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))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [g#](x0, x1) = 2x0 + x1 + 1/2,
    
    [f#](x0, x1) = 2x0 + 3x1 + 2,
    
    [cons](x0, x1) = 3x0 + x1 + 1/2,
    
    [g](x0, x1) = 2x0 + 2x1 + 7/2,
    
    [f](x0, x1) = 2x0 + 2x1 + 7/2,
    
    [empty] = 3/2
   orientation:
    f#(a,empty()) = 2a + 13/2 >= 2a + 2 = g#(a,empty())
    
    f#(a,cons(x,k)) = 2a + 3k + 9x + 7/2 >= 2a + 3k + 6x + 3 = f#(cons(x,a),k)
    
    g#(cons(x,k),d) = d + 2k + 6x + 3/2 >= d + 2k + 3x + 1 = g#(k,cons(x,d))
    
    f(a,empty()) = 2a + 13/2 >= 2a + 13/2 = g(a,empty())
    
    f(a,cons(x,k)) = 2a + 2k + 6x + 9/2 >= 2a + 2k + 6x + 9/2 = f(cons(x,a),k)
    
    g(empty(),d) = 2d + 13/2 >= d = d
    
    g(cons(x,k),d) = 2d + 2k + 6x + 9/2 >= 2d + 2k + 6x + 9/2 = g(k,cons(x,d))
   problem:
    DPs:
     
    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))
   Qed