YES

We show the termination of the TRS R:

  g(h(g(x))) -> g(x)
  g(g(x)) -> g(h(g(x)))
  h(h(x)) -> h(f(h(x),x))

-- SCC decomposition.

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

p1: g#(g(x)) -> g#(h(g(x)))
p2: g#(g(x)) -> h#(g(x))
p3: h#(h(x)) -> h#(f(h(x),x))

and R consists of:

r1: g(h(g(x))) -> g(x)
r2: g(g(x)) -> g(h(g(x)))
r3: h(h(x)) -> h(f(h(x),x))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: g#(g(x)) -> g#(h(g(x)))

and R consists of:

r1: g(h(g(x))) -> g(x)
r2: g(g(x)) -> g(h(g(x)))
r3: h(h(x)) -> h(f(h(x),x))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          g#_A(x1) = max{1, x1 - 2}
          g_A(x1) = x1 + 5
          h_A(x1) = max{0, x1 - 5}
          f_A(x1,x2) = max{x1, x2 - 11}
    
      precedence:
      
        g# > g > f > h
      
      partial status:
    
        pi(g#) = []
        pi(g) = []
        pi(h) = []
        pi(f) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          g#_A(x1) = 7
          g_A(x1) = 6
          h_A(x1) = 8
          f_A(x1,x2) = max{8, x1}
    
      precedence:
      
        g# > g = h > f
      
      partial status:
    
        pi(g#) = []
        pi(g) = []
        pi(h) = []
        pi(f) = []
    

The next rules are strictly ordered:

  p1

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