YES

Problem:
 f(empty(),l) -> l
 f(cons(x,k),l) -> g(k,l,cons(x,k))
 g(a,b,c) -> f(a,cons(b,c))

Proof:
 DP Processor:
  DPs:
   f#(cons(x,k),l) -> g#(k,l,cons(x,k))
   g#(a,b,c) -> f#(a,cons(b,c))
  TRS:
   f(empty(),l) -> l
   f(cons(x,k),l) -> g(k,l,cons(x,k))
   g(a,b,c) -> f(a,cons(b,c))
  TDG Processor:
   DPs:
    f#(cons(x,k),l) -> g#(k,l,cons(x,k))
    g#(a,b,c) -> f#(a,cons(b,c))
   TRS:
    f(empty(),l) -> l
    f(cons(x,k),l) -> g(k,l,cons(x,k))
    g(a,b,c) -> f(a,cons(b,c))
   graph:
    g#(a,b,c) -> f#(a,cons(b,c)) ->
    f#(cons(x,k),l) -> g#(k,l,cons(x,k))
    f#(cons(x,k),l) -> g#(k,l,cons(x,k)) -> g#(a,b,c) -> f#(a,cons(b,c))
   Arctic Interpretation Processor:
    dimension: 1
    interpretation:
     [g#](x0, x1, x2) = 2x0 + x2 + 0,
     
     [f#](x0, x1) = 1x0,
     
     [g](x0, x1, x2) = 6x0 + 4x2 + 3,
     
     [cons](x0, x1) = 2x1 + 0,
     
     [f](x0, x1) = 5x0 + 1x1 + 2,
     
     [empty] = 1
    orientation:
     f#(cons(x,k),l) = 3k + 1 >= 2k + 0 = g#(k,l,cons(x,k))
     
     g#(a,b,c) = 2a + c + 0 >= 1a = f#(a,cons(b,c))
     
     f(empty(),l) = 1l + 6 >= l = l
     
     f(cons(x,k),l) = 7k + 1l + 5 >= 6k + 4 = g(k,l,cons(x,k))
     
     g(a,b,c) = 6a + 4c + 3 >= 5a + 3c + 2 = f(a,cons(b,c))
    problem:
     DPs:
      
     TRS:
      f(empty(),l) -> l
      f(cons(x,k),l) -> g(k,l,cons(x,k))
      g(a,b,c) -> f(a,cons(b,c))
    Qed