YES

We show the termination of the TRS R:

  g(a()) -> g(b())
  b() -> f(a(),a())
  f(a(),a()) -> g(d())

-- SCC decomposition.

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

p1: g#(a()) -> g#(b())
p2: g#(a()) -> b#()
p3: b#() -> f#(a(),a())
p4: f#(a(),a()) -> g#(d())

and R consists of:

r1: g(a()) -> g(b())
r2: b() -> f(a(),a())
r3: f(a(),a()) -> g(d())

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: g#(a()) -> g#(b())

and R consists of:

r1: g(a()) -> g(b())
r2: b() -> f(a(),a())
r3: f(a(),a()) -> g(d())

The set of usable rules consists of

  r2, r3

Take the reduction pair:

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

The next rules are strictly ordered:

  p1

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