YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

p1: f#(a(),x) -> f#(g(x),x)
p2: f#(a(),x) -> g#(x)
p3: h#(g(x)) -> h#(a())
p4: g#(h(x)) -> g#(x)

and R consists of:

r1: f(a(),x) -> f(g(x),x)
r2: h(g(x)) -> h(a())
r3: g(h(x)) -> g(x)
r4: h(h(x)) -> x

The estimated dependency graph contains the following SCCs:

  {p1}
  {p4}


-- Reduction pair.

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

p1: f#(a(),x) -> f#(g(x),x)

and R consists of:

r1: f(a(),x) -> f(g(x),x)
r2: h(g(x)) -> h(a())
r3: g(h(x)) -> g(x)
r4: h(h(x)) -> x

The set of usable rules consists of

  r3

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      f#_A(x1,x2) = x1
      a_A() = (1,0)
      g_A(x1) = ((0,0),(1,1)) x1
      h_A(x1) = ((1,1),(1,1)) x1

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

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

and R consists of:

r1: f(a(),x) -> f(g(x),x)
r2: h(g(x)) -> h(a())
r3: g(h(x)) -> g(x)
r4: h(h(x)) -> x

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      g#_A(x1) = ((1,0),(1,0)) x1
      h_A(x1) = ((1,1),(1,1)) x1 + (1,0)

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4

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