YES

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

Proof:
 DP Processor:
  DPs:
   f#(a(),x) -> g#(a(),x)
   g#(a(),x) -> f#(b(),x)
   f#(a(),x) -> f#(b(),x)
  TRS:
   f(a(),x) -> g(a(),x)
   g(a(),x) -> f(b(),x)
   f(a(),x) -> f(b(),x)
  Usable Rule Processor:
   DPs:
    f#(a(),x) -> g#(a(),x)
    g#(a(),x) -> f#(b(),x)
    f#(a(),x) -> f#(b(),x)
   TRS:
    
   LPO Processor:
    argument filtering:
     pi(a) = []
     pi(b) = []
     pi(f#) = [0]
     pi(g#) = []
    usable rules:
     
    precedence:
     a > g# > f# ~ b
    problem:
     DPs:
      
     TRS:
      
    Qed