YES

We show the termination of the TRS R:

  a() -> g(c())
  g(a()) -> b()
  f(g(X),b()) -> f(a(),X)

-- SCC decomposition.

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

p1: a#() -> g#(c())
p2: f#(g(X),b()) -> f#(a(),X)
p3: f#(g(X),b()) -> a#()

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p2}


-- Reduction pair.

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

p1: f#(g(X),b()) -> f#(a(),X)

and R consists of:

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

The set of usable rules consists of

  r1

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      f#_A(x1,x2) = x1 + x2
      g_A(x1) = x1
      b_A() = 1
      a_A() = 0
      c_A() = 0

The next rules are strictly ordered:

  p1
  r2, r3

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