YES

We show the termination of the TRS R:

  g(|0|()) -> |0|()
  g(s(x)) -> f(g(x))
  f(|0|()) -> |0|()

-- SCC decomposition.

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

p1: g#(s(x)) -> f#(g(x))
p2: g#(s(x)) -> g#(x)

and R consists of:

r1: g(|0|()) -> |0|()
r2: g(s(x)) -> f(g(x))
r3: f(|0|()) -> |0|()

The estimated dependency graph contains the following SCCs:

  {p2}


-- Reduction pair.

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

p1: g#(s(x)) -> g#(x)

and R consists of:

r1: g(|0|()) -> |0|()
r2: g(s(x)) -> f(g(x))
r3: f(|0|()) -> |0|()

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

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

The next rules are strictly ordered:

  p1
  r1, r2, r3

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