We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^3)).
Strict Trs:
{ shuffle(Cons(x, xs)) -> Cons(x, shuffle(reverse(xs)))
, shuffle(Nil()) -> Nil()
, reverse(Cons(x, xs)) -> append(reverse(xs), Cons(x, Nil()))
, reverse(Nil()) -> Nil()
, append(Cons(x, xs), ys) -> Cons(x, append(xs, ys))
, append(Nil(), ys) -> ys
, goal(xs) -> shuffle(xs) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^3))
We add the following dependency tuples:
Strict DPs:
{ shuffle^#(Cons(x, xs)) ->
c_1(shuffle^#(reverse(xs)), reverse^#(xs))
, shuffle^#(Nil()) -> c_2()
, reverse^#(Cons(x, xs)) ->
c_3(append^#(reverse(xs), Cons(x, Nil())), reverse^#(xs))
, reverse^#(Nil()) -> c_4()
, append^#(Cons(x, xs), ys) -> c_5(append^#(xs, ys))
, append^#(Nil(), ys) -> c_6()
, goal^#(xs) -> c_7(shuffle^#(xs)) }
and mark the set of starting terms.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^3)).
Strict DPs:
{ shuffle^#(Cons(x, xs)) ->
c_1(shuffle^#(reverse(xs)), reverse^#(xs))
, shuffle^#(Nil()) -> c_2()
, reverse^#(Cons(x, xs)) ->
c_3(append^#(reverse(xs), Cons(x, Nil())), reverse^#(xs))
, reverse^#(Nil()) -> c_4()
, append^#(Cons(x, xs), ys) -> c_5(append^#(xs, ys))
, append^#(Nil(), ys) -> c_6()
, goal^#(xs) -> c_7(shuffle^#(xs)) }
Weak Trs:
{ shuffle(Cons(x, xs)) -> Cons(x, shuffle(reverse(xs)))
, shuffle(Nil()) -> Nil()
, reverse(Cons(x, xs)) -> append(reverse(xs), Cons(x, Nil()))
, reverse(Nil()) -> Nil()
, append(Cons(x, xs), ys) -> Cons(x, append(xs, ys))
, append(Nil(), ys) -> ys
, goal(xs) -> shuffle(xs) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^3))
We estimate the number of application of {2,4,6} by applications of
Pre({2,4,6}) = {1,3,5,7}. Here rules are labeled as follows:
DPs:
{ 1: shuffle^#(Cons(x, xs)) ->
c_1(shuffle^#(reverse(xs)), reverse^#(xs))
, 2: shuffle^#(Nil()) -> c_2()
, 3: reverse^#(Cons(x, xs)) ->
c_3(append^#(reverse(xs), Cons(x, Nil())), reverse^#(xs))
, 4: reverse^#(Nil()) -> c_4()
, 5: append^#(Cons(x, xs), ys) -> c_5(append^#(xs, ys))
, 6: append^#(Nil(), ys) -> c_6()
, 7: goal^#(xs) -> c_7(shuffle^#(xs)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^3)).
Strict DPs:
{ shuffle^#(Cons(x, xs)) ->
c_1(shuffle^#(reverse(xs)), reverse^#(xs))
, reverse^#(Cons(x, xs)) ->
c_3(append^#(reverse(xs), Cons(x, Nil())), reverse^#(xs))
, append^#(Cons(x, xs), ys) -> c_5(append^#(xs, ys))
, goal^#(xs) -> c_7(shuffle^#(xs)) }
Weak DPs:
{ shuffle^#(Nil()) -> c_2()
, reverse^#(Nil()) -> c_4()
, append^#(Nil(), ys) -> c_6() }
Weak Trs:
{ shuffle(Cons(x, xs)) -> Cons(x, shuffle(reverse(xs)))
, shuffle(Nil()) -> Nil()
, reverse(Cons(x, xs)) -> append(reverse(xs), Cons(x, Nil()))
, reverse(Nil()) -> Nil()
, append(Cons(x, xs), ys) -> Cons(x, append(xs, ys))
, append(Nil(), ys) -> ys
, goal(xs) -> shuffle(xs) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^3))
The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.
{ shuffle^#(Nil()) -> c_2()
, reverse^#(Nil()) -> c_4()
, append^#(Nil(), ys) -> c_6() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^3)).
Strict DPs:
{ shuffle^#(Cons(x, xs)) ->
c_1(shuffle^#(reverse(xs)), reverse^#(xs))
, reverse^#(Cons(x, xs)) ->
c_3(append^#(reverse(xs), Cons(x, Nil())), reverse^#(xs))
, append^#(Cons(x, xs), ys) -> c_5(append^#(xs, ys))
, goal^#(xs) -> c_7(shuffle^#(xs)) }
Weak Trs:
{ shuffle(Cons(x, xs)) -> Cons(x, shuffle(reverse(xs)))
, shuffle(Nil()) -> Nil()
, reverse(Cons(x, xs)) -> append(reverse(xs), Cons(x, Nil()))
, reverse(Nil()) -> Nil()
, append(Cons(x, xs), ys) -> Cons(x, append(xs, ys))
, append(Nil(), ys) -> ys
, goal(xs) -> shuffle(xs) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^3))
Consider the dependency graph
1: shuffle^#(Cons(x, xs)) ->
c_1(shuffle^#(reverse(xs)), reverse^#(xs))
-->_2 reverse^#(Cons(x, xs)) ->
c_3(append^#(reverse(xs), Cons(x, Nil())), reverse^#(xs)) :2
-->_1 shuffle^#(Cons(x, xs)) ->
c_1(shuffle^#(reverse(xs)), reverse^#(xs)) :1
2: reverse^#(Cons(x, xs)) ->
c_3(append^#(reverse(xs), Cons(x, Nil())), reverse^#(xs))
-->_1 append^#(Cons(x, xs), ys) -> c_5(append^#(xs, ys)) :3
-->_2 reverse^#(Cons(x, xs)) ->
c_3(append^#(reverse(xs), Cons(x, Nil())), reverse^#(xs)) :2
3: append^#(Cons(x, xs), ys) -> c_5(append^#(xs, ys))
-->_1 append^#(Cons(x, xs), ys) -> c_5(append^#(xs, ys)) :3
4: goal^#(xs) -> c_7(shuffle^#(xs))
-->_1 shuffle^#(Cons(x, xs)) ->
c_1(shuffle^#(reverse(xs)), reverse^#(xs)) :1
Following roots of the dependency graph are removed, as the
considered set of starting terms is closed under reduction with
respect to these rules (modulo compound contexts).
{ goal^#(xs) -> c_7(shuffle^#(xs)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^3)).
Strict DPs:
{ shuffle^#(Cons(x, xs)) ->
c_1(shuffle^#(reverse(xs)), reverse^#(xs))
, reverse^#(Cons(x, xs)) ->
c_3(append^#(reverse(xs), Cons(x, Nil())), reverse^#(xs))
, append^#(Cons(x, xs), ys) -> c_5(append^#(xs, ys)) }
Weak Trs:
{ shuffle(Cons(x, xs)) -> Cons(x, shuffle(reverse(xs)))
, shuffle(Nil()) -> Nil()
, reverse(Cons(x, xs)) -> append(reverse(xs), Cons(x, Nil()))
, reverse(Nil()) -> Nil()
, append(Cons(x, xs), ys) -> Cons(x, append(xs, ys))
, append(Nil(), ys) -> ys
, goal(xs) -> shuffle(xs) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^3))
We replace rewrite rules by usable rules:
Weak Usable Rules:
{ reverse(Cons(x, xs)) -> append(reverse(xs), Cons(x, Nil()))
, reverse(Nil()) -> Nil()
, append(Cons(x, xs), ys) -> Cons(x, append(xs, ys))
, append(Nil(), ys) -> ys }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^3)).
Strict DPs:
{ shuffle^#(Cons(x, xs)) ->
c_1(shuffle^#(reverse(xs)), reverse^#(xs))
, reverse^#(Cons(x, xs)) ->
c_3(append^#(reverse(xs), Cons(x, Nil())), reverse^#(xs))
, append^#(Cons(x, xs), ys) -> c_5(append^#(xs, ys)) }
Weak Trs:
{ reverse(Cons(x, xs)) -> append(reverse(xs), Cons(x, Nil()))
, reverse(Nil()) -> Nil()
, append(Cons(x, xs), ys) -> Cons(x, append(xs, ys))
, append(Nil(), ys) -> ys }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^3))
We decompose the input problem according to the dependency graph
into the upper component
{ shuffle^#(Cons(x, xs)) ->
c_1(shuffle^#(reverse(xs)), reverse^#(xs)) }
and lower component
{ reverse^#(Cons(x, xs)) ->
c_3(append^#(reverse(xs), Cons(x, Nil())), reverse^#(xs))
, append^#(Cons(x, xs), ys) -> c_5(append^#(xs, ys)) }
Further, following extension rules are added to the lower
component.
{ shuffle^#(Cons(x, xs)) -> shuffle^#(reverse(xs))
, shuffle^#(Cons(x, xs)) -> reverse^#(xs) }
TcT solves the upper component with certificate YES(O(1),O(n^1)).
Sub-proof:
----------
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ shuffle^#(Cons(x, xs)) ->
c_1(shuffle^#(reverse(xs)), reverse^#(xs)) }
Weak Trs:
{ reverse(Cons(x, xs)) -> append(reverse(xs), Cons(x, Nil()))
, reverse(Nil()) -> Nil()
, append(Cons(x, xs), ys) -> Cons(x, append(xs, ys))
, append(Nil(), ys) -> ys }
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: shuffle^#(Cons(x, xs)) ->
c_1(shuffle^#(reverse(xs)), reverse^#(xs)) }
Trs: { reverse(Nil()) -> Nil() }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_1) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[Cons](x1, x2) = [1] x2 + [7]
[reverse](x1) = [1] x1 + [2]
[append](x1, x2) = [1] x1 + [1] x2 + [0]
[Nil] = [0]
[shuffle^#](x1) = [1] x1 + [7]
[c_1](x1, x2) = [1] x1 + [4] x2 + [0]
[reverse^#](x1) = [1]
The order satisfies the following ordering constraints:
[reverse(Cons(x, xs))] = [1] xs + [9]
>= [1] xs + [9]
= [append(reverse(xs), Cons(x, Nil()))]
[reverse(Nil())] = [2]
> [0]
= [Nil()]
[append(Cons(x, xs), ys)] = [1] xs + [1] ys + [7]
>= [1] xs + [1] ys + [7]
= [Cons(x, append(xs, ys))]
[append(Nil(), ys)] = [1] ys + [0]
>= [1] ys + [0]
= [ys]
[shuffle^#(Cons(x, xs))] = [1] xs + [14]
> [1] xs + [13]
= [c_1(shuffle^#(reverse(xs)), reverse^#(xs))]
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:
{ shuffle^#(Cons(x, xs)) ->
c_1(shuffle^#(reverse(xs)), reverse^#(xs)) }
Weak Trs:
{ reverse(Cons(x, xs)) -> append(reverse(xs), Cons(x, Nil()))
, reverse(Nil()) -> Nil()
, append(Cons(x, xs), ys) -> Cons(x, append(xs, ys))
, append(Nil(), ys) -> ys }
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.
{ shuffle^#(Cons(x, xs)) ->
c_1(shuffle^#(reverse(xs)), reverse^#(xs)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Weak Trs:
{ reverse(Cons(x, xs)) -> append(reverse(xs), Cons(x, Nil()))
, reverse(Nil()) -> Nil()
, append(Cons(x, xs), ys) -> Cons(x, append(xs, ys))
, append(Nil(), ys) -> ys }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(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(1)).
Rules: Empty
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
Empty rules are trivially bounded
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict DPs:
{ reverse^#(Cons(x, xs)) ->
c_3(append^#(reverse(xs), Cons(x, Nil())), reverse^#(xs))
, append^#(Cons(x, xs), ys) -> c_5(append^#(xs, ys)) }
Weak DPs:
{ shuffle^#(Cons(x, xs)) -> shuffle^#(reverse(xs))
, shuffle^#(Cons(x, xs)) -> reverse^#(xs) }
Weak Trs:
{ reverse(Cons(x, xs)) -> append(reverse(xs), Cons(x, Nil()))
, reverse(Nil()) -> Nil()
, append(Cons(x, xs), ys) -> Cons(x, append(xs, ys))
, append(Nil(), ys) -> ys }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
We decompose the input problem according to the dependency graph
into the upper component
{ shuffle^#(Cons(x, xs)) -> shuffle^#(reverse(xs))
, shuffle^#(Cons(x, xs)) -> reverse^#(xs)
, reverse^#(Cons(x, xs)) ->
c_3(append^#(reverse(xs), Cons(x, Nil())), reverse^#(xs)) }
and lower component
{ append^#(Cons(x, xs), ys) -> c_5(append^#(xs, ys)) }
Further, following extension rules are added to the lower
component.
{ shuffle^#(Cons(x, xs)) -> shuffle^#(reverse(xs))
, shuffle^#(Cons(x, xs)) -> reverse^#(xs)
, reverse^#(Cons(x, xs)) -> reverse^#(xs)
, reverse^#(Cons(x, xs)) -> append^#(reverse(xs), Cons(x, Nil())) }
TcT solves the upper component with certificate YES(O(1),O(n^1)).
Sub-proof:
----------
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ reverse^#(Cons(x, xs)) ->
c_3(append^#(reverse(xs), Cons(x, Nil())), reverse^#(xs)) }
Weak DPs:
{ shuffle^#(Cons(x, xs)) -> shuffle^#(reverse(xs))
, shuffle^#(Cons(x, xs)) -> reverse^#(xs) }
Weak Trs:
{ reverse(Cons(x, xs)) -> append(reverse(xs), Cons(x, Nil()))
, reverse(Nil()) -> Nil()
, append(Cons(x, xs), ys) -> Cons(x, append(xs, ys))
, append(Nil(), ys) -> ys }
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: reverse^#(Cons(x, xs)) ->
c_3(append^#(reverse(xs), Cons(x, Nil())), reverse^#(xs))
, 2: shuffle^#(Cons(x, xs)) -> shuffle^#(reverse(xs))
, 3: shuffle^#(Cons(x, xs)) -> reverse^#(xs) }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_3) = {1, 2}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[Cons](x1, x2) = [1] x2 + [3]
[reverse](x1) = [1] x1 + [0]
[append](x1, x2) = [1] x1 + [1] x2 + [0]
[Nil] = [0]
[shuffle^#](x1) = [3] x1 + [3]
[reverse^#](x1) = [2] x1 + [1]
[c_3](x1, x2) = [1] x1 + [1] x2 + [0]
[append^#](x1, x2) = [1] x2 + [2]
The order satisfies the following ordering constraints:
[reverse(Cons(x, xs))] = [1] xs + [3]
>= [1] xs + [3]
= [append(reverse(xs), Cons(x, Nil()))]
[reverse(Nil())] = [0]
>= [0]
= [Nil()]
[append(Cons(x, xs), ys)] = [1] xs + [1] ys + [3]
>= [1] xs + [1] ys + [3]
= [Cons(x, append(xs, ys))]
[append(Nil(), ys)] = [1] ys + [0]
>= [1] ys + [0]
= [ys]
[shuffle^#(Cons(x, xs))] = [3] xs + [12]
> [3] xs + [3]
= [shuffle^#(reverse(xs))]
[shuffle^#(Cons(x, xs))] = [3] xs + [12]
> [2] xs + [1]
= [reverse^#(xs)]
[reverse^#(Cons(x, xs))] = [2] xs + [7]
> [2] xs + [6]
= [c_3(append^#(reverse(xs), Cons(x, Nil())), reverse^#(xs))]
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:
{ shuffle^#(Cons(x, xs)) -> shuffle^#(reverse(xs))
, shuffle^#(Cons(x, xs)) -> reverse^#(xs)
, reverse^#(Cons(x, xs)) ->
c_3(append^#(reverse(xs), Cons(x, Nil())), reverse^#(xs)) }
Weak Trs:
{ reverse(Cons(x, xs)) -> append(reverse(xs), Cons(x, Nil()))
, reverse(Nil()) -> Nil()
, append(Cons(x, xs), ys) -> Cons(x, append(xs, ys))
, append(Nil(), ys) -> ys }
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.
{ shuffle^#(Cons(x, xs)) -> shuffle^#(reverse(xs))
, shuffle^#(Cons(x, xs)) -> reverse^#(xs)
, reverse^#(Cons(x, xs)) ->
c_3(append^#(reverse(xs), Cons(x, Nil())), reverse^#(xs)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Weak Trs:
{ reverse(Cons(x, xs)) -> append(reverse(xs), Cons(x, Nil()))
, reverse(Nil()) -> Nil()
, append(Cons(x, xs), ys) -> Cons(x, append(xs, ys))
, append(Nil(), ys) -> ys }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(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(1)).
Rules: Empty
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
Empty rules are trivially bounded
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs: { append^#(Cons(x, xs), ys) -> c_5(append^#(xs, ys)) }
Weak DPs:
{ shuffle^#(Cons(x, xs)) -> shuffle^#(reverse(xs))
, shuffle^#(Cons(x, xs)) -> reverse^#(xs)
, reverse^#(Cons(x, xs)) -> reverse^#(xs)
, reverse^#(Cons(x, xs)) -> append^#(reverse(xs), Cons(x, Nil())) }
Weak Trs:
{ reverse(Cons(x, xs)) -> append(reverse(xs), Cons(x, Nil()))
, reverse(Nil()) -> Nil()
, append(Cons(x, xs), ys) -> Cons(x, append(xs, ys))
, append(Nil(), ys) -> ys }
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: append^#(Cons(x, xs), ys) -> c_5(append^#(xs, ys))
, 2: shuffle^#(Cons(x, xs)) -> shuffle^#(reverse(xs))
, 3: shuffle^#(Cons(x, xs)) -> reverse^#(xs)
, 4: reverse^#(Cons(x, xs)) -> reverse^#(xs)
, 5: reverse^#(Cons(x, xs)) ->
append^#(reverse(xs), Cons(x, Nil())) }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_5) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[Cons](x1, x2) = [1] x2 + [3]
[reverse](x1) = [1] x1 + [0]
[append](x1, x2) = [1] x1 + [1] x2 + [0]
[Nil] = [0]
[shuffle^#](x1) = [2] x1 + [0]
[reverse^#](x1) = [2] x1 + [0]
[append^#](x1, x2) = [2] x1 + [1]
[c_5](x1) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[reverse(Cons(x, xs))] = [1] xs + [3]
>= [1] xs + [3]
= [append(reverse(xs), Cons(x, Nil()))]
[reverse(Nil())] = [0]
>= [0]
= [Nil()]
[append(Cons(x, xs), ys)] = [1] xs + [1] ys + [3]
>= [1] xs + [1] ys + [3]
= [Cons(x, append(xs, ys))]
[append(Nil(), ys)] = [1] ys + [0]
>= [1] ys + [0]
= [ys]
[shuffle^#(Cons(x, xs))] = [2] xs + [6]
> [2] xs + [0]
= [shuffle^#(reverse(xs))]
[shuffle^#(Cons(x, xs))] = [2] xs + [6]
> [2] xs + [0]
= [reverse^#(xs)]
[reverse^#(Cons(x, xs))] = [2] xs + [6]
> [2] xs + [0]
= [reverse^#(xs)]
[reverse^#(Cons(x, xs))] = [2] xs + [6]
> [2] xs + [1]
= [append^#(reverse(xs), Cons(x, Nil()))]
[append^#(Cons(x, xs), ys)] = [2] xs + [7]
> [2] xs + [1]
= [c_5(append^#(xs, ys))]
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:
{ shuffle^#(Cons(x, xs)) -> shuffle^#(reverse(xs))
, shuffle^#(Cons(x, xs)) -> reverse^#(xs)
, reverse^#(Cons(x, xs)) -> reverse^#(xs)
, reverse^#(Cons(x, xs)) -> append^#(reverse(xs), Cons(x, Nil()))
, append^#(Cons(x, xs), ys) -> c_5(append^#(xs, ys)) }
Weak Trs:
{ reverse(Cons(x, xs)) -> append(reverse(xs), Cons(x, Nil()))
, reverse(Nil()) -> Nil()
, append(Cons(x, xs), ys) -> Cons(x, append(xs, ys))
, append(Nil(), ys) -> ys }
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.
{ shuffle^#(Cons(x, xs)) -> shuffle^#(reverse(xs))
, shuffle^#(Cons(x, xs)) -> reverse^#(xs)
, reverse^#(Cons(x, xs)) -> reverse^#(xs)
, reverse^#(Cons(x, xs)) -> append^#(reverse(xs), Cons(x, Nil()))
, append^#(Cons(x, xs), ys) -> c_5(append^#(xs, ys)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Weak Trs:
{ reverse(Cons(x, xs)) -> append(reverse(xs), Cons(x, Nil()))
, reverse(Nil()) -> Nil()
, append(Cons(x, xs), ys) -> Cons(x, append(xs, ys))
, append(Nil(), ys) -> ys }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(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(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^3))