YES

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

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