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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          f#_A(x1,x2) = ((0,1),(0,0)) x1 + ((0,1),(0,0)) x2 + (1,0)
          g_A(x1) = ((0,0),(0,1)) x1 + (4,1)
          b_A() = (5,3)
          a_A() = (4,2)
          c_A() = (3,1)
  
    precedence:
    
      b > a > f# = g = c
    
    partial status:
  
      pi(f#) = []
      pi(g) = []
      pi(b) = []
      pi(a) = []
      pi(c) = []

The next rules are strictly ordered:

  p1

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