We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ dec(Cons(Cons(x', xs'), Cons(x, xs))) -> dec(Cons(x, xs))
, dec(Cons(Cons(x, xs), Nil())) -> dec(Nil())
, dec(Cons(Nil(), Cons(x, xs))) -> dec(Cons(x, xs))
, dec(Cons(Nil(), Nil())) -> Nil()
, isNilNil(Cons(Cons(x', xs'), Cons(x, xs))) -> False()
, isNilNil(Cons(Cons(x, xs), Nil())) -> False()
, isNilNil(Cons(Nil(), Cons(x, xs))) -> False()
, isNilNil(Cons(Nil(), Nil())) -> True()
, nestdec(Cons(x, xs)) -> nestdec(dec(Cons(x, xs)))
, nestdec(Nil()) ->
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Nil())))))))))))))))))
, number17(n) ->
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Nil())))))))))))))))))
, goal(x) -> nestdec(x) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We add the following weak dependency pairs:
Strict DPs:
{ dec^#(Cons(Cons(x', xs'), Cons(x, xs))) ->
c_1(dec^#(Cons(x, xs)))
, dec^#(Cons(Cons(x, xs), Nil())) -> c_2(dec^#(Nil()))
, dec^#(Cons(Nil(), Cons(x, xs))) -> c_3(dec^#(Cons(x, xs)))
, dec^#(Cons(Nil(), Nil())) -> c_4()
, isNilNil^#(Cons(Cons(x', xs'), Cons(x, xs))) -> c_5()
, isNilNil^#(Cons(Cons(x, xs), Nil())) -> c_6()
, isNilNil^#(Cons(Nil(), Cons(x, xs))) -> c_7()
, isNilNil^#(Cons(Nil(), Nil())) -> c_8()
, nestdec^#(Cons(x, xs)) -> c_9(nestdec^#(dec(Cons(x, xs))))
, nestdec^#(Nil()) -> c_10()
, number17^#(n) -> c_11()
, goal^#(x) -> c_12(nestdec^#(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^1)).
Strict DPs:
{ dec^#(Cons(Cons(x', xs'), Cons(x, xs))) ->
c_1(dec^#(Cons(x, xs)))
, dec^#(Cons(Cons(x, xs), Nil())) -> c_2(dec^#(Nil()))
, dec^#(Cons(Nil(), Cons(x, xs))) -> c_3(dec^#(Cons(x, xs)))
, dec^#(Cons(Nil(), Nil())) -> c_4()
, isNilNil^#(Cons(Cons(x', xs'), Cons(x, xs))) -> c_5()
, isNilNil^#(Cons(Cons(x, xs), Nil())) -> c_6()
, isNilNil^#(Cons(Nil(), Cons(x, xs))) -> c_7()
, isNilNil^#(Cons(Nil(), Nil())) -> c_8()
, nestdec^#(Cons(x, xs)) -> c_9(nestdec^#(dec(Cons(x, xs))))
, nestdec^#(Nil()) -> c_10()
, number17^#(n) -> c_11()
, goal^#(x) -> c_12(nestdec^#(x)) }
Strict Trs:
{ dec(Cons(Cons(x', xs'), Cons(x, xs))) -> dec(Cons(x, xs))
, dec(Cons(Cons(x, xs), Nil())) -> dec(Nil())
, dec(Cons(Nil(), Cons(x, xs))) -> dec(Cons(x, xs))
, dec(Cons(Nil(), Nil())) -> Nil()
, isNilNil(Cons(Cons(x', xs'), Cons(x, xs))) -> False()
, isNilNil(Cons(Cons(x, xs), Nil())) -> False()
, isNilNil(Cons(Nil(), Cons(x, xs))) -> False()
, isNilNil(Cons(Nil(), Nil())) -> True()
, nestdec(Cons(x, xs)) -> nestdec(dec(Cons(x, xs)))
, nestdec(Nil()) ->
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Nil())))))))))))))))))
, number17(n) ->
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Cons(Nil(),
Nil())))))))))))))))))
, goal(x) -> nestdec(x) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We replace rewrite rules by usable rules:
Strict Usable Rules:
{ dec(Cons(Cons(x', xs'), Cons(x, xs))) -> dec(Cons(x, xs))
, dec(Cons(Cons(x, xs), Nil())) -> dec(Nil())
, dec(Cons(Nil(), Cons(x, xs))) -> dec(Cons(x, xs))
, dec(Cons(Nil(), Nil())) -> Nil() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ dec^#(Cons(Cons(x', xs'), Cons(x, xs))) ->
c_1(dec^#(Cons(x, xs)))
, dec^#(Cons(Cons(x, xs), Nil())) -> c_2(dec^#(Nil()))
, dec^#(Cons(Nil(), Cons(x, xs))) -> c_3(dec^#(Cons(x, xs)))
, dec^#(Cons(Nil(), Nil())) -> c_4()
, isNilNil^#(Cons(Cons(x', xs'), Cons(x, xs))) -> c_5()
, isNilNil^#(Cons(Cons(x, xs), Nil())) -> c_6()
, isNilNil^#(Cons(Nil(), Cons(x, xs))) -> c_7()
, isNilNil^#(Cons(Nil(), Nil())) -> c_8()
, nestdec^#(Cons(x, xs)) -> c_9(nestdec^#(dec(Cons(x, xs))))
, nestdec^#(Nil()) -> c_10()
, number17^#(n) -> c_11()
, goal^#(x) -> c_12(nestdec^#(x)) }
Strict Trs:
{ dec(Cons(Cons(x', xs'), Cons(x, xs))) -> dec(Cons(x, xs))
, dec(Cons(Cons(x, xs), Nil())) -> dec(Nil())
, dec(Cons(Nil(), Cons(x, xs))) -> dec(Cons(x, xs))
, dec(Cons(Nil(), Nil())) -> Nil() }
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(c_3) = {1}, Uargs(nestdec^#) = {1},
Uargs(c_9) = {1}, Uargs(c_12) = {1}
TcT has computed the following constructor-restricted matrix
interpretation.
[dec](x1) = [1 1] x1 + [1]
[2 0] [2]
[Cons](x1, x2) = [0 0] x2 + [1]
[0 1] [1]
[Nil] = [1]
[2]
[dec^#](x1) = [0]
[0]
[c_1](x1) = [1 0] x1 + [0]
[0 1] [0]
[c_2](x1) = [0]
[0]
[c_3](x1) = [1 0] x1 + [0]
[0 1] [0]
[c_4] = [0]
[0]
[isNilNil^#](x1) = [0]
[0]
[c_5] = [0]
[0]
[c_6] = [0]
[0]
[c_7] = [0]
[0]
[c_8] = [0]
[0]
[nestdec^#](x1) = [1 1] x1 + [0]
[0 0] [0]
[c_9](x1) = [1 0] x1 + [0]
[0 1] [0]
[c_10] = [0]
[0]
[number17^#](x1) = [0]
[0]
[c_11] = [0]
[0]
[goal^#](x1) = [2 2] x1 + [0]
[0 0] [0]
[c_12](x1) = [1 0] x1 + [0]
[0 1] [0]
The order satisfies the following ordering constraints:
[dec(Cons(Cons(x', xs'), Cons(x, xs)))] = [0 1] xs + [4]
[0 0] [4]
> [0 1] xs + [3]
[0 0] [4]
= [dec(Cons(x, xs))]
[dec(Cons(Cons(x, xs), Nil()))] = [5]
[4]
> [4]
[4]
= [dec(Nil())]
[dec(Cons(Nil(), Cons(x, xs)))] = [0 1] xs + [4]
[0 0] [4]
> [0 1] xs + [3]
[0 0] [4]
= [dec(Cons(x, xs))]
[dec(Cons(Nil(), Nil()))] = [5]
[4]
> [1]
[2]
= [Nil()]
[dec^#(Cons(Cons(x', xs'), Cons(x, xs)))] = [0]
[0]
>= [0]
[0]
= [c_1(dec^#(Cons(x, xs)))]
[dec^#(Cons(Cons(x, xs), Nil()))] = [0]
[0]
>= [0]
[0]
= [c_2(dec^#(Nil()))]
[dec^#(Cons(Nil(), Cons(x, xs)))] = [0]
[0]
>= [0]
[0]
= [c_3(dec^#(Cons(x, xs)))]
[dec^#(Cons(Nil(), Nil()))] = [0]
[0]
>= [0]
[0]
= [c_4()]
[isNilNil^#(Cons(Cons(x', xs'), Cons(x, xs)))] = [0]
[0]
>= [0]
[0]
= [c_5()]
[isNilNil^#(Cons(Cons(x, xs), Nil()))] = [0]
[0]
>= [0]
[0]
= [c_6()]
[isNilNil^#(Cons(Nil(), Cons(x, xs)))] = [0]
[0]
>= [0]
[0]
= [c_7()]
[isNilNil^#(Cons(Nil(), Nil()))] = [0]
[0]
>= [0]
[0]
= [c_8()]
[nestdec^#(Cons(x, xs))] = [0 1] xs + [2]
[0 0] [0]
? [0 1] xs + [7]
[0 0] [0]
= [c_9(nestdec^#(dec(Cons(x, xs))))]
[nestdec^#(Nil())] = [3]
[0]
> [0]
[0]
= [c_10()]
[number17^#(n)] = [0]
[0]
>= [0]
[0]
= [c_11()]
[goal^#(x)] = [2 2] x + [0]
[0 0] [0]
>= [1 1] x + [0]
[0 0] [0]
= [c_12(nestdec^#(x))]
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:
{ dec^#(Cons(Cons(x', xs'), Cons(x, xs))) ->
c_1(dec^#(Cons(x, xs)))
, dec^#(Cons(Cons(x, xs), Nil())) -> c_2(dec^#(Nil()))
, dec^#(Cons(Nil(), Cons(x, xs))) -> c_3(dec^#(Cons(x, xs)))
, dec^#(Cons(Nil(), Nil())) -> c_4()
, isNilNil^#(Cons(Cons(x', xs'), Cons(x, xs))) -> c_5()
, isNilNil^#(Cons(Cons(x, xs), Nil())) -> c_6()
, isNilNil^#(Cons(Nil(), Cons(x, xs))) -> c_7()
, isNilNil^#(Cons(Nil(), Nil())) -> c_8()
, nestdec^#(Cons(x, xs)) -> c_9(nestdec^#(dec(Cons(x, xs))))
, number17^#(n) -> c_11()
, goal^#(x) -> c_12(nestdec^#(x)) }
Weak DPs: { nestdec^#(Nil()) -> c_10() }
Weak Trs:
{ dec(Cons(Cons(x', xs'), Cons(x, xs))) -> dec(Cons(x, xs))
, dec(Cons(Cons(x, xs), Nil())) -> dec(Nil())
, dec(Cons(Nil(), Cons(x, xs))) -> dec(Cons(x, xs))
, dec(Cons(Nil(), Nil())) -> Nil() }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We estimate the number of application of {2,4,5,6,7,8,9,10} by
applications of Pre({2,4,5,6,7,8,9,10}) = {1,3,11}. Here rules are
labeled as follows:
DPs:
{ 1: dec^#(Cons(Cons(x', xs'), Cons(x, xs))) ->
c_1(dec^#(Cons(x, xs)))
, 2: dec^#(Cons(Cons(x, xs), Nil())) -> c_2(dec^#(Nil()))
, 3: dec^#(Cons(Nil(), Cons(x, xs))) -> c_3(dec^#(Cons(x, xs)))
, 4: dec^#(Cons(Nil(), Nil())) -> c_4()
, 5: isNilNil^#(Cons(Cons(x', xs'), Cons(x, xs))) -> c_5()
, 6: isNilNil^#(Cons(Cons(x, xs), Nil())) -> c_6()
, 7: isNilNil^#(Cons(Nil(), Cons(x, xs))) -> c_7()
, 8: isNilNil^#(Cons(Nil(), Nil())) -> c_8()
, 9: nestdec^#(Cons(x, xs)) -> c_9(nestdec^#(dec(Cons(x, xs))))
, 10: number17^#(n) -> c_11()
, 11: goal^#(x) -> c_12(nestdec^#(x))
, 12: nestdec^#(Nil()) -> c_10() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ dec^#(Cons(Cons(x', xs'), Cons(x, xs))) ->
c_1(dec^#(Cons(x, xs)))
, dec^#(Cons(Nil(), Cons(x, xs))) -> c_3(dec^#(Cons(x, xs)))
, goal^#(x) -> c_12(nestdec^#(x)) }
Weak DPs:
{ dec^#(Cons(Cons(x, xs), Nil())) -> c_2(dec^#(Nil()))
, dec^#(Cons(Nil(), Nil())) -> c_4()
, isNilNil^#(Cons(Cons(x', xs'), Cons(x, xs))) -> c_5()
, isNilNil^#(Cons(Cons(x, xs), Nil())) -> c_6()
, isNilNil^#(Cons(Nil(), Cons(x, xs))) -> c_7()
, isNilNil^#(Cons(Nil(), Nil())) -> c_8()
, nestdec^#(Cons(x, xs)) -> c_9(nestdec^#(dec(Cons(x, xs))))
, nestdec^#(Nil()) -> c_10()
, number17^#(n) -> c_11() }
Weak Trs:
{ dec(Cons(Cons(x', xs'), Cons(x, xs))) -> dec(Cons(x, xs))
, dec(Cons(Cons(x, xs), Nil())) -> dec(Nil())
, dec(Cons(Nil(), Cons(x, xs))) -> dec(Cons(x, xs))
, dec(Cons(Nil(), Nil())) -> Nil() }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We estimate the number of application of {3} by applications of
Pre({3}) = {}. Here rules are labeled as follows:
DPs:
{ 1: dec^#(Cons(Cons(x', xs'), Cons(x, xs))) ->
c_1(dec^#(Cons(x, xs)))
, 2: dec^#(Cons(Nil(), Cons(x, xs))) -> c_3(dec^#(Cons(x, xs)))
, 3: goal^#(x) -> c_12(nestdec^#(x))
, 4: dec^#(Cons(Cons(x, xs), Nil())) -> c_2(dec^#(Nil()))
, 5: dec^#(Cons(Nil(), Nil())) -> c_4()
, 6: isNilNil^#(Cons(Cons(x', xs'), Cons(x, xs))) -> c_5()
, 7: isNilNil^#(Cons(Cons(x, xs), Nil())) -> c_6()
, 8: isNilNil^#(Cons(Nil(), Cons(x, xs))) -> c_7()
, 9: isNilNil^#(Cons(Nil(), Nil())) -> c_8()
, 10: nestdec^#(Cons(x, xs)) -> c_9(nestdec^#(dec(Cons(x, xs))))
, 11: nestdec^#(Nil()) -> c_10()
, 12: number17^#(n) -> c_11() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ dec^#(Cons(Cons(x', xs'), Cons(x, xs))) ->
c_1(dec^#(Cons(x, xs)))
, dec^#(Cons(Nil(), Cons(x, xs))) -> c_3(dec^#(Cons(x, xs))) }
Weak DPs:
{ dec^#(Cons(Cons(x, xs), Nil())) -> c_2(dec^#(Nil()))
, dec^#(Cons(Nil(), Nil())) -> c_4()
, isNilNil^#(Cons(Cons(x', xs'), Cons(x, xs))) -> c_5()
, isNilNil^#(Cons(Cons(x, xs), Nil())) -> c_6()
, isNilNil^#(Cons(Nil(), Cons(x, xs))) -> c_7()
, isNilNil^#(Cons(Nil(), Nil())) -> c_8()
, nestdec^#(Cons(x, xs)) -> c_9(nestdec^#(dec(Cons(x, xs))))
, nestdec^#(Nil()) -> c_10()
, number17^#(n) -> c_11()
, goal^#(x) -> c_12(nestdec^#(x)) }
Weak Trs:
{ dec(Cons(Cons(x', xs'), Cons(x, xs))) -> dec(Cons(x, xs))
, dec(Cons(Cons(x, xs), Nil())) -> dec(Nil())
, dec(Cons(Nil(), Cons(x, xs))) -> dec(Cons(x, xs))
, dec(Cons(Nil(), Nil())) -> Nil() }
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.
{ dec^#(Cons(Cons(x, xs), Nil())) -> c_2(dec^#(Nil()))
, dec^#(Cons(Nil(), Nil())) -> c_4()
, isNilNil^#(Cons(Cons(x', xs'), Cons(x, xs))) -> c_5()
, isNilNil^#(Cons(Cons(x, xs), Nil())) -> c_6()
, isNilNil^#(Cons(Nil(), Cons(x, xs))) -> c_7()
, isNilNil^#(Cons(Nil(), Nil())) -> c_8()
, nestdec^#(Cons(x, xs)) -> c_9(nestdec^#(dec(Cons(x, xs))))
, nestdec^#(Nil()) -> c_10()
, number17^#(n) -> c_11()
, goal^#(x) -> c_12(nestdec^#(x)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ dec^#(Cons(Cons(x', xs'), Cons(x, xs))) ->
c_1(dec^#(Cons(x, xs)))
, dec^#(Cons(Nil(), Cons(x, xs))) -> c_3(dec^#(Cons(x, xs))) }
Weak Trs:
{ dec(Cons(Cons(x', xs'), Cons(x, xs))) -> dec(Cons(x, xs))
, dec(Cons(Cons(x, xs), Nil())) -> dec(Nil())
, dec(Cons(Nil(), Cons(x, xs))) -> dec(Cons(x, xs))
, dec(Cons(Nil(), Nil())) -> Nil() }
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:
{ dec^#(Cons(Cons(x', xs'), Cons(x, xs))) ->
c_1(dec^#(Cons(x, xs)))
, dec^#(Cons(Nil(), Cons(x, xs))) -> c_3(dec^#(Cons(x, xs))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We use the processor 'Small Polynomial Path Order (PS,1-bounded)'
to orient following rules strictly.
DPs:
{ 1: dec^#(Cons(Cons(x', xs'), Cons(x, xs))) ->
c_1(dec^#(Cons(x, xs)))
, 2: dec^#(Cons(Nil(), Cons(x, xs))) -> c_3(dec^#(Cons(x, xs))) }
Sub-proof:
----------
The input was oriented with the instance of 'Small Polynomial Path
Order (PS,1-bounded)' as induced by the safe mapping
safe(Cons) = {1, 2}, safe(Nil) = {}, safe(dec^#) = {},
safe(c_1) = {}, safe(c_3) = {}
and precedence
empty .
Following symbols are considered recursive:
{dec^#}
The recursion depth is 1.
Further, following argument filtering is employed:
pi(Cons) = [2], pi(Nil) = [], pi(dec^#) = [1], pi(c_1) = [1],
pi(c_3) = [1]
Usable defined function symbols are a subset of:
{dec^#}
For your convenience, here are the satisfied ordering constraints:
pi(dec^#(Cons(Cons(x', xs'), Cons(x, xs)))) = dec^#(Cons(; Cons(; xs));)
> c_1(dec^#(Cons(; xs););)
= pi(c_1(dec^#(Cons(x, xs))))
pi(dec^#(Cons(Nil(), Cons(x, xs)))) = dec^#(Cons(; Cons(; xs));)
> c_3(dec^#(Cons(; xs););)
= pi(c_3(dec^#(Cons(x, 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:
{ dec^#(Cons(Cons(x', xs'), Cons(x, xs))) ->
c_1(dec^#(Cons(x, xs)))
, dec^#(Cons(Nil(), Cons(x, xs))) -> c_3(dec^#(Cons(x, xs))) }
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.
{ dec^#(Cons(Cons(x', xs'), Cons(x, xs))) ->
c_1(dec^#(Cons(x, xs)))
, dec^#(Cons(Nil(), Cons(x, xs))) -> c_3(dec^#(Cons(x, xs))) }
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))