We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^5)).
Strict Trs:
{ sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
, sum(cons(0(), x), y) -> sum(x, y)
, sum(nil(), y) -> y
, weight(cons(n, cons(m, x))) ->
weight(sum(cons(n, cons(m, x)), cons(0(), x)))
, weight(cons(n, nil())) -> n }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^5))
We add the following dependency tuples:
Strict DPs:
{ sum^#(cons(s(n), x), cons(m, y)) ->
c_1(sum^#(cons(n, x), cons(s(m), y)))
, sum^#(cons(0(), x), y) -> c_2(sum^#(x, y))
, sum^#(nil(), y) -> c_3()
, weight^#(cons(n, cons(m, x))) ->
c_4(weight^#(sum(cons(n, cons(m, x)), cons(0(), x))),
sum^#(cons(n, cons(m, x)), cons(0(), x)))
, weight^#(cons(n, nil())) -> c_5() }
and mark the set of starting terms.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^5)).
Strict DPs:
{ sum^#(cons(s(n), x), cons(m, y)) ->
c_1(sum^#(cons(n, x), cons(s(m), y)))
, sum^#(cons(0(), x), y) -> c_2(sum^#(x, y))
, sum^#(nil(), y) -> c_3()
, weight^#(cons(n, cons(m, x))) ->
c_4(weight^#(sum(cons(n, cons(m, x)), cons(0(), x))),
sum^#(cons(n, cons(m, x)), cons(0(), x)))
, weight^#(cons(n, nil())) -> c_5() }
Weak Trs:
{ sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
, sum(cons(0(), x), y) -> sum(x, y)
, sum(nil(), y) -> y
, weight(cons(n, cons(m, x))) ->
weight(sum(cons(n, cons(m, x)), cons(0(), x)))
, weight(cons(n, nil())) -> n }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^5))
We estimate the number of application of {3,5} by applications of
Pre({3,5}) = {2,4}. Here rules are labeled as follows:
DPs:
{ 1: sum^#(cons(s(n), x), cons(m, y)) ->
c_1(sum^#(cons(n, x), cons(s(m), y)))
, 2: sum^#(cons(0(), x), y) -> c_2(sum^#(x, y))
, 3: sum^#(nil(), y) -> c_3()
, 4: weight^#(cons(n, cons(m, x))) ->
c_4(weight^#(sum(cons(n, cons(m, x)), cons(0(), x))),
sum^#(cons(n, cons(m, x)), cons(0(), x)))
, 5: weight^#(cons(n, nil())) -> c_5() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^5)).
Strict DPs:
{ sum^#(cons(s(n), x), cons(m, y)) ->
c_1(sum^#(cons(n, x), cons(s(m), y)))
, sum^#(cons(0(), x), y) -> c_2(sum^#(x, y))
, weight^#(cons(n, cons(m, x))) ->
c_4(weight^#(sum(cons(n, cons(m, x)), cons(0(), x))),
sum^#(cons(n, cons(m, x)), cons(0(), x))) }
Weak DPs:
{ sum^#(nil(), y) -> c_3()
, weight^#(cons(n, nil())) -> c_5() }
Weak Trs:
{ sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
, sum(cons(0(), x), y) -> sum(x, y)
, sum(nil(), y) -> y
, weight(cons(n, cons(m, x))) ->
weight(sum(cons(n, cons(m, x)), cons(0(), x)))
, weight(cons(n, nil())) -> n }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^5))
The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.
{ sum^#(nil(), y) -> c_3()
, weight^#(cons(n, nil())) -> c_5() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^5)).
Strict DPs:
{ sum^#(cons(s(n), x), cons(m, y)) ->
c_1(sum^#(cons(n, x), cons(s(m), y)))
, sum^#(cons(0(), x), y) -> c_2(sum^#(x, y))
, weight^#(cons(n, cons(m, x))) ->
c_4(weight^#(sum(cons(n, cons(m, x)), cons(0(), x))),
sum^#(cons(n, cons(m, x)), cons(0(), x))) }
Weak Trs:
{ sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
, sum(cons(0(), x), y) -> sum(x, y)
, sum(nil(), y) -> y
, weight(cons(n, cons(m, x))) ->
weight(sum(cons(n, cons(m, x)), cons(0(), x)))
, weight(cons(n, nil())) -> n }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^5))
We replace rewrite rules by usable rules:
Weak Usable Rules:
{ sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
, sum(cons(0(), x), y) -> sum(x, y)
, sum(nil(), y) -> y }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^5)).
Strict DPs:
{ sum^#(cons(s(n), x), cons(m, y)) ->
c_1(sum^#(cons(n, x), cons(s(m), y)))
, sum^#(cons(0(), x), y) -> c_2(sum^#(x, y))
, weight^#(cons(n, cons(m, x))) ->
c_4(weight^#(sum(cons(n, cons(m, x)), cons(0(), x))),
sum^#(cons(n, cons(m, x)), cons(0(), x))) }
Weak Trs:
{ sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
, sum(cons(0(), x), y) -> sum(x, y)
, sum(nil(), y) -> y }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^5))
We decompose the input problem according to the dependency graph
into the upper component
{ weight^#(cons(n, cons(m, x))) ->
c_4(weight^#(sum(cons(n, cons(m, x)), cons(0(), x))),
sum^#(cons(n, cons(m, x)), cons(0(), x))) }
and lower component
{ sum^#(cons(s(n), x), cons(m, y)) ->
c_1(sum^#(cons(n, x), cons(s(m), y)))
, sum^#(cons(0(), x), y) -> c_2(sum^#(x, y)) }
Further, following extension rules are added to the lower
component.
{ weight^#(cons(n, cons(m, x))) ->
sum^#(cons(n, cons(m, x)), cons(0(), x))
, weight^#(cons(n, cons(m, x))) ->
weight^#(sum(cons(n, cons(m, x)), cons(0(), x))) }
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:
{ weight^#(cons(n, cons(m, x))) ->
c_4(weight^#(sum(cons(n, cons(m, x)), cons(0(), x))),
sum^#(cons(n, cons(m, x)), cons(0(), x))) }
Weak Trs:
{ sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
, sum(cons(0(), x), y) -> sum(x, y)
, sum(nil(), y) -> 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: weight^#(cons(n, cons(m, x))) ->
c_4(weight^#(sum(cons(n, cons(m, x)), cons(0(), x))),
sum^#(cons(n, cons(m, x)), cons(0(), x))) }
Sub-proof:
----------
The input was oriented with the instance of 'Small Polynomial Path
Order (PS,1-bounded)' as induced by the safe mapping
safe(sum) = {}, safe(cons) = {1, 2}, safe(s) = {1}, safe(0) = {},
safe(nil) = {}, safe(sum^#) = {}, safe(weight^#) = {},
safe(c_4) = {}
and precedence
empty .
Following symbols are considered recursive:
{weight^#}
The recursion depth is 1.
Further, following argument filtering is employed:
pi(sum) = 2, pi(cons) = [2], pi(s) = [], pi(0) = [], pi(nil) = [],
pi(sum^#) = [], pi(weight^#) = [1], pi(c_4) = [1, 2]
Usable defined function symbols are a subset of:
{sum, sum^#, weight^#}
For your convenience, here are the satisfied ordering constraints:
pi(weight^#(cons(n, cons(m, x)))) = weight^#(cons(; cons(; x));)
> c_4(weight^#(cons(; x);), sum^#();)
= pi(c_4(weight^#(sum(cons(n, cons(m, x)), cons(0(), x))),
sum^#(cons(n, cons(m, x)), cons(0(), x))))
pi(sum(cons(s(n), x), cons(m, y))) = cons(; y)
>= cons(; y)
= pi(sum(cons(n, x), cons(s(m), y)))
pi(sum(cons(0(), x), y)) = y
>= y
= pi(sum(x, y))
pi(sum(nil(), y)) = y
>= y
= pi(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:
{ weight^#(cons(n, cons(m, x))) ->
c_4(weight^#(sum(cons(n, cons(m, x)), cons(0(), x))),
sum^#(cons(n, cons(m, x)), cons(0(), x))) }
Weak Trs:
{ sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
, sum(cons(0(), x), y) -> sum(x, y)
, sum(nil(), y) -> 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.
{ weight^#(cons(n, cons(m, x))) ->
c_4(weight^#(sum(cons(n, cons(m, x)), cons(0(), x))),
sum^#(cons(n, cons(m, x)), cons(0(), x))) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Weak Trs:
{ sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
, sum(cons(0(), x), y) -> sum(x, y)
, sum(nil(), y) -> 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^4)).
Strict DPs:
{ sum^#(cons(s(n), x), cons(m, y)) ->
c_1(sum^#(cons(n, x), cons(s(m), y)))
, sum^#(cons(0(), x), y) -> c_2(sum^#(x, y)) }
Weak DPs:
{ weight^#(cons(n, cons(m, x))) ->
sum^#(cons(n, cons(m, x)), cons(0(), x))
, weight^#(cons(n, cons(m, x))) ->
weight^#(sum(cons(n, cons(m, x)), cons(0(), x))) }
Weak Trs:
{ sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
, sum(cons(0(), x), y) -> sum(x, y)
, sum(nil(), y) -> y }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^4))
We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.
DPs:
{ 2: sum^#(cons(0(), x), y) -> c_2(sum^#(x, y))
, 4: weight^#(cons(n, cons(m, x))) ->
weight^#(sum(cons(n, cons(m, x)), cons(0(), x))) }
Trs: { sum(nil(), y) -> y }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_1) = {1}, Uargs(c_2) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[sum](x1, x2) = [1] x2 + [1]
[cons](x1, x2) = [1] x2 + [2]
[s](x1) = [1] x1 + [2]
[0] = [0]
[nil] = [0]
[sum^#](x1, x2) = [3] x1 + [3]
[c_1](x1) = [1] x1 + [0]
[c_2](x1) = [1] x1 + [0]
[weight^#](x1) = [3] x1 + [3]
The order satisfies the following ordering constraints:
[sum(cons(s(n), x), cons(m, y))] = [1] y + [3]
>= [1] y + [3]
= [sum(cons(n, x), cons(s(m), y))]
[sum(cons(0(), x), y)] = [1] y + [1]
>= [1] y + [1]
= [sum(x, y)]
[sum(nil(), y)] = [1] y + [1]
> [1] y + [0]
= [y]
[sum^#(cons(s(n), x), cons(m, y))] = [3] x + [9]
>= [3] x + [9]
= [c_1(sum^#(cons(n, x), cons(s(m), y)))]
[sum^#(cons(0(), x), y)] = [3] x + [9]
> [3] x + [3]
= [c_2(sum^#(x, y))]
[weight^#(cons(n, cons(m, x)))] = [3] x + [15]
>= [3] x + [15]
= [sum^#(cons(n, cons(m, x)), cons(0(), x))]
[weight^#(cons(n, cons(m, x)))] = [3] x + [15]
> [3] x + [12]
= [weight^#(sum(cons(n, cons(m, x)), cons(0(), x)))]
We return to the main proof. Consider the set of all dependency
pairs
:
{ 1: sum^#(cons(s(n), x), cons(m, y)) ->
c_1(sum^#(cons(n, x), cons(s(m), y)))
, 2: sum^#(cons(0(), x), y) -> c_2(sum^#(x, y))
, 3: weight^#(cons(n, cons(m, x))) ->
sum^#(cons(n, cons(m, x)), cons(0(), x))
, 4: weight^#(cons(n, cons(m, x))) ->
weight^#(sum(cons(n, cons(m, x)), cons(0(), x))) }
Processor 'matrix interpretation of dimension 1' induces the
complexity certificate YES(?,O(n^1)) on application of dependency
pairs {2,4}. These cover all (indirect) predecessors of dependency
pairs {2,3,4}, their number of application is equally bounded. The
dependency pairs are shifted into the weak component.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^4)).
Strict DPs:
{ sum^#(cons(s(n), x), cons(m, y)) ->
c_1(sum^#(cons(n, x), cons(s(m), y))) }
Weak DPs:
{ sum^#(cons(0(), x), y) -> c_2(sum^#(x, y))
, weight^#(cons(n, cons(m, x))) ->
sum^#(cons(n, cons(m, x)), cons(0(), x))
, weight^#(cons(n, cons(m, x))) ->
weight^#(sum(cons(n, cons(m, x)), cons(0(), x))) }
Weak Trs:
{ sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
, sum(cons(0(), x), y) -> sum(x, y)
, sum(nil(), y) -> y }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^4))
We use the processor 'matrix interpretation of dimension 4' to
orient following rules strictly.
DPs:
{ sum^#(cons(s(n), x), cons(m, y)) ->
c_1(sum^#(cons(n, x), cons(s(m), y))) }
The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^4)) . These rules are moved into the corresponding weak
component(s).
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_1) = {1}, Uargs(c_2) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[1 0 0 0] [1 0 0 0] [0]
[sum](x1, x2) = [0 0 0 0] x1 + [0 1 0 0] x2 + [0]
[1 0 0 0] [0 0 1 0] [1]
[0 0 1 0] [0 0 0 1] [1]
[0 0 0 1] [1 0 0 1] [0]
[cons](x1, x2) = [0 0 0 0] x1 + [1 0 0 1] x2 + [1]
[0 0 0 1] [0 0 1 0] [0]
[0 0 0 1] [0 0 1 0] [1]
[0 0 1 0] [0]
[s](x1) = [0 1 1 0] x1 + [0]
[0 0 0 0] [0]
[0 0 0 1] [1]
[1]
[0] = [1]
[1]
[0]
[0]
[nil] = [0]
[0]
[0]
[1 0 0 0] [0]
[sum^#](x1, x2) = [0 0 0 0] x1 + [0]
[0 0 0 0] [0]
[0 0 0 0] [0]
[1 1 1 0] [0]
[c_1](x1) = [0 0 0 0] x1 + [0]
[0 0 0 0] [0]
[0 0 0 0] [0]
[1 0 0 0] [0]
[c_2](x1) = [0 0 0 0] x1 + [0]
[0 0 0 0] [0]
[0 0 0 0] [0]
[0 1 0 1] [0]
[weight^#](x1) = [0 0 0 0] x1 + [0]
[0 0 0 0] [0]
[0 0 0 0] [0]
The order satisfies the following ordering constraints:
[sum(cons(s(n), x), cons(m, y))] = [0 0 0 1] [1 0 0 1] [0 0 0 1] [1 0 0 1] [1]
[0 0 0 0] n + [0 0 0 0] x + [0 0 0 0] m + [1 0 0 1] y + [1]
[0 0 0 1] [1 0 0 1] [0 0 0 1] [0 0 1 0] [2]
[0 0 0 1] [0 0 1 0] [0 0 0 1] [0 0 1 0] [3]
>= [0 0 0 1] [1 0 0 1] [0 0 0 1] [1 0 0 1] [1]
[0 0 0 0] n + [0 0 0 0] x + [0 0 0 0] m + [1 0 0 1] y + [1]
[0 0 0 1] [1 0 0 1] [0 0 0 1] [0 0 1 0] [2]
[0 0 0 1] [0 0 1 0] [0 0 0 1] [0 0 1 0] [3]
= [sum(cons(n, x), cons(s(m), y))]
[sum(cons(0(), x), y)] = [1 0 0 1] [1 0 0 0] [0]
[0 0 0 0] x + [0 1 0 0] y + [0]
[1 0 0 1] [0 0 1 0] [1]
[0 0 1 0] [0 0 0 1] [1]
>= [1 0 0 0] [1 0 0 0] [0]
[0 0 0 0] x + [0 1 0 0] y + [0]
[1 0 0 0] [0 0 1 0] [1]
[0 0 1 0] [0 0 0 1] [1]
= [sum(x, y)]
[sum(nil(), y)] = [1 0 0 0] [0]
[0 1 0 0] y + [0]
[0 0 1 0] [1]
[0 0 0 1] [1]
>= [1 0 0 0] [0]
[0 1 0 0] y + [0]
[0 0 1 0] [0]
[0 0 0 1] [0]
= [y]
[sum^#(cons(s(n), x), cons(m, y))] = [0 0 0 1] [1 0 0 1] [1]
[0 0 0 0] n + [0 0 0 0] x + [0]
[0 0 0 0] [0 0 0 0] [0]
[0 0 0 0] [0 0 0 0] [0]
> [0 0 0 1] [1 0 0 1] [0]
[0 0 0 0] n + [0 0 0 0] x + [0]
[0 0 0 0] [0 0 0 0] [0]
[0 0 0 0] [0 0 0 0] [0]
= [c_1(sum^#(cons(n, x), cons(s(m), y)))]
[sum^#(cons(0(), x), y)] = [1 0 0 1] [0]
[0 0 0 0] x + [0]
[0 0 0 0] [0]
[0 0 0 0] [0]
>= [1 0 0 0] [0]
[0 0 0 0] x + [0]
[0 0 0 0] [0]
[0 0 0 0] [0]
= [c_2(sum^#(x, y))]
[weight^#(cons(n, cons(m, x)))] = [0 0 0 1] [1 0 2 1] [0 0 0 3] [3]
[0 0 0 0] n + [0 0 0 0] x + [0 0 0 0] m + [0]
[0 0 0 0] [0 0 0 0] [0 0 0 0] [0]
[0 0 0 0] [0 0 0 0] [0 0 0 0] [0]
> [0 0 0 1] [1 0 1 1] [0 0 0 2] [1]
[0 0 0 0] n + [0 0 0 0] x + [0 0 0 0] m + [0]
[0 0 0 0] [0 0 0 0] [0 0 0 0] [0]
[0 0 0 0] [0 0 0 0] [0 0 0 0] [0]
= [sum^#(cons(n, cons(m, x)), cons(0(), x))]
[weight^#(cons(n, cons(m, x)))] = [0 0 0 1] [1 0 2 1] [0 0 0 3] [3]
[0 0 0 0] n + [0 0 0 0] x + [0 0 0 0] m + [0]
[0 0 0 0] [0 0 0 0] [0 0 0 0] [0]
[0 0 0 0] [0 0 0 0] [0 0 0 0] [0]
>= [0 0 0 1] [1 0 2 1] [0 0 0 1] [3]
[0 0 0 0] n + [0 0 0 0] x + [0 0 0 0] m + [0]
[0 0 0 0] [0 0 0 0] [0 0 0 0] [0]
[0 0 0 0] [0 0 0 0] [0 0 0 0] [0]
= [weight^#(sum(cons(n, cons(m, x)), cons(0(), x)))]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Weak DPs:
{ sum^#(cons(s(n), x), cons(m, y)) ->
c_1(sum^#(cons(n, x), cons(s(m), y)))
, sum^#(cons(0(), x), y) -> c_2(sum^#(x, y))
, weight^#(cons(n, cons(m, x))) ->
sum^#(cons(n, cons(m, x)), cons(0(), x))
, weight^#(cons(n, cons(m, x))) ->
weight^#(sum(cons(n, cons(m, x)), cons(0(), x))) }
Weak Trs:
{ sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
, sum(cons(0(), x), y) -> sum(x, y)
, sum(nil(), y) -> y }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
Empty rules are trivially bounded
Hurray, we answered YES(O(1),O(n^5))