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:
  { f(a, empty()) -> g(a, empty())
  , f(a, cons(x, k)) -> f(cons(x, a), k)
  , g(empty(), d) -> d
  , g(cons(x, k), d) -> g(k, cons(x, d)) }
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.

Trs: { g(empty(), d) -> d }

The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^1)) . These rules are moved into the corresponding weak
component(s).

Sub-proof:
----------
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
       [f](x1, x2) = [2] x1 + [2] x2 + [0]
                                          
           [empty] = [2]                  
                                          
       [g](x1, x2) = [2] x1 + [2] x2 + [0]
                                          
    [cons](x1, x2) = [1] x1 + [1] x2 + [0]
  
  This order satisfies the following ordering constraints:
  
       [f(a, empty())] =  [2] a + [4]                
                       >= [2] a + [4]                
                       =  [g(a, empty())]            
                                                     
    [f(a, cons(x, k))] =  [2] a + [2] x + [2] k + [0]
                       >= [2] a + [2] x + [2] k + [0]
                       =  [f(cons(x, a), k)]         
                                                     
       [g(empty(), d)] =  [2] d + [4]                
                       >  [1] d + [0]                
                       =  [d]                        
                                                     
    [g(cons(x, k), d)] =  [2] x + [2] k + [2] d + [0]
                       >= [2] x + [2] k + [2] d + [0]
                       =  [g(k, cons(x, d))]         
                                                     

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 Trs:
  { f(a, empty()) -> g(a, empty())
  , f(a, cons(x, k)) -> f(cons(x, a), k)
  , g(cons(x, k), d) -> g(k, cons(x, d)) }
Weak Trs: { g(empty(), d) -> d }
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.

Trs: { g(cons(x, k), d) -> g(k, cons(x, d)) }

The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^1)) . These rules are moved into the corresponding weak
component(s).

Sub-proof:
----------
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
       [f](x1, x2) = [2] x1 + [2] x2 + [0]
                                          
           [empty] = [0]                  
                                          
       [g](x1, x2) = [2] x1 + [1] x2 + [0]
                                          
    [cons](x1, x2) = [1] x1 + [1] x2 + [2]
  
  This order satisfies the following ordering constraints:
  
       [f(a, empty())] =  [2] a + [0]                
                       >= [2] a + [0]                
                       =  [g(a, empty())]            
                                                     
    [f(a, cons(x, k))] =  [2] a + [2] x + [2] k + [4]
                       >= [2] a + [2] x + [2] k + [4]
                       =  [f(cons(x, a), k)]         
                                                     
       [g(empty(), d)] =  [1] d + [0]                
                       >= [1] d + [0]                
                       =  [d]                        
                                                     
    [g(cons(x, k), d)] =  [2] x + [2] k + [1] d + [4]
                       >  [1] x + [2] k + [1] d + [2]
                       =  [g(k, cons(x, d))]         
                                                     

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 Trs:
  { f(a, empty()) -> g(a, empty())
  , f(a, cons(x, k)) -> f(cons(x, a), k) }
Weak Trs:
  { g(empty(), d) -> d
  , g(cons(x, k), d) -> g(k, cons(x, d)) }
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.

Trs: { f(a, empty()) -> g(a, empty()) }

The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^1)) . These rules are moved into the corresponding weak
component(s).

Sub-proof:
----------
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
       [f](x1, x2) = [1] x1 + [2] x2 + [0]
                                          
           [empty] = [2]                  
                                          
       [g](x1, x2) = [1] x1 + [1] x2 + [1]
                                          
    [cons](x1, x2) = [1] x1 + [1] x2 + [0]
  
  This order satisfies the following ordering constraints:
  
       [f(a, empty())] =  [1] a + [4]                
                       >  [1] a + [3]                
                       =  [g(a, empty())]            
                                                     
    [f(a, cons(x, k))] =  [1] a + [2] x + [2] k + [0]
                       >= [1] a + [1] x + [2] k + [0]
                       =  [f(cons(x, a), k)]         
                                                     
       [g(empty(), d)] =  [1] d + [3]                
                       >  [1] d + [0]                
                       =  [d]                        
                                                     
    [g(cons(x, k), d)] =  [1] x + [1] k + [1] d + [1]
                       >= [1] x + [1] k + [1] d + [1]
                       =  [g(k, cons(x, d))]         
                                                     

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 Trs: { f(a, cons(x, k)) -> f(cons(x, a), k) }
Weak Trs:
  { f(a, empty()) -> g(a, empty())
  , g(empty(), d) -> d
  , g(cons(x, k), d) -> g(k, cons(x, d)) }
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.

Trs: { f(a, cons(x, k)) -> f(cons(x, a), k) }

The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^1)) . These rules are moved into the corresponding weak
component(s).

Sub-proof:
----------
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
       [f](x1, x2) = [1] x1 + [2] x2 + [2]
                                          
           [empty] = [1]                  
                                          
       [g](x1, x2) = [1] x1 + [1] x2 + [0]
                                          
    [cons](x1, x2) = [1] x1 + [1] x2 + [1]
  
  This order satisfies the following ordering constraints:
  
       [f(a, empty())] =  [1] a + [4]                
                       >  [1] a + [1]                
                       =  [g(a, empty())]            
                                                     
    [f(a, cons(x, k))] =  [1] a + [2] x + [2] k + [4]
                       >  [1] a + [1] x + [2] k + [3]
                       =  [f(cons(x, a), k)]         
                                                     
       [g(empty(), d)] =  [1] d + [1]                
                       >  [1] d + [0]                
                       =  [d]                        
                                                     
    [g(cons(x, k), d)] =  [1] x + [1] k + [1] d + [1]
                       >= [1] x + [1] k + [1] d + [1]
                       =  [g(k, cons(x, d))]         
                                                     

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(a, empty()) -> g(a, empty())
  , f(a, cons(x, k)) -> f(cons(x, a), k)
  , g(empty(), d) -> d
  , g(cons(x, k), d) -> g(k, cons(x, d)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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