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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          g#_A(x1) = max{4, x1 - 2}
          a_A = 8
          b_A = 5
          f_A(x1,x2) = max{4, x1 - 4, x2 - 6}
          g_A(x1) = max{3, x1 + 1}
          d_A = 0
    
      precedence:
      
        a > g# = b = g = d > f
      
      partial status:
    
        pi(g#) = []
        pi(a) = []
        pi(b) = []
        pi(f) = []
        pi(g) = [1]
        pi(d) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          g#_A(x1) = 4
          a_A = 3
          b_A = 0
          f_A(x1,x2) = 1
          g_A(x1) = max{6, x1 + 3}
          d_A = 2
    
      precedence:
      
        g# = a = b = f = g = d
      
      partial status:
    
        pi(g#) = []
        pi(a) = []
        pi(b) = []
        pi(f) = []
        pi(g) = [1]
        pi(d) = []
    

The next rules are strictly ordered:

  p1

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