YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

r1: a(f(),a(f(),x)) -> a(x,x)
r2: a(h(),x) -> a(f(),a(g(),a(f(),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: a#(f(),a(f(),x)) -> a#(x,x)
p2: a#(h(),x) -> a#(f(),x)

and R consists of:

r1: a(f(),a(f(),x)) -> a(x,x)
r2: a(h(),x) -> a(f(),a(g(),a(f(),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:
    
        max/plus interpretations on natural numbers:
        
          a#_A(x1,x2) = max{x1 + 3, x2 + 3}
          f_A = 0
          a_A(x1,x2) = max{x1 + 1, x2 + 4}
          h_A = 0
    
      precedence:
      
        a = h > a# = f
      
      partial status:
    
        pi(a#) = [1, 2]
        pi(f) = []
        pi(a) = [2]
        pi(h) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a#_A(x1,x2) = max{x1 + 2, x2}
          f_A = 2
          a_A(x1,x2) = x2 + 2
          h_A = 1
    
      precedence:
      
        a > a# = f > h
      
      partial status:
    
        pi(a#) = [2]
        pi(f) = []
        pi(a) = [2]
        pi(h) = []
    

The next rules are strictly ordered:

  p1, p2

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