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:
  
      model-based reduction triple (=A, >=A!, >A!):
      
        finite algebra:
        
          carrier: {0,1}
          interpretations:
            f#_A(0) = 0
            f#_A(1) = 0
            s_A(0) = 0
            s_A(1) = 0
            f_A(0) = 0
            f_A(1) = 0
            f#!_A(0) = 0
            f#!_A(1) = 0
            s!_A(0) = 0
            s!_A(1) = 0
            f!_A(0) = 0
            f!_A(1) = 0
  
    partial status:
  
      pi(f#) = [1]
      pi(s) = [1]
      pi(f) = [1]

The following rules are removed:

  p1, p2

Then no dependency pair remains.