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. matrix interpretations:
    
      carrier: N^4
      order: lexicographic order
      interpretations:
        a#_A(x1,x2) = ((0,0,0,0),(0,0,0,0),(1,0,0,0),(0,0,0,0)) x1 + ((1,0,0,0),(0,1,0,0),(1,0,1,0),(1,1,1,1)) x2
        f_A() = (1,1,1,1)
        a_A(x1,x2) = ((0,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x1 + ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x2 + (1,0,1,1)
        h_A() = (2,1,1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        f > h > a > a#
      
      argument filter:
    
        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.