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 the 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 the following constructor-restricted matrix
interpretation.
[a] = [0]
[0]
[plus](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [0]
[times](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [0]
[minus](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [0]
[neg](x1) = [1 0] x1 + [0]
[0 0] [0]
[div](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [0]
[exp](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [0]
[ln](x1) = [1 0] x1 + [0]
[0 0] [0]
[dx^#](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] [0]
[c_5](x1, x2) = [1 0] x1 + [1 0] x2 + [2]
[0 1] [0 1] [0]
[c_6](x1) = [1 0] x1 + [0]
[0 1] [0]
[c_7](x1, x2) = [1 0] x1 + [1 0] x2 + [2]
[0 1] [0 1] [0]
[c_8](x1, x2) = [1 0] x1 + [1 0] x2 + [2]
[0 1] [0 1] [0]
[c_9](x1) = [1 0] x1 + [0]
[0 1] [0]
The order satisfies the following ordering constraints:
[dx^#(X)] = [1]
[0]
> [0]
[0]
= [c_1()]
[dx^#(a())] = [1]
[0]
> [0]
[0]
= [c_2()]
[dx^#(plus(ALPHA, BETA))] = [1]
[0]
? [4]
[0]
= [c_3(dx^#(ALPHA), dx^#(BETA))]
[dx^#(times(ALPHA, BETA))] = [1]
[0]
? [4]
[0]
= [c_4(dx^#(ALPHA), dx^#(BETA))]
[dx^#(minus(ALPHA, BETA))] = [1]
[0]
? [4]
[0]
= [c_5(dx^#(ALPHA), dx^#(BETA))]
[dx^#(neg(ALPHA))] = [1]
[0]
>= [1]
[0]
= [c_6(dx^#(ALPHA))]
[dx^#(div(ALPHA, BETA))] = [1]
[0]
? [4]
[0]
= [c_7(dx^#(ALPHA), dx^#(BETA))]
[dx^#(exp(ALPHA, BETA))] = [1]
[0]
? [4]
[0]
= [c_8(dx^#(ALPHA), dx^#(BETA))]
[dx^#(ln(ALPHA))] = [1]
[0]
>= [1]
[0]
= [c_9(dx^#(ALPHA))]
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:
{ 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(1),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(1),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(1),O(n^1))
We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.
DPs:
{ 4: dx^#(neg(ALPHA)) -> c_6(dx^#(ALPHA))
, 6: dx^#(exp(ALPHA, BETA)) -> c_8(dx^#(ALPHA), dx^#(BETA))
, 7: dx^#(ln(ALPHA)) -> c_9(dx^#(ALPHA)) }
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).
[plus](x1, x2) = [1] x1 + [1] x2 + [0]
[times](x1, x2) = [1] x1 + [1] x2 + [0]
[minus](x1, x2) = [1] x1 + [1] x2 + [0]
[neg](x1) = [1] x1 + [4]
[div](x1, x2) = [1] x1 + [1] x2 + [0]
[exp](x1, x2) = [1] x1 + [1] x2 + [4]
[ln](x1) = [1] x1 + [4]
[dx^#](x1) = [2] x1 + [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 + [0]
[c_6](x1) = [1] x1 + [1]
[c_7](x1, x2) = [1] x1 + [1] x2 + [0]
[c_8](x1, x2) = [1] x1 + [1] x2 + [0]
[c_9](x1) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[dx^#(plus(ALPHA, BETA))] = [2] ALPHA + [2] BETA + [0]
>= [2] ALPHA + [2] BETA + [0]
= [c_3(dx^#(ALPHA), dx^#(BETA))]
[dx^#(times(ALPHA, BETA))] = [2] ALPHA + [2] BETA + [0]
>= [2] ALPHA + [2] BETA + [0]
= [c_4(dx^#(ALPHA), dx^#(BETA))]
[dx^#(minus(ALPHA, BETA))] = [2] ALPHA + [2] BETA + [0]
>= [2] ALPHA + [2] BETA + [0]
= [c_5(dx^#(ALPHA), dx^#(BETA))]
[dx^#(neg(ALPHA))] = [2] ALPHA + [8]
> [2] ALPHA + [1]
= [c_6(dx^#(ALPHA))]
[dx^#(div(ALPHA, BETA))] = [2] ALPHA + [2] BETA + [0]
>= [2] ALPHA + [2] BETA + [0]
= [c_7(dx^#(ALPHA), dx^#(BETA))]
[dx^#(exp(ALPHA, BETA))] = [2] ALPHA + [2] BETA + [8]
> [2] ALPHA + [2] BETA + [0]
= [c_8(dx^#(ALPHA), dx^#(BETA))]
[dx^#(ln(ALPHA))] = [2] ALPHA + [8]
> [2] ALPHA + [0]
= [c_9(dx^#(ALPHA))]
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:
{ 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^#(div(ALPHA, BETA)) -> c_7(dx^#(ALPHA), dx^#(BETA)) }
Weak DPs:
{ dx^#(neg(ALPHA)) -> c_6(dx^#(ALPHA))
, 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))
We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.
DPs:
{ 3: dx^#(minus(ALPHA, BETA)) -> c_5(dx^#(ALPHA), dx^#(BETA))
, 4: dx^#(div(ALPHA, BETA)) -> c_7(dx^#(ALPHA), dx^#(BETA))
, 5: dx^#(neg(ALPHA)) -> c_6(dx^#(ALPHA))
, 6: dx^#(exp(ALPHA, BETA)) -> c_8(dx^#(ALPHA), dx^#(BETA))
, 7: dx^#(ln(ALPHA)) -> c_9(dx^#(ALPHA)) }
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).
[plus](x1, x2) = [1] x1 + [1] x2 + [0]
[times](x1, x2) = [1] x1 + [1] x2 + [0]
[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) = [4] x1 + [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 + [0]
[c_6](x1) = [1] x1 + [1]
[c_7](x1, x2) = [1] x1 + [1] x2 + [1]
[c_8](x1, x2) = [1] x1 + [1] x2 + [0]
[c_9](x1) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[dx^#(plus(ALPHA, BETA))] = [4] ALPHA + [4] BETA + [0]
>= [4] ALPHA + [4] BETA + [0]
= [c_3(dx^#(ALPHA), dx^#(BETA))]
[dx^#(times(ALPHA, BETA))] = [4] ALPHA + [4] BETA + [0]
>= [4] ALPHA + [4] BETA + [0]
= [c_4(dx^#(ALPHA), dx^#(BETA))]
[dx^#(minus(ALPHA, BETA))] = [4] ALPHA + [4] BETA + [8]
> [4] ALPHA + [4] BETA + [0]
= [c_5(dx^#(ALPHA), dx^#(BETA))]
[dx^#(neg(ALPHA))] = [4] ALPHA + [8]
> [4] ALPHA + [1]
= [c_6(dx^#(ALPHA))]
[dx^#(div(ALPHA, BETA))] = [4] ALPHA + [4] BETA + [8]
> [4] ALPHA + [4] BETA + [1]
= [c_7(dx^#(ALPHA), dx^#(BETA))]
[dx^#(exp(ALPHA, BETA))] = [4] ALPHA + [4] BETA + [8]
> [4] ALPHA + [4] BETA + [0]
= [c_8(dx^#(ALPHA), dx^#(BETA))]
[dx^#(ln(ALPHA))] = [4] ALPHA + [8]
> [4] ALPHA + [0]
= [c_9(dx^#(ALPHA))]
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:
{ dx^#(plus(ALPHA, BETA)) -> c_3(dx^#(ALPHA), dx^#(BETA))
, dx^#(times(ALPHA, BETA)) -> c_4(dx^#(ALPHA), dx^#(BETA)) }
Weak DPs:
{ 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))
We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.
DPs:
{ 1: dx^#(plus(ALPHA, BETA)) -> c_3(dx^#(ALPHA), dx^#(BETA))
, 2: dx^#(times(ALPHA, BETA)) -> c_4(dx^#(ALPHA), dx^#(BETA))
, 5: dx^#(div(ALPHA, BETA)) -> c_7(dx^#(ALPHA), dx^#(BETA))
, 6: dx^#(exp(ALPHA, BETA)) -> c_8(dx^#(ALPHA), dx^#(BETA))
, 7: dx^#(ln(ALPHA)) -> c_9(dx^#(ALPHA)) }
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).
[plus](x1, x2) = [1] x1 + [1] x2 + [2]
[times](x1, x2) = [1] x1 + [1] x2 + [2]
[minus](x1, x2) = [1] x1 + [1] x2 + [0]
[neg](x1) = [1] x1 + [0]
[div](x1, x2) = [1] x1 + [1] x2 + [2]
[exp](x1, x2) = [1] x1 + [1] x2 + [2]
[ln](x1) = [1] x1 + [2]
[dx^#](x1) = [4] x1 + [0]
[c_3](x1, x2) = [1] x1 + [1] x2 + [0]
[c_4](x1, x2) = [1] x1 + [1] x2 + [1]
[c_5](x1, x2) = [1] x1 + [1] x2 + [0]
[c_6](x1) = [1] x1 + [0]
[c_7](x1, x2) = [1] x1 + [1] x2 + [0]
[c_8](x1, x2) = [1] x1 + [1] x2 + [0]
[c_9](x1) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[dx^#(plus(ALPHA, BETA))] = [4] ALPHA + [4] BETA + [8]
> [4] ALPHA + [4] BETA + [0]
= [c_3(dx^#(ALPHA), dx^#(BETA))]
[dx^#(times(ALPHA, BETA))] = [4] ALPHA + [4] BETA + [8]
> [4] ALPHA + [4] BETA + [1]
= [c_4(dx^#(ALPHA), dx^#(BETA))]
[dx^#(minus(ALPHA, BETA))] = [4] ALPHA + [4] BETA + [0]
>= [4] ALPHA + [4] BETA + [0]
= [c_5(dx^#(ALPHA), dx^#(BETA))]
[dx^#(neg(ALPHA))] = [4] ALPHA + [0]
>= [4] ALPHA + [0]
= [c_6(dx^#(ALPHA))]
[dx^#(div(ALPHA, BETA))] = [4] ALPHA + [4] BETA + [8]
> [4] ALPHA + [4] BETA + [0]
= [c_7(dx^#(ALPHA), dx^#(BETA))]
[dx^#(exp(ALPHA, BETA))] = [4] ALPHA + [4] BETA + [8]
> [4] ALPHA + [4] BETA + [0]
= [c_8(dx^#(ALPHA), dx^#(BETA))]
[dx^#(ln(ALPHA))] = [4] ALPHA + [8]
> [4] ALPHA + [0]
= [c_9(dx^#(ALPHA))]
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:
{ 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(1))
The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.
{ 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)) }
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))