We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^3)).
Strict Trs:
{ app(nil(), y) -> y
, app(add(n, x), y) -> add(n, app(x, y))
, reverse(nil()) -> nil()
, reverse(add(n, x)) -> app(reverse(x), add(n, nil()))
, shuffle(nil()) -> nil()
, shuffle(add(n, x)) -> add(n, shuffle(reverse(x))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^3))
We add the following dependency tuples:
Strict DPs:
{ app^#(nil(), y) -> c_1()
, app^#(add(n, x), y) -> c_2(app^#(x, y))
, reverse^#(nil()) -> c_3()
, reverse^#(add(n, x)) ->
c_4(app^#(reverse(x), add(n, nil())), reverse^#(x))
, shuffle^#(nil()) -> c_5()
, shuffle^#(add(n, x)) ->
c_6(shuffle^#(reverse(x)), reverse^#(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^3)).
Strict DPs:
{ app^#(nil(), y) -> c_1()
, app^#(add(n, x), y) -> c_2(app^#(x, y))
, reverse^#(nil()) -> c_3()
, reverse^#(add(n, x)) ->
c_4(app^#(reverse(x), add(n, nil())), reverse^#(x))
, shuffle^#(nil()) -> c_5()
, shuffle^#(add(n, x)) ->
c_6(shuffle^#(reverse(x)), reverse^#(x)) }
Weak Trs:
{ app(nil(), y) -> y
, app(add(n, x), y) -> add(n, app(x, y))
, reverse(nil()) -> nil()
, reverse(add(n, x)) -> app(reverse(x), add(n, nil()))
, shuffle(nil()) -> nil()
, shuffle(add(n, x)) -> add(n, shuffle(reverse(x))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^3))
We estimate the number of application of {1,3,5} by applications of
Pre({1,3,5}) = {2,4,6}. Here rules are labeled as follows:
DPs:
{ 1: app^#(nil(), y) -> c_1()
, 2: app^#(add(n, x), y) -> c_2(app^#(x, y))
, 3: reverse^#(nil()) -> c_3()
, 4: reverse^#(add(n, x)) ->
c_4(app^#(reverse(x), add(n, nil())), reverse^#(x))
, 5: shuffle^#(nil()) -> c_5()
, 6: shuffle^#(add(n, x)) ->
c_6(shuffle^#(reverse(x)), reverse^#(x)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^3)).
Strict DPs:
{ app^#(add(n, x), y) -> c_2(app^#(x, y))
, reverse^#(add(n, x)) ->
c_4(app^#(reverse(x), add(n, nil())), reverse^#(x))
, shuffle^#(add(n, x)) ->
c_6(shuffle^#(reverse(x)), reverse^#(x)) }
Weak DPs:
{ app^#(nil(), y) -> c_1()
, reverse^#(nil()) -> c_3()
, shuffle^#(nil()) -> c_5() }
Weak Trs:
{ app(nil(), y) -> y
, app(add(n, x), y) -> add(n, app(x, y))
, reverse(nil()) -> nil()
, reverse(add(n, x)) -> app(reverse(x), add(n, nil()))
, shuffle(nil()) -> nil()
, shuffle(add(n, x)) -> add(n, shuffle(reverse(x))) }
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.
{ app^#(nil(), y) -> c_1()
, reverse^#(nil()) -> c_3()
, shuffle^#(nil()) -> c_5() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^3)).
Strict DPs:
{ app^#(add(n, x), y) -> c_2(app^#(x, y))
, reverse^#(add(n, x)) ->
c_4(app^#(reverse(x), add(n, nil())), reverse^#(x))
, shuffle^#(add(n, x)) ->
c_6(shuffle^#(reverse(x)), reverse^#(x)) }
Weak Trs:
{ app(nil(), y) -> y
, app(add(n, x), y) -> add(n, app(x, y))
, reverse(nil()) -> nil()
, reverse(add(n, x)) -> app(reverse(x), add(n, nil()))
, shuffle(nil()) -> nil()
, shuffle(add(n, x)) -> add(n, shuffle(reverse(x))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^3))
We replace rewrite rules by usable rules:
Weak Usable Rules:
{ app(nil(), y) -> y
, app(add(n, x), y) -> add(n, app(x, y))
, reverse(nil()) -> nil()
, reverse(add(n, x)) -> app(reverse(x), add(n, nil())) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^3)).
Strict DPs:
{ app^#(add(n, x), y) -> c_2(app^#(x, y))
, reverse^#(add(n, x)) ->
c_4(app^#(reverse(x), add(n, nil())), reverse^#(x))
, shuffle^#(add(n, x)) ->
c_6(shuffle^#(reverse(x)), reverse^#(x)) }
Weak Trs:
{ app(nil(), y) -> y
, app(add(n, x), y) -> add(n, app(x, y))
, reverse(nil()) -> nil()
, reverse(add(n, x)) -> app(reverse(x), add(n, nil())) }
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^#(add(n, x)) ->
c_6(shuffle^#(reverse(x)), reverse^#(x)) }
and lower component
{ app^#(add(n, x), y) -> c_2(app^#(x, y))
, reverse^#(add(n, x)) ->
c_4(app^#(reverse(x), add(n, nil())), reverse^#(x)) }
Further, following extension rules are added to the lower
component.
{ shuffle^#(add(n, x)) -> reverse^#(x)
, shuffle^#(add(n, x)) -> shuffle^#(reverse(x)) }
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^#(add(n, x)) ->
c_6(shuffle^#(reverse(x)), reverse^#(x)) }
Weak Trs:
{ app(nil(), y) -> y
, app(add(n, x), y) -> add(n, app(x, y))
, reverse(nil()) -> nil()
, reverse(add(n, x)) -> app(reverse(x), add(n, nil())) }
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^#(add(n, x)) ->
c_6(shuffle^#(reverse(x)), reverse^#(x)) }
Trs: { reverse(nil()) -> nil() }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_6) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[app](x1, x2) = [1] x1 + [1] x2 + [0]
[nil] = [0]
[add](x1, x2) = [1] x2 + [2]
[reverse](x1) = [1] x1 + [1]
[reverse^#](x1) = [0]
[shuffle^#](x1) = [1] x1 + [0]
[c_6](x1, x2) = [1] x1 + [1] x2 + [0]
The order satisfies the following ordering constraints:
[app(nil(), y)] = [1] y + [0]
>= [1] y + [0]
= [y]
[app(add(n, x), y)] = [1] y + [1] x + [2]
>= [1] y + [1] x + [2]
= [add(n, app(x, y))]
[reverse(nil())] = [1]
> [0]
= [nil()]
[reverse(add(n, x))] = [1] x + [3]
>= [1] x + [3]
= [app(reverse(x), add(n, nil()))]
[shuffle^#(add(n, x))] = [1] x + [2]
> [1] x + [1]
= [c_6(shuffle^#(reverse(x)), reverse^#(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:
{ shuffle^#(add(n, x)) ->
c_6(shuffle^#(reverse(x)), reverse^#(x)) }
Weak Trs:
{ app(nil(), y) -> y
, app(add(n, x), y) -> add(n, app(x, y))
, reverse(nil()) -> nil()
, reverse(add(n, x)) -> app(reverse(x), add(n, nil())) }
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^#(add(n, x)) ->
c_6(shuffle^#(reverse(x)), reverse^#(x)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Weak Trs:
{ app(nil(), y) -> y
, app(add(n, x), y) -> add(n, app(x, y))
, reverse(nil()) -> nil()
, reverse(add(n, x)) -> app(reverse(x), add(n, nil())) }
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:
{ app^#(add(n, x), y) -> c_2(app^#(x, y))
, reverse^#(add(n, x)) ->
c_4(app^#(reverse(x), add(n, nil())), reverse^#(x)) }
Weak DPs:
{ shuffle^#(add(n, x)) -> reverse^#(x)
, shuffle^#(add(n, x)) -> shuffle^#(reverse(x)) }
Weak Trs:
{ app(nil(), y) -> y
, app(add(n, x), y) -> add(n, app(x, y))
, reverse(nil()) -> nil()
, reverse(add(n, x)) -> app(reverse(x), add(n, nil())) }
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
{ reverse^#(add(n, x)) ->
c_4(app^#(reverse(x), add(n, nil())), reverse^#(x))
, shuffle^#(add(n, x)) -> reverse^#(x)
, shuffle^#(add(n, x)) -> shuffle^#(reverse(x)) }
and lower component
{ app^#(add(n, x), y) -> c_2(app^#(x, y)) }
Further, following extension rules are added to the lower
component.
{ reverse^#(add(n, x)) -> app^#(reverse(x), add(n, nil()))
, reverse^#(add(n, x)) -> reverse^#(x)
, shuffle^#(add(n, x)) -> reverse^#(x)
, shuffle^#(add(n, x)) -> shuffle^#(reverse(x)) }
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^#(add(n, x)) ->
c_4(app^#(reverse(x), add(n, nil())), reverse^#(x)) }
Weak DPs:
{ shuffle^#(add(n, x)) -> reverse^#(x)
, shuffle^#(add(n, x)) -> shuffle^#(reverse(x)) }
Weak Trs:
{ app(nil(), y) -> y
, app(add(n, x), y) -> add(n, app(x, y))
, reverse(nil()) -> nil()
, reverse(add(n, x)) -> app(reverse(x), add(n, nil())) }
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^#(add(n, x)) ->
c_4(app^#(reverse(x), add(n, nil())), reverse^#(x))
, 2: shuffle^#(add(n, x)) -> reverse^#(x)
, 3: shuffle^#(add(n, x)) -> shuffle^#(reverse(x)) }
Trs: { reverse(nil()) -> nil() }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_4) = {1, 2}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[app](x1, x2) = [1] x1 + [1] x2 + [0]
[nil] = [0]
[add](x1, x2) = [1] x2 + [3]
[reverse](x1) = [1] x1 + [2]
[app^#](x1, x2) = [2]
[reverse^#](x1) = [4] x1 + [0]
[c_4](x1, x2) = [4] x1 + [1] x2 + [3]
[shuffle^#](x1) = [4] x1 + [3]
The order satisfies the following ordering constraints:
[app(nil(), y)] = [1] y + [0]
>= [1] y + [0]
= [y]
[app(add(n, x), y)] = [1] y + [1] x + [3]
>= [1] y + [1] x + [3]
= [add(n, app(x, y))]
[reverse(nil())] = [2]
> [0]
= [nil()]
[reverse(add(n, x))] = [1] x + [5]
>= [1] x + [5]
= [app(reverse(x), add(n, nil()))]
[reverse^#(add(n, x))] = [4] x + [12]
> [4] x + [11]
= [c_4(app^#(reverse(x), add(n, nil())), reverse^#(x))]
[shuffle^#(add(n, x))] = [4] x + [15]
> [4] x + [0]
= [reverse^#(x)]
[shuffle^#(add(n, x))] = [4] x + [15]
> [4] x + [11]
= [shuffle^#(reverse(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:
{ reverse^#(add(n, x)) ->
c_4(app^#(reverse(x), add(n, nil())), reverse^#(x))
, shuffle^#(add(n, x)) -> reverse^#(x)
, shuffle^#(add(n, x)) -> shuffle^#(reverse(x)) }
Weak Trs:
{ app(nil(), y) -> y
, app(add(n, x), y) -> add(n, app(x, y))
, reverse(nil()) -> nil()
, reverse(add(n, x)) -> app(reverse(x), add(n, nil())) }
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.
{ reverse^#(add(n, x)) ->
c_4(app^#(reverse(x), add(n, nil())), reverse^#(x))
, shuffle^#(add(n, x)) -> reverse^#(x)
, shuffle^#(add(n, x)) -> shuffle^#(reverse(x)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Weak Trs:
{ app(nil(), y) -> y
, app(add(n, x), y) -> add(n, app(x, y))
, reverse(nil()) -> nil()
, reverse(add(n, x)) -> app(reverse(x), add(n, nil())) }
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: { app^#(add(n, x), y) -> c_2(app^#(x, y)) }
Weak DPs:
{ reverse^#(add(n, x)) -> app^#(reverse(x), add(n, nil()))
, reverse^#(add(n, x)) -> reverse^#(x)
, shuffle^#(add(n, x)) -> reverse^#(x)
, shuffle^#(add(n, x)) -> shuffle^#(reverse(x)) }
Weak Trs:
{ app(nil(), y) -> y
, app(add(n, x), y) -> add(n, app(x, y))
, reverse(nil()) -> nil()
, reverse(add(n, x)) -> app(reverse(x), add(n, nil())) }
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: app^#(add(n, x), y) -> c_2(app^#(x, y))
, 2: reverse^#(add(n, x)) -> app^#(reverse(x), add(n, nil()))
, 3: reverse^#(add(n, x)) -> reverse^#(x)
, 4: shuffle^#(add(n, x)) -> reverse^#(x)
, 5: shuffle^#(add(n, x)) -> shuffle^#(reverse(x)) }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_2) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[app](x1, x2) = [1] x1 + [1] x2 + [0]
[nil] = [0]
[add](x1, x2) = [1] x2 + [2]
[reverse](x1) = [1] x1 + [0]
[app^#](x1, x2) = [1] x1 + [5]
[c_2](x1) = [1] x1 + [0]
[reverse^#](x1) = [4] x1 + [5]
[shuffle^#](x1) = [5] x1 + [4]
The order satisfies the following ordering constraints:
[app(nil(), y)] = [1] y + [0]
>= [1] y + [0]
= [y]
[app(add(n, x), y)] = [1] y + [1] x + [2]
>= [1] y + [1] x + [2]
= [add(n, app(x, y))]
[reverse(nil())] = [0]
>= [0]
= [nil()]
[reverse(add(n, x))] = [1] x + [2]
>= [1] x + [2]
= [app(reverse(x), add(n, nil()))]
[app^#(add(n, x), y)] = [1] x + [7]
> [1] x + [5]
= [c_2(app^#(x, y))]
[reverse^#(add(n, x))] = [4] x + [13]
> [1] x + [5]
= [app^#(reverse(x), add(n, nil()))]
[reverse^#(add(n, x))] = [4] x + [13]
> [4] x + [5]
= [reverse^#(x)]
[shuffle^#(add(n, x))] = [5] x + [14]
> [4] x + [5]
= [reverse^#(x)]
[shuffle^#(add(n, x))] = [5] x + [14]
> [5] x + [4]
= [shuffle^#(reverse(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:
{ app^#(add(n, x), y) -> c_2(app^#(x, y))
, reverse^#(add(n, x)) -> app^#(reverse(x), add(n, nil()))
, reverse^#(add(n, x)) -> reverse^#(x)
, shuffle^#(add(n, x)) -> reverse^#(x)
, shuffle^#(add(n, x)) -> shuffle^#(reverse(x)) }
Weak Trs:
{ app(nil(), y) -> y
, app(add(n, x), y) -> add(n, app(x, y))
, reverse(nil()) -> nil()
, reverse(add(n, x)) -> app(reverse(x), add(n, nil())) }
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.
{ app^#(add(n, x), y) -> c_2(app^#(x, y))
, reverse^#(add(n, x)) -> app^#(reverse(x), add(n, nil()))
, reverse^#(add(n, x)) -> reverse^#(x)
, shuffle^#(add(n, x)) -> reverse^#(x)
, shuffle^#(add(n, x)) -> shuffle^#(reverse(x)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Weak Trs:
{ app(nil(), y) -> y
, app(add(n, x), y) -> add(n, app(x, y))
, reverse(nil()) -> nil()
, reverse(add(n, x)) -> app(reverse(x), add(n, nil())) }
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))