We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ perfectp(0()) -> false()
, perfectp(s(x)) -> f(x, s(0()), s(x), s(x))
, f(0(), y, 0(), u) -> true()
, f(0(), y, s(z), u) -> false()
, f(s(x), 0(), z, u) -> f(x, u, minus(z, s(x)), u)
, f(s(x), s(y), z, u) ->
if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We add the following weak dependency pairs:
Strict DPs:
{ perfectp^#(0()) -> c_1()
, perfectp^#(s(x)) -> c_2(f^#(x, s(0()), s(x), s(x)))
, f^#(0(), y, 0(), u) -> c_3()
, f^#(0(), y, s(z), u) -> c_4()
, f^#(s(x), 0(), z, u) -> c_5(f^#(x, u, minus(z, s(x)), u))
, f^#(s(x), s(y), z, u) ->
c_6(f^#(s(x), minus(y, x), z, u), f^#(x, u, z, u)) }
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:
{ perfectp^#(0()) -> c_1()
, perfectp^#(s(x)) -> c_2(f^#(x, s(0()), s(x), s(x)))
, f^#(0(), y, 0(), u) -> c_3()
, f^#(0(), y, s(z), u) -> c_4()
, f^#(s(x), 0(), z, u) -> c_5(f^#(x, u, minus(z, s(x)), u))
, f^#(s(x), s(y), z, u) ->
c_6(f^#(s(x), minus(y, x), z, u), f^#(x, u, z, u)) }
Strict Trs:
{ perfectp(0()) -> false()
, perfectp(s(x)) -> f(x, s(0()), s(x), s(x))
, f(0(), y, 0(), u) -> true()
, f(0(), y, s(z), u) -> false()
, f(s(x), 0(), z, u) -> f(x, u, minus(z, s(x)), u)
, f(s(x), s(y), z, u) ->
if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)) }
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:
{ perfectp^#(0()) -> c_1()
, perfectp^#(s(x)) -> c_2(f^#(x, s(0()), s(x), s(x)))
, f^#(0(), y, 0(), u) -> c_3()
, f^#(0(), y, s(z), u) -> c_4()
, f^#(s(x), 0(), z, u) -> c_5(f^#(x, u, minus(z, s(x)), u))
, f^#(s(x), s(y), z, u) ->
c_6(f^#(s(x), minus(y, x), z, u), f^#(x, u, z, u)) }
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_2) = {1}, Uargs(c_5) = {1}, Uargs(c_6) = {2}
TcT has computed the following constructor-restricted matrix
interpretation.
[0] = [0]
[0]
[s](x1) = [0]
[0]
[minus](x1, x2) = [0]
[0]
[perfectp^#](x1) = [0]
[0]
[c_1] = [0]
[0]
[c_2](x1) = [1 0] x1 + [0]
[0 1] [0]
[f^#](x1, x2, x3, x4) = [1]
[0]
[c_3] = [0]
[0]
[c_4] = [0]
[0]
[c_5](x1) = [1 0] x1 + [0]
[0 1] [2]
[c_6](x1, x2) = [1 0] x2 + [0]
[0 1] [0]
The order satisfies the following ordering constraints:
[perfectp^#(0())] = [0]
[0]
>= [0]
[0]
= [c_1()]
[perfectp^#(s(x))] = [0]
[0]
? [1]
[0]
= [c_2(f^#(x, s(0()), s(x), s(x)))]
[f^#(0(), y, 0(), u)] = [1]
[0]
> [0]
[0]
= [c_3()]
[f^#(0(), y, s(z), u)] = [1]
[0]
> [0]
[0]
= [c_4()]
[f^#(s(x), 0(), z, u)] = [1]
[0]
? [1]
[2]
= [c_5(f^#(x, u, minus(z, s(x)), u))]
[f^#(s(x), s(y), z, u)] = [1]
[0]
>= [1]
[0]
= [c_6(f^#(s(x), minus(y, x), z, u), f^#(x, u, z, u))]
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:
{ perfectp^#(0()) -> c_1()
, perfectp^#(s(x)) -> c_2(f^#(x, s(0()), s(x), s(x)))
, f^#(s(x), 0(), z, u) -> c_5(f^#(x, u, minus(z, s(x)), u))
, f^#(s(x), s(y), z, u) ->
c_6(f^#(s(x), minus(y, x), z, u), f^#(x, u, z, u)) }
Weak DPs:
{ f^#(0(), y, 0(), u) -> c_3()
, f^#(0(), y, s(z), u) -> c_4() }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We estimate the number of application of {1} by applications of
Pre({1}) = {}. Here rules are labeled as follows:
DPs:
{ 1: perfectp^#(0()) -> c_1()
, 2: perfectp^#(s(x)) -> c_2(f^#(x, s(0()), s(x), s(x)))
, 3: f^#(s(x), 0(), z, u) -> c_5(f^#(x, u, minus(z, s(x)), u))
, 4: f^#(s(x), s(y), z, u) ->
c_6(f^#(s(x), minus(y, x), z, u), f^#(x, u, z, u))
, 5: f^#(0(), y, 0(), u) -> c_3()
, 6: f^#(0(), y, s(z), u) -> c_4() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ perfectp^#(s(x)) -> c_2(f^#(x, s(0()), s(x), s(x)))
, f^#(s(x), 0(), z, u) -> c_5(f^#(x, u, minus(z, s(x)), u))
, f^#(s(x), s(y), z, u) ->
c_6(f^#(s(x), minus(y, x), z, u), f^#(x, u, z, u)) }
Weak DPs:
{ perfectp^#(0()) -> c_1()
, f^#(0(), y, 0(), u) -> c_3()
, f^#(0(), y, s(z), u) -> c_4() }
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.
{ perfectp^#(0()) -> c_1()
, f^#(0(), y, 0(), u) -> c_3()
, f^#(0(), y, s(z), u) -> c_4() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ perfectp^#(s(x)) -> c_2(f^#(x, s(0()), s(x), s(x)))
, f^#(s(x), 0(), z, u) -> c_5(f^#(x, u, minus(z, s(x)), u))
, f^#(s(x), s(y), z, u) ->
c_6(f^#(s(x), minus(y, x), z, u), f^#(x, u, z, u)) }
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:
{ f^#(s(x), s(y), z, u) ->
c_6(f^#(s(x), minus(y, x), z, u), f^#(x, u, z, u)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ perfectp^#(s(x)) -> c_1(f^#(x, s(0()), s(x), s(x)))
, f^#(s(x), 0(), z, u) -> c_2(f^#(x, u, minus(z, s(x)), u))
, f^#(s(x), s(y), z, u) -> c_3(f^#(x, u, z, u)) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
Consider the dependency graph
1: perfectp^#(s(x)) -> c_1(f^#(x, s(0()), s(x), s(x)))
-->_1 f^#(s(x), s(y), z, u) -> c_3(f^#(x, u, z, u)) :3
2: f^#(s(x), 0(), z, u) -> c_2(f^#(x, u, minus(z, s(x)), u))
-->_1 f^#(s(x), s(y), z, u) -> c_3(f^#(x, u, z, u)) :3
-->_1 f^#(s(x), 0(), z, u) -> c_2(f^#(x, u, minus(z, s(x)), u)) :2
3: f^#(s(x), s(y), z, u) -> c_3(f^#(x, u, z, u))
-->_1 f^#(s(x), s(y), z, u) -> c_3(f^#(x, u, z, u)) :3
-->_1 f^#(s(x), 0(), z, u) -> c_2(f^#(x, u, minus(z, s(x)), u)) :2
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).
{ perfectp^#(s(x)) -> c_1(f^#(x, s(0()), s(x), s(x))) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ f^#(s(x), 0(), z, u) -> c_2(f^#(x, u, minus(z, s(x)), u))
, f^#(s(x), s(y), z, u) -> c_3(f^#(x, u, z, u)) }
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: f^#(s(x), 0(), z, u) -> c_2(f^#(x, u, minus(z, s(x)), u))
, 2: f^#(s(x), s(y), z, u) -> c_3(f^#(x, u, z, u)) }
Sub-proof:
----------
The input was oriented with the instance of 'Small Polynomial Path
Order (PS,1-bounded)' as induced by the safe mapping
safe(0) = {}, safe(s) = {1}, safe(minus) = {1, 2},
safe(f^#) = {2, 3}, safe(c_2) = {}, safe(c_3) = {}
and precedence
empty .
Following symbols are considered recursive:
{f^#}
The recursion depth is 1.
Further, following argument filtering is employed:
pi(0) = [], pi(s) = [1], pi(minus) = 2, pi(f^#) = [1, 2, 3, 4],
pi(c_2) = [1], pi(c_3) = [1]
Usable defined function symbols are a subset of:
{f^#}
For your convenience, here are the satisfied ordering constraints:
pi(f^#(s(x), 0(), z, u)) = f^#(s(; x), u; 0(), z)
> c_2(f^#(x, u; u, s(; x));)
= pi(c_2(f^#(x, u, minus(z, s(x)), u)))
pi(f^#(s(x), s(y), z, u)) = f^#(s(; x), u; s(; y), z)
> c_3(f^#(x, u; u, z);)
= pi(c_3(f^#(x, u, z, u)))
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:
{ f^#(s(x), 0(), z, u) -> c_2(f^#(x, u, minus(z, s(x)), u))
, f^#(s(x), s(y), z, u) -> c_3(f^#(x, u, z, u)) }
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.
{ f^#(s(x), 0(), z, u) -> c_2(f^#(x, u, minus(z, s(x)), u))
, f^#(s(x), s(y), z, u) -> c_3(f^#(x, u, z, u)) }
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))