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:
 DP Processor:
  DPs:
   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))
  TRS:
   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))
  Usable Rule Processor:
   DPs:
    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))
   TRS:
    
   Matrix Interpretation Processor: dim=4
    
    usable rules:
     
    interpretation:
     [g#](x0, x1) = [0 0 1 1]x0 + [0 0 1 0]x1,
     
     [f#](x0, x1) = [0 0 1 1]x0 + [1 1 1 1]x1,
     
                      [0 0 1 0]     [1 1 0 1]     [1]
                      [1 0 1 0]     [0 1 1 1]     [0]
     [cons](x0, x1) = [1 0 0 1]x0 + [0 0 1 0]x1 + [0]
                      [1 1 0 1]     [0 0 0 1]     [1],
     
               [0]
               [0]
     [empty] = [0]
               [1]
    orientation:
     f#(a,empty()) = [0 0 1 1]a + [1] >= [0 0 1 1]a = g#(a,empty())
     
     f#(a,cons(x,k)) = [0 0 1 1]a + [1 2 2 3]k + [3 1 2 2]x + [2] >= [0 0 1 1]a + [1 1 1 1]k + [2 1 0 2]x + [1] = f#(cons(x,a),k)
     
     g#(cons(x,k),d) = [0 0 1 0]d + [0 0 1 1]k + [2 1 0 2]x + [1] >= [0 0 1 0]d + [0 0 1 1]k + [1 0 0 1]x = g#(k,cons(x,d))
    problem:
     DPs:
      
     TRS:
      
    Qed