YES

Problem:
 rec[foldr_0][3](Nil()) -> fleft_0()
 rec[foldr_0][3](Cons(x40,x14)) -> step_x_f(x40,rec[foldr_0][3](x14))
 step_x_f[1](x4,step_x_f(x3,x2),x1) -> step_x_f[1](x3,x2,rev_0[2](x1,x4))
 step_x_f[1](x11,fleft_0(),x7) -> rev_0[2](x7,x11)
 rev_0[2](x6,x10) -> Cons(x10,x6)
 main(Nil()) -> Nil()
 main(Cons(x20,x25)) -> step_x_f[1](x20,rec[foldr_0][3](x25),Nil())

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [main](x0) = 5x0 + 4,
   
   [rev_0[2]](x0, x1) = x0 + 4x1 + 4,
   
   [step_x_f[1]](x0, x1, x2) = 4x0 + 2x1 + x2 + 2,
   
   [step_x_f](x0, x1) = 2x0 + x1 + 2,
   
   [Cons](x0, x1) = x0 + x1 + 1,
   
   [fleft_0] = 1,
   
   [rec[foldr_0][3]](x0) = 2x0 + 1,
   
   [Nil] = 0
  orientation:
   rec[foldr_0][3](Nil()) = 1 >= 1 = fleft_0()
   
   rec[foldr_0][3](Cons(x40,x14)) = 2x14 + 2x40 + 3 >= 2x14 + 2x40 + 3 = step_x_f(x40,rec[foldr_0][3](x14))
   
   step_x_f[1](x4,step_x_f(x3,x2),x1) = x1 + 2x2 + 4x3 + 4x4 + 6 >= x1 + 2x2 + 4x3 + 4x4 + 6 = step_x_f[1](x3,x2,rev_0[2](x1,x4))
   
   step_x_f[1](x11,fleft_0(),x7) = 4x11 + x7 + 4 >= 4x11 + x7 + 4 = rev_0[2](x7,x11)
   
   rev_0[2](x6,x10) = 4x10 + x6 + 4 >= x10 + x6 + 1 = Cons(x10,x6)
   
   main(Nil()) = 4 >= 0 = Nil()
   
   main(Cons(x20,x25)) = 5x20 + 5x25 + 9 >= 4x20 + 4x25 + 4 = step_x_f[1](x20,rec[foldr_0][3](x25),Nil())
  problem:
   rec[foldr_0][3](Nil()) -> fleft_0()
   rec[foldr_0][3](Cons(x40,x14)) -> step_x_f(x40,rec[foldr_0][3](x14))
   step_x_f[1](x4,step_x_f(x3,x2),x1) -> step_x_f[1](x3,x2,rev_0[2](x1,x4))
   step_x_f[1](x11,fleft_0(),x7) -> rev_0[2](x7,x11)
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [rev_0[2]](x0, x1) = x0 + x1,
    
    [step_x_f[1]](x0, x1, x2) = 4x0 + 5x1 + 4x2 + 4,
    
    [step_x_f](x0, x1) = 4x0 + 5x1,
    
    [Cons](x0, x1) = 4x0 + 5x1,
    
    [fleft_0] = 1,
    
    [rec[foldr_0][3]](x0) = x0,
    
    [Nil] = 1
   orientation:
    rec[foldr_0][3](Nil()) = 1 >= 1 = fleft_0()
    
    rec[foldr_0][3](Cons(x40,x14)) = 5x14 + 4x40 >= 5x14 + 4x40 = step_x_f(x40,rec[foldr_0][3](x14))
    
    step_x_f[1](x4,step_x_f(x3,x2),x1) = 4x1 + 25x2 + 20x3 + 4x4 + 4 >= 4x1 + 5x2 + 4x3 + 4x4 + 4 = step_x_f[1](x3,x2,rev_0[2](x1,x4))
    
    step_x_f[1](x11,fleft_0(),x7) = 4x11 + 4x7 + 9 >= x11 + x7 = rev_0[2](x7,x11)
   problem:
    rec[foldr_0][3](Nil()) -> fleft_0()
    rec[foldr_0][3](Cons(x40,x14)) -> step_x_f(x40,rec[foldr_0][3](x14))
    step_x_f[1](x4,step_x_f(x3,x2),x1) -> step_x_f[1](x3,x2,rev_0[2](x1,x4))
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [rev_0[2]](x0, x1) = x0 + x1,
     
     [step_x_f[1]](x0, x1, x2) = 4x0 + x1 + 4x2,
     
     [step_x_f](x0, x1) = 4x0 + x1 + 5,
     
     [Cons](x0, x1) = 4x0 + x1 + 5,
     
     [fleft_0] = 0,
     
     [rec[foldr_0][3]](x0) = x0,
     
     [Nil] = 0
    orientation:
     rec[foldr_0][3](Nil()) = 0 >= 0 = fleft_0()
     
     rec[foldr_0][3](Cons(x40,x14)) = x14 + 4x40 + 5 >= x14 + 4x40 + 5 = step_x_f(x40,rec[foldr_0][3](x14))
     
     step_x_f[1](x4,step_x_f(x3,x2),x1) = 4x1 + x2 + 4x3 + 4x4 + 5 >= 4x1 + x2 + 4x3 + 4x4 = step_x_f[1](x3,x2,rev_0[2](x1,x4))
    problem:
     rec[foldr_0][3](Nil()) -> fleft_0()
     rec[foldr_0][3](Cons(x40,x14)) -> step_x_f(x40,rec[foldr_0][3](x14))
    Matrix Interpretation Processor: dim=3
     
     interpretation:
                           [1 0 0]     [1 0 0]  
      [step_x_f](x0, x1) = [0 0 0]x0 + [0 0 0]x1
                           [0 0 0]     [0 1 0]  ,
      
                       [1 0 0]     [1 0 1]     [0]
      [Cons](x0, x1) = [0 1 0]x0 + [0 0 1]x1 + [0]
                       [0 0 0]     [0 0 0]     [1],
      
                  [0]
      [fleft_0] = [0]
                  [0],
      
                              [1 0 1]  
      [rec[foldr_0][3]](x0) = [0 0 1]x0
                              [0 1 1]  ,
      
              [1]
      [Nil] = [0]
              [0]
     orientation:
                               [1]    [0]            
      rec[foldr_0][3](Nil()) = [0] >= [0] = fleft_0()
                               [0]    [0]            
      
                                       [1 0 1]      [1 0 0]      [1]    [1 0 1]      [1 0 0]                                        
      rec[foldr_0][3](Cons(x40,x14)) = [0 0 0]x14 + [0 0 0]x40 + [1] >= [0 0 0]x14 + [0 0 0]x40 = step_x_f(x40,rec[foldr_0][3](x14))
                                       [0 0 1]      [0 1 0]      [1]    [0 0 1]      [0 0 0]                                        
     problem:
      
     Qed