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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2) = max{7, x1 + 3, x2 + 2}
          g_A(x1) = max{2, x1 + 1}
          b_A = 6
          a_A = 3
          c_A = 1
    
      precedence:
      
        f# > g = b = a = c
      
      partial status:
    
        pi(f#) = [1, 2]
        pi(g) = [1]
        pi(b) = []
        pi(a) = []
        pi(c) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2) = max{x1 - 1, x2 + 2}
          g_A(x1) = x1
          b_A = 2
          a_A = 3
          c_A = 4
    
      precedence:
      
        f# = g = b = a = c
      
      partial status:
    
        pi(f#) = []
        pi(g) = [1]
        pi(b) = []
        pi(a) = []
        pi(c) = []
    

The next rules are strictly ordered:

  p1

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