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

  r2, r3

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          f#_A(x1,x2) = x1
          a_A() = 4
          g_A(x1) = 2
          h_A(x1) = 1
  
    base order 2:
  
      precedence:
      
        a > f# = g > h
      
  
    partial status:
  
      pi(f#) = [1]
      pi(a) = []
      pi(g) = []
      pi(h) = []

The following rules are removed:

  p1

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

  r2, r4

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          g#_A(x1) = x1
          h_A(x1) = x1 + 1
          g_A(x1) = x1 + 3
          a_A() = 1
  
    base order 2:
  
      precedence:
      
        g > h > g# = a
      
  
    partial status:
  
      pi(g#) = [1]
      pi(h) = [1]
      pi(g) = [1]
      pi(a) = []

The following rules are removed:

  p1

Then no dependency pair remains.