YES(O(1),O(n^1))

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 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 following constructor-restricted matrix
interpretation.

                    [0] = [0]         
                                      
                [s](x1) = [0]         
                                      
        [minus](x1, x2) = [1] x1 + [2]
                                      
       [perfectp^#](x1) = [0]         
                                      
                  [c_1] = [1]         
                                      
              [c_2](x1) = [1] x1 + [0]
                                      
  [f^#](x1, x2, x3, x4) = [1]         
                                      
                  [c_3] = [0]         
                                      
                  [c_4] = [0]         
                                      
              [c_5](x1) = [1] x1 + [0]
                                      
          [c_6](x1, x2) = [1] x2 + [0]

This order satisfies following ordering constraints:


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(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(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(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(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(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(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(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(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(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(n^1))

The following argument positions are usable:
  Uargs(c_2) = {1}, Uargs(c_3) = {1}

TcT has computed following constructor-based matrix interpretation
satisfying not(EDA).

                    [0] = [3]                  
                                               
                [s](x1) = [1] x1 + [2]         
                                               
        [minus](x1, x2) = [1] x1 + [3]         
                                               
  [f^#](x1, x2, x3, x4) = [2] x1 + [3] x4 + [0]
                                               
              [c_2](x1) = [1] x1 + [3]         
                                               
              [c_3](x1) = [1] x1 + [1]         

This order satisfies following ordering constraints:

   [f^#(s(x), 0(), z, u)] = [2] x + [3] u + [4]                
                          > [2] x + [3] u + [3]                
                          = [c_2(f^#(x, u, minus(z, s(x)), u))]
                                                               
  [f^#(s(x), s(y), z, u)] = [2] x + [3] u + [4]                
                          > [2] x + [3] u + [1]                
                          = [c_3(f^#(x, u, z, u))]             
                                                               

Hurray, we answered YES(O(1),O(n^1))