We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ is_empty(nil()) -> true()
, is_empty(cons(x, l)) -> false()
, hd(cons(x, l)) -> x
, tl(cons(x, l)) -> l
, append(l1, l2) -> ifappend(l1, l2, l1)
, ifappend(l1, l2, nil()) -> l2
, ifappend(l1, l2, cons(x, l)) -> cons(x, append(l, l2)) }
Obligation:
runtime complexity
Answer:
YES(O(1),O(n^1))
We add the following weak dependency pairs:
Strict DPs:
{ is_empty^#(nil()) -> c_1()
, is_empty^#(cons(x, l)) -> c_2()
, hd^#(cons(x, l)) -> c_3(x)
, tl^#(cons(x, l)) -> c_4(l)
, append^#(l1, l2) -> c_5(ifappend^#(l1, l2, l1))
, ifappend^#(l1, l2, nil()) -> c_6(l2)
, ifappend^#(l1, l2, cons(x, l)) -> c_7(x, append^#(l, l2)) }
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:
{ is_empty^#(nil()) -> c_1()
, is_empty^#(cons(x, l)) -> c_2()
, hd^#(cons(x, l)) -> c_3(x)
, tl^#(cons(x, l)) -> c_4(l)
, append^#(l1, l2) -> c_5(ifappend^#(l1, l2, l1))
, ifappend^#(l1, l2, nil()) -> c_6(l2)
, ifappend^#(l1, l2, cons(x, l)) -> c_7(x, append^#(l, l2)) }
Strict Trs:
{ is_empty(nil()) -> true()
, is_empty(cons(x, l)) -> false()
, hd(cons(x, l)) -> x
, tl(cons(x, l)) -> l
, append(l1, l2) -> ifappend(l1, l2, l1)
, ifappend(l1, l2, nil()) -> l2
, ifappend(l1, l2, cons(x, l)) -> cons(x, append(l, l2)) }
Obligation:
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:
{ is_empty^#(nil()) -> c_1()
, is_empty^#(cons(x, l)) -> c_2()
, hd^#(cons(x, l)) -> c_3(x)
, tl^#(cons(x, l)) -> c_4(l)
, append^#(l1, l2) -> c_5(ifappend^#(l1, l2, l1))
, ifappend^#(l1, l2, nil()) -> c_6(l2)
, ifappend^#(l1, l2, cons(x, l)) -> c_7(x, append^#(l, l2)) }
Obligation:
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_5) = {1}, Uargs(c_7) = {2}
TcT has computed the following constructor-restricted matrix
interpretation.
[nil] = [0]
[0]
[cons](x1, x2) = [1 0] x2 + [0]
[0 1] [0]
[is_empty^#](x1) = [0]
[0]
[c_1] = [0]
[0]
[c_2] = [0]
[0]
[hd^#](x1) = [1]
[0]
[c_3](x1) = [0]
[0]
[tl^#](x1) = [0]
[0]
[c_4](x1) = [0]
[0]
[append^#](x1, x2) = [2]
[2]
[c_5](x1) = [1 0] x1 + [0]
[0 1] [0]
[ifappend^#](x1, x2, x3) = [1]
[0]
[c_6](x1) = [0]
[0]
[c_7](x1, x2) = [1 0] x2 + [2]
[0 1] [2]
The order satisfies the following ordering constraints:
[is_empty^#(nil())] = [0]
[0]
>= [0]
[0]
= [c_1()]
[is_empty^#(cons(x, l))] = [0]
[0]
>= [0]
[0]
= [c_2()]
[hd^#(cons(x, l))] = [1]
[0]
> [0]
[0]
= [c_3(x)]
[tl^#(cons(x, l))] = [0]
[0]
>= [0]
[0]
= [c_4(l)]
[append^#(l1, l2)] = [2]
[2]
> [1]
[0]
= [c_5(ifappend^#(l1, l2, l1))]
[ifappend^#(l1, l2, nil())] = [1]
[0]
> [0]
[0]
= [c_6(l2)]
[ifappend^#(l1, l2, cons(x, l))] = [1]
[0]
? [4]
[4]
= [c_7(x, append^#(l, l2))]
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(n^1)).
Strict DPs:
{ is_empty^#(nil()) -> c_1()
, is_empty^#(cons(x, l)) -> c_2()
, tl^#(cons(x, l)) -> c_4(l)
, ifappend^#(l1, l2, cons(x, l)) -> c_7(x, append^#(l, l2)) }
Weak DPs:
{ hd^#(cons(x, l)) -> c_3(x)
, append^#(l1, l2) -> c_5(ifappend^#(l1, l2, l1))
, ifappend^#(l1, l2, nil()) -> c_6(l2) }
Obligation:
runtime complexity
Answer:
YES(?,O(n^1))
We employ 'linear path analysis' using the following approximated
dependency graph:
->{3,7,6,5,4} [ ? ]
|
|->{1} [ YES(O(1),O(n^1)) ]
|
`->{2} [ YES(O(1),O(n^1)) ]
Here dependency-pairs are as follows:
Strict DPs:
{ 1: is_empty^#(nil()) -> c_1()
, 2: is_empty^#(cons(x, l)) -> c_2()
, 3: tl^#(cons(x, l)) -> c_4(l)
, 4: ifappend^#(l1, l2, cons(x, l)) -> c_7(x, append^#(l, l2)) }
Weak DPs:
{ 5: hd^#(cons(x, l)) -> c_3(x)
, 6: append^#(l1, l2) -> c_5(ifappend^#(l1, l2, l1))
, 7: ifappend^#(l1, l2, nil()) -> c_6(l2) }
* Path {3,7,6,5,4}->{1}: YES(O(1),O(n^1))
---------------------------------------
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ is_empty^#(nil()) -> c_1()
, tl^#(cons(x, l)) -> c_4(l)
, ifappend^#(l1, l2, cons(x, l)) -> c_7(x, append^#(l, l2)) }
Weak DPs:
{ hd^#(cons(x, l)) -> c_3(x)
, append^#(l1, l2) -> c_5(ifappend^#(l1, l2, l1))
, ifappend^#(l1, l2, nil()) -> c_6(l2) }
Obligation:
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: is_empty^#(nil()) -> c_1()
, 2: tl^#(cons(x, l)) -> c_4(l)
, 4: hd^#(cons(x, l)) -> c_3(x) }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_5) = {1}, Uargs(c_7) = {2}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[nil] = [0]
[cons](x1, x2) = [1] x2 + [0]
[is_empty^#](x1) = [5]
[c_1] = [0]
[hd^#](x1) = [1] x1 + [4]
[c_3](x1) = [0]
[tl^#](x1) = [1] x1 + [4]
[c_4](x1) = [1] x1 + [0]
[append^#](x1, x2) = [4] x1 + [0]
[c_5](x1) = [1] x1 + [0]
[ifappend^#](x1, x2, x3) = [4] x3 + [0]
[c_6](x1) = [0]
[c_7](x1, x2) = [1] x2 + [0]
The order satisfies the following ordering constraints:
[is_empty^#(nil())] = [5]
> [0]
= [c_1()]
[hd^#(cons(x, l))] = [1] l + [4]
> [0]
= [c_3(x)]
[tl^#(cons(x, l))] = [1] l + [4]
> [1] l + [0]
= [c_4(l)]
[append^#(l1, l2)] = [4] l1 + [0]
>= [4] l1 + [0]
= [c_5(ifappend^#(l1, l2, l1))]
[ifappend^#(l1, l2, nil())] = [0]
>= [0]
= [c_6(l2)]
[ifappend^#(l1, l2, cons(x, l))] = [4] l + [0]
>= [4] l + [0]
= [c_7(x, append^#(l, l2))]
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:
{ ifappend^#(l1, l2, cons(x, l)) -> c_7(x, append^#(l, l2)) }
Weak DPs:
{ is_empty^#(nil()) -> c_1()
, hd^#(cons(x, l)) -> c_3(x)
, tl^#(cons(x, l)) -> c_4(l)
, append^#(l1, l2) -> c_5(ifappend^#(l1, l2, l1))
, ifappend^#(l1, l2, nil()) -> c_6(l2) }
Obligation:
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.
{ is_empty^#(nil()) -> c_1() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ ifappend^#(l1, l2, cons(x, l)) -> c_7(x, append^#(l, l2)) }
Weak DPs:
{ hd^#(cons(x, l)) -> c_3(x)
, tl^#(cons(x, l)) -> c_4(l)
, append^#(l1, l2) -> c_5(ifappend^#(l1, l2, l1))
, ifappend^#(l1, l2, nil()) -> c_6(l2) }
Obligation:
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: ifappend^#(l1, l2, cons(x, l)) -> c_7(x, append^#(l, l2))
, 2: hd^#(cons(x, l)) -> c_3(x)
, 5: ifappend^#(l1, l2, nil()) -> c_6(l2) }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_5) = {1}, Uargs(c_7) = {2}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[nil] = [0]
[cons](x1, x2) = [1] x2 + [1]
[is_empty^#](x1) = [0]
[c_1] = [0]
[hd^#](x1) = [1] x1 + [0]
[c_3](x1) = [0]
[tl^#](x1) = [0]
[c_4](x1) = [0]
[append^#](x1, x2) = [4] x1 + [1]
[c_5](x1) = [1] x1 + [0]
[ifappend^#](x1, x2, x3) = [4] x3 + [1]
[c_6](x1) = [0]
[c_7](x1, x2) = [1] x2 + [2]
The order satisfies the following ordering constraints:
[hd^#(cons(x, l))] = [1] l + [1]
> [0]
= [c_3(x)]
[tl^#(cons(x, l))] = [0]
>= [0]
= [c_4(l)]
[append^#(l1, l2)] = [4] l1 + [1]
>= [4] l1 + [1]
= [c_5(ifappend^#(l1, l2, l1))]
[ifappend^#(l1, l2, nil())] = [1]
> [0]
= [c_6(l2)]
[ifappend^#(l1, l2, cons(x, l))] = [4] l + [5]
> [4] l + [3]
= [c_7(x, append^#(l, l2))]
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:
{ hd^#(cons(x, l)) -> c_3(x)
, tl^#(cons(x, l)) -> c_4(l)
, append^#(l1, l2) -> c_5(ifappend^#(l1, l2, l1))
, ifappend^#(l1, l2, nil()) -> c_6(l2)
, ifappend^#(l1, l2, cons(x, l)) -> c_7(x, append^#(l, l2)) }
Obligation:
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.
{ hd^#(cons(x, l)) -> c_3(x)
, tl^#(cons(x, l)) -> c_4(l)
, append^#(l1, l2) -> c_5(ifappend^#(l1, l2, l1))
, ifappend^#(l1, l2, nil()) -> c_6(l2)
, ifappend^#(l1, l2, cons(x, l)) -> c_7(x, append^#(l, l2)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Rules: Empty
Obligation:
runtime complexity
Answer:
YES(O(1),O(1))
Empty rules are trivially bounded
* Path {3,7,6,5,4}->{2}: YES(O(1),O(n^1))
---------------------------------------
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ is_empty^#(cons(x, l)) -> c_2()
, tl^#(cons(x, l)) -> c_4(l)
, ifappend^#(l1, l2, cons(x, l)) -> c_7(x, append^#(l, l2)) }
Weak DPs:
{ hd^#(cons(x, l)) -> c_3(x)
, append^#(l1, l2) -> c_5(ifappend^#(l1, l2, l1))
, ifappend^#(l1, l2, nil()) -> c_6(l2) }
Obligation:
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: is_empty^#(cons(x, l)) -> c_2()
, 2: tl^#(cons(x, l)) -> c_4(l)
, 4: hd^#(cons(x, l)) -> c_3(x) }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_5) = {1}, Uargs(c_7) = {2}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[nil] = [0]
[cons](x1, x2) = [1] x2 + [0]
[is_empty^#](x1) = [5]
[c_2] = [1]
[hd^#](x1) = [1] x1 + [4]
[c_3](x1) = [0]
[tl^#](x1) = [1] x1 + [2]
[c_4](x1) = [1] x1 + [0]
[append^#](x1, x2) = [4] x1 + [0]
[c_5](x1) = [1] x1 + [0]
[ifappend^#](x1, x2, x3) = [4] x3 + [0]
[c_6](x1) = [0]
[c_7](x1, x2) = [1] x2 + [0]
The order satisfies the following ordering constraints:
[is_empty^#(cons(x, l))] = [5]
> [1]
= [c_2()]
[hd^#(cons(x, l))] = [1] l + [4]
> [0]
= [c_3(x)]
[tl^#(cons(x, l))] = [1] l + [2]
> [1] l + [0]
= [c_4(l)]
[append^#(l1, l2)] = [4] l1 + [0]
>= [4] l1 + [0]
= [c_5(ifappend^#(l1, l2, l1))]
[ifappend^#(l1, l2, nil())] = [0]
>= [0]
= [c_6(l2)]
[ifappend^#(l1, l2, cons(x, l))] = [4] l + [0]
>= [4] l + [0]
= [c_7(x, append^#(l, l2))]
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:
{ ifappend^#(l1, l2, cons(x, l)) -> c_7(x, append^#(l, l2)) }
Weak DPs:
{ is_empty^#(cons(x, l)) -> c_2()
, hd^#(cons(x, l)) -> c_3(x)
, tl^#(cons(x, l)) -> c_4(l)
, append^#(l1, l2) -> c_5(ifappend^#(l1, l2, l1))
, ifappend^#(l1, l2, nil()) -> c_6(l2) }
Obligation:
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.
{ is_empty^#(cons(x, l)) -> c_2() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ ifappend^#(l1, l2, cons(x, l)) -> c_7(x, append^#(l, l2)) }
Weak DPs:
{ hd^#(cons(x, l)) -> c_3(x)
, tl^#(cons(x, l)) -> c_4(l)
, append^#(l1, l2) -> c_5(ifappend^#(l1, l2, l1))
, ifappend^#(l1, l2, nil()) -> c_6(l2) }
Obligation:
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: ifappend^#(l1, l2, cons(x, l)) -> c_7(x, append^#(l, l2))
, 2: hd^#(cons(x, l)) -> c_3(x)
, 5: ifappend^#(l1, l2, nil()) -> c_6(l2) }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_5) = {1}, Uargs(c_7) = {2}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[nil] = [0]
[cons](x1, x2) = [1] x2 + [1]
[is_empty^#](x1) = [0]
[c_2] = [0]
[hd^#](x1) = [1] x1 + [0]
[c_3](x1) = [0]
[tl^#](x1) = [0]
[c_4](x1) = [0]
[append^#](x1, x2) = [4] x1 + [1]
[c_5](x1) = [1] x1 + [0]
[ifappend^#](x1, x2, x3) = [4] x3 + [1]
[c_6](x1) = [0]
[c_7](x1, x2) = [1] x2 + [2]
The order satisfies the following ordering constraints:
[hd^#(cons(x, l))] = [1] l + [1]
> [0]
= [c_3(x)]
[tl^#(cons(x, l))] = [0]
>= [0]
= [c_4(l)]
[append^#(l1, l2)] = [4] l1 + [1]
>= [4] l1 + [1]
= [c_5(ifappend^#(l1, l2, l1))]
[ifappend^#(l1, l2, nil())] = [1]
> [0]
= [c_6(l2)]
[ifappend^#(l1, l2, cons(x, l))] = [4] l + [5]
> [4] l + [3]
= [c_7(x, append^#(l, l2))]
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:
{ hd^#(cons(x, l)) -> c_3(x)
, tl^#(cons(x, l)) -> c_4(l)
, append^#(l1, l2) -> c_5(ifappend^#(l1, l2, l1))
, ifappend^#(l1, l2, nil()) -> c_6(l2)
, ifappend^#(l1, l2, cons(x, l)) -> c_7(x, append^#(l, l2)) }
Obligation:
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.
{ hd^#(cons(x, l)) -> c_3(x)
, tl^#(cons(x, l)) -> c_4(l)
, append^#(l1, l2) -> c_5(ifappend^#(l1, l2, l1))
, ifappend^#(l1, l2, nil()) -> c_6(l2)
, ifappend^#(l1, l2, cons(x, l)) -> c_7(x, append^#(l, l2)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Rules: Empty
Obligation:
runtime complexity
Answer:
YES(O(1),O(1))
Empty rules are trivially bounded
Hurray, we answered YES(O(1),O(n^1))