We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict Trs:
{ app(l, nil()) -> l
, app(nil(), k) -> k
, app(cons(x, l), k) -> cons(x, app(l, k))
, sum(app(l, cons(x, cons(y, k)))) ->
sum(app(l, sum(cons(x, cons(y, k)))))
, sum(cons(x, nil())) -> cons(x, nil())
, sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
, sum(plus(cons(0(), x), cons(y, l))) ->
pred(sum(cons(s(x), cons(y, l))))
, plus(0(), y) -> y
, plus(s(x), y) -> s(plus(x, y))
, pred(cons(s(x), nil())) -> cons(x, nil()) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
We add the following dependency tuples:
Strict DPs:
{ app^#(l, nil()) -> c_1()
, app^#(nil(), k) -> c_2()
, app^#(cons(x, l), k) -> c_3(app^#(l, k))
, sum^#(app(l, cons(x, cons(y, k)))) ->
c_4(sum^#(app(l, sum(cons(x, cons(y, k))))),
app^#(l, sum(cons(x, cons(y, k)))),
sum^#(cons(x, cons(y, k))))
, sum^#(cons(x, nil())) -> c_5()
, sum^#(cons(x, cons(y, l))) ->
c_6(sum^#(cons(plus(x, y), l)), plus^#(x, y))
, sum^#(plus(cons(0(), x), cons(y, l))) ->
c_7(pred^#(sum(cons(s(x), cons(y, l)))),
sum^#(cons(s(x), cons(y, l))))
, plus^#(0(), y) -> c_8()
, plus^#(s(x), y) -> c_9(plus^#(x, y))
, pred^#(cons(s(x), nil())) -> c_10() }
and mark the set of starting terms.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict DPs:
{ app^#(l, nil()) -> c_1()
, app^#(nil(), k) -> c_2()
, app^#(cons(x, l), k) -> c_3(app^#(l, k))
, sum^#(app(l, cons(x, cons(y, k)))) ->
c_4(sum^#(app(l, sum(cons(x, cons(y, k))))),
app^#(l, sum(cons(x, cons(y, k)))),
sum^#(cons(x, cons(y, k))))
, sum^#(cons(x, nil())) -> c_5()
, sum^#(cons(x, cons(y, l))) ->
c_6(sum^#(cons(plus(x, y), l)), plus^#(x, y))
, sum^#(plus(cons(0(), x), cons(y, l))) ->
c_7(pred^#(sum(cons(s(x), cons(y, l)))),
sum^#(cons(s(x), cons(y, l))))
, plus^#(0(), y) -> c_8()
, plus^#(s(x), y) -> c_9(plus^#(x, y))
, pred^#(cons(s(x), nil())) -> c_10() }
Weak Trs:
{ app(l, nil()) -> l
, app(nil(), k) -> k
, app(cons(x, l), k) -> cons(x, app(l, k))
, sum(app(l, cons(x, cons(y, k)))) ->
sum(app(l, sum(cons(x, cons(y, k)))))
, sum(cons(x, nil())) -> cons(x, nil())
, sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
, sum(plus(cons(0(), x), cons(y, l))) ->
pred(sum(cons(s(x), cons(y, l))))
, plus(0(), y) -> y
, plus(s(x), y) -> s(plus(x, y))
, pred(cons(s(x), nil())) -> cons(x, nil()) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
Consider the dependency graph:
1: app^#(l, nil()) -> c_1()
2: app^#(nil(), k) -> c_2()
3: app^#(cons(x, l), k) -> c_3(app^#(l, k))
-->_1 app^#(cons(x, l), k) -> c_3(app^#(l, k)) :3
-->_1 app^#(nil(), k) -> c_2() :2
-->_1 app^#(l, nil()) -> c_1() :1
4: sum^#(app(l, cons(x, cons(y, k)))) ->
c_4(sum^#(app(l, sum(cons(x, cons(y, k))))),
app^#(l, sum(cons(x, cons(y, k)))),
sum^#(cons(x, cons(y, k))))
-->_1 sum^#(plus(cons(0(), x), cons(y, l))) ->
c_7(pred^#(sum(cons(s(x), cons(y, l)))),
sum^#(cons(s(x), cons(y, l)))) :7
-->_3 sum^#(cons(x, cons(y, l))) ->
c_6(sum^#(cons(plus(x, y), l)), plus^#(x, y)) :6
-->_1 sum^#(cons(x, cons(y, l))) ->
c_6(sum^#(cons(plus(x, y), l)), plus^#(x, y)) :6
-->_1 sum^#(cons(x, nil())) -> c_5() :5
-->_1 sum^#(app(l, cons(x, cons(y, k)))) ->
c_4(sum^#(app(l, sum(cons(x, cons(y, k))))),
app^#(l, sum(cons(x, cons(y, k)))),
sum^#(cons(x, cons(y, k)))) :4
-->_2 app^#(l, nil()) -> c_1() :1
5: sum^#(cons(x, nil())) -> c_5()
6: sum^#(cons(x, cons(y, l))) ->
c_6(sum^#(cons(plus(x, y), l)), plus^#(x, y))
-->_2 plus^#(s(x), y) -> c_9(plus^#(x, y)) :9
-->_2 plus^#(0(), y) -> c_8() :8
-->_1 sum^#(cons(x, cons(y, l))) ->
c_6(sum^#(cons(plus(x, y), l)), plus^#(x, y)) :6
-->_1 sum^#(cons(x, nil())) -> c_5() :5
7: sum^#(plus(cons(0(), x), cons(y, l))) ->
c_7(pred^#(sum(cons(s(x), cons(y, l)))),
sum^#(cons(s(x), cons(y, l))))
-->_1 pred^#(cons(s(x), nil())) -> c_10() :10
-->_2 sum^#(cons(x, cons(y, l))) ->
c_6(sum^#(cons(plus(x, y), l)), plus^#(x, y)) :6
8: plus^#(0(), y) -> c_8()
9: plus^#(s(x), y) -> c_9(plus^#(x, y))
-->_1 plus^#(s(x), y) -> c_9(plus^#(x, y)) :9
-->_1 plus^#(0(), y) -> c_8() :8
10: pred^#(cons(s(x), nil())) -> c_10()
Only the nodes {1,2,3,5,6,9,8,10} are reachable from nodes
{1,2,3,5,6,8,9,10} that start derivation from marked basic terms.
The nodes not reachable are removed from the problem.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict DPs:
{ app^#(l, nil()) -> c_1()
, app^#(nil(), k) -> c_2()
, app^#(cons(x, l), k) -> c_3(app^#(l, k))
, sum^#(cons(x, nil())) -> c_5()
, sum^#(cons(x, cons(y, l))) ->
c_6(sum^#(cons(plus(x, y), l)), plus^#(x, y))
, plus^#(0(), y) -> c_8()
, plus^#(s(x), y) -> c_9(plus^#(x, y))
, pred^#(cons(s(x), nil())) -> c_10() }
Weak Trs:
{ app(l, nil()) -> l
, app(nil(), k) -> k
, app(cons(x, l), k) -> cons(x, app(l, k))
, sum(app(l, cons(x, cons(y, k)))) ->
sum(app(l, sum(cons(x, cons(y, k)))))
, sum(cons(x, nil())) -> cons(x, nil())
, sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
, sum(plus(cons(0(), x), cons(y, l))) ->
pred(sum(cons(s(x), cons(y, l))))
, plus(0(), y) -> y
, plus(s(x), y) -> s(plus(x, y))
, pred(cons(s(x), nil())) -> cons(x, nil()) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
We estimate the number of application of {1,2,4,6,8} by
applications of Pre({1,2,4,6,8}) = {3,5,7}. Here rules are labeled
as follows:
DPs:
{ 1: app^#(l, nil()) -> c_1()
, 2: app^#(nil(), k) -> c_2()
, 3: app^#(cons(x, l), k) -> c_3(app^#(l, k))
, 4: sum^#(cons(x, nil())) -> c_5()
, 5: sum^#(cons(x, cons(y, l))) ->
c_6(sum^#(cons(plus(x, y), l)), plus^#(x, y))
, 6: plus^#(0(), y) -> c_8()
, 7: plus^#(s(x), y) -> c_9(plus^#(x, y))
, 8: pred^#(cons(s(x), nil())) -> c_10() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict DPs:
{ app^#(cons(x, l), k) -> c_3(app^#(l, k))
, sum^#(cons(x, cons(y, l))) ->
c_6(sum^#(cons(plus(x, y), l)), plus^#(x, y))
, plus^#(s(x), y) -> c_9(plus^#(x, y)) }
Weak DPs:
{ app^#(l, nil()) -> c_1()
, app^#(nil(), k) -> c_2()
, sum^#(cons(x, nil())) -> c_5()
, plus^#(0(), y) -> c_8()
, pred^#(cons(s(x), nil())) -> c_10() }
Weak Trs:
{ app(l, nil()) -> l
, app(nil(), k) -> k
, app(cons(x, l), k) -> cons(x, app(l, k))
, sum(app(l, cons(x, cons(y, k)))) ->
sum(app(l, sum(cons(x, cons(y, k)))))
, sum(cons(x, nil())) -> cons(x, nil())
, sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
, sum(plus(cons(0(), x), cons(y, l))) ->
pred(sum(cons(s(x), cons(y, l))))
, plus(0(), y) -> y
, plus(s(x), y) -> s(plus(x, y))
, pred(cons(s(x), nil())) -> cons(x, nil()) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.
{ app^#(l, nil()) -> c_1()
, app^#(nil(), k) -> c_2()
, sum^#(cons(x, nil())) -> c_5()
, plus^#(0(), y) -> c_8()
, pred^#(cons(s(x), nil())) -> c_10() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict DPs:
{ app^#(cons(x, l), k) -> c_3(app^#(l, k))
, sum^#(cons(x, cons(y, l))) ->
c_6(sum^#(cons(plus(x, y), l)), plus^#(x, y))
, plus^#(s(x), y) -> c_9(plus^#(x, y)) }
Weak Trs:
{ app(l, nil()) -> l
, app(nil(), k) -> k
, app(cons(x, l), k) -> cons(x, app(l, k))
, sum(app(l, cons(x, cons(y, k)))) ->
sum(app(l, sum(cons(x, cons(y, k)))))
, sum(cons(x, nil())) -> cons(x, nil())
, sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
, sum(plus(cons(0(), x), cons(y, l))) ->
pred(sum(cons(s(x), cons(y, l))))
, plus(0(), y) -> y
, plus(s(x), y) -> s(plus(x, y))
, pred(cons(s(x), nil())) -> cons(x, nil()) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
We replace rewrite rules by usable rules:
Weak Usable Rules:
{ plus(0(), y) -> y
, plus(s(x), y) -> s(plus(x, y)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict DPs:
{ app^#(cons(x, l), k) -> c_3(app^#(l, k))
, sum^#(cons(x, cons(y, l))) ->
c_6(sum^#(cons(plus(x, y), l)), plus^#(x, y))
, plus^#(s(x), y) -> c_9(plus^#(x, y)) }
Weak Trs:
{ plus(0(), y) -> y
, plus(s(x), y) -> s(plus(x, y)) }
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.
DPs:
{ 1: app^#(cons(x, l), k) -> c_3(app^#(l, k)) }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_3) = {1}, Uargs(c_6) = {1, 2}, Uargs(c_9) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[cons](x1, x2) = [1] x1 + [1] x2 + [1]
[plus](x1, x2) = [4]
[0] = [0]
[s](x1) = [1] x1 + [3]
[app^#](x1, x2) = [1] x1 + [0]
[c_3](x1) = [1] x1 + [0]
[sum^#](x1) = [0]
[c_6](x1, x2) = [1] x1 + [2] x2 + [0]
[plus^#](x1, x2) = [0]
[c_9](x1) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[plus(0(), y)] = [4]
? [1] y + [0]
= [y]
[plus(s(x), y)] = [4]
? [7]
= [s(plus(x, y))]
[app^#(cons(x, l), k)] = [1] l + [1] x + [1]
> [1] l + [0]
= [c_3(app^#(l, k))]
[sum^#(cons(x, cons(y, l)))] = [0]
>= [0]
= [c_6(sum^#(cons(plus(x, y), l)), plus^#(x, y))]
[plus^#(s(x), y)] = [0]
>= [0]
= [c_9(plus^#(x, y))]
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^2)).
Strict DPs:
{ sum^#(cons(x, cons(y, l))) ->
c_6(sum^#(cons(plus(x, y), l)), plus^#(x, y))
, plus^#(s(x), y) -> c_9(plus^#(x, y)) }
Weak DPs: { app^#(cons(x, l), k) -> c_3(app^#(l, k)) }
Weak Trs:
{ plus(0(), y) -> y
, plus(s(x), y) -> s(plus(x, y)) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.
{ app^#(cons(x, l), k) -> c_3(app^#(l, k)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict DPs:
{ sum^#(cons(x, cons(y, l))) ->
c_6(sum^#(cons(plus(x, y), l)), plus^#(x, y))
, plus^#(s(x), y) -> c_9(plus^#(x, y)) }
Weak Trs:
{ plus(0(), y) -> y
, plus(s(x), y) -> s(plus(x, y)) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
We decompose the input problem according to the dependency graph
into the upper component
{ sum^#(cons(x, cons(y, l))) ->
c_6(sum^#(cons(plus(x, y), l)), plus^#(x, y)) }
and lower component
{ plus^#(s(x), y) -> c_9(plus^#(x, y)) }
Further, following extension rules are added to the lower
component.
{ sum^#(cons(x, cons(y, l))) -> sum^#(cons(plus(x, y), l))
, sum^#(cons(x, cons(y, l))) -> plus^#(x, y) }
TcT solves the upper component with certificate YES(O(1),O(n^1)).
Sub-proof:
----------
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ sum^#(cons(x, cons(y, l))) ->
c_6(sum^#(cons(plus(x, y), l)), plus^#(x, y)) }
Weak Trs:
{ plus(0(), y) -> y
, plus(s(x), y) -> s(plus(x, y)) }
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: sum^#(cons(x, cons(y, l))) ->
c_6(sum^#(cons(plus(x, y), l)), plus^#(x, y)) }
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(plus) = {}, safe(0) = {}, safe(s) = {1},
safe(sum^#) = {}, safe(c_6) = {}, safe(plus^#) = {}
and precedence
empty .
Following symbols are considered recursive:
{sum^#}
The recursion depth is 1.
Further, following argument filtering is employed:
pi(cons) = [1, 2], pi(plus) = 2, pi(0) = [], pi(s) = 1,
pi(sum^#) = [1], pi(c_6) = [1, 2], pi(plus^#) = []
Usable defined function symbols are a subset of:
{plus, sum^#, plus^#}
For your convenience, here are the satisfied ordering constraints:
pi(sum^#(cons(x, cons(y, l)))) = sum^#(cons(; x, cons(; y, l));)
> c_6(sum^#(cons(; y, l);), plus^#();)
= pi(c_6(sum^#(cons(plus(x, y), l)), plus^#(x, y)))
pi(plus(0(), y)) = y
>= y
= pi(y)
pi(plus(s(x), y)) = y
>= y
= pi(s(plus(x, y)))
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:
{ sum^#(cons(x, cons(y, l))) ->
c_6(sum^#(cons(plus(x, y), l)), plus^#(x, y)) }
Weak Trs:
{ plus(0(), y) -> y
, plus(s(x), y) -> s(plus(x, 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.
{ sum^#(cons(x, cons(y, l))) ->
c_6(sum^#(cons(plus(x, y), l)), plus^#(x, y)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Weak Trs:
{ plus(0(), y) -> y
, plus(s(x), y) -> s(plus(x, 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
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs: { plus^#(s(x), y) -> c_9(plus^#(x, y)) }
Weak DPs:
{ sum^#(cons(x, cons(y, l))) -> sum^#(cons(plus(x, y), l))
, sum^#(cons(x, cons(y, l))) -> plus^#(x, y) }
Weak Trs:
{ plus(0(), y) -> y
, plus(s(x), y) -> s(plus(x, y)) }
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: plus^#(s(x), y) -> c_9(plus^#(x, y)) }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_9) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[cons](x1, x2) = [1] x1 + [1] x2 + [0]
[plus](x1, x2) = [1] x1 + [1] x2 + [0]
[0] = [0]
[s](x1) = [1] x1 + [1]
[sum^#](x1) = [2] x1 + [0]
[plus^#](x1, x2) = [1] x1 + [0]
[c_9](x1) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[plus(0(), y)] = [1] y + [0]
>= [1] y + [0]
= [y]
[plus(s(x), y)] = [1] x + [1] y + [1]
>= [1] x + [1] y + [1]
= [s(plus(x, y))]
[sum^#(cons(x, cons(y, l)))] = [2] l + [2] x + [2] y + [0]
>= [2] l + [2] x + [2] y + [0]
= [sum^#(cons(plus(x, y), l))]
[sum^#(cons(x, cons(y, l)))] = [2] l + [2] x + [2] y + [0]
>= [1] x + [0]
= [plus^#(x, y)]
[plus^#(s(x), y)] = [1] x + [1]
> [1] x + [0]
= [c_9(plus^#(x, y))]
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:
{ sum^#(cons(x, cons(y, l))) -> sum^#(cons(plus(x, y), l))
, sum^#(cons(x, cons(y, l))) -> plus^#(x, y)
, plus^#(s(x), y) -> c_9(plus^#(x, y)) }
Weak Trs:
{ plus(0(), y) -> y
, plus(s(x), y) -> s(plus(x, 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.
{ sum^#(cons(x, cons(y, l))) -> sum^#(cons(plus(x, y), l))
, sum^#(cons(x, cons(y, l))) -> plus^#(x, y)
, plus^#(s(x), y) -> c_9(plus^#(x, y)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Weak Trs:
{ plus(0(), y) -> y
, plus(s(x), y) -> s(plus(x, 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
Hurray, we answered YES(O(1),O(n^2))