YES(O(1),O(n^3)) We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^3)). Strict Trs: { f(nil()) -> nil() , f(.(nil(), y)) -> .(nil(), f(y)) , f(.(.(x, y), z)) -> f(.(x, .(y, z))) , g(nil()) -> nil() , g(.(x, nil())) -> .(g(x), nil()) , g(.(x, .(y, z))) -> g(.(.(x, y), z)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^3)) We add following weak dependency pairs: Strict DPs: { f^#(nil()) -> c_1() , f^#(.(nil(), y)) -> c_2(f^#(y)) , f^#(.(.(x, y), z)) -> c_3(f^#(.(x, .(y, z)))) , g^#(nil()) -> c_4() , g^#(.(x, nil())) -> c_5(g^#(x)) , g^#(.(x, .(y, z))) -> c_6(g^#(.(.(x, y), z))) } and mark the set of starting terms. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^3)). Strict DPs: { f^#(nil()) -> c_1() , f^#(.(nil(), y)) -> c_2(f^#(y)) , f^#(.(.(x, y), z)) -> c_3(f^#(.(x, .(y, z)))) , g^#(nil()) -> c_4() , g^#(.(x, nil())) -> c_5(g^#(x)) , g^#(.(x, .(y, z))) -> c_6(g^#(.(.(x, y), z))) } Strict Trs: { f(nil()) -> nil() , f(.(nil(), y)) -> .(nil(), f(y)) , f(.(.(x, y), z)) -> f(.(x, .(y, z))) , g(nil()) -> nil() , g(.(x, nil())) -> .(g(x), nil()) , g(.(x, .(y, z))) -> g(.(.(x, y), z)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^3)) No rule is usable, rules are removed from the input problem. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^3)). Strict DPs: { f^#(nil()) -> c_1() , f^#(.(nil(), y)) -> c_2(f^#(y)) , f^#(.(.(x, y), z)) -> c_3(f^#(.(x, .(y, z)))) , g^#(nil()) -> c_4() , g^#(.(x, nil())) -> c_5(g^#(x)) , g^#(.(x, .(y, z))) -> c_6(g^#(.(.(x, y), z))) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^3)) The weightgap principle applies (using the following constant growth matrix-interpretation) The following argument positions are usable: Uargs(c_2) = {1}, Uargs(c_3) = {1}, Uargs(c_5) = {1}, Uargs(c_6) = {1} TcT has computed following constructor-restricted matrix interpretation. [nil] = [1] [.](x1, x2) = [1] x2 + [0] [f^#](x1) = [1] x1 + [0] [c_1] = [0] [c_2](x1) = [1] x1 + [2] [c_3](x1) = [1] x1 + [1] [g^#](x1) = [0] [c_4] = [1] [c_5](x1) = [1] x1 + [2] [c_6](x1) = [1] x1 + [1] This order satisfies following ordering constraints: 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(n^3)). Strict DPs: { f^#(.(nil(), y)) -> c_2(f^#(y)) , f^#(.(.(x, y), z)) -> c_3(f^#(.(x, .(y, z)))) , g^#(nil()) -> c_4() , g^#(.(x, nil())) -> c_5(g^#(x)) , g^#(.(x, .(y, z))) -> c_6(g^#(.(.(x, y), z))) } Weak DPs: { f^#(nil()) -> c_1() } Obligation: innermost runtime complexity Answer: YES(?,O(n^3)) We estimate the number of application of {3} by applications of Pre({3}) = {4}. Here rules are labeled as follows: DPs: { 1: f^#(.(nil(), y)) -> c_2(f^#(y)) , 2: f^#(.(.(x, y), z)) -> c_3(f^#(.(x, .(y, z)))) , 3: g^#(nil()) -> c_4() , 4: g^#(.(x, nil())) -> c_5(g^#(x)) , 5: g^#(.(x, .(y, z))) -> c_6(g^#(.(.(x, y), z))) , 6: f^#(nil()) -> c_1() } We are left with following problem, upon which TcT provides the certificate YES(?,O(n^3)). Strict DPs: { f^#(.(nil(), y)) -> c_2(f^#(y)) , f^#(.(.(x, y), z)) -> c_3(f^#(.(x, .(y, z)))) , g^#(.(x, nil())) -> c_5(g^#(x)) , g^#(.(x, .(y, z))) -> c_6(g^#(.(.(x, y), z))) } Weak DPs: { f^#(nil()) -> c_1() , g^#(nil()) -> c_4() } Obligation: innermost runtime complexity Answer: YES(?,O(n^3)) The following weak DPs constitute a sub-graph of the DG that is closed under successors. The DPs are removed. { f^#(nil()) -> c_1() , g^#(nil()) -> c_4() } We are left with following problem, upon which TcT provides the certificate YES(?,O(n^3)). Strict DPs: { f^#(.(nil(), y)) -> c_2(f^#(y)) , f^#(.(.(x, y), z)) -> c_3(f^#(.(x, .(y, z)))) , g^#(.(x, nil())) -> c_5(g^#(x)) , g^#(.(x, .(y, z))) -> c_6(g^#(.(.(x, y), z))) } Obligation: innermost runtime complexity Answer: YES(?,O(n^3)) The following argument positions are usable: Uargs(c_2) = {1}, Uargs(c_3) = {1}, Uargs(c_5) = {1}, Uargs(c_6) = {1} TcT has computed following constructor-based matrix interpretation satisfying not(EDA). [0] [nil] = [0] [0] [1 0 0] [1 0 0] [1] [.](x1, x2) = [1 0 0] x1 + [0 1 0] x2 + [0] [0 0 1] [1 0 0] [0] [1 1 0] [0] [f^#](x1) = [0 0 0] x1 + [0] [0 0 0] [0] [1 0 0] [0] [c_2](x1) = [0 0 0] x1 + [0] [0 0 0] [0] [1 0 0] [0] [c_3](x1) = [0 0 0] x1 + [0] [0 0 0] [0] [1 0 1] [0] [g^#](x1) = [0 0 0] x1 + [0] [0 0 0] [0] [1 0 0] [0] [c_5](x1) = [0 0 0] x1 + [0] [0 0 0] [0] [1 0 0] [0] [c_6](x1) = [0 0 0] x1 + [0] [0 0 0] [0] This order satisfies following ordering constraints: [f^#(.(nil(), y))] = [1 1 0] [1] [0 0 0] y + [0] [0 0 0] [0] > [1 1 0] [0] [0 0 0] y + [0] [0 0 0] [0] = [c_2(f^#(y))] [f^#(.(.(x, y), z))] = [2 0 0] [2 0 0] [1 1 0] [3] [0 0 0] y + [0 0 0] x + [0 0 0] z + [0] [0 0 0] [0 0 0] [0 0 0] [0] > [2 0 0] [2 0 0] [1 1 0] [2] [0 0 0] y + [0 0 0] x + [0 0 0] z + [0] [0 0 0] [0 0 0] [0 0 0] [0] = [c_3(f^#(.(x, .(y, z))))] [g^#(.(x, nil()))] = [1 0 1] [1] [0 0 0] x + [0] [0 0 0] [0] > [1 0 1] [0] [0 0 0] x + [0] [0 0 0] [0] = [c_5(g^#(x))] [g^#(.(x, .(y, z)))] = [2 0 0] [1 0 1] [2 0 0] [3] [0 0 0] y + [0 0 0] x + [0 0 0] z + [0] [0 0 0] [0 0 0] [0 0 0] [0] > [2 0 0] [1 0 1] [2 0 0] [2] [0 0 0] y + [0 0 0] x + [0 0 0] z + [0] [0 0 0] [0 0 0] [0 0 0] [0] = [c_6(g^#(.(.(x, y), z)))] Hurray, we answered YES(O(1),O(n^3))