YES

Problem:
 f(0(),1(),x) -> f(x,x,x)
 f(x,y,z) -> 2()
 0() -> 2()
 1() -> 2()

Proof:
 DP Processor:
  DPs:
   f#(0(),1(),x) -> f#(x,x,x)
  TRS:
   f(0(),1(),x) -> f(x,x,x)
   f(x,y,z) -> 2()
   0() -> 2()
   1() -> 2()
  EDG Processor:
   DPs:
    f#(0(),1(),x) -> f#(x,x,x)
   TRS:
    f(0(),1(),x) -> f(x,x,x)
    f(x,y,z) -> 2()
    0() -> 2()
    1() -> 2()
   graph:
    
   SCC Processor:
    #sccs: 0
    #rules: 0
    #arcs: 0/1