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:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        a > g > h > f#
      
      argument filter:
    
        pi(f#) = 1
        pi(a) = []
        pi(g) = []
        pi(h) = [1]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        g > h > a > f#
      
      argument filter:
    
        pi(f#) = [1]
        pi(a) = []
        pi(g) = []
        pi(h) = 1
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        g > h > a > f#
      
      argument filter:
    
        pi(f#) = []
        pi(a) = []
        pi(g) = []
        pi(h) = [1]
    

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:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        h > g#
      
      argument filter:
    
        pi(g#) = [1]
        pi(h) = [1]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        h > g#
      
      argument filter:
    
        pi(g#) = [1]
        pi(h) = [1]
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        h > g#
      
      argument filter:
    
        pi(g#) = [1]
        pi(h) = [1]
    

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4

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