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))
  Usable Rule Processor:
   DPs:
    f#(cons(x,k),l) -> g#(k,l,cons(x,k))
    g#(a,b,c) -> f#(a,cons(b,c))
   TRS:
    
   Arctic Interpretation Processor:
    dimension: 1
    usable rules:
     
    interpretation:
     [g#](x0, x1, x2) = 1x0 + -8x2,
     
     [f#](x0, x1) = x0,
     
     [cons](x0, x1) = x0 + 3x1 + 0
    orientation:
     f#(cons(x,k),l) = 3k + x + 0 >= 1k + -8x + -8 = g#(k,l,cons(x,k))
     
     g#(a,b,c) = 1a + -8c >= a = f#(a,cons(b,c))
    problem:
     DPs:
      
     TRS:
      
    Qed