YES

Problem:
 rev(ls) -> r1(ls,empty())
 r1(empty(),a) -> a
 r1(cons(x,k),a) -> r1(k,cons(x,a))

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [cons](x0, x1) = x0 + x1 + 4,
   
   [r1](x0, x1) = x0 + x1 + 3,
   
   [empty] = 4,
   
   [rev](x0) = 4x0 + 7
  orientation:
   rev(ls) = 4ls + 7 >= ls + 7 = r1(ls,empty())
   
   r1(empty(),a) = a + 7 >= a = a
   
   r1(cons(x,k),a) = a + k + x + 7 >= a + k + x + 7 = r1(k,cons(x,a))
  problem:
   rev(ls) -> r1(ls,empty())
   r1(cons(x,k),a) -> r1(k,cons(x,a))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [cons](x0, x1) = x0 + x1 + 1,
    
    [r1](x0, x1) = 5x0 + 4x1,
    
    [empty] = 0,
    
    [rev](x0) = 5x0
   orientation:
    rev(ls) = 5ls >= 5ls = r1(ls,empty())
    
    r1(cons(x,k),a) = 4a + 5k + 5x + 5 >= 4a + 5k + 4x + 4 = r1(k,cons(x,a))
   problem:
    rev(ls) -> r1(ls,empty())
   Matrix Interpretation Processor: dim=3
    
    interpretation:
                    [1 0 0]     [1 0 0]  
     [r1](x0, x1) = [0 0 0]x0 + [0 0 0]x1
                    [0 0 0]     [0 0 0]  ,
     
               [0]
     [empty] = [0]
               [0],
     
                 [1 0 0]     [1]
     [rev](x0) = [0 0 0]x0 + [0]
                 [0 0 0]     [0]
    orientation:
               [1 0 0]     [1]    [1 0 0]                   
     rev(ls) = [0 0 0]ls + [0] >= [0 0 0]ls = r1(ls,empty())
               [0 0 0]     [0]    [0 0 0]                   
    problem:
     
    Qed