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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          f#_A(x1) = x1
          s_A(x1) = x1 + 3
          f_A(x1) = 5
          |0|_A() = 1
  
    precedence:
    
      f# = f > s = |0|
    
    partial status:
  
      pi(f#) = [1]
      pi(s) = [1]
      pi(f) = []
      pi(|0|) = []

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

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


-- Reduction pair.

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

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

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          f#_A(x1) = x1
          s_A(x1) = x1 + 3
          f_A(x1) = 5
          |0|_A() = 1
  
    precedence:
    
      f# = f > s = |0|
    
    partial status:
  
      pi(f#) = [1]
      pi(s) = [1]
      pi(f) = []
      pi(|0|) = []

The next rules are strictly ordered:

  p1

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