YES(O(1),O(n^2))

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

Strict Trs:
  { f(empty(), l) -> l
  , f(cons(x, k), l) -> g(k, l, cons(x, k))
  , g(a, b, c) -> f(a, cons(b, c)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

We use the processor 'custom shape polynomial interpretation' to
orient following rules strictly.

Trs:
  { f(empty(), l) -> l
  , f(cons(x, k), l) -> g(k, l, cons(x, k)) }

The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^2)) . These rules are moved into the corresponding weak
component(s).

Sub-proof:
----------
  TcT has computed the following constructor-restricted polynomial
  interpretation.
      [f](x1, x2) = 2 + x1 + 2*x1^2 + x2       
                                               
        [empty]() = 0                          
                                               
   [cons](x1, x2) = 1 + x1 + x2                
                                               
  [g](x1, x2, x3) = 3 + 3*x1 + 2*x1^2 + x2 + x3
                                               
  
  This order satisfies the following ordering constraints.
  
       [f(empty(), l)] =  2 + l                                            
                       >  l                                                
                       =  [l]                                              
                                                                           
    [f(cons(x, k), l)] =  5 + 5*x + 5*k + 2*x^2 + 2*x*k + 2*k*x + 2*k^2 + l
                       >  4 + 4*k + 2*k^2 + l + x                          
                       =  [g(k, l, cons(x, k))]                            
                                                                           
          [g(a, b, c)] =  3 + 3*a + 2*a^2 + b + c                          
                       >= 3 + a + 2*a^2 + b + c                            
                       =  [f(a, cons(b, c))]                               
                                                                           

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 Trs: { g(a, b, c) -> f(a, cons(b, c)) }
Weak Trs:
  { f(empty(), l) -> l
  , f(cons(x, k), l) -> g(k, l, cons(x, k)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

We use the processor 'custom shape polynomial interpretation' to
orient following rules strictly.

Trs: { g(a, b, c) -> f(a, cons(b, c)) }

The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^2)) . These rules are moved into the corresponding weak
component(s).

Sub-proof:
----------
  TcT has computed the following constructor-restricted polynomial
  interpretation.
      [f](x1, x2) = 2*x1 + 2*x1^2 + x2           
                                                 
        [empty]() = 0                            
                                                 
   [cons](x1, x2) = 1 + x1 + x2                  
                                                 
  [g](x1, x2, x3) = 2 + 2*x1 + 2*x1^2 + x2 + 2*x3
                                                 
  
  This order satisfies the following ordering constraints.
  
       [f(empty(), l)] =  l                                                
                       >= l                                                
                       =  [l]                                              
                                                                           
    [f(cons(x, k), l)] =  4 + 6*x + 6*k + 2*x^2 + 2*x*k + 2*k*x + 2*k^2 + l
                       >= 4 + 4*k + 2*k^2 + l + 2*x                        
                       =  [g(k, l, cons(x, k))]                            
                                                                           
          [g(a, b, c)] =  2 + 2*a + 2*a^2 + b + 2*c                        
                       >  2*a + 2*a^2 + 1 + b + c                          
                       =  [f(a, cons(b, c))]                               
                                                                           

We return to the main proof.

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

Weak Trs:
  { f(empty(), l) -> l
  , f(cons(x, k), l) -> g(k, l, cons(x, k))
  , g(a, b, c) -> f(a, cons(b, c)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

Hurray, we answered YES(O(1),O(n^2))