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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            f#_A(x1) = max{0, x1 - 23}
            s_A(x1) = max{22, x1 + 7}
            f_A(x1) = max{22, x1 + 7}
            f#!_A(x1) = max{3, x1 - 10}
            s!_A(x1) = max{1, x1 - 16}
            f!_A(x1) = max{2, x1 - 15}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              f#_A(x1) = 0
              s_A(x1) = 1
              f_A(x1) = 2
              f#!_A(x1) = x1 + 1
              s!_A(x1) = 0
              f!_A(x1) = 1
        
  
    partial status:
  
      pi(f#) = []
      pi(s) = []
      pi(f) = []

The following rules are removed:

  p1, p2

Then no dependency pair remains.