YES

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

Proof:
 DP Processor:
  DPs:
   a#() -> g#(c())
   f#(g(X),b()) -> a#()
   f#(g(X),b()) -> f#(a(),X)
  TRS:
   a() -> g(c())
   g(a()) -> b()
   f(g(X),b()) -> f(a(),X)
  TDG Processor:
   DPs:
    a#() -> g#(c())
    f#(g(X),b()) -> a#()
    f#(g(X),b()) -> f#(a(),X)
   TRS:
    a() -> g(c())
    g(a()) -> b()
    f(g(X),b()) -> f(a(),X)
   graph:
    f#(g(X),b()) -> f#(a(),X) -> f#(g(X),b()) -> f#(a(),X)
    f#(g(X),b()) -> f#(a(),X) -> f#(g(X),b()) -> a#()
    f#(g(X),b()) -> a#() -> a#() -> g#(c())
   SCC Processor:
    #sccs: 1
    #rules: 1
    #arcs: 3/9
    DPs:
     f#(g(X),b()) -> f#(a(),X)
    TRS:
     a() -> g(c())
     g(a()) -> b()
     f(g(X),b()) -> f(a(),X)
    CDG Processor:
     DPs:
      f#(g(X),b()) -> f#(a(),X)
     TRS:
      a() -> g(c())
      g(a()) -> b()
      f(g(X),b()) -> f(a(),X)
     graph:
      
     Qed