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^2
      order: standard order
      interpretations:
        g#_A(x1) = ((0,1),(0,0)) x1
        a_A() = (3,2)
        b_A() = (4,1)
        f_A(x1,x2) = x1
        g_A(x1) = ((1,1),(0,0)) x1
        d_A() = (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        g#_A(x1) = (0,0)
        a_A() = (1,1)
        b_A() = (1,0)
        f_A(x1,x2) = (2,1)
        g_A(x1) = x1 + (0,1)
        d_A() = (1,1)
    

The next rules are strictly ordered:

  p1

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