YES

Problem:
 rec[fold_0][3](Nil()) -> 0()
 rec[fold_0][3](Cons(x10,x6)) -> rec[plus_0][2](x10,rec[fold_0][3](x6))
 rec[plus_0][2](0(),x18) -> x18
 rec[plus_0][2](S(x10),x6) -> S(rec[plus_0][2](x10,x6))
 main(x3) -> rec[fold_0][3](x3)

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [main](x0) = x0,
   
   [S](x0) = x0,
   
   [rec[plus_0][2]](x0, x1) = x0 + x1 + 4,
   
   [Cons](x0, x1) = x0 + x1 + 4,
   
   [0] = 3,
   
   [rec[fold_0][3]](x0) = x0,
   
   [Nil] = 3
  orientation:
   rec[fold_0][3](Nil()) = 3 >= 3 = 0()
   
   rec[fold_0][3](Cons(x10,x6)) = x10 + x6 + 4 >= x10 + x6 + 4 = rec[plus_0][2](x10,rec[fold_0][3](x6))
   
   rec[plus_0][2](0(),x18) = x18 + 7 >= x18 = x18
   
   rec[plus_0][2](S(x10),x6) = x10 + x6 + 4 >= x10 + x6 + 4 = S(rec[plus_0][2](x10,x6))
   
   main(x3) = x3 >= x3 = rec[fold_0][3](x3)
  problem:
   rec[fold_0][3](Nil()) -> 0()
   rec[fold_0][3](Cons(x10,x6)) -> rec[plus_0][2](x10,rec[fold_0][3](x6))
   rec[plus_0][2](S(x10),x6) -> S(rec[plus_0][2](x10,x6))
   main(x3) -> rec[fold_0][3](x3)
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [main](x0) = 5x0 + 6,
    
    [S](x0) = x0 + 6,
    
    [rec[plus_0][2]](x0, x1) = 2x0 + 4x1,
    
    [Cons](x0, x1) = 2x0 + 4x1 + 4,
    
    [0] = 0,
    
    [rec[fold_0][3]](x0) = 5x0 + 5,
    
    [Nil] = 0
   orientation:
    rec[fold_0][3](Nil()) = 5 >= 0 = 0()
    
    rec[fold_0][3](Cons(x10,x6)) = 10x10 + 20x6 + 25 >= 2x10 + 20x6 + 20 = rec[plus_0][2](x10,rec[fold_0][3](x6))
    
    rec[plus_0][2](S(x10),x6) = 2x10 + 4x6 + 12 >= 2x10 + 4x6 + 6 = S(rec[plus_0][2](x10,x6))
    
    main(x3) = 5x3 + 6 >= 5x3 + 5 = rec[fold_0][3](x3)
   problem:
    
   Qed