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: { first(X1, X2) -> n__first(X1, X2) , first(0(), X) -> nil() , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z))) , activate(X) -> X , activate(n__first(X1, X2)) -> first(X1, X2) , activate(n__from(X)) -> from(X) , from(X) -> cons(X, n__from(s(X))) , from(X) -> n__from(X) } Obligation: runtime complexity Answer: YES(O(1),O(n^1)) The weightgap principle applies (using the following nonconstant growth matrix-interpretation) The following argument positions are usable: Uargs(cons) = {2}, Uargs(n__first) = {2} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [first](x1, x2) = [1] x2 + [0] [0] = [7] [nil] = [7] [s](x1) = [0] [cons](x1, x2) = [1] x2 + [0] [n__first](x1, x2) = [1] x2 + [0] [activate](x1) = [1] x1 + [0] [from](x1) = [1] x1 + [5] [n__from](x1) = [1] x1 + [0] The following symbols are considered usable {first, activate, from} The order satisfies the following ordering constraints: [first(X1, X2)] = [1] X2 + [0] >= [1] X2 + [0] = [n__first(X1, X2)] [first(0(), X)] = [1] X + [0] ? [7] = [nil()] [first(s(X), cons(Y, Z))] = [1] Z + [0] >= [1] Z + [0] = [cons(Y, n__first(X, activate(Z)))] [activate(X)] = [1] X + [0] >= [1] X + [0] = [X] [activate(n__first(X1, X2))] = [1] X2 + [0] >= [1] X2 + [0] = [first(X1, X2)] [activate(n__from(X))] = [1] X + [0] ? [1] X + [5] = [from(X)] [from(X)] = [1] X + [5] > [0] = [cons(X, n__from(s(X)))] [from(X)] = [1] X + [5] > [1] X + [0] = [n__from(X)] 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 Trs: { first(X1, X2) -> n__first(X1, X2) , first(0(), X) -> nil() , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z))) , activate(X) -> X , activate(n__first(X1, X2)) -> first(X1, X2) , activate(n__from(X)) -> from(X) } Weak Trs: { from(X) -> cons(X, n__from(s(X))) , from(X) -> n__from(X) } Obligation: runtime complexity Answer: YES(O(1),O(n^1)) The weightgap principle applies (using the following nonconstant growth matrix-interpretation) The following argument positions are usable: Uargs(cons) = {2}, Uargs(n__first) = {2} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [first](x1, x2) = [1] x2 + [1] [0] = [7] [nil] = [0] [s](x1) = [0] [cons](x1, x2) = [1] x2 + [0] [n__first](x1, x2) = [1] x2 + [0] [activate](x1) = [1] x1 + [0] [from](x1) = [1] x1 + [5] [n__from](x1) = [1] x1 + [0] The following symbols are considered usable {first, activate, from} The order satisfies the following ordering constraints: [first(X1, X2)] = [1] X2 + [1] > [1] X2 + [0] = [n__first(X1, X2)] [first(0(), X)] = [1] X + [1] > [0] = [nil()] [first(s(X), cons(Y, Z))] = [1] Z + [1] > [1] Z + [0] = [cons(Y, n__first(X, activate(Z)))] [activate(X)] = [1] X + [0] >= [1] X + [0] = [X] [activate(n__first(X1, X2))] = [1] X2 + [0] ? [1] X2 + [1] = [first(X1, X2)] [activate(n__from(X))] = [1] X + [0] ? [1] X + [5] = [from(X)] [from(X)] = [1] X + [5] > [0] = [cons(X, n__from(s(X)))] [from(X)] = [1] X + [5] > [1] X + [0] = [n__from(X)] 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 Trs: { activate(X) -> X , activate(n__first(X1, X2)) -> first(X1, X2) , activate(n__from(X)) -> from(X) } Weak Trs: { first(X1, X2) -> n__first(X1, X2) , first(0(), X) -> nil() , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z))) , from(X) -> cons(X, n__from(s(X))) , from(X) -> n__from(X) } Obligation: runtime complexity Answer: YES(O(1),O(n^1)) The weightgap principle applies (using the following nonconstant growth matrix-interpretation) The following argument positions are usable: Uargs(cons) = {2}, Uargs(n__first) = {2} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [first](x1, x2) = [1] x1 + [1] x2 + [4] [0] = [7] [nil] = [3] [s](x1) = [1] x1 + [4] [cons](x1, x2) = [1] x2 + [0] [n__first](x1, x2) = [1] x1 + [1] x2 + [0] [activate](x1) = [1] x1 + [4] [from](x1) = [5] [n__from](x1) = [0] The following symbols are considered usable {first, activate, from} The order satisfies the following ordering constraints: [first(X1, X2)] = [1] X1 + [1] X2 + [4] > [1] X1 + [1] X2 + [0] = [n__first(X1, X2)] [first(0(), X)] = [1] X + [11] > [3] = [nil()] [first(s(X), cons(Y, Z))] = [1] X + [1] Z + [8] > [1] X + [1] Z + [4] = [cons(Y, n__first(X, activate(Z)))] [activate(X)] = [1] X + [4] > [1] X + [0] = [X] [activate(n__first(X1, X2))] = [1] X1 + [1] X2 + [4] >= [1] X1 + [1] X2 + [4] = [first(X1, X2)] [activate(n__from(X))] = [4] ? [5] = [from(X)] [from(X)] = [5] > [0] = [cons(X, n__from(s(X)))] [from(X)] = [5] > [0] = [n__from(X)] 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 Trs: { activate(n__first(X1, X2)) -> first(X1, X2) , activate(n__from(X)) -> from(X) } Weak Trs: { first(X1, X2) -> n__first(X1, X2) , first(0(), X) -> nil() , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z))) , activate(X) -> X , from(X) -> cons(X, n__from(s(X))) , from(X) -> n__from(X) } Obligation: runtime complexity Answer: YES(O(1),O(n^1)) The weightgap principle applies (using the following nonconstant growth matrix-interpretation) The following argument positions are usable: Uargs(cons) = {2}, Uargs(n__first) = {2} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [first](x1, x2) = [1] x1 + [1] x2 + [0] [0] = [7] [nil] = [7] [s](x1) = [1] x1 + [1] [cons](x1, x2) = [1] x2 + [0] [n__first](x1, x2) = [1] x1 + [1] x2 + [0] [activate](x1) = [1] x1 + [1] [from](x1) = [0] [n__from](x1) = [0] The following symbols are considered usable {first, activate, from} The order satisfies the following ordering constraints: [first(X1, X2)] = [1] X1 + [1] X2 + [0] >= [1] X1 + [1] X2 + [0] = [n__first(X1, X2)] [first(0(), X)] = [1] X + [7] >= [7] = [nil()] [first(s(X), cons(Y, Z))] = [1] X + [1] Z + [1] >= [1] X + [1] Z + [1] = [cons(Y, n__first(X, activate(Z)))] [activate(X)] = [1] X + [1] > [1] X + [0] = [X] [activate(n__first(X1, X2))] = [1] X1 + [1] X2 + [1] > [1] X1 + [1] X2 + [0] = [first(X1, X2)] [activate(n__from(X))] = [1] > [0] = [from(X)] [from(X)] = [0] >= [0] = [cons(X, n__from(s(X)))] [from(X)] = [0] >= [0] = [n__from(X)] Further, it can be verified that all rules not oriented are covered by the weightgap condition. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak Trs: { first(X1, X2) -> n__first(X1, X2) , first(0(), X) -> nil() , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z))) , activate(X) -> X , activate(n__first(X1, X2)) -> first(X1, X2) , activate(n__from(X)) -> from(X) , from(X) -> cons(X, n__from(s(X))) , from(X) -> n__from(X) } Obligation: runtime complexity Answer: YES(O(1),O(1)) Empty rules are trivially bounded Hurray, we answered YES(O(1),O(n^1))