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. matrix interpretations:
    
      carrier: N^4
      order: lexicographic order
      interpretations:
        g#_A(x1) = ((1,0,0,0),(0,1,0,0),(0,1,1,0),(0,1,1,1)) x1
        a_A() = (3,5,1,1)
        b_A() = (2,4,1,1)
        f_A(x1,x2) = ((0,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x1 + (1,0,0,0)
        g_A(x1) = (0,2,1,1)
        d_A() = (1,1,1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        f > a > b > d > g > g#
      
      argument filter:
    
        pi(g#) = 1
        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.