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:
innermost 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()
, tl^#(cons(x, l)) -> c_4()
, append^#(l1, l2) -> c_5(ifappend^#(l1, l2, l1))
, ifappend^#(l1, l2, nil()) -> c_6()
, ifappend^#(l1, l2, cons(x, l)) -> c_7(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()
, tl^#(cons(x, l)) -> c_4()
, append^#(l1, l2) -> c_5(ifappend^#(l1, l2, l1))
, ifappend^#(l1, l2, nil()) -> c_6()
, ifappend^#(l1, l2, cons(x, l)) -> c_7(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:
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:
{ is_empty^#(nil()) -> c_1()
, is_empty^#(cons(x, l)) -> c_2()
, hd^#(cons(x, l)) -> c_3()
, tl^#(cons(x, l)) -> c_4()
, append^#(l1, l2) -> c_5(ifappend^#(l1, l2, l1))
, ifappend^#(l1, l2, nil()) -> c_6()
, ifappend^#(l1, l2, cons(x, l)) -> c_7(append^#(l, l2)) }
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_5) = {1}, Uargs(c_7) = {1}
TcT has computed the following constructor-restricted matrix
interpretation.
[nil] = [0]
[0]
[cons](x1, x2) = [1 0] x2 + [0]
[0 0] [0]
[is_empty^#](x1) = [0]
[0]
[c_1] = [0]
[0]
[c_2] = [0]
[0]
[hd^#](x1) = [1]
[0]
[c_3] = [0]
[0]
[tl^#](x1) = [0]
[0]
[c_4] = [0]
[0]
[append^#](x1, x2) = [0]
[2]
[c_5](x1) = [1 0] x1 + [0]
[0 1] [0]
[ifappend^#](x1, x2, x3) = [0 0] x1 + [0]
[0 2] [0]
[c_6] = [0]
[0]
[c_7](x1) = [1 0] x1 + [0]
[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()]
[tl^#(cons(x, l))] = [0]
[0]
>= [0]
[0]
= [c_4()]
[append^#(l1, l2)] = [0]
[2]
? [0 0] l1 + [0]
[0 2] [0]
= [c_5(ifappend^#(l1, l2, l1))]
[ifappend^#(l1, l2, nil())] = [0 0] l1 + [0]
[0 2] [0]
>= [0]
[0]
= [c_6()]
[ifappend^#(l1, l2, cons(x, l))] = [0 0] l1 + [0]
[0 2] [0]
? [0]
[4]
= [c_7(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(1),O(n^1)).
Strict DPs:
{ is_empty^#(nil()) -> c_1()
, is_empty^#(cons(x, l)) -> c_2()
, tl^#(cons(x, l)) -> c_4()
, append^#(l1, l2) -> c_5(ifappend^#(l1, l2, l1))
, ifappend^#(l1, l2, nil()) -> c_6()
, ifappend^#(l1, l2, cons(x, l)) -> c_7(append^#(l, l2)) }
Weak DPs: { hd^#(cons(x, l)) -> c_3() }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We estimate the number of application of {1,2,3,5} by applications
of Pre({1,2,3,5}) = {4}. Here rules are labeled as follows:
DPs:
{ 1: is_empty^#(nil()) -> c_1()
, 2: is_empty^#(cons(x, l)) -> c_2()
, 3: tl^#(cons(x, l)) -> c_4()
, 4: append^#(l1, l2) -> c_5(ifappend^#(l1, l2, l1))
, 5: ifappend^#(l1, l2, nil()) -> c_6()
, 6: ifappend^#(l1, l2, cons(x, l)) -> c_7(append^#(l, l2))
, 7: hd^#(cons(x, l)) -> c_3() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ append^#(l1, l2) -> c_5(ifappend^#(l1, l2, l1))
, ifappend^#(l1, l2, cons(x, l)) -> c_7(append^#(l, l2)) }
Weak DPs:
{ is_empty^#(nil()) -> c_1()
, is_empty^#(cons(x, l)) -> c_2()
, hd^#(cons(x, l)) -> c_3()
, tl^#(cons(x, l)) -> c_4()
, ifappend^#(l1, l2, nil()) -> c_6() }
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.
{ is_empty^#(nil()) -> c_1()
, is_empty^#(cons(x, l)) -> c_2()
, hd^#(cons(x, l)) -> c_3()
, tl^#(cons(x, l)) -> c_4()
, ifappend^#(l1, l2, nil()) -> c_6() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ append^#(l1, l2) -> c_5(ifappend^#(l1, l2, l1))
, ifappend^#(l1, l2, cons(x, l)) -> c_7(append^#(l, l2)) }
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: append^#(l1, l2) -> c_5(ifappend^#(l1, l2, l1))
, 2: ifappend^#(l1, l2, cons(x, l)) -> c_7(append^#(l, l2)) }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_5) = {1}, Uargs(c_7) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[cons](x1, x2) = [1] x2 + [2]
[append^#](x1, x2) = [4] x1 + [4]
[c_5](x1) = [1] x1 + [1]
[ifappend^#](x1, x2, x3) = [4] x3 + [0]
[c_7](x1) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[append^#(l1, l2)] = [4] l1 + [4]
> [4] l1 + [1]
= [c_5(ifappend^#(l1, l2, l1))]
[ifappend^#(l1, l2, cons(x, l))] = [4] l + [8]
> [4] l + [4]
= [c_7(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:
{ append^#(l1, l2) -> c_5(ifappend^#(l1, l2, l1))
, ifappend^#(l1, l2, cons(x, l)) -> c_7(append^#(l, l2)) }
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.
{ append^#(l1, l2) -> c_5(ifappend^#(l1, l2, l1))
, ifappend^#(l1, l2, cons(x, l)) -> c_7(append^#(l, l2)) }
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))