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: { D(t()) -> 1() , D(constant()) -> 0() , D(+(x, y)) -> +(D(x), D(y)) , D(*(x, y)) -> +(*(y, D(x)), *(x, D(y))) , D(-(x, y)) -> -(D(x), D(y)) , D(minus(x)) -> minus(D(x)) , D(div(x, y)) -> -(div(D(x), y), div(*(x, D(y)), pow(y, 2()))) , D(pow(x, y)) -> +(*(*(y, pow(x, -(y, 1()))), D(x)), *(*(pow(x, y), ln(x)), D(y))) , D(ln(x)) -> div(D(x), x) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We add the following innermost weak dependency pairs: Strict DPs: { D^#(t()) -> c_1() , D^#(constant()) -> c_2() , D^#(+(x, y)) -> c_3(D^#(x), D^#(y)) , D^#(*(x, y)) -> c_4(D^#(x), D^#(y)) , D^#(-(x, y)) -> c_5(D^#(x), D^#(y)) , D^#(minus(x)) -> c_6(D^#(x)) , D^#(div(x, y)) -> c_7(D^#(x), D^#(y)) , D^#(pow(x, y)) -> c_8(D^#(x), D^#(y)) , D^#(ln(x)) -> c_9(D^#(x)) } 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: { D^#(t()) -> c_1() , D^#(constant()) -> c_2() , D^#(+(x, y)) -> c_3(D^#(x), D^#(y)) , D^#(*(x, y)) -> c_4(D^#(x), D^#(y)) , D^#(-(x, y)) -> c_5(D^#(x), D^#(y)) , D^#(minus(x)) -> c_6(D^#(x)) , D^#(div(x, y)) -> c_7(D^#(x), D^#(y)) , D^#(pow(x, y)) -> c_8(D^#(x), D^#(y)) , D^#(ln(x)) -> c_9(D^#(x)) } Strict Trs: { D(t()) -> 1() , D(constant()) -> 0() , D(+(x, y)) -> +(D(x), D(y)) , D(*(x, y)) -> +(*(y, D(x)), *(x, D(y))) , D(-(x, y)) -> -(D(x), D(y)) , D(minus(x)) -> minus(D(x)) , D(div(x, y)) -> -(div(D(x), y), div(*(x, D(y)), pow(y, 2()))) , D(pow(x, y)) -> +(*(*(y, pow(x, -(y, 1()))), D(x)), *(*(pow(x, y), ln(x)), D(y))) , D(ln(x)) -> div(D(x), x) } 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: { D^#(t()) -> c_1() , D^#(constant()) -> c_2() , D^#(+(x, y)) -> c_3(D^#(x), D^#(y)) , D^#(*(x, y)) -> c_4(D^#(x), D^#(y)) , D^#(-(x, y)) -> c_5(D^#(x), D^#(y)) , D^#(minus(x)) -> c_6(D^#(x)) , D^#(div(x, y)) -> c_7(D^#(x), D^#(y)) , D^#(pow(x, y)) -> c_8(D^#(x), D^#(y)) , D^#(ln(x)) -> c_9(D^#(x)) } Obligation: innermost 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_3) = {1, 2}, Uargs(c_4) = {1, 2}, Uargs(c_5) = {1, 2}, Uargs(c_6) = {1}, Uargs(c_7) = {1, 2}, Uargs(c_8) = {1, 2}, Uargs(c_9) = {1} TcT has computed the following constructor-restricted matrix interpretation. [t] = [0] [0] [constant] = [0] [0] [+](x1, x2) = [1 0] x1 + [1 0] x2 + [0] [0 0] [0 0] [0] [*](x1, x2) = [1 0] x1 + [1 0] x2 + [0] [0 0] [0 0] [0] [-](x1, x2) = [1 0] x1 + [1 0] x2 + [0] [0 0] [0 0] [0] [minus](x1) = [1 0] x1 + [0] [0 0] [0] [div](x1, x2) = [1 0] x1 + [1 0] x2 + [0] [0 0] [0 0] [0] [pow](x1, x2) = [1 0] x1 + [1 0] x2 + [0] [0 0] [0 0] [0] [ln](x1) = [1 0] x1 + [0] [0 0] [0] [D^#](x1) = [1] [0] [c_1] = [0] [0] [c_2] = [0] [0] [c_3](x1, x2) = [1 0] x1 + [1 0] x2 + [2] [0 1] [0 1] [0] [c_4](x1, x2) = [1 0] x1 + [1 0] x2 + [2] [0 1] [0 1] [2] [c_5](x1, x2) = [1 0] x1 + [1 0] x2 + [2] [0 1] [0 1] [2] [c_6](x1) = [1 0] x1 + [1] [0 1] [0] [c_7](x1, x2) = [1 0] x1 + [1 0] x2 + [2] [0 1] [0 1] [1] [c_8](x1, x2) = [1 0] x1 + [1 0] x2 + [2] [0 1] [0 1] [2] [c_9](x1) = [1 0] x1 + [1] [0 1] [0] The following symbols are considered usable {D^#} The order satisfies the following ordering constraints: [D^#(t())] = [1] [0] > [0] [0] = [c_1()] [D^#(constant())] = [1] [0] > [0] [0] = [c_2()] [D^#(+(x, y))] = [1] [0] ? [4] [0] = [c_3(D^#(x), D^#(y))] [D^#(*(x, y))] = [1] [0] ? [4] [2] = [c_4(D^#(x), D^#(y))] [D^#(-(x, y))] = [1] [0] ? [4] [2] = [c_5(D^#(x), D^#(y))] [D^#(minus(x))] = [1] [0] ? [2] [0] = [c_6(D^#(x))] [D^#(div(x, y))] = [1] [0] ? [4] [1] = [c_7(D^#(x), D^#(y))] [D^#(pow(x, y))] = [1] [0] ? [4] [2] = [c_8(D^#(x), D^#(y))] [D^#(ln(x))] = [1] [0] ? [2] [0] = [c_9(D^#(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 DPs: { D^#(+(x, y)) -> c_3(D^#(x), D^#(y)) , D^#(*(x, y)) -> c_4(D^#(x), D^#(y)) , D^#(-(x, y)) -> c_5(D^#(x), D^#(y)) , D^#(minus(x)) -> c_6(D^#(x)) , D^#(div(x, y)) -> c_7(D^#(x), D^#(y)) , D^#(pow(x, y)) -> c_8(D^#(x), D^#(y)) , D^#(ln(x)) -> c_9(D^#(x)) } Weak DPs: { D^#(t()) -> c_1() , D^#(constant()) -> c_2() } 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. { D^#(t()) -> c_1() , D^#(constant()) -> c_2() } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { D^#(+(x, y)) -> c_3(D^#(x), D^#(y)) , D^#(*(x, y)) -> c_4(D^#(x), D^#(y)) , D^#(-(x, y)) -> c_5(D^#(x), D^#(y)) , D^#(minus(x)) -> c_6(D^#(x)) , D^#(div(x, y)) -> c_7(D^#(x), D^#(y)) , D^#(pow(x, y)) -> c_8(D^#(x), D^#(y)) , D^#(ln(x)) -> c_9(D^#(x)) } 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: { 3: D^#(-(x, y)) -> c_5(D^#(x), D^#(y)) , 4: D^#(minus(x)) -> c_6(D^#(x)) , 5: D^#(div(x, y)) -> c_7(D^#(x), D^#(y)) , 6: D^#(pow(x, y)) -> c_8(D^#(x), D^#(y)) , 7: D^#(ln(x)) -> c_9(D^#(x)) } Sub-proof: ---------- The following argument positions are usable: Uargs(c_3) = {1, 2}, Uargs(c_4) = {1, 2}, Uargs(c_5) = {1, 2}, Uargs(c_6) = {1}, Uargs(c_7) = {1, 2}, Uargs(c_8) = {1, 2}, Uargs(c_9) = {1} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [D](x1) = [7] x1 + [0] [t] = [0] [1] = [0] [constant] = [0] [0] = [0] [+](x1, x2) = [1] x1 + [1] x2 + [0] [*](x1, x2) = [1] x1 + [1] x2 + [0] [-](x1, x2) = [1] x1 + [1] x2 + [4] [minus](x1) = [1] x1 + [4] [div](x1, x2) = [1] x1 + [1] x2 + [4] [pow](x1, x2) = [1] x1 + [1] x2 + [4] [2] = [0] [ln](x1) = [1] x1 + [4] [D^#](x1) = [2] x1 + [0] [c_1] = [0] [c_2] = [0] [c_3](x1, x2) = [1] x1 + [1] x2 + [0] [c_4](x1, x2) = [1] x1 + [1] x2 + [0] [c_5](x1, x2) = [1] x1 + [1] x2 + [3] [c_6](x1) = [1] x1 + [7] [c_7](x1, x2) = [1] x1 + [1] x2 + [3] [c_8](x1, x2) = [1] x1 + [1] x2 + [3] [c_9](x1) = [1] x1 + [1] The following symbols are considered usable {D^#} The order satisfies the following ordering constraints: [D^#(+(x, y))] = [2] x + [2] y + [0] >= [2] x + [2] y + [0] = [c_3(D^#(x), D^#(y))] [D^#(*(x, y))] = [2] x + [2] y + [0] >= [2] x + [2] y + [0] = [c_4(D^#(x), D^#(y))] [D^#(-(x, y))] = [2] x + [2] y + [8] > [2] x + [2] y + [3] = [c_5(D^#(x), D^#(y))] [D^#(minus(x))] = [2] x + [8] > [2] x + [7] = [c_6(D^#(x))] [D^#(div(x, y))] = [2] x + [2] y + [8] > [2] x + [2] y + [3] = [c_7(D^#(x), D^#(y))] [D^#(pow(x, y))] = [2] x + [2] y + [8] > [2] x + [2] y + [3] = [c_8(D^#(x), D^#(y))] [D^#(ln(x))] = [2] x + [8] > [2] x + [1] = [c_9(D^#(x))] 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: { D^#(+(x, y)) -> c_3(D^#(x), D^#(y)) , D^#(*(x, y)) -> c_4(D^#(x), D^#(y)) } Weak DPs: { D^#(-(x, y)) -> c_5(D^#(x), D^#(y)) , D^#(minus(x)) -> c_6(D^#(x)) , D^#(div(x, y)) -> c_7(D^#(x), D^#(y)) , D^#(pow(x, y)) -> c_8(D^#(x), D^#(y)) , D^#(ln(x)) -> c_9(D^#(x)) } 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: D^#(+(x, y)) -> c_3(D^#(x), D^#(y)) , 2: D^#(*(x, y)) -> c_4(D^#(x), D^#(y)) , 5: D^#(div(x, y)) -> c_7(D^#(x), D^#(y)) , 6: D^#(pow(x, y)) -> c_8(D^#(x), D^#(y)) , 7: D^#(ln(x)) -> c_9(D^#(x)) } Sub-proof: ---------- The following argument positions are usable: Uargs(c_3) = {1, 2}, Uargs(c_4) = {1, 2}, Uargs(c_5) = {1, 2}, Uargs(c_6) = {1}, Uargs(c_7) = {1, 2}, Uargs(c_8) = {1, 2}, Uargs(c_9) = {1} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [D](x1) = [7] x1 + [0] [t] = [0] [1] = [0] [constant] = [0] [0] = [0] [+](x1, x2) = [1] x1 + [1] x2 + [4] [*](x1, x2) = [1] x1 + [1] x2 + [4] [-](x1, x2) = [1] x1 + [1] x2 + [0] [minus](x1) = [1] x1 + [0] [div](x1, x2) = [1] x1 + [1] x2 + [4] [pow](x1, x2) = [1] x1 + [1] x2 + [4] [2] = [0] [ln](x1) = [1] x1 + [4] [D^#](x1) = [2] x1 + [0] [c_1] = [0] [c_2] = [0] [c_3](x1, x2) = [1] x1 + [1] x2 + [7] [c_4](x1, x2) = [1] x1 + [1] x2 + [5] [c_5](x1, x2) = [1] x1 + [1] x2 + [0] [c_6](x1) = [1] x1 + [0] [c_7](x1, x2) = [1] x1 + [1] x2 + [7] [c_8](x1, x2) = [1] x1 + [1] x2 + [3] [c_9](x1) = [1] x1 + [5] The following symbols are considered usable {D^#} The order satisfies the following ordering constraints: [D^#(+(x, y))] = [2] x + [2] y + [8] > [2] x + [2] y + [7] = [c_3(D^#(x), D^#(y))] [D^#(*(x, y))] = [2] x + [2] y + [8] > [2] x + [2] y + [5] = [c_4(D^#(x), D^#(y))] [D^#(-(x, y))] = [2] x + [2] y + [0] >= [2] x + [2] y + [0] = [c_5(D^#(x), D^#(y))] [D^#(minus(x))] = [2] x + [0] >= [2] x + [0] = [c_6(D^#(x))] [D^#(div(x, y))] = [2] x + [2] y + [8] > [2] x + [2] y + [7] = [c_7(D^#(x), D^#(y))] [D^#(pow(x, y))] = [2] x + [2] y + [8] > [2] x + [2] y + [3] = [c_8(D^#(x), D^#(y))] [D^#(ln(x))] = [2] x + [8] > [2] x + [5] = [c_9(D^#(x))] 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: { D^#(+(x, y)) -> c_3(D^#(x), D^#(y)) , D^#(*(x, y)) -> c_4(D^#(x), D^#(y)) , D^#(-(x, y)) -> c_5(D^#(x), D^#(y)) , D^#(minus(x)) -> c_6(D^#(x)) , D^#(div(x, y)) -> c_7(D^#(x), D^#(y)) , D^#(pow(x, y)) -> c_8(D^#(x), D^#(y)) , D^#(ln(x)) -> c_9(D^#(x)) } 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. { D^#(+(x, y)) -> c_3(D^#(x), D^#(y)) , D^#(*(x, y)) -> c_4(D^#(x), D^#(y)) , D^#(-(x, y)) -> c_5(D^#(x), D^#(y)) , D^#(minus(x)) -> c_6(D^#(x)) , D^#(div(x, y)) -> c_7(D^#(x), D^#(y)) , D^#(pow(x, y)) -> c_8(D^#(x), D^#(y)) , D^#(ln(x)) -> c_9(D^#(x)) } 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))