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:
    
   CDG 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:
     
    graph:
     
    Qed