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()
  TDG Processor:
   DPs:
    g#(b()) -> f#(b())
    f#(a()) -> g#(a())
   TRS:
    g(b()) -> f(b())
    f(a()) -> g(a())
    b() -> a()
   graph:
    f#(a()) -> g#(a()) -> g#(b()) -> f#(b())
    g#(b()) -> f#(b()) -> f#(a()) -> g#(a())
   EDG Processor:
    DPs:
     g#(b()) -> f#(b())
     f#(a()) -> g#(a())
    TRS:
     g(b()) -> f(b())
     f(a()) -> g(a())
     b() -> a()
    graph:
     g#(b()) -> f#(b()) -> f#(a()) -> g#(a())
    SCC Processor:
     #sccs: 0
     #rules: 0
     #arcs: 1/4