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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            g#_A(x1) = ((1,0),(1,0)) x1
            g_A(x1) = (3,5)
            h_A(x1) = (2,4)
            f_A(x1,x2) = ((0,0),(1,0)) x1 + ((0,0),(1,0)) x2 + (1,1)
    
      precedence:
      
        g# = g > h = f
      
      partial status:
    
        pi(g#) = []
        pi(g) = []
        pi(h) = []
        pi(f) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            g#_A(x1) = (0,0)
            g_A(x1) = (0,0)
            h_A(x1) = (2,1)
            f_A(x1,x2) = (1,2)
    
      precedence:
      
        h > g = f > g#
      
      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.