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))
   Subterm Criterion Processor:
    simple projection:
     pi(f#) = 0
     pi(g#) = 0
    problem:
     DPs:
      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))
    SCC Processor:
     #sccs: 0
     #rules: 0
     #arcs: 2/1