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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          g#_A(x1) = ((0,0),(1,0)) x1 + (4,1)
          a_A() = (5,5)
          b_A() = (3,2)
          f_A(x1,x2) = (2,1)
          g_A(x1) = (0,0)
          d_A() = (1,6)
  
    precedence:
    
      g# = a = b > f = g = d
    
    partial status:
  
      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.