MAYBE

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

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