We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ admit(x, nil()) -> nil()
, admit(x, .(u, .(v, .(w(), z)))) ->
cond(=(sum(x, u, v), w()),
.(u, .(v, .(w(), admit(carry(x, u, v), z)))))
, cond(true(), y) -> y }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We add the following dependency tuples:
Strict DPs:
{ admit^#(x, nil()) -> c_1()
, admit^#(x, .(u, .(v, .(w(), z)))) ->
c_2(cond^#(=(sum(x, u, v), w()),
.(u, .(v, .(w(), admit(carry(x, u, v), z))))),
admit^#(carry(x, u, v), z))
, cond^#(true(), y) -> c_3() }
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:
{ admit^#(x, nil()) -> c_1()
, admit^#(x, .(u, .(v, .(w(), z)))) ->
c_2(cond^#(=(sum(x, u, v), w()),
.(u, .(v, .(w(), admit(carry(x, u, v), z))))),
admit^#(carry(x, u, v), z))
, cond^#(true(), y) -> c_3() }
Weak Trs:
{ admit(x, nil()) -> nil()
, admit(x, .(u, .(v, .(w(), z)))) ->
cond(=(sum(x, u, v), w()),
.(u, .(v, .(w(), admit(carry(x, u, v), z)))))
, cond(true(), y) -> y }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We estimate the number of application of {1,3} by applications of
Pre({1,3}) = {2}. Here rules are labeled as follows:
DPs:
{ 1: admit^#(x, nil()) -> c_1()
, 2: admit^#(x, .(u, .(v, .(w(), z)))) ->
c_2(cond^#(=(sum(x, u, v), w()),
.(u, .(v, .(w(), admit(carry(x, u, v), z))))),
admit^#(carry(x, u, v), z))
, 3: cond^#(true(), y) -> c_3() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ admit^#(x, .(u, .(v, .(w(), z)))) ->
c_2(cond^#(=(sum(x, u, v), w()),
.(u, .(v, .(w(), admit(carry(x, u, v), z))))),
admit^#(carry(x, u, v), z)) }
Weak DPs:
{ admit^#(x, nil()) -> c_1()
, cond^#(true(), y) -> c_3() }
Weak Trs:
{ admit(x, nil()) -> nil()
, admit(x, .(u, .(v, .(w(), z)))) ->
cond(=(sum(x, u, v), w()),
.(u, .(v, .(w(), admit(carry(x, u, v), z)))))
, cond(true(), y) -> 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.
{ admit^#(x, nil()) -> c_1()
, cond^#(true(), y) -> c_3() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ admit^#(x, .(u, .(v, .(w(), z)))) ->
c_2(cond^#(=(sum(x, u, v), w()),
.(u, .(v, .(w(), admit(carry(x, u, v), z))))),
admit^#(carry(x, u, v), z)) }
Weak Trs:
{ admit(x, nil()) -> nil()
, admit(x, .(u, .(v, .(w(), z)))) ->
cond(=(sum(x, u, v), w()),
.(u, .(v, .(w(), admit(carry(x, u, v), z)))))
, cond(true(), y) -> y }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
Due to missing edges in the dependency-graph, the right-hand sides
of following rules could be simplified:
{ admit^#(x, .(u, .(v, .(w(), z)))) ->
c_2(cond^#(=(sum(x, u, v), w()),
.(u, .(v, .(w(), admit(carry(x, u, v), z))))),
admit^#(carry(x, u, v), z)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ admit^#(x, .(u, .(v, .(w(), z)))) ->
c_1(admit^#(carry(x, u, v), z)) }
Weak Trs:
{ admit(x, nil()) -> nil()
, admit(x, .(u, .(v, .(w(), z)))) ->
cond(=(sum(x, u, v), w()),
.(u, .(v, .(w(), admit(carry(x, u, v), z)))))
, cond(true(), y) -> 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:
{ admit^#(x, .(u, .(v, .(w(), z)))) ->
c_1(admit^#(carry(x, u, v), z)) }
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: admit^#(x, .(u, .(v, .(w(), z)))) ->
c_1(admit^#(carry(x, u, v), z)) }
Sub-proof:
----------
The input was oriented with the instance of 'Small Polynomial Path
Order (PS,1-bounded)' as induced by the safe mapping
safe(.) = {1, 2}, safe(w) = {}, safe(carry) = {1, 2, 3},
safe(admit^#) = {1}, safe(c_1) = {}
and precedence
empty .
Following symbols are considered recursive:
{admit^#}
The recursion depth is 1.
Further, following argument filtering is employed:
pi(.) = [2], pi(w) = [], pi(carry) = [], pi(admit^#) = [1, 2],
pi(c_1) = [1]
Usable defined function symbols are a subset of:
{admit^#}
For your convenience, here are the satisfied ordering constraints:
pi(admit^#(x, .(u, .(v, .(w(), z))))) = admit^#(.(; .(; .(; z))); x)
> c_1(admit^#(z; carry());)
= pi(c_1(admit^#(carry(x, u, v), z)))
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:
{ admit^#(x, .(u, .(v, .(w(), z)))) ->
c_1(admit^#(carry(x, u, v), z)) }
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.
{ admit^#(x, .(u, .(v, .(w(), z)))) ->
c_1(admit^#(carry(x, u, v), z)) }
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))