YES

Problem:
 rec[take_l_0][2](0()) -> Nil()
 rec[take_l_0][2](S(x6)) -> Cons(rec[take_l_0][2](x6))
 main(x3) -> rec[take_l_0][2](x3)

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
                [1 0 0]  
   [main](x0) = [0 0 0]x0
                [0 0 0]  ,
   
                [1 0 0]  
   [Cons](x0) = [0 0 0]x0
                [0 0 0]  ,
   
             [1 0 0]     [1]
   [S](x0) = [0 0 0]x0 + [0]
             [0 0 0]     [0],
   
           [0]
   [Nil] = [0]
           [0],
   
                            [1 0 0]  
   [rec[take_l_0][2]](x0) = [0 0 0]x0
                            [0 0 0]  ,
   
         [1]
   [0] = [0]
         [0]
  orientation:
                           [1]    [0]        
   rec[take_l_0][2](0()) = [0] >= [0] = Nil()
                           [0]    [0]        
   
                             [1 0 0]     [1]    [1 0 0]                               
   rec[take_l_0][2](S(x6)) = [0 0 0]x6 + [0] >= [0 0 0]x6 = Cons(rec[take_l_0][2](x6))
                             [0 0 0]     [0]    [0 0 0]                               
   
              [1 0 0]      [1 0 0]                         
   main(x3) = [0 0 0]x3 >= [0 0 0]x3 = rec[take_l_0][2](x3)
              [0 0 0]      [0 0 0]                         
  problem:
   main(x3) -> rec[take_l_0][2](x3)
  Matrix Interpretation Processor: dim=3
   
   interpretation:
                 [1 0 0]     [1]
    [main](x0) = [0 0 0]x0 + [0]
                 [0 0 1]     [0],
    
                             [1 0 0]  
    [rec[take_l_0][2]](x0) = [0 0 0]x0
                             [0 0 0]  
   orientation:
               [1 0 0]     [1]    [1 0 0]                         
    main(x3) = [0 0 0]x3 + [0] >= [0 0 0]x3 = rec[take_l_0][2](x3)
               [0 0 1]     [0]    [0 0 0]                         
   problem:
    
   Qed