We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).

Strict Trs:
  { rev(nil()) -> nil()
  , rev(.(x, y)) -> ++(rev(y), .(x, nil()))
  , ++(nil(), y) -> y
  , ++(.(x, y), z) -> .(x, ++(y, z))
  , car(.(x, y)) -> x
  , cdr(.(x, y)) -> y
  , null(nil()) -> true()
  , null(.(x, y)) -> false() }
Obligation:
  runtime complexity
Answer:
  YES(O(1),O(n^2))

The input is overlay and right-linear. Switching to innermost
rewriting.

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).

Strict Trs:
  { rev(nil()) -> nil()
  , rev(.(x, y)) -> ++(rev(y), .(x, nil()))
  , ++(nil(), y) -> y
  , ++(.(x, y), z) -> .(x, ++(y, z))
  , car(.(x, y)) -> x
  , cdr(.(x, y)) -> y
  , null(nil()) -> true()
  , null(.(x, y)) -> false() }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

We add the following dependency tuples:

Strict DPs:
  { rev^#(nil()) -> c_1()
  , rev^#(.(x, y)) -> c_2(++^#(rev(y), .(x, nil())), rev^#(y))
  , ++^#(nil(), y) -> c_3()
  , ++^#(.(x, y), z) -> c_4(++^#(y, z))
  , car^#(.(x, y)) -> c_5()
  , cdr^#(.(x, y)) -> c_6()
  , null^#(nil()) -> c_7()
  , null^#(.(x, y)) -> c_8() }

and mark the set of starting terms.

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).

Strict DPs:
  { rev^#(nil()) -> c_1()
  , rev^#(.(x, y)) -> c_2(++^#(rev(y), .(x, nil())), rev^#(y))
  , ++^#(nil(), y) -> c_3()
  , ++^#(.(x, y), z) -> c_4(++^#(y, z))
  , car^#(.(x, y)) -> c_5()
  , cdr^#(.(x, y)) -> c_6()
  , null^#(nil()) -> c_7()
  , null^#(.(x, y)) -> c_8() }
Weak Trs:
  { rev(nil()) -> nil()
  , rev(.(x, y)) -> ++(rev(y), .(x, nil()))
  , ++(nil(), y) -> y
  , ++(.(x, y), z) -> .(x, ++(y, z))
  , car(.(x, y)) -> x
  , cdr(.(x, y)) -> y
  , null(nil()) -> true()
  , null(.(x, y)) -> false() }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

We estimate the number of application of {1,3,5,6,7,8} by
applications of Pre({1,3,5,6,7,8}) = {2,4}. Here rules are labeled
as follows:

  DPs:
    { 1: rev^#(nil()) -> c_1()
    , 2: rev^#(.(x, y)) -> c_2(++^#(rev(y), .(x, nil())), rev^#(y))
    , 3: ++^#(nil(), y) -> c_3()
    , 4: ++^#(.(x, y), z) -> c_4(++^#(y, z))
    , 5: car^#(.(x, y)) -> c_5()
    , 6: cdr^#(.(x, y)) -> c_6()
    , 7: null^#(nil()) -> c_7()
    , 8: null^#(.(x, y)) -> c_8() }

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).

Strict DPs:
  { rev^#(.(x, y)) -> c_2(++^#(rev(y), .(x, nil())), rev^#(y))
  , ++^#(.(x, y), z) -> c_4(++^#(y, z)) }
Weak DPs:
  { rev^#(nil()) -> c_1()
  , ++^#(nil(), y) -> c_3()
  , car^#(.(x, y)) -> c_5()
  , cdr^#(.(x, y)) -> c_6()
  , null^#(nil()) -> c_7()
  , null^#(.(x, y)) -> c_8() }
Weak Trs:
  { rev(nil()) -> nil()
  , rev(.(x, y)) -> ++(rev(y), .(x, nil()))
  , ++(nil(), y) -> y
  , ++(.(x, y), z) -> .(x, ++(y, z))
  , car(.(x, y)) -> x
  , cdr(.(x, y)) -> y
  , null(nil()) -> true()
  , null(.(x, y)) -> false() }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

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

{ rev^#(nil()) -> c_1()
, ++^#(nil(), y) -> c_3()
, car^#(.(x, y)) -> c_5()
, cdr^#(.(x, y)) -> c_6()
, null^#(nil()) -> c_7()
, null^#(.(x, y)) -> c_8() }

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).

Strict DPs:
  { rev^#(.(x, y)) -> c_2(++^#(rev(y), .(x, nil())), rev^#(y))
  , ++^#(.(x, y), z) -> c_4(++^#(y, z)) }
Weak Trs:
  { rev(nil()) -> nil()
  , rev(.(x, y)) -> ++(rev(y), .(x, nil()))
  , ++(nil(), y) -> y
  , ++(.(x, y), z) -> .(x, ++(y, z))
  , car(.(x, y)) -> x
  , cdr(.(x, y)) -> y
  , null(nil()) -> true()
  , null(.(x, y)) -> false() }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

We replace rewrite rules by usable rules:

  Weak Usable Rules:
    { rev(nil()) -> nil()
    , rev(.(x, y)) -> ++(rev(y), .(x, nil()))
    , ++(nil(), y) -> y
    , ++(.(x, y), z) -> .(x, ++(y, z)) }

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).

Strict DPs:
  { rev^#(.(x, y)) -> c_2(++^#(rev(y), .(x, nil())), rev^#(y))
  , ++^#(.(x, y), z) -> c_4(++^#(y, z)) }
Weak Trs:
  { rev(nil()) -> nil()
  , rev(.(x, y)) -> ++(rev(y), .(x, nil()))
  , ++(nil(), y) -> y
  , ++(.(x, y), z) -> .(x, ++(y, z)) }
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

  { rev^#(.(x, y)) -> c_2(++^#(rev(y), .(x, nil())), rev^#(y)) }

and lower component

  { ++^#(.(x, y), z) -> c_4(++^#(y, z)) }

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

{ rev^#(.(x, y)) -> rev^#(y)
, rev^#(.(x, y)) -> ++^#(rev(y), .(x, nil())) }

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:
    { rev^#(.(x, y)) -> c_2(++^#(rev(y), .(x, nil())), rev^#(y)) }
  Weak Trs:
    { rev(nil()) -> nil()
    , rev(.(x, y)) -> ++(rev(y), .(x, nil()))
    , ++(nil(), y) -> y
    , ++(.(x, y), z) -> .(x, ++(y, 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: rev^#(.(x, y)) -> c_2(++^#(rev(y), .(x, nil())), rev^#(y)) }
  
  Sub-proof:
  ----------
    The input was oriented with the instance of 'Small Polynomial Path
    Order (PS,1-bounded)' as induced by the safe mapping
    
     safe(rev) = {}, safe(nil) = {}, safe(.) = {1, 2}, safe(++) = {},
     safe(rev^#) = {}, safe(c_2) = {}, safe(++^#) = {}
    
    and precedence
    
     empty .
    
    Following symbols are considered recursive:
    
     {rev^#}
    
    The recursion depth is 1.
    
    Further, following argument filtering is employed:
    
     pi(rev) = [], pi(nil) = [], pi(.) = [2], pi(++) = 2,
     pi(rev^#) = [1], pi(c_2) = [1, 2], pi(++^#) = []
    
    Usable defined function symbols are a subset of:
    
     {rev^#, ++^#}
    
    For your convenience, here are the satisfied ordering constraints:
    
      pi(rev^#(.(x, y))) = rev^#(.(; y);)                              
                         > c_2(++^#(),  rev^#(y;);)                    
                         = pi(c_2(++^#(rev(y), .(x, nil())), rev^#(y)))
                                                                       
  
  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:
    { rev^#(.(x, y)) -> c_2(++^#(rev(y), .(x, nil())), rev^#(y)) }
  Weak Trs:
    { rev(nil()) -> nil()
    , rev(.(x, y)) -> ++(rev(y), .(x, nil()))
    , ++(nil(), y) -> y
    , ++(.(x, y), z) -> .(x, ++(y, 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.
  
  { rev^#(.(x, y)) -> c_2(++^#(rev(y), .(x, nil())), rev^#(y)) }
  
  We are left with following problem, upon which TcT provides the
  certificate YES(O(1),O(1)).
  
  Weak Trs:
    { rev(nil()) -> nil()
    , rev(.(x, y)) -> ++(rev(y), .(x, nil()))
    , ++(nil(), y) -> y
    , ++(.(x, y), z) -> .(x, ++(y, z)) }
  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: { ++^#(.(x, y), z) -> c_4(++^#(y, z)) }
Weak DPs:
  { rev^#(.(x, y)) -> rev^#(y)
  , rev^#(.(x, y)) -> ++^#(rev(y), .(x, nil())) }
Weak Trs:
  { rev(nil()) -> nil()
  , rev(.(x, y)) -> ++(rev(y), .(x, nil()))
  , ++(nil(), y) -> y
  , ++(.(x, y), z) -> .(x, ++(y, z)) }
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: ++^#(.(x, y), z) -> c_4(++^#(y, z))
  , 2: rev^#(.(x, y)) -> rev^#(y)
  , 3: rev^#(.(x, y)) -> ++^#(rev(y), .(x, nil())) }
Trs: { rev(nil()) -> nil() }

Sub-proof:
----------
  The following argument positions are usable:
    Uargs(c_4) = {1}
  
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
         [rev](x1) = [1] x1 + [1]         
                                          
             [nil] = [0]                  
                                          
       [.](x1, x2) = [1] x2 + [2]         
                                          
      [++](x1, x2) = [1] x1 + [1] x2 + [0]
                                          
       [rev^#](x1) = [4] x1 + [3]         
                                          
    [++^#](x1, x2) = [4] x1 + [3] x2 + [0]
                                          
         [c_4](x1) = [1] x1 + [7]         
  
  The order satisfies the following ordering constraints:
  
          [rev(nil())] =  [1]                        
                       >  [0]                        
                       =  [nil()]                    
                                                     
        [rev(.(x, y))] =  [1] y + [3]                
                       >= [1] y + [3]                
                       =  [++(rev(y), .(x, nil()))]  
                                                     
        [++(nil(), y)] =  [1] y + [0]                
                       >= [1] y + [0]                
                       =  [y]                        
                                                     
      [++(.(x, y), z)] =  [1] y + [1] z + [2]        
                       >= [1] y + [1] z + [2]        
                       =  [.(x, ++(y, z))]           
                                                     
      [rev^#(.(x, y))] =  [4] y + [11]               
                       >  [4] y + [3]                
                       =  [rev^#(y)]                 
                                                     
      [rev^#(.(x, y))] =  [4] y + [11]               
                       >  [4] y + [10]               
                       =  [++^#(rev(y), .(x, nil()))]
                                                     
    [++^#(.(x, y), z)] =  [4] y + [3] z + [8]        
                       >  [4] y + [3] z + [7]        
                       =  [c_4(++^#(y, 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:
  { rev^#(.(x, y)) -> rev^#(y)
  , rev^#(.(x, y)) -> ++^#(rev(y), .(x, nil()))
  , ++^#(.(x, y), z) -> c_4(++^#(y, z)) }
Weak Trs:
  { rev(nil()) -> nil()
  , rev(.(x, y)) -> ++(rev(y), .(x, nil()))
  , ++(nil(), y) -> y
  , ++(.(x, y), z) -> .(x, ++(y, 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.

{ rev^#(.(x, y)) -> rev^#(y)
, rev^#(.(x, y)) -> ++^#(rev(y), .(x, nil()))
, ++^#(.(x, y), z) -> c_4(++^#(y, z)) }

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

Weak Trs:
  { rev(nil()) -> nil()
  , rev(.(x, y)) -> ++(rev(y), .(x, nil()))
  , ++(nil(), y) -> y
  , ++(.(x, y), z) -> .(x, ++(y, z)) }
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^2))