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 reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        f#_A(x1,x2) = ((1,1),(0,0)) x1 + x2
        g_A(x1) = x1 + (1,1)
        b_A() = (4,1)
        a_A() = (3,2)
        c_A() = (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        f#_A(x1,x2) = ((1,1),(0,1)) x1 + ((0,1),(0,0)) x2
        g_A(x1) = ((0,1),(0,1)) x1 + (1,1)
        b_A() = (1,1)
        a_A() = (2,2)
        c_A() = (1,2)
    

The next rules are strictly ordered:

  p1

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