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)
  Usable Rule Processor:
   DPs:
    a#() -> g#(c())
    f#(g(X),b()) -> a#()
    f#(g(X),b()) -> f#(a(),X)
   TRS:
    a() -> g(c())
   Arctic Interpretation Processor:
    dimension: 1
    usable rules:
     a() -> g(c())
    interpretation:
     [f#](x0, x1) = -6x0 + x1 + 4,
     
     [g#](x0) = 2x0 + 1,
     
     [a#] = 4,
     
     [b] = 5,
     
     [g](x0) = 7x0 + 2,
     
     [c] = 0,
     
     [a] = 8
    orientation:
     a#() = 4 >= 2 = g#(c())
     
     f#(g(X),b()) = 1X + 5 >= 4 = a#()
     
     f#(g(X),b()) = 1X + 5 >= X + 4 = f#(a(),X)
     
     a() = 8 >= 7 = g(c())
    problem:
     DPs:
      
     TRS:
      a() -> g(c())
    Qed