YES

We show the termination of the TRS R:

  f(f(x)) -> g(f(x))
  g(g(x)) -> f(x)

-- SCC decomposition.

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

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

and R consists of:

r1: f(f(x)) -> g(f(x))
r2: g(g(x)) -> 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#(f(x)) -> g#(f(x))
p2: g#(g(x)) -> f#(x)

and R consists of:

r1: f(f(x)) -> g(f(x))
r2: g(g(x)) -> f(x)

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = x1 + 6
          f_A(x1) = x1 + 8
          g#_A(x1) = x1 + 6
          g_A(x1) = x1 + 7
    
      precedence:
      
        f# = f = g# = g
      
      partial status:
    
        pi(f#) = []
        pi(f) = []
        pi(g#) = []
        pi(g) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = x1 + 10
          f_A(x1) = x1 + 4
          g#_A(x1) = x1 + 9
          g_A(x1) = max{9, x1 + 3}
    
      precedence:
      
        f# = f = g# = g
      
      partial status:
    
        pi(f#) = [1]
        pi(f) = [1]
        pi(g#) = [1]
        pi(g) = []
    

The next rules are strictly ordered:

  p1, p2

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