MAYBE

Problem:
 f(x) -> s(x)
 f(s(s(x))) -> s(f(f(x)))

Proof:
 DP Processor:
  DPs:
   f#(s(s(x))) -> f#(x)
   f#(s(s(x))) -> f#(f(x))
  TRS:
   f(x) -> s(x)
   f(s(s(x))) -> s(f(f(x)))
  SCC Processor:
   #sccs: 1
   #rules: 2
   #arcs: 4/4
   DPs:
    f#(s(s(x))) -> f#(x)
    f#(s(s(x))) -> f#(f(x))
   TRS:
    f(x) -> s(x)
    f(s(s(x))) -> s(f(f(x)))
   Open