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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        f#_A(x1) = max{5, x1 + 3}
        s_A(x1) = max{4, x1 + 3}
        f_A(x1) = max{4, x1 + 3}
  
    precedence:
    
      f# = s = f
    
    partial status:
  
      pi(f#) = [1]
      pi(s) = [1]
      pi(f) = [1]

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(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 monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        f#_A(x1) = max{4, x1 + 1}
        s_A(x1) = x1 + 4
        f_A(x1) = x1 + 4
  
    precedence:
    
      f# = s = f
    
    partial status:
  
      pi(f#) = [1]
      pi(s) = [1]
      pi(f) = [1]

The next rules are strictly ordered:

  p1

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