YES
We show the termination of the TRS R:

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


-- SCC decomposition.

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

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

and R consists of:

r1: f(|0|()) -> s(|0|())
r2: f(s(|0|())) -> s(|0|())
r3: f(s(s(x))) -> f(f(s(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(s(x)))
p2: f#(s(s(x))) -> f#(s(x))

and R consists of:

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

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        f#_A(x1) = 0
        s_A(x1) = max{0, x1 - 3}
        f_A(x1) = max{1, x1 - 3}
        |0|_A = 2
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          f#_A(x1) = max{0, x1 - 9}
          s_A(x1) = x1 + 4
          f_A(x1) = max{5, x1 - 9}
          |0|_A = 1
          f#!_A(x1) = max{1, x1 - 6}
          s!_A(x1) = 3
          f!_A(x1) = max{4, x1 - 9}
          |0|!_A = 5
  
    partial status:
  
      pi(f#) = []
      pi(s) = []
      pi(f) = []
      pi(|0|) = []

The following rules are removed:

  p1, p2

Then no dependency pair remains.