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:
  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(x, y, 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(x, y, 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:
  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(x, y, f^#(s(x), minus(y, x), z, u), f^#(x, u, z, u)) }
Obligation:
  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) = {4}

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]      [0]           
                                                   
  [c_6](x1, x2, x3, x4) = [0 0] x2 + [1 0] x4 + [0]
                          [2 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]                                                       
                             [0]                                                       
                          =  [c_5(f^#(x, u, minus(z, s(x)), u))]                       
                                                                                       
  [f^#(s(x), s(y), z, u)] =  [1]                                                       
                             [0]                                                       
                          ?  [0 0] y + [1]                                             
                             [2 0]     [0]                                             
                          =  [c_6(x, y, 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(x, y, 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:
  runtime complexity
Answer:
  YES(O(1),O(n^1))

We estimate the number of application of {1} by applications of
Pre({1}) = {4}. 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(x, y, 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(x, y, 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:
  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(x, y, f^#(s(x), minus(y, x), z, u), f^#(x, u, z, u)) }
Obligation:
  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(x, y, 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(x, y, f^#(x, u, z, u)) }
Obligation:
  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: perfectp^#(s(x)) -> c_1(f^#(x, s(0()), s(x), s(x))) }

Sub-proof:
----------
  The following argument positions are usable:
    Uargs(c_1) = {1}, Uargs(c_2) = {1}, Uargs(c_3) = {3}
  
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
                      [0] = [0]         
                                        
                  [s](x1) = [1] x1 + [0]
                                        
          [minus](x1, x2) = [0]         
                                        
         [perfectp^#](x1) = [4] x1 + [4]
                                        
    [f^#](x1, x2, x3, x4) = [0]         
                                        
                [c_1](x1) = [1] x1 + [1]
                                        
                [c_2](x1) = [4] x1 + [0]
                                        
        [c_3](x1, x2, x3) = [1] x3 + [0]
  
  The order satisfies the following ordering constraints:
  
         [perfectp^#(s(x))] =  [4] x + [4]                        
                            >  [1]                                
                            =  [c_1(f^#(x, s(0()), s(x), s(x)))]  
                                                                  
     [f^#(s(x), 0(), z, u)] =  [0]                                
                            >= [0]                                
                            =  [c_2(f^#(x, u, minus(z, s(x)), u))]
                                                                  
    [f^#(s(x), s(y), z, u)] =  [0]                                
                            >= [0]                                
                            =  [c_3(x, y, 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(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(x, y, f^#(x, u, z, u)) }
Weak DPs: { perfectp^#(s(x)) -> c_1(f^#(x, s(0()), s(x), s(x))) }
Obligation:
  runtime complexity
Answer:
  YES(O(1),O(n^1))

We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.

DPs:
  { 2: f^#(s(x), s(y), z, u) -> c_3(x, y, f^#(x, u, z, u))
  , 3: perfectp^#(s(x)) -> c_1(f^#(x, s(0()), s(x), s(x))) }

Sub-proof:
----------
  The following argument positions are usable:
    Uargs(c_1) = {1}, Uargs(c_2) = {1}, Uargs(c_3) = {3}
  
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
                      [0] = [0]         
                                        
                  [s](x1) = [1] x1 + [4]
                                        
          [minus](x1, x2) = [0]         
                                        
         [perfectp^#](x1) = [2] x1 + [6]
                                        
    [f^#](x1, x2, x3, x4) = [1] x1 + [5]
                                        
                [c_1](x1) = [1] x1 + [5]
                                        
                [c_2](x1) = [1] x1 + [4]
                                        
        [c_3](x1, x2, x3) = [1] x3 + [0]
  
  The order satisfies the following ordering constraints:
  
         [perfectp^#(s(x))] =  [2] x + [14]                       
                            >  [1] x + [10]                       
                            =  [c_1(f^#(x, s(0()), s(x), s(x)))]  
                                                                  
     [f^#(s(x), 0(), z, u)] =  [1] x + [9]                        
                            >= [1] x + [9]                        
                            =  [c_2(f^#(x, u, minus(z, s(x)), u))]
                                                                  
    [f^#(s(x), s(y), z, u)] =  [1] x + [9]                        
                            >  [1] x + [5]                        
                            =  [c_3(x, y, 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(n^1)).

Strict DPs:
  { f^#(s(x), 0(), z, u) -> c_2(f^#(x, u, minus(z, s(x)), u)) }
Weak DPs:
  { perfectp^#(s(x)) -> c_1(f^#(x, s(0()), s(x), s(x)))
  , f^#(s(x), s(y), z, u) -> c_3(x, y, f^#(x, u, z, u)) }
Obligation:
  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: f^#(s(x), 0(), z, u) -> c_2(f^#(x, u, minus(z, s(x)), u))
  , 2: perfectp^#(s(x)) -> c_1(f^#(x, s(0()), s(x), s(x)))
  , 3: f^#(s(x), s(y), z, u) -> c_3(x, y, f^#(x, u, z, u)) }

Sub-proof:
----------
  The following argument positions are usable:
    Uargs(c_1) = {1}, Uargs(c_2) = {1}, Uargs(c_3) = {3}
  
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
                      [0] = [0]         
                                        
                  [s](x1) = [1] x1 + [4]
                                        
          [minus](x1, x2) = [0]         
                                        
         [perfectp^#](x1) = [1] x1 + [1]
                                        
    [f^#](x1, x2, x3, x4) = [1] x1 + [0]
                                        
                [c_1](x1) = [1] x1 + [0]
                                        
                [c_2](x1) = [1] x1 + [1]
                                        
        [c_3](x1, x2, x3) = [1] x3 + [0]
  
  The order satisfies the following ordering constraints:
  
         [perfectp^#(s(x))] = [1] x + [5]                        
                            > [1] x + [0]                        
                            = [c_1(f^#(x, s(0()), s(x), s(x)))]  
                                                                 
     [f^#(s(x), 0(), z, u)] = [1] x + [4]                        
                            > [1] x + [1]                        
                            = [c_2(f^#(x, u, minus(z, s(x)), u))]
                                                                 
    [f^#(s(x), s(y), z, u)] = [1] x + [4]                        
                            > [1] x + [0]                        
                            = [c_3(x, y, 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:
  { 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(x, y, f^#(x, u, z, u)) }
Obligation:
  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.

{ 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(x, y, 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:
  runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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