YES

We show the termination of the TRS R:

  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 decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(cons(x,k),l) -> g#(k,l,cons(x,k))
p2: g#(a,b,c) -> f#(a,cons(b,c))

and R consists of:

r1: f(empty(),l) -> l
r2: f(cons(x,k),l) -> g(k,l,cons(x,k))
r3: g(a,b,c) -> f(a,cons(b,c))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(cons(x,k),l) -> g#(k,l,cons(x,k))
p2: g#(a,b,c) -> f#(a,cons(b,c))

and R consists of:

r1: f(empty(),l) -> l
r2: f(cons(x,k),l) -> g(k,l,cons(x,k))
r3: g(a,b,c) -> f(a,cons(b,c))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        cons > f# > g#
      
      argument filter:
    
        pi(f#) = 1
        pi(cons) = 2
        pi(g#) = 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        g# > cons > f#
      
      argument filter:
    
        pi(f#) = 1
        pi(cons) = 2
        pi(g#) = 1
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        f# > cons > g#
      
      argument filter:
    
        pi(f#) = 1
        pi(cons) = [2]
        pi(g#) = [1]
    

The next rules are strictly ordered:

  p1, p2

We remove them from the problem.  Then no dependency pair remains.