MAYBE

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

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