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:
  { merge(x, nil()) -> x
  , merge(nil(), y) -> y
  , merge(++(x, y), ++(u(), v())) -> ++(x, merge(y, ++(u(), v())))
  , merge(++(x, y), ++(u(), v())) -> ++(u(), merge(++(x, y), v())) }
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:
  { merge(++(x, y), ++(u(), v())) -> ++(x, merge(y, ++(u(), v()))) }

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).
  
    [merge](x1, x2) = [2] x1 + [1] x2 + [0]
                                           
              [nil] = [0]                  
                                           
       [++](x1, x2) = [1] x1 + [1] x2 + [2]
                                           
                [u] = [0]                  
                                           
                [v] = [0]                  
  
  This order satisfies the following ordering constraints:
  
                  [merge(x, nil())] =  [2] x + [0]                    
                                    >= [1] x + [0]                    
                                    =  [x]                            
                                                                      
                  [merge(nil(), y)] =  [1] y + [0]                    
                                    >= [1] y + [0]                    
                                    =  [y]                            
                                                                      
    [merge(++(x, y), ++(u(), v()))] =  [2] x + [2] y + [6]            
                                    >  [1] x + [2] y + [4]            
                                    =  [++(x, merge(y, ++(u(), v())))]
                                                                      
    [merge(++(x, y), ++(u(), v()))] =  [2] x + [2] y + [6]            
                                    >= [2] x + [2] y + [6]            
                                    =  [++(u(), merge(++(x, y), v()))]
                                                                      

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:
  { merge(x, nil()) -> x
  , merge(nil(), y) -> y
  , merge(++(x, y), ++(u(), v())) -> ++(u(), merge(++(x, y), v())) }
Weak Trs:
  { merge(++(x, y), ++(u(), v())) -> ++(x, merge(y, ++(u(), v()))) }
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:
  { merge(++(x, y), ++(u(), v())) -> ++(u(), merge(++(x, y), v())) }

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).
  
    [merge](x1, x2) = [1] x1 + [3] x2 + [0]
                                           
              [nil] = [0]                  
                                           
       [++](x1, x2) = [1] x1 + [1] x2 + [1]
                                           
                [u] = [0]                  
                                           
                [v] = [0]                  
  
  This order satisfies the following ordering constraints:
  
                  [merge(x, nil())] =  [1] x + [0]                    
                                    >= [1] x + [0]                    
                                    =  [x]                            
                                                                      
                  [merge(nil(), y)] =  [3] y + [0]                    
                                    >= [1] y + [0]                    
                                    =  [y]                            
                                                                      
    [merge(++(x, y), ++(u(), v()))] =  [1] x + [1] y + [4]            
                                    >= [1] x + [1] y + [4]            
                                    =  [++(x, merge(y, ++(u(), v())))]
                                                                      
    [merge(++(x, y), ++(u(), v()))] =  [1] x + [1] y + [4]            
                                    >  [1] x + [1] y + [2]            
                                    =  [++(u(), merge(++(x, y), v()))]
                                                                      

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:
  { merge(x, nil()) -> x
  , merge(nil(), y) -> y }
Weak Trs:
  { merge(++(x, y), ++(u(), v())) -> ++(x, merge(y, ++(u(), v())))
  , merge(++(x, y), ++(u(), v())) -> ++(u(), merge(++(x, y), v())) }
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:
  { merge(x, nil()) -> x
  , merge(nil(), y) -> y }

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).
  
    [merge](x1, x2) = [2] x1 + [2] x2 + [1]
                                           
              [nil] = [0]                  
                                           
       [++](x1, x2) = [1] x1 + [1] x2 + [0]
                                           
                [u] = [0]                  
                                           
                [v] = [0]                  
  
  This order satisfies the following ordering constraints:
  
                  [merge(x, nil())] =  [2] x + [1]                    
                                    >  [1] x + [0]                    
                                    =  [x]                            
                                                                      
                  [merge(nil(), y)] =  [2] y + [1]                    
                                    >  [1] y + [0]                    
                                    =  [y]                            
                                                                      
    [merge(++(x, y), ++(u(), v()))] =  [2] x + [2] y + [1]            
                                    >= [1] x + [2] y + [1]            
                                    =  [++(x, merge(y, ++(u(), v())))]
                                                                      
    [merge(++(x, y), ++(u(), v()))] =  [2] x + [2] y + [1]            
                                    >= [2] x + [2] y + [1]            
                                    =  [++(u(), merge(++(x, y), v()))]
                                                                      

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:
  { merge(x, nil()) -> x
  , merge(nil(), y) -> y
  , merge(++(x, y), ++(u(), v())) -> ++(x, merge(y, ++(u(), v())))
  , merge(++(x, y), ++(u(), v())) -> ++(u(), merge(++(x, y), v())) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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