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: { dx(X) -> one() , dx(a()) -> zero() , dx(plus(ALPHA, BETA)) -> plus(dx(ALPHA), dx(BETA)) , dx(times(ALPHA, BETA)) -> plus(times(BETA, dx(ALPHA)), times(ALPHA, dx(BETA))) , dx(minus(ALPHA, BETA)) -> minus(dx(ALPHA), dx(BETA)) , dx(neg(ALPHA)) -> neg(dx(ALPHA)) , dx(div(ALPHA, BETA)) -> minus(div(dx(ALPHA), BETA), times(ALPHA, div(dx(BETA), exp(BETA, two())))) , dx(exp(ALPHA, BETA)) -> plus(times(BETA, times(exp(ALPHA, minus(BETA, one())), dx(ALPHA))), times(exp(ALPHA, BETA), times(ln(ALPHA), dx(BETA)))) , dx(ln(ALPHA)) -> div(dx(ALPHA), ALPHA) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We add following weak dependency pairs: Strict DPs: { dx^#(X) -> c_1() , dx^#(a()) -> c_2() , dx^#(plus(ALPHA, BETA)) -> c_3(dx^#(ALPHA), dx^#(BETA)) , dx^#(times(ALPHA, BETA)) -> c_4(dx^#(ALPHA), dx^#(BETA)) , dx^#(minus(ALPHA, BETA)) -> c_5(dx^#(ALPHA), dx^#(BETA)) , dx^#(neg(ALPHA)) -> c_6(dx^#(ALPHA)) , dx^#(div(ALPHA, BETA)) -> c_7(dx^#(ALPHA), dx^#(BETA)) , dx^#(exp(ALPHA, BETA)) -> c_8(dx^#(ALPHA), dx^#(BETA)) , dx^#(ln(ALPHA)) -> c_9(dx^#(ALPHA)) } 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: { dx^#(X) -> c_1() , dx^#(a()) -> c_2() , dx^#(plus(ALPHA, BETA)) -> c_3(dx^#(ALPHA), dx^#(BETA)) , dx^#(times(ALPHA, BETA)) -> c_4(dx^#(ALPHA), dx^#(BETA)) , dx^#(minus(ALPHA, BETA)) -> c_5(dx^#(ALPHA), dx^#(BETA)) , dx^#(neg(ALPHA)) -> c_6(dx^#(ALPHA)) , dx^#(div(ALPHA, BETA)) -> c_7(dx^#(ALPHA), dx^#(BETA)) , dx^#(exp(ALPHA, BETA)) -> c_8(dx^#(ALPHA), dx^#(BETA)) , dx^#(ln(ALPHA)) -> c_9(dx^#(ALPHA)) } Strict Trs: { dx(X) -> one() , dx(a()) -> zero() , dx(plus(ALPHA, BETA)) -> plus(dx(ALPHA), dx(BETA)) , dx(times(ALPHA, BETA)) -> plus(times(BETA, dx(ALPHA)), times(ALPHA, dx(BETA))) , dx(minus(ALPHA, BETA)) -> minus(dx(ALPHA), dx(BETA)) , dx(neg(ALPHA)) -> neg(dx(ALPHA)) , dx(div(ALPHA, BETA)) -> minus(div(dx(ALPHA), BETA), times(ALPHA, div(dx(BETA), exp(BETA, two())))) , dx(exp(ALPHA, BETA)) -> plus(times(BETA, times(exp(ALPHA, minus(BETA, one())), dx(ALPHA))), times(exp(ALPHA, BETA), times(ln(ALPHA), dx(BETA)))) , dx(ln(ALPHA)) -> div(dx(ALPHA), ALPHA) } 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: { dx^#(X) -> c_1() , dx^#(a()) -> c_2() , dx^#(plus(ALPHA, BETA)) -> c_3(dx^#(ALPHA), dx^#(BETA)) , dx^#(times(ALPHA, BETA)) -> c_4(dx^#(ALPHA), dx^#(BETA)) , dx^#(minus(ALPHA, BETA)) -> c_5(dx^#(ALPHA), dx^#(BETA)) , dx^#(neg(ALPHA)) -> c_6(dx^#(ALPHA)) , dx^#(div(ALPHA, BETA)) -> c_7(dx^#(ALPHA), dx^#(BETA)) , dx^#(exp(ALPHA, BETA)) -> c_8(dx^#(ALPHA), dx^#(BETA)) , dx^#(ln(ALPHA)) -> c_9(dx^#(ALPHA)) } 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 following constructor-restricted matrix interpretation. [a] = [1] [plus](x1, x2) = [1] x1 + [1] x2 + [1] [times](x1, x2) = [1] x1 + [1] x2 + [1] [minus](x1, x2) = [1] x1 + [1] x2 + [2] [neg](x1) = [1] x1 + [2] [div](x1, x2) = [1] x1 + [1] x2 + [2] [exp](x1, x2) = [1] x1 + [1] x2 + [2] [ln](x1) = [1] x1 + [2] [dx^#](x1) = [1] [c_1] = [0] [c_2] = [0] [c_3](x1, x2) = [1] x1 + [1] x2 + [2] [c_4](x1, x2) = [1] x1 + [1] x2 + [2] [c_5](x1, x2) = [1] x1 + [1] x2 + [2] [c_6](x1) = [1] x1 + [0] [c_7](x1, x2) = [1] x1 + [1] x2 + [2] [c_8](x1, x2) = [1] x1 + [1] x2 + [2] [c_9](x1) = [1] x1 + [0] 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^1)). Strict DPs: { dx^#(plus(ALPHA, BETA)) -> c_3(dx^#(ALPHA), dx^#(BETA)) , dx^#(times(ALPHA, BETA)) -> c_4(dx^#(ALPHA), dx^#(BETA)) , dx^#(minus(ALPHA, BETA)) -> c_5(dx^#(ALPHA), dx^#(BETA)) , dx^#(neg(ALPHA)) -> c_6(dx^#(ALPHA)) , dx^#(div(ALPHA, BETA)) -> c_7(dx^#(ALPHA), dx^#(BETA)) , dx^#(exp(ALPHA, BETA)) -> c_8(dx^#(ALPHA), dx^#(BETA)) , dx^#(ln(ALPHA)) -> c_9(dx^#(ALPHA)) } Weak DPs: { dx^#(X) -> c_1() , dx^#(a()) -> c_2() } Obligation: innermost runtime complexity Answer: YES(?,O(n^1)) The following weak DPs constitute a sub-graph of the DG that is closed under successors. The DPs are removed. { dx^#(X) -> c_1() , dx^#(a()) -> c_2() } We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict DPs: { dx^#(plus(ALPHA, BETA)) -> c_3(dx^#(ALPHA), dx^#(BETA)) , dx^#(times(ALPHA, BETA)) -> c_4(dx^#(ALPHA), dx^#(BETA)) , dx^#(minus(ALPHA, BETA)) -> c_5(dx^#(ALPHA), dx^#(BETA)) , dx^#(neg(ALPHA)) -> c_6(dx^#(ALPHA)) , dx^#(div(ALPHA, BETA)) -> c_7(dx^#(ALPHA), dx^#(BETA)) , dx^#(exp(ALPHA, BETA)) -> c_8(dx^#(ALPHA), dx^#(BETA)) , dx^#(ln(ALPHA)) -> c_9(dx^#(ALPHA)) } Obligation: innermost runtime complexity Answer: YES(?,O(n^1)) 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 following constructor-based matrix interpretation satisfying not(EDA). [plus](x1, x2) = [1] x1 + [1] x2 + [2] [times](x1, x2) = [1] x1 + [1] x2 + [2] [minus](x1, x2) = [1] x1 + [1] x2 + [2] [neg](x1) = [1] x1 + [2] [div](x1, x2) = [1] x1 + [1] x2 + [2] [exp](x1, x2) = [1] x1 + [1] x2 + [2] [ln](x1) = [1] x1 + [2] [dx^#](x1) = [2] x1 + [0] [c_3](x1, x2) = [1] x1 + [1] x2 + [3] [c_4](x1, x2) = [1] x1 + [1] x2 + [3] [c_5](x1, x2) = [1] x1 + [1] x2 + [3] [c_6](x1) = [1] x1 + [1] [c_7](x1, x2) = [1] x1 + [1] x2 + [3] [c_8](x1, x2) = [1] x1 + [1] x2 + [3] [c_9](x1) = [1] x1 + [1] This order satisfies following ordering constraints: [dx^#(plus(ALPHA, BETA))] = [2] ALPHA + [2] BETA + [4] > [2] ALPHA + [2] BETA + [3] = [c_3(dx^#(ALPHA), dx^#(BETA))] [dx^#(times(ALPHA, BETA))] = [2] ALPHA + [2] BETA + [4] > [2] ALPHA + [2] BETA + [3] = [c_4(dx^#(ALPHA), dx^#(BETA))] [dx^#(minus(ALPHA, BETA))] = [2] ALPHA + [2] BETA + [4] > [2] ALPHA + [2] BETA + [3] = [c_5(dx^#(ALPHA), dx^#(BETA))] [dx^#(neg(ALPHA))] = [2] ALPHA + [4] > [2] ALPHA + [1] = [c_6(dx^#(ALPHA))] [dx^#(div(ALPHA, BETA))] = [2] ALPHA + [2] BETA + [4] > [2] ALPHA + [2] BETA + [3] = [c_7(dx^#(ALPHA), dx^#(BETA))] [dx^#(exp(ALPHA, BETA))] = [2] ALPHA + [2] BETA + [4] > [2] ALPHA + [2] BETA + [3] = [c_8(dx^#(ALPHA), dx^#(BETA))] [dx^#(ln(ALPHA))] = [2] ALPHA + [4] > [2] ALPHA + [1] = [c_9(dx^#(ALPHA))] Hurray, we answered YES(O(1),O(n^1))