We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { f(x, 0()) -> x , f(f(x, y), z) -> f(x, f(y, z)) , f(0(), y) -> y , f(i(x), y) -> i(x) , f(g(x, y), z) -> g(f(x, z), f(y, z)) , f(1(), g(x, y)) -> x , f(2(), g(x, y)) -> y } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We add the following dependency tuples: Strict DPs: { f^#(x, 0()) -> c_1() , f^#(f(x, y), z) -> c_2(f^#(x, f(y, z)), f^#(y, z)) , f^#(0(), y) -> c_3() , f^#(i(x), y) -> c_4() , f^#(g(x, y), z) -> c_5(f^#(x, z), f^#(y, z)) , f^#(1(), g(x, y)) -> c_6() , f^#(2(), g(x, y)) -> c_7() } and mark the set of starting terms. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { f^#(x, 0()) -> c_1() , f^#(f(x, y), z) -> c_2(f^#(x, f(y, z)), f^#(y, z)) , f^#(0(), y) -> c_3() , f^#(i(x), y) -> c_4() , f^#(g(x, y), z) -> c_5(f^#(x, z), f^#(y, z)) , f^#(1(), g(x, y)) -> c_6() , f^#(2(), g(x, y)) -> c_7() } Weak Trs: { f(x, 0()) -> x , f(f(x, y), z) -> f(x, f(y, z)) , f(0(), y) -> y , f(i(x), y) -> i(x) , f(g(x, y), z) -> g(f(x, z), f(y, z)) , f(1(), g(x, y)) -> x , f(2(), g(x, y)) -> y } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We estimate the number of application of {1,3,4,6,7} by applications of Pre({1,3,4,6,7}) = {2,5}. Here rules are labeled as follows: DPs: { 1: f^#(x, 0()) -> c_1() , 2: f^#(f(x, y), z) -> c_2(f^#(x, f(y, z)), f^#(y, z)) , 3: f^#(0(), y) -> c_3() , 4: f^#(i(x), y) -> c_4() , 5: f^#(g(x, y), z) -> c_5(f^#(x, z), f^#(y, z)) , 6: f^#(1(), g(x, y)) -> c_6() , 7: f^#(2(), g(x, y)) -> c_7() } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { f^#(f(x, y), z) -> c_2(f^#(x, f(y, z)), f^#(y, z)) , f^#(g(x, y), z) -> c_5(f^#(x, z), f^#(y, z)) } Weak DPs: { f^#(x, 0()) -> c_1() , f^#(0(), y) -> c_3() , f^#(i(x), y) -> c_4() , f^#(1(), g(x, y)) -> c_6() , f^#(2(), g(x, y)) -> c_7() } Weak Trs: { f(x, 0()) -> x , f(f(x, y), z) -> f(x, f(y, z)) , f(0(), y) -> y , f(i(x), y) -> i(x) , f(g(x, y), z) -> g(f(x, z), f(y, z)) , f(1(), g(x, y)) -> x , f(2(), g(x, y)) -> y } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) The following weak DPs constitute a sub-graph of the DG that is closed under successors. The DPs are removed. { f^#(x, 0()) -> c_1() , f^#(0(), y) -> c_3() , f^#(i(x), y) -> c_4() , f^#(1(), g(x, y)) -> c_6() , f^#(2(), g(x, y)) -> c_7() } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { f^#(f(x, y), z) -> c_2(f^#(x, f(y, z)), f^#(y, z)) , f^#(g(x, y), z) -> c_5(f^#(x, z), f^#(y, z)) } Weak Trs: { f(x, 0()) -> x , f(f(x, y), z) -> f(x, f(y, z)) , f(0(), y) -> y , f(i(x), y) -> i(x) , f(g(x, y), z) -> g(f(x, z), f(y, z)) , f(1(), g(x, y)) -> x , f(2(), g(x, y)) -> y } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) Due to missing edges in the dependency-graph, the right-hand sides of following rules could be simplified: { f^#(f(x, y), z) -> c_2(f^#(x, f(y, z)), f^#(y, z)) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { f^#(f(x, y), z) -> c_1(f^#(y, z)) , f^#(g(x, y), z) -> c_2(f^#(x, z), f^#(y, z)) } Weak Trs: { f(x, 0()) -> x , f(f(x, y), z) -> f(x, f(y, z)) , f(0(), y) -> y , f(i(x), y) -> i(x) , f(g(x, y), z) -> g(f(x, z), f(y, z)) , f(1(), g(x, y)) -> x , f(2(), g(x, y)) -> y } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) 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^1)). Strict DPs: { f^#(f(x, y), z) -> c_1(f^#(y, z)) , f^#(g(x, y), z) -> c_2(f^#(x, z), f^#(y, z)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. DPs: { 1: f^#(f(x, y), z) -> c_1(f^#(y, z)) , 2: f^#(g(x, y), z) -> c_2(f^#(x, z), f^#(y, z)) } Sub-proof: ---------- The following argument positions are usable: Uargs(c_1) = {1}, Uargs(c_2) = {1, 2} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [f](x1, x2) = [2] x1 + [2] x2 + [2] [g](x1, x2) = [1] x1 + [1] x2 + [2] [f^#](x1, x2) = [4] x1 + [0] [c_1](x1) = [1] x1 + [1] [c_2](x1, x2) = [1] x1 + [1] x2 + [0] The order satisfies the following ordering constraints: [f^#(f(x, y), z)] = [8] y + [8] x + [8] > [4] y + [1] = [c_1(f^#(y, z))] [f^#(g(x, y), z)] = [4] y + [4] x + [8] > [4] y + [4] x + [0] = [c_2(f^#(x, z), f^#(y, z))] The strictly oriented rules are moved into the weak component. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak DPs: { f^#(f(x, y), z) -> c_1(f^#(y, z)) , f^#(g(x, y), z) -> c_2(f^#(x, z), f^#(y, z)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) The following weak DPs constitute a sub-graph of the DG that is closed under successors. The DPs are removed. { f^#(f(x, y), z) -> c_1(f^#(y, z)) , f^#(g(x, y), z) -> c_2(f^#(x, z), f^#(y, z)) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Rules: Empty Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) Empty rules are trivially bounded Hurray, we answered YES(O(1),O(n^1))