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: { perfectp(0()) -> false() , perfectp(s(x)) -> f(x, s(0()), s(x), s(x)) , f(0(), y, 0(), u) -> true() , f(0(), y, s(z), u) -> false() , f(s(x), 0(), z, u) -> f(x, u, minus(z, s(x)), u) , f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)) } Obligation: runtime complexity Answer: YES(O(1),O(n^1)) We add the following weak dependency pairs: Strict DPs: { perfectp^#(0()) -> c_1() , perfectp^#(s(x)) -> c_2(f^#(x, s(0()), s(x), s(x))) , f^#(0(), y, 0(), u) -> c_3() , f^#(0(), y, s(z), u) -> c_4() , f^#(s(x), 0(), z, u) -> c_5(f^#(x, u, minus(z, s(x)), u)) , f^#(s(x), s(y), z, u) -> c_6(x, y, f^#(s(x), minus(y, x), z, u), f^#(x, u, z, u)) } 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: { perfectp^#(0()) -> c_1() , perfectp^#(s(x)) -> c_2(f^#(x, s(0()), s(x), s(x))) , f^#(0(), y, 0(), u) -> c_3() , f^#(0(), y, s(z), u) -> c_4() , f^#(s(x), 0(), z, u) -> c_5(f^#(x, u, minus(z, s(x)), u)) , f^#(s(x), s(y), z, u) -> c_6(x, y, f^#(s(x), minus(y, x), z, u), f^#(x, u, z, u)) } Strict Trs: { perfectp(0()) -> false() , perfectp(s(x)) -> f(x, s(0()), s(x), s(x)) , f(0(), y, 0(), u) -> true() , f(0(), y, s(z), u) -> false() , f(s(x), 0(), z, u) -> f(x, u, minus(z, s(x)), u) , f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)) } Obligation: 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: { perfectp^#(0()) -> c_1() , perfectp^#(s(x)) -> c_2(f^#(x, s(0()), s(x), s(x))) , f^#(0(), y, 0(), u) -> c_3() , f^#(0(), y, s(z), u) -> c_4() , f^#(s(x), 0(), z, u) -> c_5(f^#(x, u, minus(z, s(x)), u)) , f^#(s(x), s(y), z, u) -> c_6(x, y, f^#(s(x), minus(y, x), z, u), f^#(x, u, z, u)) } Obligation: runtime complexity Answer: YES(O(1),O(n^1)) The weightgap principle applies (using the following constant growth matrix-interpretation) The following argument positions are usable: Uargs(c_2) = {1}, Uargs(c_5) = {1}, Uargs(c_6) = {4} TcT has computed the following constructor-restricted matrix interpretation. [0] = [0] [0] [s](x1) = [0] [0] [minus](x1, x2) = [1 0] x1 + [0] [0 0] [0] [perfectp^#](x1) = [0] [0] [c_1] = [1] [0] [c_2](x1) = [1 0] x1 + [0] [0 1] [0] [f^#](x1, x2, x3, x4) = [1] [0] [c_3] = [0] [0] [c_4] = [0] [0] [c_5](x1) = [1 0] x1 + [0] [0 1] [0] [c_6](x1, x2, x3, x4) = [0 0] x1 + [0 0] x2 + [1 0] x4 + [0] [1 2] [2 0] [0 1] [0] The following symbols are considered usable {perfectp^#, f^#} The order satisfies the following ordering constraints: [perfectp^#(0())] = [0] [0] ? [1] [0] = [c_1()] [perfectp^#(s(x))] = [0] [0] ? [1] [0] = [c_2(f^#(x, s(0()), s(x), s(x)))] [f^#(0(), y, 0(), u)] = [1] [0] > [0] [0] = [c_3()] [f^#(0(), y, s(z), u)] = [1] [0] > [0] [0] = [c_4()] [f^#(s(x), 0(), z, u)] = [1] [0] >= [1] [0] = [c_5(f^#(x, u, minus(z, s(x)), u))] [f^#(s(x), s(y), z, u)] = [1] [0] ? [0 0] x + [0 0] y + [1] [1 2] [2 0] [0] = [c_6(x, y, f^#(s(x), minus(y, x), z, u), f^#(x, u, z, u))] 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 DPs: { perfectp^#(0()) -> c_1() , perfectp^#(s(x)) -> c_2(f^#(x, s(0()), s(x), s(x))) , f^#(s(x), 0(), z, u) -> c_5(f^#(x, u, minus(z, s(x)), u)) , f^#(s(x), s(y), z, u) -> c_6(x, y, f^#(s(x), minus(y, x), z, u), f^#(x, u, z, u)) } Weak DPs: { f^#(0(), y, 0(), u) -> c_3() , f^#(0(), y, s(z), u) -> c_4() } Obligation: runtime complexity Answer: YES(O(1),O(n^1)) We estimate the number of application of {1} by applications of Pre({1}) = {4}. Here rules are labeled as follows: DPs: { 1: perfectp^#(0()) -> c_1() , 2: perfectp^#(s(x)) -> c_2(f^#(x, s(0()), s(x), s(x))) , 3: f^#(s(x), 0(), z, u) -> c_5(f^#(x, u, minus(z, s(x)), u)) , 4: f^#(s(x), s(y), z, u) -> c_6(x, y, f^#(s(x), minus(y, x), z, u), f^#(x, u, z, u)) , 5: f^#(0(), y, 0(), u) -> c_3() , 6: f^#(0(), y, s(z), u) -> c_4() } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { perfectp^#(s(x)) -> c_2(f^#(x, s(0()), s(x), s(x))) , f^#(s(x), 0(), z, u) -> c_5(f^#(x, u, minus(z, s(x)), u)) , f^#(s(x), s(y), z, u) -> c_6(x, y, f^#(s(x), minus(y, x), z, u), f^#(x, u, z, u)) } Weak DPs: { perfectp^#(0()) -> c_1() , f^#(0(), y, 0(), u) -> c_3() , f^#(0(), y, s(z), u) -> c_4() } Obligation: 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. { perfectp^#(0()) -> c_1() , f^#(0(), y, 0(), u) -> c_3() , f^#(0(), y, s(z), u) -> c_4() } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { perfectp^#(s(x)) -> c_2(f^#(x, s(0()), s(x), s(x))) , f^#(s(x), 0(), z, u) -> c_5(f^#(x, u, minus(z, s(x)), u)) , f^#(s(x), s(y), z, u) -> c_6(x, y, f^#(s(x), minus(y, x), z, u), f^#(x, u, z, u)) } Obligation: 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^#(s(x), s(y), z, u) -> c_6(x, y, f^#(s(x), minus(y, x), z, u), f^#(x, u, z, u)) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { perfectp^#(s(x)) -> c_1(f^#(x, s(0()), s(x), s(x))) , f^#(s(x), 0(), z, u) -> c_2(f^#(x, u, minus(z, s(x)), u)) , f^#(s(x), s(y), z, u) -> c_3(x, y, f^#(x, u, z, u)) } Obligation: 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: perfectp^#(s(x)) -> c_1(f^#(x, s(0()), s(x), s(x))) } Sub-proof: ---------- The following argument positions are usable: Uargs(c_1) = {1}, Uargs(c_2) = {1}, Uargs(c_3) = {3} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [perfectp](x1) = [7] x1 + [0] [0] = [0] [false] = [0] [s](x1) = [1] x1 + [0] [f](x1, x2, x3, x4) = [7] x1 + [7] x2 + [7] x3 + [7] x4 + [0] [true] = [0] [minus](x1, x2) = [1] x1 + [4] [if](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [0] [le](x1, x2) = [1] x1 + [1] x2 + [0] [perfectp^#](x1) = [4] x1 + [7] [c_1] = [0] [c_2](x1) = [7] x1 + [0] [f^#](x1, x2, x3, x4) = [0] [c_3] = [0] [c_4] = [0] [c_5](x1) = [7] x1 + [0] [c_6](x1, x2, x3, x4) = [7] x1 + [7] x2 + [7] x3 + [7] x4 + [0] [c] = [0] [c_1](x1) = [4] x1 + [2] [c_2](x1) = [4] x1 + [0] [c_3](x1, x2, x3) = [2] x3 + [0] The following symbols are considered usable {perfectp^#, f^#} The order satisfies the following ordering constraints: [perfectp^#(s(x))] = [4] x + [7] > [2] = [c_1(f^#(x, s(0()), s(x), s(x)))] [f^#(s(x), 0(), z, u)] = [0] >= [0] = [c_2(f^#(x, u, minus(z, s(x)), u))] [f^#(s(x), s(y), z, u)] = [0] >= [0] = [c_3(x, y, f^#(x, u, z, u))] 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(n^1)). Strict DPs: { f^#(s(x), 0(), z, u) -> c_2(f^#(x, u, minus(z, s(x)), u)) , f^#(s(x), s(y), z, u) -> c_3(x, y, f^#(x, u, z, u)) } Weak DPs: { perfectp^#(s(x)) -> c_1(f^#(x, s(0()), s(x), s(x))) } Obligation: 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^#(s(x), 0(), z, u) -> c_2(f^#(x, u, minus(z, s(x)), u)) , 2: f^#(s(x), s(y), z, u) -> c_3(x, y, f^#(x, u, z, u)) , 3: perfectp^#(s(x)) -> c_1(f^#(x, s(0()), s(x), s(x))) } Sub-proof: ---------- The following argument positions are usable: Uargs(c_1) = {1}, Uargs(c_2) = {1}, Uargs(c_3) = {3} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [perfectp](x1) = [7] x1 + [0] [0] = [0] [false] = [0] [s](x1) = [1] x1 + [4] [f](x1, x2, x3, x4) = [7] x1 + [7] x2 + [7] x3 + [7] x4 + [0] [true] = [0] [minus](x1, x2) = [1] x1 + [4] [if](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [0] [le](x1, x2) = [1] x1 + [1] x2 + [0] [perfectp^#](x1) = [1] x1 + [3] [c_1] = [0] [c_2](x1) = [7] x1 + [0] [f^#](x1, x2, x3, x4) = [1] x1 + [0] [c_3] = [0] [c_4] = [0] [c_5](x1) = [7] x1 + [0] [c_6](x1, x2, x3, x4) = [7] x1 + [7] x2 + [7] x3 + [7] x4 + [0] [c] = [0] [c_1](x1) = [1] x1 + [3] [c_2](x1) = [1] x1 + [3] [c_3](x1, x2, x3) = [1] x3 + [1] The following symbols are considered usable {perfectp^#, f^#} The order satisfies the following ordering constraints: [perfectp^#(s(x))] = [1] x + [7] > [1] x + [3] = [c_1(f^#(x, s(0()), s(x), s(x)))] [f^#(s(x), 0(), z, u)] = [1] x + [4] > [1] x + [3] = [c_2(f^#(x, u, minus(z, s(x)), u))] [f^#(s(x), s(y), z, u)] = [1] x + [4] > [1] x + [1] = [c_3(x, y, f^#(x, u, z, u))] 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: { perfectp^#(s(x)) -> c_1(f^#(x, s(0()), s(x), s(x))) , f^#(s(x), 0(), z, u) -> c_2(f^#(x, u, minus(z, s(x)), u)) , f^#(s(x), s(y), z, u) -> c_3(x, y, f^#(x, u, z, u)) } Obligation: 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. { perfectp^#(s(x)) -> c_1(f^#(x, s(0()), s(x), s(x))) , f^#(s(x), 0(), z, u) -> c_2(f^#(x, u, minus(z, s(x)), u)) , f^#(s(x), s(y), z, u) -> c_3(x, y, f^#(x, u, z, u)) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Rules: Empty Obligation: runtime complexity Answer: YES(O(1),O(1)) Empty rules are trivially bounded Hurray, we answered YES(O(1),O(n^1))