YES(O(1),O(n^2))
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict Trs:
{ rec[append_0][2](Nil(), Cons(x6, Nil())) -> Cons(x6, Nil())
, rec[append_0][2](Cons(x14, x10), Cons(x6, Nil())) ->
Cons(x14, rec[append_0][2](x10, Cons(x6, Nil())))
, rec[rev_0][1](Nil()) -> Nil()
, rec[rev_0][1](Cons(x5, x3)) ->
rec[append_0][2](rec[rev_0][1](x3), Cons(x5, Nil()))
, main(x1) -> rec[rev_0][1](x1) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(Cons) = {2}, Uargs(rec[append_0][2]) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[Nil] = [0]
[Cons](x1, x2) = [1] x2 + [0]
[rec[append_0][2]](x1, x2) = [1] x1 + [0]
[rec[rev_0][1]](x1) = [0]
[main](x1) = [1] x1 + [7]
The following symbols are considered usable
{rec[append_0][2], rec[rev_0][1], main}
The order satisfies the following ordering constraints:
[rec[append_0][2](Nil(), Cons(x6, Nil()))] = [0]
>= [0]
= [Cons(x6, Nil())]
[rec[append_0][2](Cons(x14, x10), Cons(x6, Nil()))] = [1] x10 + [0]
>= [1] x10 + [0]
= [Cons(x14, rec[append_0][2](x10, Cons(x6, Nil())))]
[rec[rev_0][1](Nil())] = [0]
>= [0]
= [Nil()]
[rec[rev_0][1](Cons(x5, x3))] = [0]
>= [0]
= [rec[append_0][2](rec[rev_0][1](x3), Cons(x5, Nil()))]
[main(x1)] = [1] x1 + [7]
> [0]
= [rec[rev_0][1](x1)]
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^2)).
Strict Trs:
{ rec[append_0][2](Nil(), Cons(x6, Nil())) -> Cons(x6, Nil())
, rec[append_0][2](Cons(x14, x10), Cons(x6, Nil())) ->
Cons(x14, rec[append_0][2](x10, Cons(x6, Nil())))
, rec[rev_0][1](Nil()) -> Nil()
, rec[rev_0][1](Cons(x5, x3)) ->
rec[append_0][2](rec[rev_0][1](x3), Cons(x5, Nil())) }
Weak Trs: { main(x1) -> rec[rev_0][1](x1) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(Cons) = {2}, Uargs(rec[append_0][2]) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[Nil] = [0]
[Cons](x1, x2) = [1] x2 + [0]
[rec[append_0][2]](x1, x2) = [1] x1 + [0]
[rec[rev_0][1]](x1) = [1]
[main](x1) = [1] x1 + [7]
The following symbols are considered usable
{rec[append_0][2], rec[rev_0][1], main}
The order satisfies the following ordering constraints:
[rec[append_0][2](Nil(), Cons(x6, Nil()))] = [0]
>= [0]
= [Cons(x6, Nil())]
[rec[append_0][2](Cons(x14, x10), Cons(x6, Nil()))] = [1] x10 + [0]
>= [1] x10 + [0]
= [Cons(x14, rec[append_0][2](x10, Cons(x6, Nil())))]
[rec[rev_0][1](Nil())] = [1]
> [0]
= [Nil()]
[rec[rev_0][1](Cons(x5, x3))] = [1]
>= [1]
= [rec[append_0][2](rec[rev_0][1](x3), Cons(x5, Nil()))]
[main(x1)] = [1] x1 + [7]
> [1]
= [rec[rev_0][1](x1)]
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^2)).
Strict Trs:
{ rec[append_0][2](Nil(), Cons(x6, Nil())) -> Cons(x6, Nil())
, rec[append_0][2](Cons(x14, x10), Cons(x6, Nil())) ->
Cons(x14, rec[append_0][2](x10, Cons(x6, Nil())))
, rec[rev_0][1](Cons(x5, x3)) ->
rec[append_0][2](rec[rev_0][1](x3), Cons(x5, Nil())) }
Weak Trs:
{ rec[rev_0][1](Nil()) -> Nil()
, main(x1) -> rec[rev_0][1](x1) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(Cons) = {2}, Uargs(rec[append_0][2]) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[Nil] = [0]
[Cons](x1, x2) = [1] x2 + [0]
[rec[append_0][2]](x1, x2) = [1] x1 + [1]
[rec[rev_0][1]](x1) = [0]
[main](x1) = [1] x1 + [7]
The following symbols are considered usable
{rec[append_0][2], rec[rev_0][1], main}
The order satisfies the following ordering constraints:
[rec[append_0][2](Nil(), Cons(x6, Nil()))] = [1]
> [0]
= [Cons(x6, Nil())]
[rec[append_0][2](Cons(x14, x10), Cons(x6, Nil()))] = [1] x10 + [1]
>= [1] x10 + [1]
= [Cons(x14, rec[append_0][2](x10, Cons(x6, Nil())))]
[rec[rev_0][1](Nil())] = [0]
>= [0]
= [Nil()]
[rec[rev_0][1](Cons(x5, x3))] = [0]
? [1]
= [rec[append_0][2](rec[rev_0][1](x3), Cons(x5, Nil()))]
[main(x1)] = [1] x1 + [7]
> [0]
= [rec[rev_0][1](x1)]
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^2)).
Strict Trs:
{ rec[append_0][2](Cons(x14, x10), Cons(x6, Nil())) ->
Cons(x14, rec[append_0][2](x10, Cons(x6, Nil())))
, rec[rev_0][1](Cons(x5, x3)) ->
rec[append_0][2](rec[rev_0][1](x3), Cons(x5, Nil())) }
Weak Trs:
{ rec[append_0][2](Nil(), Cons(x6, Nil())) -> Cons(x6, Nil())
, rec[rev_0][1](Nil()) -> Nil()
, main(x1) -> rec[rev_0][1](x1) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.
Trs:
{ rec[rev_0][1](Cons(x5, x3)) ->
rec[append_0][2](rec[rev_0][1](x3), Cons(x5, Nil())) }
The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^1)) . These rules are moved into the corresponding weak
component(s).
Sub-proof:
----------
The following argument positions are usable:
Uargs(Cons) = {2}, Uargs(rec[append_0][2]) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[Nil] = [4]
[Cons](x1, x2) = [1] x2 + [4]
[rec[append_0][2]](x1, x2) = [1] x1 + [4]
[rec[rev_0][1]](x1) = [2] x1 + [0]
[main](x1) = [7] x1 + [7]
The following symbols are considered usable
{rec[append_0][2], rec[rev_0][1], main}
The order satisfies the following ordering constraints:
[rec[append_0][2](Nil(), Cons(x6, Nil()))] = [8]
>= [8]
= [Cons(x6, Nil())]
[rec[append_0][2](Cons(x14, x10), Cons(x6, Nil()))] = [1] x10 + [8]
>= [1] x10 + [8]
= [Cons(x14, rec[append_0][2](x10, Cons(x6, Nil())))]
[rec[rev_0][1](Nil())] = [8]
> [4]
= [Nil()]
[rec[rev_0][1](Cons(x5, x3))] = [2] x3 + [8]
> [2] x3 + [4]
= [rec[append_0][2](rec[rev_0][1](x3), Cons(x5, Nil()))]
[main(x1)] = [7] x1 + [7]
> [2] x1 + [0]
= [rec[rev_0][1](x1)]
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 Trs:
{ rec[append_0][2](Cons(x14, x10), Cons(x6, Nil())) ->
Cons(x14, rec[append_0][2](x10, Cons(x6, Nil()))) }
Weak Trs:
{ rec[append_0][2](Nil(), Cons(x6, Nil())) -> Cons(x6, Nil())
, rec[rev_0][1](Nil()) -> Nil()
, rec[rev_0][1](Cons(x5, x3)) ->
rec[append_0][2](rec[rev_0][1](x3), Cons(x5, Nil()))
, main(x1) -> rec[rev_0][1](x1) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
We use the processor 'matrix interpretation of dimension 2' to
orient following rules strictly.
Trs:
{ rec[append_0][2](Cons(x14, x10), Cons(x6, Nil())) ->
Cons(x14, rec[append_0][2](x10, Cons(x6, Nil()))) }
The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^2)) . These rules are moved into the corresponding weak
component(s).
Sub-proof:
----------
The following argument positions are usable:
Uargs(Cons) = {2}, Uargs(rec[append_0][2]) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[Nil] = [0]
[1]
[Cons](x1, x2) = [1 1] x2 + [0]
[0 1] [6]
[rec[append_0][2]](x1, x2) = [1 2] x1 + [0 0] x2 + [0]
[0 1] [0 1] [0]
[rec[rev_0][1]](x1) = [4 0] x1 + [0]
[0 2] [0]
[main](x1) = [7 7] x1 + [7]
[7 7] [7]
The following symbols are considered usable
{rec[append_0][2], rec[rev_0][1], main}
The order satisfies the following ordering constraints:
[rec[append_0][2](Nil(), Cons(x6, Nil()))] = [2]
[8]
> [1]
[7]
= [Cons(x6, Nil())]
[rec[append_0][2](Cons(x14, x10), Cons(x6, Nil()))] = [1 3] x10 + [12]
[0 1] [13]
> [1 3] x10 + [7]
[0 1] [13]
= [Cons(x14, rec[append_0][2](x10, Cons(x6, Nil())))]
[rec[rev_0][1](Nil())] = [0]
[2]
>= [0]
[1]
= [Nil()]
[rec[rev_0][1](Cons(x5, x3))] = [4 4] x3 + [0]
[0 2] [12]
>= [4 4] x3 + [0]
[0 2] [7]
= [rec[append_0][2](rec[rev_0][1](x3), Cons(x5, Nil()))]
[main(x1)] = [7 7] x1 + [7]
[7 7] [7]
> [4 0] x1 + [0]
[0 2] [0]
= [rec[rev_0][1](x1)]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Weak Trs:
{ rec[append_0][2](Nil(), Cons(x6, Nil())) -> Cons(x6, Nil())
, rec[append_0][2](Cons(x14, x10), Cons(x6, Nil())) ->
Cons(x14, rec[append_0][2](x10, Cons(x6, Nil())))
, rec[rev_0][1](Nil()) -> Nil()
, rec[rev_0][1](Cons(x5, x3)) ->
rec[append_0][2](rec[rev_0][1](x3), Cons(x5, Nil()))
, main(x1) -> rec[rev_0][1](x1) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
Empty rules are trivially bounded
Hurray, we answered YES(O(1),O(n^2))