YES

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

Proof:
 DP Processor:
  DPs:
   f#(a()) -> f#(b())
   g#(b()) -> g#(a())
   f#(x) -> g#(x)
  TRS:
   f(a()) -> f(b())
   g(b()) -> g(a())
   f(x) -> g(x)
  EDG Processor:
   DPs:
    f#(a()) -> f#(b())
    g#(b()) -> g#(a())
    f#(x) -> g#(x)
   TRS:
    f(a()) -> f(b())
    g(b()) -> g(a())
    f(x) -> g(x)
   graph:
    f#(a()) -> f#(b()) -> f#(x) -> g#(x)
    f#(x) -> g#(x) -> g#(b()) -> g#(a())
   SCC Processor:
    #sccs: 0
    #rules: 0
    #arcs: 2/9