YES

We show the termination of the TRS R:

  f(x) -> s(x)
  f(s(s(x))) -> s(f(f(x)))

-- SCC decomposition.

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

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

and R consists of:

r1: f(x) -> s(x)
r2: f(s(s(x))) -> s(f(f(x)))

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

and R consists of:

r1: f(x) -> s(x)
r2: f(s(s(x))) -> s(f(f(x)))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = x1 + 4
          s_A(x1) = x1 + 3
          f_A(x1) = x1 + 3
    
      precedence:
      
        f# = f > s
      
      partial status:
    
        pi(f#) = [1]
        pi(s) = [1]
        pi(f) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = x1 + 13
          s_A(x1) = x1 + 6
          f_A(x1) = x1 + 3
    
      precedence:
      
        f# = s = f
      
      partial status:
    
        pi(f#) = [1]
        pi(s) = [1]
        pi(f) = []
    

The next rules are strictly ordered:

  p1, p2

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