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 monotone reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          f#_A(x1) = x1 + 1
          s_A(x1) = x1 + 2
          f_A(x1) = x1 + 2
  
    base order 2:
  
      precedence:
      
        f > s > f#
      
  
    partial status:
  
      pi(f#) = [1]
      pi(s) = [1]
      pi(f) = [1]

The following rules are removed:

  p2



-- SCC decomposition.

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

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


-- Reduction pair.

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

p1: f#(s(s(x))) -> f#(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, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          f#_A(x1) = x1
          s_A(x1) = x1 + 1
          f_A(x1) = x1 + 1
  
    base order 2:
  
      precedence:
      
        f# = s = f
      
  
    partial status:
  
      pi(f#) = []
      pi(s) = [1]
      pi(f) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.