MAYBE

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

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