YES

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

Proof:
 DP Processor:
  DPs:
   f#(a()) -> f#(b())
   g#(b()) -> g#(a())
  TRS:
   f(a()) -> f(b())
   g(b()) -> g(a())
  TDG Processor:
   DPs:
    f#(a()) -> f#(b())
    g#(b()) -> g#(a())
   TRS:
    f(a()) -> f(b())
    g(b()) -> g(a())
   graph:
    g#(b()) -> g#(a()) -> g#(b()) -> g#(a())
    f#(a()) -> f#(b()) -> f#(a()) -> f#(b())
   SCC Processor:
    #sccs: 2
    #rules: 2
    #arcs: 2/4
    DPs:
     f#(a()) -> f#(b())
    TRS:
     f(a()) -> f(b())
     g(b()) -> g(a())
    EDG Processor:
     DPs:
      f#(a()) -> f#(b())
     TRS:
      f(a()) -> f(b())
      g(b()) -> g(a())
     graph:
      
     Qed
    
    DPs:
     g#(b()) -> g#(a())
    TRS:
     f(a()) -> f(b())
     g(b()) -> g(a())
    EDG Processor:
     DPs:
      g#(b()) -> g#(a())
     TRS:
      f(a()) -> f(b())
      g(b()) -> g(a())
     graph:
      
     Qed