YES

Problem:
 g(b()) -> f(b())
 f(a()) -> g(a())
 b() -> a()

Proof:
 DP Processor:
  DPs:
   g#(b()) -> f#(b())
   f#(a()) -> g#(a())
  TRS:
   g(b()) -> f(b())
   f(a()) -> g(a())
   b() -> a()
  Usable Rule Processor:
   DPs:
    g#(b()) -> f#(b())
    f#(a()) -> g#(a())
   TRS:
    b() -> a()
   Arctic Interpretation Processor:
    dimension: 1
    usable rules:
     
    interpretation:
     [f#](x0) = 1,
     
     [g#](x0) = -8x0 + 0,
     
     [a] = 6,
     
     [b] = 10
    orientation:
     g#(b()) = 2 >= 1 = f#(b())
     
     f#(a()) = 1 >= 0 = g#(a())
     
     b() = 10 >= 6 = a()
    problem:
     DPs:
      
     TRS:
      b() -> a()
    Qed