We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ foldl(x, Cons(S(0()), xs)) -> foldl(S(x), xs)
, foldl(a, Nil()) -> a
, foldl(S(0()), Cons(x, xs)) -> foldl(S(x), xs)
, foldr(a, Cons(x, xs)) -> op(x, foldr(a, xs))
, foldr(a, Nil()) -> a
, op(x, S(0())) -> S(x)
, op(S(0()), y) -> S(y)
, notEmpty(Cons(x, xs)) -> True()
, notEmpty(Nil()) -> False()
, fold(a, xs) -> Cons(foldl(a, xs), Cons(foldr(a, xs), Nil())) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We add the following weak dependency pairs:
Strict DPs:
{ foldl^#(x, Cons(S(0()), xs)) -> c_1(foldl^#(S(x), xs))
, foldl^#(a, Nil()) -> c_2()
, foldl^#(S(0()), Cons(x, xs)) -> c_3(foldl^#(S(x), xs))
, foldr^#(a, Cons(x, xs)) -> c_4(op^#(x, foldr(a, xs)))
, foldr^#(a, Nil()) -> c_5()
, op^#(x, S(0())) -> c_6()
, op^#(S(0()), y) -> c_7()
, notEmpty^#(Cons(x, xs)) -> c_8()
, notEmpty^#(Nil()) -> c_9()
, fold^#(a, xs) -> c_10(foldl^#(a, xs), foldr^#(a, xs)) }
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:
{ foldl^#(x, Cons(S(0()), xs)) -> c_1(foldl^#(S(x), xs))
, foldl^#(a, Nil()) -> c_2()
, foldl^#(S(0()), Cons(x, xs)) -> c_3(foldl^#(S(x), xs))
, foldr^#(a, Cons(x, xs)) -> c_4(op^#(x, foldr(a, xs)))
, foldr^#(a, Nil()) -> c_5()
, op^#(x, S(0())) -> c_6()
, op^#(S(0()), y) -> c_7()
, notEmpty^#(Cons(x, xs)) -> c_8()
, notEmpty^#(Nil()) -> c_9()
, fold^#(a, xs) -> c_10(foldl^#(a, xs), foldr^#(a, xs)) }
Strict Trs:
{ foldl(x, Cons(S(0()), xs)) -> foldl(S(x), xs)
, foldl(a, Nil()) -> a
, foldl(S(0()), Cons(x, xs)) -> foldl(S(x), xs)
, foldr(a, Cons(x, xs)) -> op(x, foldr(a, xs))
, foldr(a, Nil()) -> a
, op(x, S(0())) -> S(x)
, op(S(0()), y) -> S(y)
, notEmpty(Cons(x, xs)) -> True()
, notEmpty(Nil()) -> False()
, fold(a, xs) -> Cons(foldl(a, xs), Cons(foldr(a, xs), Nil())) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We replace rewrite rules by usable rules:
Strict Usable Rules:
{ foldr(a, Cons(x, xs)) -> op(x, foldr(a, xs))
, foldr(a, Nil()) -> a
, op(x, S(0())) -> S(x)
, op(S(0()), y) -> S(y) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ foldl^#(x, Cons(S(0()), xs)) -> c_1(foldl^#(S(x), xs))
, foldl^#(a, Nil()) -> c_2()
, foldl^#(S(0()), Cons(x, xs)) -> c_3(foldl^#(S(x), xs))
, foldr^#(a, Cons(x, xs)) -> c_4(op^#(x, foldr(a, xs)))
, foldr^#(a, Nil()) -> c_5()
, op^#(x, S(0())) -> c_6()
, op^#(S(0()), y) -> c_7()
, notEmpty^#(Cons(x, xs)) -> c_8()
, notEmpty^#(Nil()) -> c_9()
, fold^#(a, xs) -> c_10(foldl^#(a, xs), foldr^#(a, xs)) }
Strict Trs:
{ foldr(a, Cons(x, xs)) -> op(x, foldr(a, xs))
, foldr(a, Nil()) -> a
, op(x, S(0())) -> S(x)
, op(S(0()), y) -> S(y) }
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(op) = {2}, Uargs(c_1) = {1}, Uargs(c_3) = {1},
Uargs(c_4) = {1}, Uargs(op^#) = {2}, Uargs(c_10) = {1, 2}
TcT has computed the following constructor-restricted matrix
interpretation.
[Cons](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 1] [0 1] [2]
[S](x1) = [1 0] x1 + [0]
[0 1] [0]
[0] = [1]
[0]
[foldr](x1, x2) = [2 0] x1 + [2 1] x2 + [1]
[0 2] [0 1] [2]
[op](x1, x2) = [1 1] x1 + [1 0] x2 + [0]
[0 1] [0 1] [0]
[Nil] = [0]
[0]
[foldl^#](x1, x2) = [0]
[0]
[c_1](x1) = [1 0] x1 + [0]
[0 1] [0]
[c_2] = [0]
[0]
[c_3](x1) = [1 0] x1 + [0]
[0 1] [0]
[foldr^#](x1, x2) = [2 0] x1 + [2 1] x2 + [0]
[0 0] [0 0] [0]
[c_4](x1) = [1 0] x1 + [0]
[0 1] [0]
[op^#](x1, x2) = [1 0] x2 + [0]
[0 0] [0]
[c_5] = [0]
[0]
[c_6] = [0]
[0]
[c_7] = [0]
[0]
[notEmpty^#](x1) = [0]
[0]
[c_8] = [0]
[0]
[c_9] = [0]
[0]
[fold^#](x1, x2) = [2 1] x1 + [2 2] x2 + [0]
[0 0] [0 0] [0]
[c_10](x1, x2) = [1 0] x1 + [1 0] x2 + [2]
[0 1] [0 1] [2]
The order satisfies the following ordering constraints:
[foldr(a, Cons(x, xs))] = [2 1] x + [2 1] xs + [2 0] a + [3]
[0 1] [0 1] [0 2] [4]
> [1 1] x + [2 1] xs + [2 0] a + [1]
[0 1] [0 1] [0 2] [2]
= [op(x, foldr(a, xs))]
[foldr(a, Nil())] = [2 0] a + [1]
[0 2] [2]
> [1 0] a + [0]
[0 1] [0]
= [a]
[op(x, S(0()))] = [1 1] x + [1]
[0 1] [0]
> [1 0] x + [0]
[0 1] [0]
= [S(x)]
[op(S(0()), y)] = [1 0] y + [1]
[0 1] [0]
> [1 0] y + [0]
[0 1] [0]
= [S(y)]
[foldl^#(x, Cons(S(0()), xs))] = [0]
[0]
>= [0]
[0]
= [c_1(foldl^#(S(x), xs))]
[foldl^#(a, Nil())] = [0]
[0]
>= [0]
[0]
= [c_2()]
[foldl^#(S(0()), Cons(x, xs))] = [0]
[0]
>= [0]
[0]
= [c_3(foldl^#(S(x), xs))]
[foldr^#(a, Cons(x, xs))] = [2 1] x + [2 1] xs + [2 0] a + [2]
[0 0] [0 0] [0 0] [0]
> [2 1] xs + [2 0] a + [1]
[0 0] [0 0] [0]
= [c_4(op^#(x, foldr(a, xs)))]
[foldr^#(a, Nil())] = [2 0] a + [0]
[0 0] [0]
>= [0]
[0]
= [c_5()]
[op^#(x, S(0()))] = [1]
[0]
> [0]
[0]
= [c_6()]
[op^#(S(0()), y)] = [1 0] y + [0]
[0 0] [0]
>= [0]
[0]
= [c_7()]
[notEmpty^#(Cons(x, xs))] = [0]
[0]
>= [0]
[0]
= [c_8()]
[notEmpty^#(Nil())] = [0]
[0]
>= [0]
[0]
= [c_9()]
[fold^#(a, xs)] = [2 2] xs + [2 1] a + [0]
[0 0] [0 0] [0]
? [2 1] xs + [2 0] a + [2]
[0 0] [0 0] [2]
= [c_10(foldl^#(a, xs), foldr^#(a, xs))]
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:
{ foldl^#(x, Cons(S(0()), xs)) -> c_1(foldl^#(S(x), xs))
, foldl^#(a, Nil()) -> c_2()
, foldl^#(S(0()), Cons(x, xs)) -> c_3(foldl^#(S(x), xs))
, foldr^#(a, Nil()) -> c_5()
, op^#(S(0()), y) -> c_7()
, notEmpty^#(Cons(x, xs)) -> c_8()
, notEmpty^#(Nil()) -> c_9()
, fold^#(a, xs) -> c_10(foldl^#(a, xs), foldr^#(a, xs)) }
Weak DPs:
{ foldr^#(a, Cons(x, xs)) -> c_4(op^#(x, foldr(a, xs)))
, op^#(x, S(0())) -> c_6() }
Weak Trs:
{ foldr(a, Cons(x, xs)) -> op(x, foldr(a, xs))
, foldr(a, Nil()) -> a
, op(x, S(0())) -> S(x)
, op(S(0()), y) -> S(y) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We estimate the number of application of {2,4,6,7} by applications
of Pre({2,4,6,7}) = {1,3,8}. Here rules are labeled as follows:
DPs:
{ 1: foldl^#(x, Cons(S(0()), xs)) -> c_1(foldl^#(S(x), xs))
, 2: foldl^#(a, Nil()) -> c_2()
, 3: foldl^#(S(0()), Cons(x, xs)) -> c_3(foldl^#(S(x), xs))
, 4: foldr^#(a, Nil()) -> c_5()
, 5: op^#(S(0()), y) -> c_7()
, 6: notEmpty^#(Cons(x, xs)) -> c_8()
, 7: notEmpty^#(Nil()) -> c_9()
, 8: fold^#(a, xs) -> c_10(foldl^#(a, xs), foldr^#(a, xs))
, 9: foldr^#(a, Cons(x, xs)) -> c_4(op^#(x, foldr(a, xs)))
, 10: op^#(x, S(0())) -> c_6() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ foldl^#(x, Cons(S(0()), xs)) -> c_1(foldl^#(S(x), xs))
, foldl^#(S(0()), Cons(x, xs)) -> c_3(foldl^#(S(x), xs))
, op^#(S(0()), y) -> c_7()
, fold^#(a, xs) -> c_10(foldl^#(a, xs), foldr^#(a, xs)) }
Weak DPs:
{ foldl^#(a, Nil()) -> c_2()
, foldr^#(a, Cons(x, xs)) -> c_4(op^#(x, foldr(a, xs)))
, foldr^#(a, Nil()) -> c_5()
, op^#(x, S(0())) -> c_6()
, notEmpty^#(Cons(x, xs)) -> c_8()
, notEmpty^#(Nil()) -> c_9() }
Weak Trs:
{ foldr(a, Cons(x, xs)) -> op(x, foldr(a, xs))
, foldr(a, Nil()) -> a
, op(x, S(0())) -> S(x)
, op(S(0()), y) -> S(y) }
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.
{ foldl^#(a, Nil()) -> c_2()
, foldr^#(a, Nil()) -> c_5()
, op^#(x, S(0())) -> c_6()
, notEmpty^#(Cons(x, xs)) -> c_8()
, notEmpty^#(Nil()) -> c_9() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ foldl^#(x, Cons(S(0()), xs)) -> c_1(foldl^#(S(x), xs))
, foldl^#(S(0()), Cons(x, xs)) -> c_3(foldl^#(S(x), xs))
, op^#(S(0()), y) -> c_7()
, fold^#(a, xs) -> c_10(foldl^#(a, xs), foldr^#(a, xs)) }
Weak DPs: { foldr^#(a, Cons(x, xs)) -> c_4(op^#(x, foldr(a, xs))) }
Weak Trs:
{ foldr(a, Cons(x, xs)) -> op(x, foldr(a, xs))
, foldr(a, Nil()) -> a
, op(x, S(0())) -> S(x)
, op(S(0()), y) -> S(y) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
Consider the dependency graph
1: foldl^#(x, Cons(S(0()), xs)) -> c_1(foldl^#(S(x), xs))
-->_1 foldl^#(S(0()), Cons(x, xs)) -> c_3(foldl^#(S(x), xs)) :2
-->_1 foldl^#(x, Cons(S(0()), xs)) -> c_1(foldl^#(S(x), xs)) :1
2: foldl^#(S(0()), Cons(x, xs)) -> c_3(foldl^#(S(x), xs))
-->_1 foldl^#(S(0()), Cons(x, xs)) -> c_3(foldl^#(S(x), xs)) :2
-->_1 foldl^#(x, Cons(S(0()), xs)) -> c_1(foldl^#(S(x), xs)) :1
3: op^#(S(0()), y) -> c_7()
4: fold^#(a, xs) -> c_10(foldl^#(a, xs), foldr^#(a, xs))
-->_2 foldr^#(a, Cons(x, xs)) -> c_4(op^#(x, foldr(a, xs))) :5
-->_1 foldl^#(S(0()), Cons(x, xs)) -> c_3(foldl^#(S(x), xs)) :2
-->_1 foldl^#(x, Cons(S(0()), xs)) -> c_1(foldl^#(S(x), xs)) :1
5: foldr^#(a, Cons(x, xs)) -> c_4(op^#(x, foldr(a, xs)))
-->_1 op^#(S(0()), y) -> c_7() :3
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).
{ fold^#(a, xs) -> c_10(foldl^#(a, xs), foldr^#(a, xs)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ foldl^#(x, Cons(S(0()), xs)) -> c_1(foldl^#(S(x), xs))
, foldl^#(S(0()), Cons(x, xs)) -> c_3(foldl^#(S(x), xs))
, op^#(S(0()), y) -> c_7() }
Weak DPs: { foldr^#(a, Cons(x, xs)) -> c_4(op^#(x, foldr(a, xs))) }
Weak Trs:
{ foldr(a, Cons(x, xs)) -> op(x, foldr(a, xs))
, foldr(a, Nil()) -> a
, op(x, S(0())) -> S(x)
, op(S(0()), y) -> S(y) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We analyse the complexity of following sub-problems (R) and (S).
Problem (S) is obtained from the input problem by shifting strict
rules from (R) into the weak component:
Problem (R):
------------
Strict DPs: { op^#(S(0()), y) -> c_7() }
Weak DPs:
{ foldl^#(x, Cons(S(0()), xs)) -> c_1(foldl^#(S(x), xs))
, foldl^#(S(0()), Cons(x, xs)) -> c_3(foldl^#(S(x), xs))
, foldr^#(a, Cons(x, xs)) -> c_4(op^#(x, foldr(a, xs))) }
Weak Trs:
{ foldr(a, Cons(x, xs)) -> op(x, foldr(a, xs))
, foldr(a, Nil()) -> a
, op(x, S(0())) -> S(x)
, op(S(0()), y) -> S(y) }
StartTerms: basic terms
Strategy: innermost
Problem (S):
------------
Strict DPs:
{ foldl^#(x, Cons(S(0()), xs)) -> c_1(foldl^#(S(x), xs))
, foldl^#(S(0()), Cons(x, xs)) -> c_3(foldl^#(S(x), xs)) }
Weak DPs:
{ foldr^#(a, Cons(x, xs)) -> c_4(op^#(x, foldr(a, xs)))
, op^#(S(0()), y) -> c_7() }
Weak Trs:
{ foldr(a, Cons(x, xs)) -> op(x, foldr(a, xs))
, foldr(a, Nil()) -> a
, op(x, S(0())) -> S(x)
, op(S(0()), y) -> S(y) }
StartTerms: basic terms
Strategy: innermost
Overall, the transformation results in the following sub-problem(s):
Generated new problems:
-----------------------
R) Strict DPs: { op^#(S(0()), y) -> c_7() }
Weak DPs:
{ foldl^#(x, Cons(S(0()), xs)) -> c_1(foldl^#(S(x), xs))
, foldl^#(S(0()), Cons(x, xs)) -> c_3(foldl^#(S(x), xs))
, foldr^#(a, Cons(x, xs)) -> c_4(op^#(x, foldr(a, xs))) }
Weak Trs:
{ foldr(a, Cons(x, xs)) -> op(x, foldr(a, xs))
, foldr(a, Nil()) -> a
, op(x, S(0())) -> S(x)
, op(S(0()), y) -> S(y) }
StartTerms: basic terms
Strategy: innermost
This problem was proven YES(O(1),O(1)).
S) Strict DPs:
{ foldl^#(x, Cons(S(0()), xs)) -> c_1(foldl^#(S(x), xs))
, foldl^#(S(0()), Cons(x, xs)) -> c_3(foldl^#(S(x), xs)) }
Weak DPs:
{ foldr^#(a, Cons(x, xs)) -> c_4(op^#(x, foldr(a, xs)))
, op^#(S(0()), y) -> c_7() }
Weak Trs:
{ foldr(a, Cons(x, xs)) -> op(x, foldr(a, xs))
, foldr(a, Nil()) -> a
, op(x, S(0())) -> S(x)
, op(S(0()), y) -> S(y) }
StartTerms: basic terms
Strategy: innermost
This problem was proven YES(O(1),O(n^1)).
Proofs for generated problems:
------------------------------
R) We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Strict DPs: { op^#(S(0()), y) -> c_7() }
Weak DPs:
{ foldl^#(x, Cons(S(0()), xs)) -> c_1(foldl^#(S(x), xs))
, foldl^#(S(0()), Cons(x, xs)) -> c_3(foldl^#(S(x), xs))
, foldr^#(a, Cons(x, xs)) -> c_4(op^#(x, foldr(a, xs))) }
Weak Trs:
{ foldr(a, Cons(x, xs)) -> op(x, foldr(a, xs))
, foldr(a, Nil()) -> a
, op(x, S(0())) -> S(x)
, op(S(0()), y) -> S(y) }
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.
{ foldl^#(x, Cons(S(0()), xs)) -> c_1(foldl^#(S(x), xs))
, foldl^#(S(0()), Cons(x, xs)) -> c_3(foldl^#(S(x), xs)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Strict DPs: { op^#(S(0()), y) -> c_7() }
Weak DPs: { foldr^#(a, Cons(x, xs)) -> c_4(op^#(x, foldr(a, xs))) }
Weak Trs:
{ foldr(a, Cons(x, xs)) -> op(x, foldr(a, xs))
, foldr(a, Nil()) -> a
, op(x, S(0())) -> S(x)
, op(S(0()), y) -> S(y) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
The dependency graph contains no loops, we remove all dependency
pairs.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Weak Trs:
{ foldr(a, Cons(x, xs)) -> op(x, foldr(a, xs))
, foldr(a, Nil()) -> a
, op(x, S(0())) -> S(x)
, op(S(0()), y) -> S(y) }
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
S) We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ foldl^#(x, Cons(S(0()), xs)) -> c_1(foldl^#(S(x), xs))
, foldl^#(S(0()), Cons(x, xs)) -> c_3(foldl^#(S(x), xs)) }
Weak DPs:
{ foldr^#(a, Cons(x, xs)) -> c_4(op^#(x, foldr(a, xs)))
, op^#(S(0()), y) -> c_7() }
Weak Trs:
{ foldr(a, Cons(x, xs)) -> op(x, foldr(a, xs))
, foldr(a, Nil()) -> a
, op(x, S(0())) -> S(x)
, op(S(0()), y) -> S(y) }
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.
{ foldr^#(a, Cons(x, xs)) -> c_4(op^#(x, foldr(a, xs)))
, op^#(S(0()), y) -> c_7() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ foldl^#(x, Cons(S(0()), xs)) -> c_1(foldl^#(S(x), xs))
, foldl^#(S(0()), Cons(x, xs)) -> c_3(foldl^#(S(x), xs)) }
Weak Trs:
{ foldr(a, Cons(x, xs)) -> op(x, foldr(a, xs))
, foldr(a, Nil()) -> a
, op(x, S(0())) -> S(x)
, op(S(0()), y) -> S(y) }
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:
{ foldl^#(x, Cons(S(0()), xs)) -> c_1(foldl^#(S(x), xs))
, foldl^#(S(0()), Cons(x, xs)) -> c_3(foldl^#(S(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: foldl^#(x, Cons(S(0()), xs)) -> c_1(foldl^#(S(x), xs))
, 2: foldl^#(S(0()), Cons(x, xs)) -> c_3(foldl^#(S(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(S) = {1}, safe(0) = {},
safe(foldl^#) = {1}, safe(c_1) = {}, safe(c_3) = {}
and precedence
empty .
Following symbols are considered recursive:
{foldl^#}
The recursion depth is 1.
Further, following argument filtering is employed:
pi(Cons) = [1, 2], pi(S) = [1], pi(0) = [], pi(foldl^#) = [2],
pi(c_1) = [1], pi(c_3) = [1]
Usable defined function symbols are a subset of:
{foldl^#}
For your convenience, here are the satisfied ordering constraints:
pi(foldl^#(x, Cons(S(0()), xs))) = foldl^#(Cons(; S(; 0()), xs);)
> c_1(foldl^#(xs;);)
= pi(c_1(foldl^#(S(x), xs)))
pi(foldl^#(S(0()), Cons(x, xs))) = foldl^#(Cons(; x, xs);)
> c_3(foldl^#(xs;);)
= pi(c_3(foldl^#(S(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:
{ foldl^#(x, Cons(S(0()), xs)) -> c_1(foldl^#(S(x), xs))
, foldl^#(S(0()), Cons(x, xs)) -> c_3(foldl^#(S(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.
{ foldl^#(x, Cons(S(0()), xs)) -> c_1(foldl^#(S(x), xs))
, foldl^#(S(0()), Cons(x, xs)) -> c_3(foldl^#(S(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))