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))
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [g#](x0, x1, x2) = 1x0 + -11x1 + -1x2 + -2,
    
    [f#](x0, x1) = x0 + -8x1 + -4,
    
    [g](x0, x1, x2) = 6x0 + 4x2 + 4,
    
    [cons](x0, x1) = 2x1 + 0,
    
    [f](x0, x1) = 5x0 + 1x1 + -16,
    
    [empty] = 0
   orientation:
    f#(cons(x,k),l) = 2k + -8l + 0 >= 1k + -11l + -1 = g#(k,l,cons(x,k))
    
    g#(a,b,c) = 1a + -11b + -1c + -2 >= a + -6c + -4 = f#(a,cons(b,c))
    
    f(empty(),l) = 1l + 5 >= l = l
    
    f(cons(x,k),l) = 7k + 1l + 5 >= 6k + 4 = g(k,l,cons(x,k))
    
    g(a,b,c) = 6a + 4c + 4 >= 5a + 3c + 1 = 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