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. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            f#_A(x1,x2) = x1 + (0,1)
            a_A() = (1,3)
            h_A(x1) = ((1,0),(0,0)) x1 + (2,0)
            g_A(x1) = (1,1)
    
      precedence:
      
        f# = a = h = g
      
      partial status:
    
        pi(f#) = []
        pi(a) = []
        pi(h) = []
        pi(g) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            f#_A(x1,x2) = ((1,0),(1,0)) x1
            a_A() = (3,3)
            h_A(x1) = (2,2)
            g_A(x1) = (1,1)
    
      precedence:
      
        a > f# = h = g
      
      partial status:
    
        pi(f#) = []
        pi(a) = []
        pi(h) = []
        pi(g) = []
    

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. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            g#_A(x1) = ((1,0),(1,1)) x1 + (2,2)
            h_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        g# = h
      
      partial status:
    
        pi(g#) = []
        pi(h) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            g#_A(x1) = ((1,0),(0,0)) x1
            h_A(x1) = (1,1)
    
      precedence:
      
        h > g#
      
      partial status:
    
        pi(g#) = []
        pi(h) = []
    

The next rules are strictly ordered:

  p1

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