We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ ordered(Cons(x', Cons(x, xs))) ->
ordered[Ite](<(x', x), Cons(x', Cons(x, xs)))
, ordered(Cons(x, Nil())) -> True()
, ordered(Nil()) -> True()
, notEmpty(Cons(x, xs)) -> True()
, notEmpty(Nil()) -> False()
, goal(xs) -> ordered(xs) }
Weak Trs:
{ ordered[Ite](True(), Cons(x', Cons(x, xs))) -> ordered(xs)
, ordered[Ite](False(), xs) -> False()
, <(x, 0()) -> False()
, <(S(x), S(y)) -> <(x, y)
, <(0(), S(y)) -> True() }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We add the following weak dependency pairs:
Strict DPs:
{ ordered^#(Cons(x', Cons(x, xs))) ->
c_1(ordered[Ite]^#(<(x', x), Cons(x', Cons(x, xs))))
, ordered^#(Cons(x, Nil())) -> c_2()
, ordered^#(Nil()) -> c_3()
, notEmpty^#(Cons(x, xs)) -> c_4()
, notEmpty^#(Nil()) -> c_5()
, goal^#(xs) -> c_6(ordered^#(xs)) }
Weak DPs:
{ ordered[Ite]^#(True(), Cons(x', Cons(x, xs))) ->
c_7(ordered^#(xs))
, ordered[Ite]^#(False(), xs) -> c_8()
, <^#(x, 0()) -> c_9()
, <^#(S(x), S(y)) -> c_10(<^#(x, y))
, <^#(0(), S(y)) -> c_11() }
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:
{ ordered^#(Cons(x', Cons(x, xs))) ->
c_1(ordered[Ite]^#(<(x', x), Cons(x', Cons(x, xs))))
, ordered^#(Cons(x, Nil())) -> c_2()
, ordered^#(Nil()) -> c_3()
, notEmpty^#(Cons(x, xs)) -> c_4()
, notEmpty^#(Nil()) -> c_5()
, goal^#(xs) -> c_6(ordered^#(xs)) }
Strict Trs:
{ ordered(Cons(x', Cons(x, xs))) ->
ordered[Ite](<(x', x), Cons(x', Cons(x, xs)))
, ordered(Cons(x, Nil())) -> True()
, ordered(Nil()) -> True()
, notEmpty(Cons(x, xs)) -> True()
, notEmpty(Nil()) -> False()
, goal(xs) -> ordered(xs) }
Weak DPs:
{ ordered[Ite]^#(True(), Cons(x', Cons(x, xs))) ->
c_7(ordered^#(xs))
, ordered[Ite]^#(False(), xs) -> c_8()
, <^#(x, 0()) -> c_9()
, <^#(S(x), S(y)) -> c_10(<^#(x, y))
, <^#(0(), S(y)) -> c_11() }
Weak Trs:
{ ordered[Ite](True(), Cons(x', Cons(x, xs))) -> ordered(xs)
, ordered[Ite](False(), xs) -> False()
, <(x, 0()) -> False()
, <(S(x), S(y)) -> <(x, y)
, <(0(), S(y)) -> True() }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We replace rewrite rules by usable rules:
Weak Usable Rules:
{ <(x, 0()) -> False()
, <(S(x), S(y)) -> <(x, y)
, <(0(), S(y)) -> True() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ ordered^#(Cons(x', Cons(x, xs))) ->
c_1(ordered[Ite]^#(<(x', x), Cons(x', Cons(x, xs))))
, ordered^#(Cons(x, Nil())) -> c_2()
, ordered^#(Nil()) -> c_3()
, notEmpty^#(Cons(x, xs)) -> c_4()
, notEmpty^#(Nil()) -> c_5()
, goal^#(xs) -> c_6(ordered^#(xs)) }
Weak DPs:
{ ordered[Ite]^#(True(), Cons(x', Cons(x, xs))) ->
c_7(ordered^#(xs))
, ordered[Ite]^#(False(), xs) -> c_8()
, <^#(x, 0()) -> c_9()
, <^#(S(x), S(y)) -> c_10(<^#(x, y))
, <^#(0(), S(y)) -> c_11() }
Weak Trs:
{ <(x, 0()) -> False()
, <(S(x), S(y)) -> <(x, y)
, <(0(), S(y)) -> True() }
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_1) = {1}, Uargs(ordered[Ite]^#) = {1}, Uargs(c_6) = {1},
Uargs(c_7) = {1}, Uargs(c_10) = {1}
TcT has computed the following constructor-restricted matrix
interpretation.
[True] = [0]
[0]
[<](x1, x2) = [0]
[0]
[S](x1) = [1 0] x1 + [0]
[0 0] [0]
[Cons](x1, x2) = [0]
[0]
[Nil] = [0]
[0]
[0] = [0]
[0]
[False] = [0]
[0]
[ordered^#](x1) = [0]
[0]
[c_1](x1) = [1 0] x1 + [2]
[0 1] [1]
[ordered[Ite]^#](x1, x2) = [2 0] x1 + [0]
[0 0] [0]
[c_2] = [0]
[0]
[c_3] = [0]
[0]
[notEmpty^#](x1) = [0]
[0]
[c_4] = [0]
[0]
[c_5] = [0]
[0]
[goal^#](x1) = [1 1] x1 + [2]
[1 2] [2]
[c_6](x1) = [1 0] x1 + [0]
[0 1] [0]
[c_7](x1) = [1 0] x1 + [0]
[0 1] [0]
[c_8] = [0]
[0]
[<^#](x1, x2) = [0]
[0]
[c_9] = [0]
[0]
[c_10](x1) = [1 0] x1 + [0]
[0 1] [0]
[c_11] = [0]
[0]
The order satisfies the following ordering constraints:
[<(x, 0())] = [0]
[0]
>= [0]
[0]
= [False()]
[<(S(x), S(y))] = [0]
[0]
>= [0]
[0]
= [<(x, y)]
[<(0(), S(y))] = [0]
[0]
>= [0]
[0]
= [True()]
[ordered^#(Cons(x', Cons(x, xs)))] = [0]
[0]
? [2]
[1]
= [c_1(ordered[Ite]^#(<(x', x), Cons(x', Cons(x, xs))))]
[ordered^#(Cons(x, Nil()))] = [0]
[0]
>= [0]
[0]
= [c_2()]
[ordered^#(Nil())] = [0]
[0]
>= [0]
[0]
= [c_3()]
[ordered[Ite]^#(True(), Cons(x', Cons(x, xs)))] = [0]
[0]
>= [0]
[0]
= [c_7(ordered^#(xs))]
[ordered[Ite]^#(False(), xs)] = [0]
[0]
>= [0]
[0]
= [c_8()]
[notEmpty^#(Cons(x, xs))] = [0]
[0]
>= [0]
[0]
= [c_4()]
[notEmpty^#(Nil())] = [0]
[0]
>= [0]
[0]
= [c_5()]
[goal^#(xs)] = [1 1] xs + [2]
[1 2] [2]
> [0]
[0]
= [c_6(ordered^#(xs))]
[<^#(x, 0())] = [0]
[0]
>= [0]
[0]
= [c_9()]
[<^#(S(x), S(y))] = [0]
[0]
>= [0]
[0]
= [c_10(<^#(x, y))]
[<^#(0(), S(y))] = [0]
[0]
>= [0]
[0]
= [c_11()]
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:
{ ordered^#(Cons(x', Cons(x, xs))) ->
c_1(ordered[Ite]^#(<(x', x), Cons(x', Cons(x, xs))))
, ordered^#(Cons(x, Nil())) -> c_2()
, ordered^#(Nil()) -> c_3()
, notEmpty^#(Cons(x, xs)) -> c_4()
, notEmpty^#(Nil()) -> c_5() }
Weak DPs:
{ ordered[Ite]^#(True(), Cons(x', Cons(x, xs))) ->
c_7(ordered^#(xs))
, ordered[Ite]^#(False(), xs) -> c_8()
, goal^#(xs) -> c_6(ordered^#(xs))
, <^#(x, 0()) -> c_9()
, <^#(S(x), S(y)) -> c_10(<^#(x, y))
, <^#(0(), S(y)) -> c_11() }
Weak Trs:
{ <(x, 0()) -> False()
, <(S(x), S(y)) -> <(x, y)
, <(0(), S(y)) -> True() }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We estimate the number of application of {4,5} by applications of
Pre({4,5}) = {}. Here rules are labeled as follows:
DPs:
{ 1: ordered^#(Cons(x', Cons(x, xs))) ->
c_1(ordered[Ite]^#(<(x', x), Cons(x', Cons(x, xs))))
, 2: ordered^#(Cons(x, Nil())) -> c_2()
, 3: ordered^#(Nil()) -> c_3()
, 4: notEmpty^#(Cons(x, xs)) -> c_4()
, 5: notEmpty^#(Nil()) -> c_5()
, 6: ordered[Ite]^#(True(), Cons(x', Cons(x, xs))) ->
c_7(ordered^#(xs))
, 7: ordered[Ite]^#(False(), xs) -> c_8()
, 8: goal^#(xs) -> c_6(ordered^#(xs))
, 9: <^#(x, 0()) -> c_9()
, 10: <^#(S(x), S(y)) -> c_10(<^#(x, y))
, 11: <^#(0(), S(y)) -> c_11() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ ordered^#(Cons(x', Cons(x, xs))) ->
c_1(ordered[Ite]^#(<(x', x), Cons(x', Cons(x, xs))))
, ordered^#(Cons(x, Nil())) -> c_2()
, ordered^#(Nil()) -> c_3() }
Weak DPs:
{ ordered[Ite]^#(True(), Cons(x', Cons(x, xs))) ->
c_7(ordered^#(xs))
, ordered[Ite]^#(False(), xs) -> c_8()
, notEmpty^#(Cons(x, xs)) -> c_4()
, notEmpty^#(Nil()) -> c_5()
, goal^#(xs) -> c_6(ordered^#(xs))
, <^#(x, 0()) -> c_9()
, <^#(S(x), S(y)) -> c_10(<^#(x, y))
, <^#(0(), S(y)) -> c_11() }
Weak Trs:
{ <(x, 0()) -> False()
, <(S(x), S(y)) -> <(x, y)
, <(0(), S(y)) -> True() }
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.
{ ordered[Ite]^#(False(), xs) -> c_8()
, notEmpty^#(Cons(x, xs)) -> c_4()
, notEmpty^#(Nil()) -> c_5()
, <^#(x, 0()) -> c_9()
, <^#(S(x), S(y)) -> c_10(<^#(x, y))
, <^#(0(), S(y)) -> c_11() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ ordered^#(Cons(x', Cons(x, xs))) ->
c_1(ordered[Ite]^#(<(x', x), Cons(x', Cons(x, xs))))
, ordered^#(Cons(x, Nil())) -> c_2()
, ordered^#(Nil()) -> c_3() }
Weak DPs:
{ ordered[Ite]^#(True(), Cons(x', Cons(x, xs))) ->
c_7(ordered^#(xs))
, goal^#(xs) -> c_6(ordered^#(xs)) }
Weak Trs:
{ <(x, 0()) -> False()
, <(S(x), S(y)) -> <(x, y)
, <(0(), S(y)) -> True() }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
Consider the dependency graph
1: ordered^#(Cons(x', Cons(x, xs))) ->
c_1(ordered[Ite]^#(<(x', x), Cons(x', Cons(x, xs))))
-->_1 ordered[Ite]^#(True(), Cons(x', Cons(x, xs))) ->
c_7(ordered^#(xs)) :4
2: ordered^#(Cons(x, Nil())) -> c_2()
3: ordered^#(Nil()) -> c_3()
4: ordered[Ite]^#(True(), Cons(x', Cons(x, xs))) ->
c_7(ordered^#(xs))
-->_1 ordered^#(Nil()) -> c_3() :3
-->_1 ordered^#(Cons(x, Nil())) -> c_2() :2
-->_1 ordered^#(Cons(x', Cons(x, xs))) ->
c_1(ordered[Ite]^#(<(x', x), Cons(x', Cons(x, xs)))) :1
5: goal^#(xs) -> c_6(ordered^#(xs))
-->_1 ordered^#(Nil()) -> c_3() :3
-->_1 ordered^#(Cons(x, Nil())) -> c_2() :2
-->_1 ordered^#(Cons(x', Cons(x, xs))) ->
c_1(ordered[Ite]^#(<(x', x), Cons(x', Cons(x, 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_6(ordered^#(xs)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ ordered^#(Cons(x', Cons(x, xs))) ->
c_1(ordered[Ite]^#(<(x', x), Cons(x', Cons(x, xs))))
, ordered^#(Cons(x, Nil())) -> c_2()
, ordered^#(Nil()) -> c_3() }
Weak DPs:
{ ordered[Ite]^#(True(), Cons(x', Cons(x, xs))) ->
c_7(ordered^#(xs)) }
Weak Trs:
{ <(x, 0()) -> False()
, <(S(x), S(y)) -> <(x, y)
, <(0(), S(y)) -> True() }
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:
{ 2: ordered^#(Cons(x, Nil())) -> c_2()
, 3: ordered^#(Nil()) -> c_3() }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_1) = {1}, Uargs(c_7) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[True] = [0]
[<](x1, x2) = [0]
[S](x1) = [1] x1 + [0]
[Cons](x1, x2) = [0]
[Nil] = [0]
[0] = [0]
[False] = [0]
[ordered^#](x1) = [1]
[c_1](x1) = [1] x1 + [0]
[ordered[Ite]^#](x1, x2) = [1]
[c_2] = [0]
[c_3] = [0]
[c_7](x1) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[<(x, 0())] = [0]
>= [0]
= [False()]
[<(S(x), S(y))] = [0]
>= [0]
= [<(x, y)]
[<(0(), S(y))] = [0]
>= [0]
= [True()]
[ordered^#(Cons(x', Cons(x, xs)))] = [1]
>= [1]
= [c_1(ordered[Ite]^#(<(x', x), Cons(x', Cons(x, xs))))]
[ordered^#(Cons(x, Nil()))] = [1]
> [0]
= [c_2()]
[ordered^#(Nil())] = [1]
> [0]
= [c_3()]
[ordered[Ite]^#(True(), Cons(x', Cons(x, xs)))] = [1]
>= [1]
= [c_7(ordered^#(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(n^1)).
Strict DPs:
{ ordered^#(Cons(x', Cons(x, xs))) ->
c_1(ordered[Ite]^#(<(x', x), Cons(x', Cons(x, xs)))) }
Weak DPs:
{ ordered^#(Cons(x, Nil())) -> c_2()
, ordered^#(Nil()) -> c_3()
, ordered[Ite]^#(True(), Cons(x', Cons(x, xs))) ->
c_7(ordered^#(xs)) }
Weak Trs:
{ <(x, 0()) -> False()
, <(S(x), S(y)) -> <(x, y)
, <(0(), S(y)) -> True() }
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.
{ ordered^#(Cons(x, Nil())) -> c_2()
, ordered^#(Nil()) -> c_3() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ ordered^#(Cons(x', Cons(x, xs))) ->
c_1(ordered[Ite]^#(<(x', x), Cons(x', Cons(x, xs)))) }
Weak DPs:
{ ordered[Ite]^#(True(), Cons(x', Cons(x, xs))) ->
c_7(ordered^#(xs)) }
Weak Trs:
{ <(x, 0()) -> False()
, <(S(x), S(y)) -> <(x, y)
, <(0(), S(y)) -> True() }
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: ordered^#(Cons(x', Cons(x, xs))) ->
c_1(ordered[Ite]^#(<(x', x), Cons(x', Cons(x, xs))))
, 2: ordered[Ite]^#(True(), Cons(x', Cons(x, xs))) ->
c_7(ordered^#(xs)) }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_1) = {1}, Uargs(c_7) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[True] = [0]
[<](x1, x2) = [0]
[S](x1) = [1] x1 + [0]
[Cons](x1, x2) = [1] x2 + [4]
[Nil] = [0]
[0] = [0]
[False] = [0]
[ordered^#](x1) = [1] x1 + [4]
[c_1](x1) = [1] x1 + [1]
[ordered[Ite]^#](x1, x2) = [1] x2 + [1]
[c_2] = [0]
[c_3] = [0]
[c_7](x1) = [1] x1 + [4]
The order satisfies the following ordering constraints:
[<(x, 0())] = [0]
>= [0]
= [False()]
[<(S(x), S(y))] = [0]
>= [0]
= [<(x, y)]
[<(0(), S(y))] = [0]
>= [0]
= [True()]
[ordered^#(Cons(x', Cons(x, xs)))] = [1] xs + [12]
> [1] xs + [10]
= [c_1(ordered[Ite]^#(<(x', x), Cons(x', Cons(x, xs))))]
[ordered[Ite]^#(True(), Cons(x', Cons(x, xs)))] = [1] xs + [9]
> [1] xs + [8]
= [c_7(ordered^#(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:
{ ordered^#(Cons(x', Cons(x, xs))) ->
c_1(ordered[Ite]^#(<(x', x), Cons(x', Cons(x, xs))))
, ordered[Ite]^#(True(), Cons(x', Cons(x, xs))) ->
c_7(ordered^#(xs)) }
Weak Trs:
{ <(x, 0()) -> False()
, <(S(x), S(y)) -> <(x, y)
, <(0(), S(y)) -> True() }
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.
{ ordered^#(Cons(x', Cons(x, xs))) ->
c_1(ordered[Ite]^#(<(x', x), Cons(x', Cons(x, xs))))
, ordered[Ite]^#(True(), Cons(x', Cons(x, xs))) ->
c_7(ordered^#(xs)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Weak Trs:
{ <(x, 0()) -> False()
, <(S(x), S(y)) -> <(x, y)
, <(0(), S(y)) -> True() }
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^1))