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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      g#_A(x1) = x1
      g_A(x1) = ((1,0,0,0),(1,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + (1,3,3,2)
      h_A(x1) = ((1,0,0,0),(1,0,0,0),(1,0,0,0),(0,0,0,0)) x1 + (0,1,1,1)
      f_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,0,1,0),(1,1,0,1)) x1

The next rules are strictly ordered:

  p1

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