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) = 2x0 + x1 + 1,
   
   [r1](x0, x1) = 3x0 + 2x1,
   
   [empty] = 0,
   
   [rev](x0) = 4x0
  orientation:
   rev(ls) = 4ls >= 3ls = r1(ls,empty())
   
   r1(empty(),a) = 2a >= a = a
   
   r1(cons(x,k),a) = 2a + 3k + 6x + 3 >= 2a + 3k + 4x + 2 = r1(k,cons(x,a))
  problem:
   rev(ls) -> r1(ls,empty())
   r1(empty(),a) -> a
  Matrix Interpretation Processor: dim=3
   
   interpretation:
                   [1 0 1]       
    [r1](x0, x1) = [0 0 0]x0 + x1
                   [0 0 0]       ,
    
              [0]
    [empty] = [0]
              [1],
    
                [1 1 1]     [0]
    [rev](x0) = [0 0 0]x0 + [0]
                [0 0 1]     [1]
   orientation:
              [1 1 1]     [0]    [1 0 1]     [0]                 
    rev(ls) = [0 0 0]ls + [0] >= [0 0 0]ls + [0] = r1(ls,empty())
              [0 0 1]     [1]    [0 0 0]     [1]                 
    
                        [1]         
    r1(empty(),a) = a + [0] >= a = a
                        [0]         
   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