MAYBE

Problem:
 f(a()) -> f(a())
 a() -> b()

Proof:
 DP Processor:
  DPs:
   f#(a()) -> f#(a())
  TRS:
   f(a()) -> f(a())
   a() -> b()
  SCC Processor:
   #sccs: 1
   #rules: 1
   #arcs: 1/1
   DPs:
    f#(a()) -> f#(a())
   TRS:
    f(a()) -> f(a())
    a() -> b()
   Open