YES

Problem:
 f(a,empty()) -> g(a,empty())
 f(a,cons(x,k)) -> f(cons(x,a),k)
 g(empty(),d) -> d
 g(cons(x,k),d) -> g(k,cons(x,d))

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [cons](x0, x1) = x0 + x1 + 2,
   
   [g](x0, x1) = x0 + x1,
   
   [f](x0, x1) = x0 + x1,
   
   [empty] = 5
  orientation:
   f(a,empty()) = a + 5 >= a + 5 = g(a,empty())
   
   f(a,cons(x,k)) = a + k + x + 2 >= a + k + x + 2 = f(cons(x,a),k)
   
   g(empty(),d) = d + 5 >= d = d
   
   g(cons(x,k),d) = d + k + x + 2 >= d + k + x + 2 = g(k,cons(x,d))
  problem:
   f(a,empty()) -> g(a,empty())
   f(a,cons(x,k)) -> f(cons(x,a),k)
   g(cons(x,k),d) -> g(k,cons(x,d))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [cons](x0, x1) = x0 + x1 + 1,
    
    [g](x0, x1) = 3x0 + 2x1 + 4,
    
    [f](x0, x1) = 3x0 + 3x1 + 3,
    
    [empty] = 1
   orientation:
    f(a,empty()) = 3a + 6 >= 3a + 6 = g(a,empty())
    
    f(a,cons(x,k)) = 3a + 3k + 3x + 6 >= 3a + 3k + 3x + 6 = f(cons(x,a),k)
    
    g(cons(x,k),d) = 2d + 3k + 3x + 7 >= 2d + 3k + 2x + 6 = g(k,cons(x,d))
   problem:
    f(a,empty()) -> g(a,empty())
    f(a,cons(x,k)) -> f(cons(x,a),k)
   Matrix Interpretation Processor: dim=3
    
    interpretation:
                      [1 0 0]     [1 0 0]     [0]
     [cons](x0, x1) = [0 0 0]x0 + [0 0 0]x1 + [0]
                      [0 0 0]     [0 0 1]     [1],
     
                   [1 0 0]     [1 0 0]  
     [g](x0, x1) = [0 0 0]x0 + [0 0 0]x1
                   [0 0 0]     [0 0 0]  ,
     
                   [1 0 0]     [1 0 1]  
     [f](x0, x1) = [0 0 0]x0 + [0 0 0]x1
                   [0 0 0]     [0 0 0]  ,
     
               [0]
     [empty] = [0]
               [1]
    orientation:
                    [1 0 0]    [1]    [1 0 0]                
     f(a,empty()) = [0 0 0]a + [0] >= [0 0 0]a = g(a,empty())
                    [0 0 0]    [0]    [0 0 0]                
     
                      [1 0 0]    [1 0 1]    [1 0 0]    [1]    [1 0 0]    [1 0 1]    [1 0 0]                  
     f(a,cons(x,k)) = [0 0 0]a + [0 0 0]k + [0 0 0]x + [0] >= [0 0 0]a + [0 0 0]k + [0 0 0]x = f(cons(x,a),k)
                      [0 0 0]    [0 0 0]    [0 0 0]    [0]    [0 0 0]    [0 0 0]    [0 0 0]                  
    problem:
     
    Qed