YES

We show the termination of the TRS R:

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

and R consists of:

r1: f(a(),h(x)) -> f(g(x),h(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(),h(x)) -> f#(g(x),h(x))

and R consists of:

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

  r2, r3, r4

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        f#_A(x1,x2) = x1 + x2
        a_A() = 0
        h_A(x1) = x1 + 1
        g_A(x1) = 0
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        f#_A(x1,x2) = x1 + x2
        a_A() = 1
        h_A(x1) = 0
        g_A(x1) = 0
    

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(),h(x)) -> f(g(x),h(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 reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        g#_A(x1) = x1
        h_A(x1) = x1 + 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        g#_A(x1) = 0
        h_A(x1) = x1 + 1
    

The next rules are strictly ordered:

  p1

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