00/0.96 > [1] X + [1] Y + [0] 0.00/0.96 = [s(n__add(activate(X), Y))] 0.00/0.96 0.00/0.96 [len(X)] = [1] X + [0] 0.00/0.96 >= [1] X + [0] 0.00/0.96 = [n__len(X)] 0.00/0.96 0.00/0.96 [len(nil())] = [0] 0.00/0.96 >= [0] 0.00/0.96 = [0()] 0.00/0.96 0.00/0.96 [len(cons(X, Z))] = [1] Z + [0] 0.00/0.96 >= [1] Z + [0] 0.00/0.96 = [s(n__len(activate(Z)))] 0.00/0.96 0.00/0.96 0.00/0.96 Further, it can be verified that all rules not oriented are covered by the weightgap condition. 0.00/0.96 0.00/0.96 We are left with following problem, upon which TcT provides the 0.00/0.96 certificate YES(O(1),O(n^1)). 0.00/0.96 0.00/0.96 Strict Trs: 0.00/0.96 { fst(X1, X2) -> n__fst(X1, X2) 0.00/0.96 , fst(s(X), cons(Y, Z)) -> 0.00/0.96 cons(Y, n__fst(activate(X), activate(Z))) 0.00/0.96 , activate(n__from(X)) -> from(X) 0.00/0.96 , activate(n__add(X1, X2)) -> add(X1, X2) 0.00/0.96 , len(X) -> n__len(X) 0.00/0.96 , len(cons(X, Z)) -> s(n__len(activate(Z))) } 0.00/0.96 Weak Trs: 0.00/0.96 { fst(0(), Z) -> nil() 0.00/0.96 , activate(X) -> X 0.00/0.96 , activate(n__fst(X1, X2)) -> fst(X1, X2) 0.00/0.96 , activate(n__len(X)) -> len(X) 0.00/0.96 , from(X) -> cons(X, n__from(s(X))) 0.00/0.96 , from(X) -> n__from(X) 0.00/0.96 , add(X1, X2) -> n__add(X1, X2) 0.00/0.96 , add(0(), X) -> X 0.00/0.96 , add(s(X), Y) -> s(n__add(activate(X), Y)) 0.00/0.96 , len(nil()) -> 0() } 0.00/0.96 Obligation: 0.00/0.96 innermost runtime complexity 0.00/0.96 Answer: 0.00/0.96 YES(O(1),O(n^1)) 0.00/0.96 0.00/0.96 The weightgap principle applies (using the following nonconstant 0.00/0.96 growth matrix-interpretation) 0.00/0.96 0.00/0.96 The following argument positions are usable: 0.00/0.96 Uargs(s) = {1}, Uargs(cons) = {2}, Uargs(n__fst) = {1, 2}, 0.00/0.96 Uargs(n__add) = {1}, Uargs(n__len) = {1} 0.00/0.96 0.00/0.96 TcT has computed the following matrix interpretation satisfying 0.00/0.96 not(EDA) and not(IDA(1)). 0.00/0.96 0.00/0.96 [fst](x1, x2) = [1] x1 + [1] x2 + [4] 0.00/0.96 0.00/0.96 [0] = [1] 0.00/0.96 0.00/0.96 [nil] = [5] 0.00/0.96 0.00/0.96 [s](x1) = [1] x1 + [4] 0.00/0.96 0.00/0.96 [cons](x1, x2) = [1] x2 + [0] 0.00/0.96 0.00/0.96 [n__fst](x1, x2) = [1] x1 + [1] x2 + [4] 0.00/0.96 0.00/0.96 [activate](x1) = [1] x1 + [0] 0.00/0.96 0.00/0.96 [from](x1) = [0] 0.00/0.96 0.00/0.96 [n__from](x1) = [0] 0.00/0.96 0.00/0.96 [add](x1, x2) = [1] x1 + [1] x2 + [4] 0.00/0.96 0.00/0.96 [n__add](x1, x2) = [1] x1 + [1] x2 + [0] 0.00/0.96 0.00/0.96 [len](x1) = [1] x1 + [0] 0.00/0.96 0.00/0.96 [n__len](x1) = [1] x1 + [4] 0.00/0.96 0.00/0.96 The order satisfies the following ordering constraints: 0.00/0.96 0.00/0.96 [fst(X1, X2)] = [1] X1 + [1] X2 + [4] 0.00/0.96 >= [1] X1 + [1] X2 + [4] 0.00/0.96 = [n__fst(X1, X2)] 0.00/0.96 0.00/0.96 [fst(0(), Z)] = [1] Z + [5] 0.00/0.96 >= [5] 0.00/0.96 = [nil()] 0.00/0.96 0.00/0.96 [fst(s(X), cons(Y, Z))] = [1] Z + [1] X + [8] 0.00/0.96 > [1] Z + [1] X + [4] 0.00/0.96 = [cons(Y, n__fst(activate(X), activate(Z)))] 0.00/0.96 0.00/0.96 [activate(X)] = [1] X + [0] 0.00/0.96 >= [1] X + [0] 0.00/0.96 = [X] 0.00/0.96 0.00/0.96 [activate(n__fst(X1, X2))] = [1] X1 + [1] X2 + [4] 0.00/0.96 >= [1] X1 + [1] X2 + [4] 0.00/0.96 = [fst(X1, X2)] 0.00/0.96 0.00/0.96 [activate(n__from(X))] = [0] 0.00/0.96 >= [0] 0.00/0.96 = [from(X)] 0.00/0.96 0.00/0.96 [activate(n__add(X1, X2))] = [1] X1 + [1] X2 + [0] 0.00/0.96 ? [1] X1 + [1] X2 + [4] 0.00/0.96 = [add(X1, X2)] 0.00/0.96 0.00/0.96 [activate(n__len(X))] = [1] X + [4] 0.00/0.96 > [1] X + [0] 0.00/0.96 = [len(X)] 0.00/0.96 0.00/0.96 [from(X)] = [0] 0.00/0.96 >= [0] 0.00/0.96 = [cons(X, n__from(s(X)))] 0.00/0.96 0.00/0.96 [from(X)] = [0] 0.00/0.96 >= [0] 0.00/0.96 = [n__from(X)] 0.00/0.96 0.00/0.96 [add(X1, X2)] = [1] X1 + [1] X2 + [4] 0.00/0.96 > [1] X1 + [1] X2 + [0] 0.00/0.96 = [n__add(X1, X2)] 0.00/0.96 0.00/0.96 [add(0(), X)] = [1] X + [5] 0.00/0.96 > [1] X + [0] 0.00/0.96 = [X] 0.00/0.96 0.00/0.96 [add(s(X), Y)] = [1] X + [1] Y + [8] 0.00/0.96 > [1] X + [1] Y + [4] 0.00/0.96 = [s(n__add(activate(X), Y))] 0.00/0.96 0.00/0.96 [len(X)] = [1] X + [0] 0.00/0.96 ? [1] X + [4] 0.00/0.96 = [n__len(X)] 0.00/0.96 0.00/0.96 [len(nil())] = [5] 0.00/0.96 > [1] 0.00/0.96 = [0()] 0.00/0.96 0.00/0.96 [len(cons(X, Z))] = [1] Z + [0] 0.00/0.96 ? [1] Z + [8] 0.00/0.96 = [s(n__len(activate(Z)))] 0.00/0.96 0.00/0.96 0.00/0.96 Further, it can be verified that all rules not oriented are covered by the weightgap condition. 0.00/0.96 0.00/0.96 We are left with following problem, upon which TcT provides the 0.00/0.96 certificate YES(O(1),O(n^1)). 0.00/0.96 0.00/0.96 Strict Trs: 0.00/0.96 { fst(X1, X2) -> n__fst(X1, X2) 0.00/0.96 , activate(n__from(X)) -> from(X) 0.00/0.96 , activate(n__add(X1, X2)) -> add(X1, X2) 0.00/0.96 , len(X) -> n__len(X) 0.00/0.96 , len(cons(X, Z)) -> s(n__len(activate(Z))) } 0.00/0.96 Weak Trs: 0.00/0.96 { fst(0(), Z) -> nil() 0.00/0.96 , fst(s(X), cons(Y, Z)) -> 0.00/0.96 cons(Y, n__fst(activate(X), activate(Z))) 0.00/0.96 , activate(X) -> X 0.00/0.96 , activate(n__fst(X1, X2)) -> fst(X1, X2) 0.00/0.96 , activate(n__len(X)) -> len(X) 0.00/0.96 , from(X) -> cons(X, n__from(s(X))) 0.00/0.96 , from(X) -> n__from(X) 0.00/0.96 , add(X1, X2) -> n__add(X1, X2) 0.00/0.96 , add(0(), X) -> X 0.00/0.96 , add(s(X), Y) -> s(n__add(activate(X), Y)) 0.00/0.96 , len(nil()) -> 0() } 0.00/0.96 Obligation: 0.00/0.96 innermost runtime complexity 0.00/0.96 Answer: 0.00/0.96 YES(O(1),O(n^1)) 0.00/0.96 0.00/0.96 The weightgap principle applies (using the following nonconstant 0.00/0.96 growth matrix-interpretation) 0.00/0.96 0.00/0.96 The following argument positions are usable: 0.00/0.96 Uargs(s) = {1}, Uargs(cons) = {2}, Uargs(n__fst) = {1, 2}, 0.00/0.96 Uargs(n__add) = {1}, Uargs(n__len) = {1} 0.00/0.96 0.00/0.96 TcT has computed the following matrix interpretation satisfying 0.00/0.96 not(EDA) and not(IDA(1)). 0.00/0.96 0.00/0.96 [fst](x1, x2) = [1] x1 + [1] x2 + [1] 0.00/0.96 0.00/0.96 [0] = [3] 0.00/0.96 0.00/0.96 [nil] = [4] 0.00/0.96 0.00/0.96 [s](x1) = [1] x1 + [7] 0.00/0.96 0.00/0.96 [cons](x1, x2) = [1] x2 + [0] 0.00/0.96 0.00/0.96 [n__fst](x1, x2) = [1] x1 + [1] x2 + [3] 0.00/0.96 0.00/0.96 [activate](x1) = [1] x1 + [1] 0.00/0.96 0.00/0.96 [from](x1) = [0] 0.00/0.96 0.00/0.96 [n__from](x1) = [0] 0.00/0.96 0.00/0.96 [add](x1, x2) = [1] x1 + [1] x2 + [1] 0.00/0.96 0.00/0.96 [n__add](x1, x2) = [1] x1 + [1] x2 + [0] 0.00/0.96 0.00/0.96 [len](x1) = [1] x1 + [0] 0.00/0.96 0.00/0.96 [n__len](x1) = [1] x1 + [0] 0.00/0.96 0.00/0.96 The order satisfies the following ordering constraints: 0.00/0.96 0.00/0.96 [fst(X1, X2)] = [1] X1 + [1] X2 + [1] 0.00/0.96 ? [1] X1 + [1] X2 + [3] 0.00/0.96 = [n__fst(X1, X2)] 0.00/0.96 0.00/0.96 [fst(0(), Z)] = [1] Z + [4] 0.00/0.96 >= [4] 0.00/0.96 = [nil()] 0.00/0.96 0.00/0.96 [fst(s(X), cons(Y, Z))] = [1] Z + [1] X + [8] 0.00/0.96 > [1] Z + [1] X + [5] 0.00/0.96 = [cons(Y, n__fst(activate(X), activate(Z)))] 0.00/0.96 0.00/0.96 [activate(X)] = [1] X + [1] 0.00/0.96 > [1] X + [0] 0.00/0.96 = [X] 0.00/0.96 0.00/0.96 [activate(n__fst(X1, X2))] = [1] X1 + [1] X2 + [4] 0.00/0.96 > [1] X1 + [1] X2 + [1] 0.00/0.96 = [fst(X1, X2)] 0.00/0.96 0.00/0.96 [activate(n__from(X))] = [1] 0.00/0.96 > [0] 0.00/0.96 = [from(X)] 0.00/0.96 0.00/0.96 [activate(n__add(X1, X2))] = [1] X1 + [1] X2 + [1] 0.00/0.96 >= [1] X1 + [1] X2 + [1] 0.00/0.96 = [add(X1, X2)] 0.00/0.96 0.00/0.96 [activate(n__len(X))] = [1] X + [1] 0.00/0.96 > [1] X + [0] 0.00/0.96 = [len(X)] 0.00/0.96 0.00/0.96 [from(X)] = [0] 0.00/0.96 >= [0] 0.00/0.96 = [cons(X, n__from(s(X)))] 0.00/0.96 0.00/0.96 [from(X)] = [0] 0.00/0.96 >= [0] 0.00/0.96 = [n__from(X)] 0.00/0.96 0.00/0.96 [add(X1, X2)] = [1] X1 + [1] X2 + [1] 0.00/0.96 > [1] X1 + [1] X2 + [0] 0.00/0.97 = [n__add(X1, X2)] 0.00/0.97 0.00/0.97 [add(0(), X)] = [1] X + [4] 0.00/0.97 > [1] X + [0] 0.00/0.97 = [X] 0.00/0.97 0.00/0.97 [add(s(X), Y)] = [1] X + [1] Y + [8] 0.00/0.97 >= [1] X + [1] Y + [8] 0.00/0.97 = [s(n__add(activate(X), Y))] 0.00/0.97 0.00/0.97 [len(X)] = [1] X + [0] 0.00/0.97 >= [1] X + [0] 0.00/0.97 = [n__len(X)] 0.00/0.97 0.00/0.97 [len(nil())] = [4] 0.00/0.97 > [3] 0.00/0.97 = [0()] 0.00/0.97 0.00/0.97 [len(cons(X, Z))] = [1] Z + [0] 0.00/0.97 ? [1] Z + [8] 0.00/0.97 = [s(n__len(activate(Z)))] 0.00/0.97 0.00/0.97 0.00/0.97 Further, it can be verified that all rules not oriented are covered by the weightgap condition. 0.00/0.97 0.00/0.97 We are left with following problem, upon which TcT provides the 0.00/0.97 certificate YES(O(1),O(n^1)). 0.00/0.97 0.00/0.97 Strict Trs: 0.00/0.97 { fst(X1, X2) -> n__fst(X1, X2) 0.00/0.97 , activate(n__add(X1, X2)) -> add(X1, X2) 0.00/0.97 , len(X) -> n__len(X) 0.00/0.97 , len(cons(X, Z)) -> s(n__len(activate(Z))) } 0.00/0.97 Weak Trs: 0.00/0.97 { fst(0(), Z) -> nil() 0.00/0.97 , fst(s(X), cons(Y, Z)) -> 0.00/0.97 cons(Y, n__fst(activate(X), activate(Z))) 0.00/0.97 , activate(X) -> X 0.00/0.97 , activate(n__fst(X1, X2)) -> fst(X1, X2) 0.00/0.97 , activate(n__from(X)) -> from(X) 0.00/0.97 , activate(n__len(X)) -> len(X) 0.00/0.97 , from(X) -> cons(X, n__from(s(X))) 0.00/0.97 , from(X) -> n__from(X) 0.00/0.97 , add(X1, X2) -> n__add(X1, X2) 0.00/0.97 , add(0(), X) -> X 0.00/0.97 , add(s(X), Y) -> s(n__add(activate(X), Y)) 0.00/0.97 , len(nil()) -> 0() } 0.00/0.97 Obligation: 0.00/0.97 innermost runtime complexity 0.00/0.97 Answer: 0.00/0.97 YES(O(1),O(n^1)) 0.00/0.97 0.00/0.97 The weightgap principle applies (using the following nonconstant 0.00/0.97 growth matrix-interpretation) 0.00/0.97 0.00/0.97 The following argument positions are usable: 0.00/0.97 Uargs(s) = {1}, Uargs(cons) = {2}, Uargs(n__fst) = {1, 2}, 0.00/0.97 Uargs(n__add) = {1}, Uargs(n__len) = {1} 0.00/0.97 0.00/0.97 TcT has computed the following matrix interpretation satisfying 0.00/0.97 not(EDA) and not(IDA(1)). 0.00/0.97 0.00/0.97 [fst](x1, x2) = [1] x1 + [1] x2 + [3] 0.00/0.97 0.00/0.97 [0] = [6] 0.00/0.97 0.00/0.97 [nil] = [5] 0.00/0.97 0.00/0.97 [s](x1) = [1] x1 + [5] 0.00/0.97 0.00/0.97 [cons](x1, x2) = [1] x2 + [0] 0.00/0.97 0.00/0.97 [n__fst](x1, x2) = [1] x1 + [1] x2 + [0] 0.00/0.97 0.00/0.97 [activate](x1) = [1] x1 + [4] 0.00/0.97 0.00/0.97 [from](x1) = [0] 0.00/0.97 0.00/0.97 [n__from](x1) = [0] 0.00/0.97 0.00/0.97 [add](x1, x2) = [1] x1 + [1] x2 + [4] 0.00/0.97 0.00/0.97 [n__add](x1, x2) = [1] x1 + [1] x2 + [0] 0.00/0.97 0.00/0.97 [len](x1) = [1] x1 + [4] 0.00/0.97 0.00/0.97 [n__len](x1) = [1] x1 + [0] 0.00/0.97 0.00/0.97 The order satisfies the following ordering constraints: 0.00/0.97 0.00/0.97 [fst(X1, X2)] = [1] X1 + [1] X2 + [3] 0.00/0.97 > [1] X1 + [1] X2 + [0] 0.00/0.97 = [n__fst(X1, X2)] 0.00/0.97 0.00/0.97 [fst(0(), Z)] = [1] Z + [9] 0.00/0.97 > [5] 0.00/0.97 = [nil()] 0.00/0.97 0.00/0.97 [fst(s(X), cons(Y, Z))] = [1] Z + [1] X + [8] 0.00/0.97 >= [1] Z + [1] X + [8] 0.00/0.97 = [cons(Y, n__fst(activate(X), activate(Z)))] 0.00/0.97 0.00/0.97 [activate(X)] = [1] X + [4] 0.00/0.97 > [1] X + [0] 0.00/0.97 = [X] 0.00/0.97 0.00/0.97 [activate(n__fst(X1, X2))] = [1] X1 + [1] X2 + [4] 0.00/0.97 > [1] X1 + [1] X2 + [3] 0.00/0.97 = [fst(X1, X2)] 0.00/0.97 0.00/0.97 [activate(n__from(X))] = [4] 0.00/0.97 > [0] 0.00/0.97 = [from(X)] 0.00/0.97 0.00/0.97 [activate(n__add(X1, X2))] = [1] X1 + [1] X2 + [4] 0.00/0.97 >= [1] X1 + [1] X2 + [4] 0.00/0.97 = [add(X1, X2)] 0.00/0.97 0.00/0.97 [activate(n__len(X))] = [1] X + [4] 0.00/0.97 >= [1] X + [4] 0.00/0.97 = [len(X)] 0.00/0.97 0.00/0.97 [from(X)] = [0] 0.00/0.97 >= [0] 0.00/0.97 = [cons(X, n__from(s(X)))] 0.00/0.97 0.00/0.97 [from(X)] = [0] 0.00/0.97 >= [0] 0.00/0.97 = [n__from(X)] 0.00/0.97 0.00/0.97 [add(X1, X2)] = [1] X1 + [1] X2 + [4] 0.00/0.97 > [1] X1 + [1] X2 + [0] 0.00/0.97 = [n__add(X1, X2)] 0.00/0.97 0.00/0.97 [add(0(), X)] = [1] X + [10] 0.00/0.97 > [1] X + [0] 0.00/0.97 = [X] 0.00/0.97 0.00/0.97 [add(s(X), Y)] = [1] X + [1] Y + [9] 0.00/0.97 >= [1] X + [1] Y + [9] 0.00/0.97 = [s(n__add(activate(X), Y))] 0.00/0.97 0.00/0.97 [len(X)] = [1] X + [4] 0.00/0.97 > [1] X + [0] 0.00/0.97 = [n__len(X)] 0.00/0.97 0.00/0.97 [len(nil())] = [9] 0.00/0.97 > [6] 0.00/0.97 = [0()] 0.00/0.97 0.00/0.97 [len(cons(X, Z))] = [1] Z + [4] 0.00/0.97 ? [1] Z + [9] 0.00/0.97 = [s(n__len(activate(Z)))] 0.00/0.97 0.00/0.97 0.00/0.97 Further, it can be verified that all rules not oriented are covered by the weightgap condition. 0.00/0.97 0.00/0.97 We are left with following problem, upon which TcT provides the 0.00/0.97 certificate YES(O(1),O(n^1)). 0.00/0.97 0.00/0.97 Strict Trs: 0.00/0.97 { activate(n__add(X1, X2)) -> add(X1, X2) 0.00/0.97 , len(cons(X, Z)) -> s(n__len(activate(Z))) } 0.00/0.97 Weak Trs: 0.00/0.97 { fst(X1, X2) -> n__fst(X1, X2) 0.00/0.97 , fst(0(), Z) -> nil() 0.00/0.97 , fst(s(X), cons(Y, Z)) -> 0.00/0.97 cons(Y, n__fst(activate(X), activate(Z))) 0.00/0.97 , activate(X) -> X 0.00/0.97 , activate(n__fst(X1, X2)) -> fst(X1, X2) 0.00/0.97 , activate(n__from(X)) -> from(X) 0.00/0.97 , activate(n__len(X)) -> len(X) 0.00/0.97 , from(X) -> cons(X, n__from(s(X))) 0.00/0.97 , from(X) -> n__from(X) 0.00/0.97 , add(X1, X2) -> n__add(X1, X2) 0.00/0.97 , add(0(), X) -> X 0.00/0.97 , add(s(X), Y) -> s(n__add(activate(X), Y)) 0.00/0.97 , len(X) -> n__len(X) 0.00/0.97 , len(nil()) -> 0() } 0.00/0.97 Obligation: 0.00/0.97 innermost runtime complexity 0.00/0.97 Answer: 0.00/0.97 YES(O(1),O(n^1)) 0.00/0.97 0.00/0.97 We use the processor 'matrix interpretation of dimension 1' to 0.00/0.97 orient following rules strictly. 0.00/0.97 0.00/0.97 Trs: 0.00/0.97 { activate(n__add(X1, X2)) -> add(X1, X2) 0.00/0.97 , len(cons(X, Z)) -> s(n__len(activate(Z))) } 0.00/0.97 0.00/0.97 The induced complexity on above rules (modulo remaining rules) is 0.00/0.97 YES(?,O(n^1)) . These rules are moved into the corresponding weak 0.00/0.97 component(s). 0.00/0.97 0.00/0.97 Sub-proof: 0.00/0.97 ---------- 0.00/0.97 The following argument positions are usable: 0.00/0.97 Uargs(s) = {1}, Uargs(cons) = {2}, Uargs(n__fst) = {1, 2}, 0.00/0.97 Uargs(n__add) = {1}, Uargs(n__len) = {1} 0.00/0.97 0.00/0.97 TcT has computed the following constructor-based matrix 0.00/0.97 interpretation satisfying not(EDA). 0.00/0.97 0.00/0.97 [fst](x1, x2) = [2] x1 + [2] x2 + [0] 0.00/0.97 0.00/0.97 [0] = [0] 0.00/0.97 0.00/0.97 [nil] = [0] 0.00/0.97 0.00/0.97 [s](x1) = [1] x1 + [1] 0.00/0.97 0.00/0.97 [cons](x1, x2) = [1] x2 + [3] 0.00/0.97 0.00/0.97 [n__fst](x1, x2) = [1] x1 + [1] x2 + [0] 0.00/0.97 0.00/0.97 [activate](x1) = [2] x1 + [0] 0.00/0.97 0.00/0.97 [from](x1) = [7] 0.00/0.97 0.00/0.97 [n__from](x1) = [4] 0.00/0.97 0.00/0.97 [add](x1, x2) = [2] x1 + [1] x2 + [2] 0.00/0.97 0.00/0.97 [n__add](x1, x2) = [1] x1 + [1] x2 + [2] 0.00/0.97 0.00/0.97 [len](x1) = [2] x1 + [4] 0.00/0.97 0.00/0.97 [n__len](x1) = [1] x1 + [4] 0.00/0.97 0.00/0.97 The order satisfies the following ordering constraints: 0.00/0.97 0.00/0.97 [fst(X1, X2)] = [2] X1 + [2] X2 + [0] 0.00/0.97 >= [1] X1 + [1] X2 + [0] 0.00/0.97 = [n__fst(X1, X2)] 0.00/0.97 0.00/0.97 [fst(0(), Z)] = [2] Z + [0] 0.00/0.97 >= [0] 0.00/0.97 = [nil()] 0.00/0.97 0.00/0.97 [fst(s(X), cons(Y, Z))] = [2] Z + [2] X + [8] 0.00/0.97 > [2] Z + [2] X + [3] 0.00/0.97 = [cons(Y, n__fst(activate(X), activate(Z)))] 0.00/0.97 0.00/0.97 [activate(X)] = [2] X + [0] 0.00/0.97 >= [1] X + [0] 0.00/0.97 = [X] 0.00/0.97 0.00/0.97 [activate(n__fst(X1, X2))] = [2] X1 + [2] X2 + [0] 0.00/0.97 >= [2] X1 + [2] X2 + [0] 0.00/0.97 = [fst(X1, X2)] 0.00/0.97 0.00/0.97 [activate(n__from(X))] = [8] 0.00/0.97 > [7] 0.00/0.97 = [from(X)] 0.00/0.97 0.00/0.97 [activate(n__add(X1, X2))] = [2] X1 + [2] X2 + [4] 0.00/0.97 > [2] X1 + [1] X2 + [2] 0.00/0.97 = [add(X1, X2)] 0.00/0.97 0.00/0.97 [activate(n__len(X))] = [2] X + [8] 0.00/0.97 > [2] X + [4] 0.00/0.97 = [len(X)] 0.00/0.97 0.00/0.97 [from(X)] = [7] 0.00/0.97 >= [7] 0.00/0.97 = [cons(X, n__from(s(X)))] 0.00/0.97 0.00/0.97 [from(X)] = [7] 0.00/0.97 > [4] 0.00/0.97 = [n__from(X)] 0.00/0.97 0.00/0.97 [add(X1, X2)] = [2] X1 + [1] X2 + [2] 0.00/0.97 >= [1] X1 + [1] X2 + [2] 0.00/0.97 = [n__add(X1, X2)] 0.00/0.97 0.00/0.97 [add(0(), X)] = [1] X + [2] 0.00/0.97 > [1] X + [0] 0.00/0.97 = [X] 0.00/0.97 0.00/0.97 [add(s(X), Y)] = [2] X + [1] Y + [4] 0.00/0.97 > [2] X + [1] Y + [3] 0.00/0.97 = [s(n__add(activate(X), Y))] 0.00/0.97 0.00/0.97 [len(X)] = [2] X + [4] 0.00/0.97 >= [1] X + [4] 0.00/0.97 = [n__len(X)] 0.00/0.97 0.00/0.97 [len(nil())] = [4] 0.00/0.97 > [0] 0.00/0.97 = [0()] 0.00/0.97 0.00/0.97 [len(cons(X, Z))] = [2] Z + [10] 0.00/0.97 > [2] Z + [5] 0.00/0.97 = [s(n__len(activate(Z)))] 0.00/0.97 0.00/0.97 0.00/0.97 We return to the main proof. 0.00/0.97 0.00/0.97 We are left with following problem, upon which TcT provides the 0.00/0.97 certificate YES(O(1),O(1)). 0.00/0.97 0.00/0.97 Weak Trs: 0.00/0.97 { fst(X1, X2) -> n__fst(X1, X2) 0.00/0.97 , fst(0(), Z) -> nil() 0.00/0.97 , fst(s(X), cons(Y, Z)) -> 0.00/0.97 cons(Y, n__fst(activate(X), activate(Z))) 0.00/0.97 , activate(X) -> X 0.00/0.97 , activate(n__fst(X1, X2)) -> fst(X1, X2) 0.00/0.97 , activate(n__from(X)) -> from(X) 0.00/0.97 , activate(n__add(X1, X2)) -> add(X1, X2) 0.00/0.97 , activate(n__len(X)) -> len(X) 0.00/0.97 , from(X) -> cons(X, n__from(s(X))) 0.00/0.97 , from(X) -> n__from(X) 0.00/0.97 , add(X1, X2) -> n__add(X1, X2) 0.00/0.97 , add(0(), X) -> X 0.00/0.97 , add(s(X), Y) -> s(n__add(activate(X), Y)) 0.00/0.97 , len(X) -> n__len(X) 0.00/0.97 , len(nil()) -> 0() 0.00/0.97 , len(cons(X, Z)) -> s(n__len(activate(Z))) } 0.00/0.97 Obligation: 0.00/0.97 innermost runtime complexity 0.00/0.97 Answer: 0.00/0.97 YES(O(1),O(1)) 0.00/0.97 0.00/0.97 Empty rules are trivially bounded 0.00/0.97 0.00/0.97 Hurray, we answered YES(O(1),O(n^1)) 0.00/0.97 EOF