We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^2)). Strict Trs: { a__from(X) -> cons(mark(X), from(s(X))) , a__from(X) -> from(X) , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(from(X)) -> a__from(mark(X)) , mark(s(X)) -> s(mark(X)) , mark(nil()) -> nil() , mark(0()) -> 0() , mark(length(X)) -> a__length(X) , mark(length1(X)) -> a__length1(X) , a__length(X) -> length(X) , a__length(cons(X, Y)) -> s(a__length1(Y)) , a__length(nil()) -> 0() , a__length1(X) -> a__length(X) , a__length1(X) -> length1(X) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) The weightgap principle applies (using the following nonconstant growth matrix-interpretation) The following argument positions are usable: Uargs(a__from) = {1}, Uargs(cons) = {1}, Uargs(s) = {1} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [a__from](x1) = [1] x1 + [4] [cons](x1, x2) = [1] x1 + [0] [mark](x1) = [0] [from](x1) = [1] x1 + [0] [s](x1) = [1] x1 + [0] [a__length](x1) = [0] [nil] = [0] [0] = [0] [a__length1](x1) = [0] [length](x1) = [0] [length1](x1) = [0] The order satisfies the following ordering constraints: [a__from(X)] = [1] X + [4] > [0] = [cons(mark(X), from(s(X)))] [a__from(X)] = [1] X + [4] > [1] X + [0] = [from(X)] [mark(cons(X1, X2))] = [0] >= [0] = [cons(mark(X1), X2)] [mark(from(X))] = [0] ? [4] = [a__from(mark(X))] [mark(s(X))] = [0] >= [0] = [s(mark(X))] [mark(nil())] = [0] >= [0] = [nil()] [mark(0())] = [0] >= [0] = [0()] [mark(length(X))] = [0] >= [0] = [a__length(X)] [mark(length1(X))] = [0] >= [0] = [a__length1(X)] [a__length(X)] = [0] >= [0] = [length(X)] [a__length(cons(X, Y))] = [0] >= [0] = [s(a__length1(Y))] [a__length(nil())] = [0] >= [0] = [0()] [a__length1(X)] = [0] >= [0] = [a__length(X)] [a__length1(X)] = [0] >= [0] = [length1(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^2)). Strict Trs: { mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(from(X)) -> a__from(mark(X)) , mark(s(X)) -> s(mark(X)) , mark(nil()) -> nil() , mark(0()) -> 0() , mark(length(X)) -> a__length(X) , mark(length1(X)) -> a__length1(X) , a__length(X) -> length(X) , a__length(cons(X, Y)) -> s(a__length1(Y)) , a__length(nil()) -> 0() , a__length1(X) -> a__length(X) , a__length1(X) -> length1(X) } Weak Trs: { a__from(X) -> cons(mark(X), from(s(X))) , a__from(X) -> from(X) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) The weightgap principle applies (using the following nonconstant growth matrix-interpretation) The following argument positions are usable: Uargs(a__from) = {1}, Uargs(cons) = {1}, Uargs(s) = {1} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [a__from](x1) = [1] x1 + [6] [cons](x1, x2) = [1] x1 + [0] [mark](x1) = [1] x1 + [1] [from](x1) = [1] x1 + [0] [s](x1) = [1] x1 + [1] [a__length](x1) = [1] [nil] = [0] [0] = [0] [a__length1](x1) = [4] [length](x1) = [0] [length1](x1) = [0] The order satisfies the following ordering constraints: [a__from(X)] = [1] X + [6] > [1] X + [1] = [cons(mark(X), from(s(X)))] [a__from(X)] = [1] X + [6] > [1] X + [0] = [from(X)] [mark(cons(X1, X2))] = [1] X1 + [1] >= [1] X1 + [1] = [cons(mark(X1), X2)] [mark(from(X))] = [1] X + [1] ? [1] X + [7] = [a__from(mark(X))] [mark(s(X))] = [1] X + [2] >= [1] X + [2] = [s(mark(X))] [mark(nil())] = [1] > [0] = [nil()] [mark(0())] = [1] > [0] = [0()] [mark(length(X))] = [1] >= [1] = [a__length(X)] [mark(length1(X))] = [1] ? [4] = [a__length1(X)] [a__length(X)] = [1] > [0] = [length(X)] [a__length(cons(X, Y))] = [1] ? [5] = [s(a__length1(Y))] [a__length(nil())] = [1] > [0] = [0()] [a__length1(X)] = [4] > [1] = [a__length(X)] [a__length1(X)] = [4] > [0] = [length1(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^2)). Strict Trs: { mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(from(X)) -> a__from(mark(X)) , mark(s(X)) -> s(mark(X)) , mark(length(X)) -> a__length(X) , mark(length1(X)) -> a__length1(X) , a__length(cons(X, Y)) -> s(a__length1(Y)) } Weak Trs: { a__from(X) -> cons(mark(X), from(s(X))) , a__from(X) -> from(X) , mark(nil()) -> nil() , mark(0()) -> 0() , a__length(X) -> length(X) , a__length(nil()) -> 0() , a__length1(X) -> a__length(X) , a__length1(X) -> length1(X) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) The weightgap principle applies (using the following nonconstant growth matrix-interpretation) The following argument positions are usable: Uargs(a__from) = {1}, Uargs(cons) = {1}, Uargs(s) = {1} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [a__from](x1) = [1] x1 + [7] [cons](x1, x2) = [1] x1 + [0] [mark](x1) = [1] x1 + [1] [from](x1) = [1] x1 + [0] [s](x1) = [1] x1 + [0] [a__length](x1) = [3] [nil] = [0] [0] = [0] [a__length1](x1) = [4] [length](x1) = [3] [length1](x1) = [0] The order satisfies the following ordering constraints: [a__from(X)] = [1] X + [7] > [1] X + [1] = [cons(mark(X), from(s(X)))] [a__from(X)] = [1] X + [7] > [1] X + [0] = [from(X)] [mark(cons(X1, X2))] = [1] X1 + [1] >= [1] X1 + [1] = [cons(mark(X1), X2)] [mark(from(X))] = [1] X + [1] ? [1] X + [8] = [a__from(mark(X))] [mark(s(X))] = [1] X + [1] >= [1] X + [1] = [s(mark(X))] [mark(nil())] = [1] > [0] = [nil()] [mark(0())] = [1] > [0] = [0()] [mark(length(X))] = [4] > [3] = [a__length(X)] [mark(length1(X))] = [1] ? [4] = [a__length1(X)] [a__length(X)] = [3] >= [3] = [length(X)] [a__length(cons(X, Y))] = [3] ? [4] = [s(a__length1(Y))] [a__length(nil())] = [3] > [0] = [0()] [a__length1(X)] = [4] > [3] = [a__length(X)] [a__length1(X)] = [4] > [0] = [length1(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^2)). Strict Trs: { mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(from(X)) -> a__from(mark(X)) , mark(s(X)) -> s(mark(X)) , mark(length1(X)) -> a__length1(X) , a__length(cons(X, Y)) -> s(a__length1(Y)) } Weak Trs: { a__from(X) -> cons(mark(X), from(s(X))) , a__from(X) -> from(X) , mark(nil()) -> nil() , mark(0()) -> 0() , mark(length(X)) -> a__length(X) , a__length(X) -> length(X) , a__length(nil()) -> 0() , a__length1(X) -> a__length(X) , a__length1(X) -> length1(X) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) The weightgap principle applies (using the following nonconstant growth matrix-interpretation) The following argument positions are usable: Uargs(a__from) = {1}, Uargs(cons) = {1}, Uargs(s) = {1} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [a__from](x1) = [1] x1 + [7] [cons](x1, x2) = [1] x1 + [0] [mark](x1) = [1] x1 + [1] [from](x1) = [1] x1 + [0] [s](x1) = [1] x1 + [0] [a__length](x1) = [4] [nil] = [0] [0] = [0] [a__length1](x1) = [4] [length](x1) = [4] [length1](x1) = [4] The order satisfies the following ordering constraints: [a__from(X)] = [1] X + [7] > [1] X + [1] = [cons(mark(X), from(s(X)))] [a__from(X)] = [1] X + [7] > [1] X + [0] = [from(X)] [mark(cons(X1, X2))] = [1] X1 + [1] >= [1] X1 + [1] = [cons(mark(X1), X2)] [mark(from(X))] = [1] X + [1] ? [1] X + [8] = [a__from(mark(X))] [mark(s(X))] = [1] X + [1] >= [1] X + [1] = [s(mark(X))] [mark(nil())] = [1] > [0] = [nil()] [mark(0())] = [1] > [0] = [0()] [mark(length(X))] = [5] > [4] = [a__length(X)] [mark(length1(X))] = [5] > [4] = [a__length1(X)] [a__length(X)] = [4] >= [4] = [length(X)] [a__length(cons(X, Y))] = [4] >= [4] = [s(a__length1(Y))] [a__length(nil())] = [4] > [0] = [0()] [a__length1(X)] = [4] >= [4] = [a__length(X)] [a__length1(X)] = [4] >= [4] = [length1(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^2)). Strict Trs: { mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(from(X)) -> a__from(mark(X)) , mark(s(X)) -> s(mark(X)) , a__length(cons(X, Y)) -> s(a__length1(Y)) } Weak Trs: { a__from(X) -> cons(mark(X), from(s(X))) , a__from(X) -> from(X) , mark(nil()) -> nil() , mark(0()) -> 0() , mark(length(X)) -> a__length(X) , mark(length1(X)) -> a__length1(X) , a__length(X) -> length(X) , a__length(nil()) -> 0() , a__length1(X) -> a__length(X) , a__length1(X) -> length1(X) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) We use the processor 'matrix interpretation of dimension 2' to orient following rules strictly. Trs: { a__length(cons(X, Y)) -> s(a__length1(Y)) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- The following argument positions are usable: Uargs(a__from) = {1}, Uargs(cons) = {1}, Uargs(s) = {1} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA) and not(IDA(1)). [a__from](x1) = [1 0] x1 + [6] [0 0] [7] [cons](x1, x2) = [1 0] x1 + [0 1] x2 + [1] [0 0] [0 1] [2] [mark](x1) = [1 0] x1 + [1] [0 4] [2] [from](x1) = [1 0] x1 + [6] [0 0] [3] [s](x1) = [1 0] x1 + [0] [0 0] [1] [a__length](x1) = [0 4] x1 + [1] [0 0] [1] [nil] = [0] [0] [0] = [0] [0] [a__length1](x1) = [0 4] x1 + [3] [0 0] [1] [length](x1) = [0 4] x1 + [0] [0 0] [0] [length1](x1) = [0 4] x1 + [2] [0 0] [0] The order satisfies the following ordering constraints: [a__from(X)] = [1 0] X + [6] [0 0] [7] > [1 0] X + [5] [0 0] [5] = [cons(mark(X), from(s(X)))] [a__from(X)] = [1 0] X + [6] [0 0] [7] >= [1 0] X + [6] [0 0] [3] = [from(X)] [mark(cons(X1, X2))] = [1 0] X1 + [0 1] X2 + [2] [0 0] [0 4] [10] >= [1 0] X1 + [0 1] X2 + [2] [0 0] [0 1] [2] = [cons(mark(X1), X2)] [mark(from(X))] = [1 0] X + [7] [0 0] [14] >= [1 0] X + [7] [0 0] [7] = [a__from(mark(X))] [mark(s(X))] = [1 0] X + [1] [0 0] [6] >= [1 0] X + [1] [0 0] [1] = [s(mark(X))] [mark(nil())] = [1] [2] > [0] [0] = [nil()] [mark(0())] = [1] [2] > [0] [0] = [0()] [mark(length(X))] = [0 4] X + [1] [0 0] [2] >= [0 4] X + [1] [0 0] [1] = [a__length(X)] [mark(length1(X))] = [0 4] X + [3] [0 0] [2] >= [0 4] X + [3] [0 0] [1] = [a__length1(X)] [a__length(X)] = [0 4] X + [1] [0 0] [1] > [0 4] X + [0] [0 0] [0] = [length(X)] [a__length(cons(X, Y))] = [0 4] Y + [9] [0 0] [1] > [0 4] Y + [3] [0 0] [1] = [s(a__length1(Y))] [a__length(nil())] = [1] [1] > [0] [0] = [0()] [a__length1(X)] = [0 4] X + [3] [0 0] [1] > [0 4] X + [1] [0 0] [1] = [a__length(X)] [a__length1(X)] = [0 4] X + [3] [0 0] [1] > [0 4] X + [2] [0 0] [0] = [length1(X)] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^2)). Strict Trs: { mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(from(X)) -> a__from(mark(X)) , mark(s(X)) -> s(mark(X)) } Weak Trs: { a__from(X) -> cons(mark(X), from(s(X))) , a__from(X) -> from(X) , mark(nil()) -> nil() , mark(0()) -> 0() , mark(length(X)) -> a__length(X) , mark(length1(X)) -> a__length1(X) , a__length(X) -> length(X) , a__length(cons(X, Y)) -> s(a__length1(Y)) , a__length(nil()) -> 0() , a__length1(X) -> a__length(X) , a__length1(X) -> length1(X) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) We use the processor 'matrix interpretation of dimension 2' to orient following rules strictly. Trs: { mark(from(X)) -> a__from(mark(X)) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^2)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- The following argument positions are usable: Uargs(a__from) = {1}, Uargs(cons) = {1}, Uargs(s) = {1} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [a__from](x1) = [1 7] x1 + [1] [0 1] [4] [cons](x1, x2) = [1 0] x1 + [0] [0 1] [0] [mark](x1) = [1 2] x1 + [1] [0 1] [0] [from](x1) = [1 7] x1 + [0] [0 1] [4] [s](x1) = [1 5] x1 + [0] [0 1] [0] [a__length](x1) = [1] [0] [nil] = [0] [0] [0] = [0] [0] [a__length1](x1) = [1] [0] [length](x1) = [0] [0] [length1](x1) = [0] [0] The order satisfies the following ordering constraints: [a__from(X)] = [1 7] X + [1] [0 1] [4] >= [1 2] X + [1] [0 1] [0] = [cons(mark(X), from(s(X)))] [a__from(X)] = [1 7] X + [1] [0 1] [4] > [1 7] X + [0] [0 1] [4] = [from(X)] [mark(cons(X1, X2))] = [1 2] X1 + [1] [0 1] [0] >= [1 2] X1 + [1] [0 1] [0] = [cons(mark(X1), X2)] [mark(from(X))] = [1 9] X + [9] [0 1] [4] > [1 9] X + [2] [0 1] [4] = [a__from(mark(X))] [mark(s(X))] = [1 7] X + [1] [0 1] [0] >= [1 7] X + [1] [0 1] [0] = [s(mark(X))] [mark(nil())] = [1] [0] > [0] [0] = [nil()] [mark(0())] = [1] [0] > [0] [0] = [0()] [mark(length(X))] = [1] [0] >= [1] [0] = [a__length(X)] [mark(length1(X))] = [1] [0] >= [1] [0] = [a__length1(X)] [a__length(X)] = [1] [0] > [0] [0] = [length(X)] [a__length(cons(X, Y))] = [1] [0] >= [1] [0] = [s(a__length1(Y))] [a__length(nil())] = [1] [0] > [0] [0] = [0()] [a__length1(X)] = [1] [0] >= [1] [0] = [a__length(X)] [a__length1(X)] = [1] [0] > [0] [0] = [length1(X)] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^2)). Strict Trs: { mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(s(X)) -> s(mark(X)) } Weak Trs: { a__from(X) -> cons(mark(X), from(s(X))) , a__from(X) -> from(X) , mark(from(X)) -> a__from(mark(X)) , mark(nil()) -> nil() , mark(0()) -> 0() , mark(length(X)) -> a__length(X) , mark(length1(X)) -> a__length1(X) , a__length(X) -> length(X) , a__length(cons(X, Y)) -> s(a__length1(Y)) , a__length(nil()) -> 0() , a__length1(X) -> a__length(X) , a__length1(X) -> length1(X) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) We use the processor 'matrix interpretation of dimension 2' to orient following rules strictly. Trs: { mark(cons(X1, X2)) -> cons(mark(X1), X2) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^2)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- The following argument positions are usable: Uargs(a__from) = {1}, Uargs(cons) = {1}, Uargs(s) = {1} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [a__from](x1) = [1 7] x1 + [7] [0 1] [3] [cons](x1, x2) = [1 6] x1 + [0] [0 1] [3] [mark](x1) = [1 1] x1 + [3] [0 1] [0] [from](x1) = [1 7] x1 + [6] [0 1] [3] [s](x1) = [1 0] x1 + [0] [0 1] [0] [a__length](x1) = [7] [2] [nil] = [1] [3] [0] = [0] [0] [a__length1](x1) = [7] [2] [length](x1) = [4] [2] [length1](x1) = [4] [2] The order satisfies the following ordering constraints: [a__from(X)] = [1 7] X + [7] [0 1] [3] > [1 7] X + [3] [0 1] [3] = [cons(mark(X), from(s(X)))] [a__from(X)] = [1 7] X + [7] [0 1] [3] > [1 7] X + [6] [0 1] [3] = [from(X)] [mark(cons(X1, X2))] = [1 7] X1 + [6] [0 1] [3] > [1 7] X1 + [3] [0 1] [3] = [cons(mark(X1), X2)] [mark(from(X))] = [1 8] X + [12] [0 1] [3] > [1 8] X + [10] [0 1] [3] = [a__from(mark(X))] [mark(s(X))] = [1 1] X + [3] [0 1] [0] >= [1 1] X + [3] [0 1] [0] = [s(mark(X))] [mark(nil())] = [7] [3] > [1] [3] = [nil()] [mark(0())] = [3] [0] > [0] [0] = [0()] [mark(length(X))] = [9] [2] > [7] [2] = [a__length(X)] [mark(length1(X))] = [9] [2] > [7] [2] = [a__length1(X)] [a__length(X)] = [7] [2] > [4] [2] = [length(X)] [a__length(cons(X, Y))] = [7] [2] >= [7] [2] = [s(a__length1(Y))] [a__length(nil())] = [7] [2] > [0] [0] = [0()] [a__length1(X)] = [7] [2] >= [7] [2] = [a__length(X)] [a__length1(X)] = [7] [2] > [4] [2] = [length1(X)] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^2)). Strict Trs: { mark(s(X)) -> s(mark(X)) } Weak Trs: { a__from(X) -> cons(mark(X), from(s(X))) , a__from(X) -> from(X) , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(from(X)) -> a__from(mark(X)) , mark(nil()) -> nil() , mark(0()) -> 0() , mark(length(X)) -> a__length(X) , mark(length1(X)) -> a__length1(X) , a__length(X) -> length(X) , a__length(cons(X, Y)) -> s(a__length1(Y)) , a__length(nil()) -> 0() , a__length1(X) -> a__length(X) , a__length1(X) -> length1(X) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) We use the processor 'matrix interpretation of dimension 3' to orient following rules strictly. Trs: { mark(s(X)) -> s(mark(X)) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^2)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- The following argument positions are usable: Uargs(a__from) = {1}, Uargs(cons) = {1}, Uargs(s) = {1} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA) and not(IDA(2)). [1 4 0] [4] [a__from](x1) = [0 1 0] x1 + [5] [0 0 0] [1] [1 0 0] [0 0 1] [0] [cons](x1, x2) = [0 1 0] x1 + [0 0 0] x2 + [4] [0 0 0] [0 0 1] [1] [1 2 0] [4] [mark](x1) = [0 1 0] x1 + [1] [2 0 1] [1] [1 4 0] [0] [from](x1) = [0 1 0] x1 + [5] [0 0 0] [0] [1 0 0] [0] [s](x1) = [0 1 0] x1 + [1] [0 0 1] [5] [0 0 4] [0] [a__length](x1) = [0 0 4] x1 + [3] [0 0 6] [0] [0] [nil] = [0] [0] [0] [0] = [0] [0] [0 0 4] [0] [a__length1](x1) = [0 0 4] x1 + [6] [0 0 6] [0] [0 0 3] [0] [length](x1) = [0 0 4] x1 + [2] [0 0 0] [0] [0 0 4] [0] [length1](x1) = [0 0 4] x1 + [5] [0 0 0] [0] The order satisfies the following ordering constraints: [a__from(X)] = [1 4 0] [4] [0 1 0] X + [5] [0 0 0] [1] >= [1 2 0] [4] [0 1 0] X + [5] [0 0 0] [1] = [cons(mark(X), from(s(X)))] [a__from(X)] = [1 4 0] [4] [0 1 0] X + [5] [0 0 0] [1] > [1 4 0] [0] [0 1 0] X + [5] [0 0 0] [0] = [from(X)] [mark(cons(X1, X2))] = [1 2 0] [0 0 1] [12] [0 1 0] X1 + [0 0 0] X2 + [5] [2 0 0] [0 0 3] [2] > [1 2 0] [0 0 1] [4] [0 1 0] X1 + [0 0 0] X2 + [5] [0 0 0] [0 0 1] [1] = [cons(mark(X1), X2)] [mark(from(X))] = [1 6 0] [14] [0 1 0] X + [6] [2 8 0] [1] > [1 6 0] [12] [0 1 0] X + [6] [0 0 0] [1] = [a__from(mark(X))] [mark(s(X))] = [1 2 0] [6] [0 1 0] X + [2] [2 0 1] [6] > [1 2 0] [4] [0 1 0] X + [2] [2 0 1] [6] = [s(mark(X))] [mark(nil())] = [4] [1] [1] > [0] [0] [0] = [nil()] [mark(0())] = [4] [1] [1] > [0] [0] [0] = [0()] [mark(length(X))] = [0 0 11] [8] [0 0 4] X + [3] [0 0 6] [1] > [0 0 4] [0] [0 0 4] X + [3] [0 0 6] [0] = [a__length(X)] [mark(length1(X))] = [0 0 12] [14] [0 0 4] X + [6] [0 0 8] [1] > [0 0 4] [0] [0 0 4] X + [6] [0 0 6] [0] = [a__length1(X)] [a__length(X)] = [0 0 4] [0] [0 0 4] X + [3] [0 0 6] [0] >= [0 0 3] [0] [0 0 4] X + [2] [0 0 0] [0] = [length(X)] [a__length(cons(X, Y))] = [0 0 4] [4] [0 0 4] Y + [7] [0 0 6] [6] > [0 0 4] [0] [0 0 4] Y + [7] [0 0 6] [5] = [s(a__length1(Y))] [a__length(nil())] = [0] [3] [0] >= [0] [0] [0] = [0()] [a__length1(X)] = [0 0 4] [0] [0 0 4] X + [6] [0 0 6] [0] >= [0 0 4] [0] [0 0 4] X + [3] [0 0 6] [0] = [a__length(X)] [a__length1(X)] = [0 0 4] [0] [0 0 4] X + [6] [0 0 6] [0] >= [0 0 4] [0] [0 0 4] X + [5] [0 0 0] [0] = [length1(X)] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak Trs: { a__from(X) -> cons(mark(X), from(s(X))) , a__from(X) -> from(X) , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(from(X)) -> a__from(mark(X)) , mark(s(X)) -> s(mark(X)) , mark(nil()) -> nil() , mark(0()) -> 0() , mark(length(X)) -> a__length(X) , mark(length1(X)) -> a__length1(X) , a__length(X) -> length(X) , a__length(cons(X, Y)) -> s(a__length1(Y)) , a__length(nil()) -> 0() , a__length1(X) -> a__length(X) , a__length1(X) -> length1(X) } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) Empty rules are trivially bounded Hurray, we answered YES(O(1),O(n^2))