We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^3)).

Strict Trs:
  { app(nil(), y) -> y
  , app(add(n, x), y) -> add(n, app(x, y))
  , reverse(nil()) -> nil()
  , reverse(add(n, x)) -> app(reverse(x), add(n, nil()))
  , shuffle(nil()) -> nil()
  , shuffle(add(n, x)) -> add(n, shuffle(reverse(x))) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^3))

We add the following dependency tuples:

Strict DPs:
  { app^#(nil(), y) -> c_1()
  , app^#(add(n, x), y) -> c_2(app^#(x, y))
  , reverse^#(nil()) -> c_3()
  , reverse^#(add(n, x)) ->
    c_4(app^#(reverse(x), add(n, nil())), reverse^#(x))
  , shuffle^#(nil()) -> c_5()
  , shuffle^#(add(n, x)) ->
    c_6(shuffle^#(reverse(x)), reverse^#(x)) }

and mark the set of starting terms.

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^3)).

Strict DPs:
  { app^#(nil(), y) -> c_1()
  , app^#(add(n, x), y) -> c_2(app^#(x, y))
  , reverse^#(nil()) -> c_3()
  , reverse^#(add(n, x)) ->
    c_4(app^#(reverse(x), add(n, nil())), reverse^#(x))
  , shuffle^#(nil()) -> c_5()
  , shuffle^#(add(n, x)) ->
    c_6(shuffle^#(reverse(x)), reverse^#(x)) }
Weak Trs:
  { app(nil(), y) -> y
  , app(add(n, x), y) -> add(n, app(x, y))
  , reverse(nil()) -> nil()
  , reverse(add(n, x)) -> app(reverse(x), add(n, nil()))
  , shuffle(nil()) -> nil()
  , shuffle(add(n, x)) -> add(n, shuffle(reverse(x))) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^3))

We estimate the number of application of {1,3,5} by applications of
Pre({1,3,5}) = {2,4,6}. Here rules are labeled as follows:

  DPs:
    { 1: app^#(nil(), y) -> c_1()
    , 2: app^#(add(n, x), y) -> c_2(app^#(x, y))
    , 3: reverse^#(nil()) -> c_3()
    , 4: reverse^#(add(n, x)) ->
         c_4(app^#(reverse(x), add(n, nil())), reverse^#(x))
    , 5: shuffle^#(nil()) -> c_5()
    , 6: shuffle^#(add(n, x)) ->
         c_6(shuffle^#(reverse(x)), reverse^#(x)) }

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^3)).

Strict DPs:
  { app^#(add(n, x), y) -> c_2(app^#(x, y))
  , reverse^#(add(n, x)) ->
    c_4(app^#(reverse(x), add(n, nil())), reverse^#(x))
  , shuffle^#(add(n, x)) ->
    c_6(shuffle^#(reverse(x)), reverse^#(x)) }
Weak DPs:
  { app^#(nil(), y) -> c_1()
  , reverse^#(nil()) -> c_3()
  , shuffle^#(nil()) -> c_5() }
Weak Trs:
  { app(nil(), y) -> y
  , app(add(n, x), y) -> add(n, app(x, y))
  , reverse(nil()) -> nil()
  , reverse(add(n, x)) -> app(reverse(x), add(n, nil()))
  , shuffle(nil()) -> nil()
  , shuffle(add(n, x)) -> add(n, shuffle(reverse(x))) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^3))

The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.

{ app^#(nil(), y) -> c_1()
, reverse^#(nil()) -> c_3()
, shuffle^#(nil()) -> c_5() }

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^3)).

Strict DPs:
  { app^#(add(n, x), y) -> c_2(app^#(x, y))
  , reverse^#(add(n, x)) ->
    c_4(app^#(reverse(x), add(n, nil())), reverse^#(x))
  , shuffle^#(add(n, x)) ->
    c_6(shuffle^#(reverse(x)), reverse^#(x)) }
Weak Trs:
  { app(nil(), y) -> y
  , app(add(n, x), y) -> add(n, app(x, y))
  , reverse(nil()) -> nil()
  , reverse(add(n, x)) -> app(reverse(x), add(n, nil()))
  , shuffle(nil()) -> nil()
  , shuffle(add(n, x)) -> add(n, shuffle(reverse(x))) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^3))

We replace rewrite rules by usable rules:

  Weak Usable Rules:
    { app(nil(), y) -> y
    , app(add(n, x), y) -> add(n, app(x, y))
    , reverse(nil()) -> nil()
    , reverse(add(n, x)) -> app(reverse(x), add(n, nil())) }

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^3)).

Strict DPs:
  { app^#(add(n, x), y) -> c_2(app^#(x, y))
  , reverse^#(add(n, x)) ->
    c_4(app^#(reverse(x), add(n, nil())), reverse^#(x))
  , shuffle^#(add(n, x)) ->
    c_6(shuffle^#(reverse(x)), reverse^#(x)) }
Weak Trs:
  { app(nil(), y) -> y
  , app(add(n, x), y) -> add(n, app(x, y))
  , reverse(nil()) -> nil()
  , reverse(add(n, x)) -> app(reverse(x), add(n, nil())) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^3))

We decompose the input problem according to the dependency graph
into the upper component

  { shuffle^#(add(n, x)) ->
    c_6(shuffle^#(reverse(x)), reverse^#(x)) }

and lower component

  { app^#(add(n, x), y) -> c_2(app^#(x, y))
  , reverse^#(add(n, x)) ->
    c_4(app^#(reverse(x), add(n, nil())), reverse^#(x)) }

Further, following extension rules are added to the lower
component.

{ shuffle^#(add(n, x)) -> reverse^#(x)
, shuffle^#(add(n, x)) -> shuffle^#(reverse(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:
    { shuffle^#(add(n, x)) ->
      c_6(shuffle^#(reverse(x)), reverse^#(x)) }
  Weak Trs:
    { app(nil(), y) -> y
    , app(add(n, x), y) -> add(n, app(x, y))
    , reverse(nil()) -> nil()
    , reverse(add(n, x)) -> app(reverse(x), add(n, nil())) }
  Obligation:
    innermost 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: shuffle^#(add(n, x)) ->
         c_6(shuffle^#(reverse(x)), reverse^#(x)) }
  Trs: { reverse(nil()) -> nil() }
  
  Sub-proof:
  ----------
    The following argument positions are usable:
      Uargs(c_6) = {1}
    
    TcT has computed the following constructor-based matrix
    interpretation satisfying not(EDA).
    
        [app](x1, x2) = [1] x1 + [1] x2 + [0]
                                             
                [nil] = [0]                  
                                             
        [add](x1, x2) = [1] x2 + [2]         
                                             
        [reverse](x1) = [1] x1 + [1]         
                                             
      [reverse^#](x1) = [0]                  
                                             
      [shuffle^#](x1) = [1] x1 + [0]         
                                             
        [c_6](x1, x2) = [1] x1 + [1] x2 + [0]
    
    The order satisfies the following ordering constraints:
    
             [app(nil(), y)] =  [1] y + [0]                               
                             >= [1] y + [0]                               
                             =  [y]                                       
                                                                          
         [app(add(n, x), y)] =  [1] y + [1] x + [2]                       
                             >= [1] y + [1] x + [2]                       
                             =  [add(n, app(x, y))]                       
                                                                          
            [reverse(nil())] =  [1]                                       
                             >  [0]                                       
                             =  [nil()]                                   
                                                                          
        [reverse(add(n, x))] =  [1] x + [3]                               
                             >= [1] x + [3]                               
                             =  [app(reverse(x), add(n, nil()))]          
                                                                          
      [shuffle^#(add(n, x))] =  [1] x + [2]                               
                             >  [1] x + [1]                               
                             =  [c_6(shuffle^#(reverse(x)), reverse^#(x))]
                                                                          
  
  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:
    { shuffle^#(add(n, x)) ->
      c_6(shuffle^#(reverse(x)), reverse^#(x)) }
  Weak Trs:
    { app(nil(), y) -> y
    , app(add(n, x), y) -> add(n, app(x, y))
    , reverse(nil()) -> nil()
    , reverse(add(n, x)) -> app(reverse(x), add(n, nil())) }
  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.
  
  { shuffle^#(add(n, x)) ->
    c_6(shuffle^#(reverse(x)), reverse^#(x)) }
  
  We are left with following problem, upon which TcT provides the
  certificate YES(O(1),O(1)).
  
  Weak Trs:
    { app(nil(), y) -> y
    , app(add(n, x), y) -> add(n, app(x, y))
    , reverse(nil()) -> nil()
    , reverse(add(n, x)) -> app(reverse(x), add(n, nil())) }
  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^2)).

Strict DPs:
  { app^#(add(n, x), y) -> c_2(app^#(x, y))
  , reverse^#(add(n, x)) ->
    c_4(app^#(reverse(x), add(n, nil())), reverse^#(x)) }
Weak DPs:
  { shuffle^#(add(n, x)) -> reverse^#(x)
  , shuffle^#(add(n, x)) -> shuffle^#(reverse(x)) }
Weak Trs:
  { app(nil(), y) -> y
  , app(add(n, x), y) -> add(n, app(x, y))
  , reverse(nil()) -> nil()
  , reverse(add(n, x)) -> app(reverse(x), add(n, nil())) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

We decompose the input problem according to the dependency graph
into the upper component

  { reverse^#(add(n, x)) ->
    c_4(app^#(reverse(x), add(n, nil())), reverse^#(x))
  , shuffle^#(add(n, x)) -> reverse^#(x)
  , shuffle^#(add(n, x)) -> shuffle^#(reverse(x)) }

and lower component

  { app^#(add(n, x), y) -> c_2(app^#(x, y)) }

Further, following extension rules are added to the lower
component.

{ reverse^#(add(n, x)) -> app^#(reverse(x), add(n, nil()))
, reverse^#(add(n, x)) -> reverse^#(x)
, shuffle^#(add(n, x)) -> reverse^#(x)
, shuffle^#(add(n, x)) -> shuffle^#(reverse(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:
    { reverse^#(add(n, x)) ->
      c_4(app^#(reverse(x), add(n, nil())), reverse^#(x)) }
  Weak DPs:
    { shuffle^#(add(n, x)) -> reverse^#(x)
    , shuffle^#(add(n, x)) -> shuffle^#(reverse(x)) }
  Weak Trs:
    { app(nil(), y) -> y
    , app(add(n, x), y) -> add(n, app(x, y))
    , reverse(nil()) -> nil()
    , reverse(add(n, x)) -> app(reverse(x), add(n, nil())) }
  Obligation:
    innermost 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: reverse^#(add(n, x)) ->
         c_4(app^#(reverse(x), add(n, nil())), reverse^#(x))
    , 2: shuffle^#(add(n, x)) -> reverse^#(x)
    , 3: shuffle^#(add(n, x)) -> shuffle^#(reverse(x)) }
  Trs: { reverse(nil()) -> nil() }
  
  Sub-proof:
  ----------
    The following argument positions are usable:
      Uargs(c_4) = {1, 2}
    
    TcT has computed the following constructor-based matrix
    interpretation satisfying not(EDA).
    
        [app](x1, x2) = [1] x1 + [1] x2 + [0]
                                             
                [nil] = [0]                  
                                             
        [add](x1, x2) = [1] x2 + [3]         
                                             
        [reverse](x1) = [1] x1 + [2]         
                                             
      [app^#](x1, x2) = [2]                  
                                             
      [reverse^#](x1) = [4] x1 + [0]         
                                             
        [c_4](x1, x2) = [4] x1 + [1] x2 + [3]
                                             
      [shuffle^#](x1) = [4] x1 + [3]         
    
    The order satisfies the following ordering constraints:
    
             [app(nil(), y)] =  [1] y + [0]                                          
                             >= [1] y + [0]                                          
                             =  [y]                                                  
                                                                                     
         [app(add(n, x), y)] =  [1] y + [1] x + [3]                                  
                             >= [1] y + [1] x + [3]                                  
                             =  [add(n, app(x, y))]                                  
                                                                                     
            [reverse(nil())] =  [2]                                                  
                             >  [0]                                                  
                             =  [nil()]                                              
                                                                                     
        [reverse(add(n, x))] =  [1] x + [5]                                          
                             >= [1] x + [5]                                          
                             =  [app(reverse(x), add(n, nil()))]                     
                                                                                     
      [reverse^#(add(n, x))] =  [4] x + [12]                                         
                             >  [4] x + [11]                                         
                             =  [c_4(app^#(reverse(x), add(n, nil())), reverse^#(x))]
                                                                                     
      [shuffle^#(add(n, x))] =  [4] x + [15]                                         
                             >  [4] x + [0]                                          
                             =  [reverse^#(x)]                                       
                                                                                     
      [shuffle^#(add(n, x))] =  [4] x + [15]                                         
                             >  [4] x + [11]                                         
                             =  [shuffle^#(reverse(x))]                              
                                                                                     
  
  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:
    { reverse^#(add(n, x)) ->
      c_4(app^#(reverse(x), add(n, nil())), reverse^#(x))
    , shuffle^#(add(n, x)) -> reverse^#(x)
    , shuffle^#(add(n, x)) -> shuffle^#(reverse(x)) }
  Weak Trs:
    { app(nil(), y) -> y
    , app(add(n, x), y) -> add(n, app(x, y))
    , reverse(nil()) -> nil()
    , reverse(add(n, x)) -> app(reverse(x), add(n, nil())) }
  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.
  
  { reverse^#(add(n, x)) ->
    c_4(app^#(reverse(x), add(n, nil())), reverse^#(x))
  , shuffle^#(add(n, x)) -> reverse^#(x)
  , shuffle^#(add(n, x)) -> shuffle^#(reverse(x)) }
  
  We are left with following problem, upon which TcT provides the
  certificate YES(O(1),O(1)).
  
  Weak Trs:
    { app(nil(), y) -> y
    , app(add(n, x), y) -> add(n, app(x, y))
    , reverse(nil()) -> nil()
    , reverse(add(n, x)) -> app(reverse(x), add(n, nil())) }
  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^1)).

Strict DPs: { app^#(add(n, x), y) -> c_2(app^#(x, y)) }
Weak DPs:
  { reverse^#(add(n, x)) -> app^#(reverse(x), add(n, nil()))
  , reverse^#(add(n, x)) -> reverse^#(x)
  , shuffle^#(add(n, x)) -> reverse^#(x)
  , shuffle^#(add(n, x)) -> shuffle^#(reverse(x)) }
Weak Trs:
  { app(nil(), y) -> y
  , app(add(n, x), y) -> add(n, app(x, y))
  , reverse(nil()) -> nil()
  , reverse(add(n, x)) -> app(reverse(x), add(n, nil())) }
Obligation:
  innermost 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: app^#(add(n, x), y) -> c_2(app^#(x, y))
  , 2: reverse^#(add(n, x)) -> app^#(reverse(x), add(n, nil()))
  , 3: reverse^#(add(n, x)) -> reverse^#(x)
  , 4: shuffle^#(add(n, x)) -> reverse^#(x)
  , 5: shuffle^#(add(n, x)) -> shuffle^#(reverse(x)) }

Sub-proof:
----------
  The following argument positions are usable:
    Uargs(c_2) = {1}
  
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
      [app](x1, x2) = [1] x1 + [1] x2 + [0]
                                           
              [nil] = [0]                  
                                           
      [add](x1, x2) = [1] x2 + [2]         
                                           
      [reverse](x1) = [1] x1 + [0]         
                                           
    [app^#](x1, x2) = [1] x1 + [5]         
                                           
          [c_2](x1) = [1] x1 + [0]         
                                           
    [reverse^#](x1) = [4] x1 + [5]         
                                           
    [shuffle^#](x1) = [5] x1 + [4]         
  
  The order satisfies the following ordering constraints:
  
           [app(nil(), y)] =  [1] y + [0]                       
                           >= [1] y + [0]                       
                           =  [y]                               
                                                                
       [app(add(n, x), y)] =  [1] y + [1] x + [2]               
                           >= [1] y + [1] x + [2]               
                           =  [add(n, app(x, y))]               
                                                                
          [reverse(nil())] =  [0]                               
                           >= [0]                               
                           =  [nil()]                           
                                                                
      [reverse(add(n, x))] =  [1] x + [2]                       
                           >= [1] x + [2]                       
                           =  [app(reverse(x), add(n, nil()))]  
                                                                
     [app^#(add(n, x), y)] =  [1] x + [7]                       
                           >  [1] x + [5]                       
                           =  [c_2(app^#(x, y))]                
                                                                
    [reverse^#(add(n, x))] =  [4] x + [13]                      
                           >  [1] x + [5]                       
                           =  [app^#(reverse(x), add(n, nil()))]
                                                                
    [reverse^#(add(n, x))] =  [4] x + [13]                      
                           >  [4] x + [5]                       
                           =  [reverse^#(x)]                    
                                                                
    [shuffle^#(add(n, x))] =  [5] x + [14]                      
                           >  [4] x + [5]                       
                           =  [reverse^#(x)]                    
                                                                
    [shuffle^#(add(n, x))] =  [5] x + [14]                      
                           >  [5] x + [4]                       
                           =  [shuffle^#(reverse(x))]           
                                                                

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:
  { app^#(add(n, x), y) -> c_2(app^#(x, y))
  , reverse^#(add(n, x)) -> app^#(reverse(x), add(n, nil()))
  , reverse^#(add(n, x)) -> reverse^#(x)
  , shuffle^#(add(n, x)) -> reverse^#(x)
  , shuffle^#(add(n, x)) -> shuffle^#(reverse(x)) }
Weak Trs:
  { app(nil(), y) -> y
  , app(add(n, x), y) -> add(n, app(x, y))
  , reverse(nil()) -> nil()
  , reverse(add(n, x)) -> app(reverse(x), add(n, nil())) }
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.

{ app^#(add(n, x), y) -> c_2(app^#(x, y))
, reverse^#(add(n, x)) -> app^#(reverse(x), add(n, nil()))
, reverse^#(add(n, x)) -> reverse^#(x)
, shuffle^#(add(n, x)) -> reverse^#(x)
, shuffle^#(add(n, x)) -> shuffle^#(reverse(x)) }

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).

Weak Trs:
  { app(nil(), y) -> y
  , app(add(n, x), y) -> add(n, app(x, y))
  , reverse(nil()) -> nil()
  , reverse(add(n, x)) -> app(reverse(x), add(n, nil())) }
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

Hurray, we answered YES(O(1),O(n^3))