YES

Problem:
 f(f(X,Y),Z) -> f(X,f(Y,Z))
 f(X,f(Y,Z)) -> f(Y,Y)

Proof:
 DP Processor:
  DPs:
   f#(f(X,Y),Z) -> f#(Y,Z)
   f#(f(X,Y),Z) -> f#(X,f(Y,Z))
   f#(X,f(Y,Z)) -> f#(Y,Y)
  TRS:
   f(f(X,Y),Z) -> f(X,f(Y,Z))
   f(X,f(Y,Z)) -> f(Y,Y)
  EDG Processor:
   DPs:
    f#(f(X,Y),Z) -> f#(Y,Z)
    f#(f(X,Y),Z) -> f#(X,f(Y,Z))
    f#(X,f(Y,Z)) -> f#(Y,Y)
   TRS:
    f(f(X,Y),Z) -> f(X,f(Y,Z))
    f(X,f(Y,Z)) -> f(Y,Y)
   graph:
    f#(f(X,Y),Z) -> f#(Y,Z) -> f#(X,f(Y,Z)) -> f#(Y,Y)
    f#(f(X,Y),Z) -> f#(X,f(Y,Z)) -> f#(X,f(Y,Z)) -> f#(Y,Y)
   SCC Processor:
    #sccs: 0
    #rules: 0
    #arcs: 2/9