YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

r1: g(s(x)) -> f(x)
r2: f(|0|()) -> s(|0|())
r3: f(s(x)) -> s(s(g(x)))
r4: g(|0|()) -> |0|()

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

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

and R consists of:

r1: g(s(x)) -> f(x)
r2: f(|0|()) -> s(|0|())
r3: f(s(x)) -> s(s(g(x)))
r4: g(|0|()) -> |0|()

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          g#_A(x1) = x1 + 4
          s_A(x1) = x1 + 4
          f#_A(x1) = x1 + 3
    
      precedence:
      
        g# = s = f#
      
      partial status:
    
        pi(g#) = []
        pi(s) = []
        pi(f#) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          g#_A(x1) = max{3, x1 + 2}
          s_A(x1) = x1 + 1
          f#_A(x1) = x1 + 2
    
      precedence:
      
        g# = s = f#
      
      partial status:
    
        pi(g#) = []
        pi(s) = [1]
        pi(f#) = [1]
    

The next rules are strictly ordered:

  p1, p2

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