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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^4
      order: lexicographic order
      interpretations:
        f#_A(x1) = ((1,0,0,0),(0,1,0,0),(0,1,1,0),(0,0,0,0)) x1
        s_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,0,0),(1,1,1,0)) x1 + (3,1,5,1)
        f_A(x1) = ((0,0,0,0),(0,0,0,0),(1,0,0,0),(0,0,0,0)) x1 + (5,5,0,1)
        |0|_A() = (1,1,1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        |0| > f > s > f#
      
      argument filter:
    
        pi(f#) = []
        pi(s) = []
        pi(f) = []
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1, p2

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