MAYBE

Problem:
 f(0(),1(),x) -> f(x,x,x)

Proof:
 DP Processor:
  DPs:
   f#(0(),1(),x) -> f#(x,x,x)
  TRS:
   f(0(),1(),x) -> f(x,x,x)
  SCC Processor:
   #sccs: 1
   #rules: 1
   #arcs: 1/1
   DPs:
    f#(0(),1(),x) -> f#(x,x,x)
   TRS:
    f(0(),1(),x) -> f(x,x,x)
   Open