MAYBE

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

Proof:
 DP Processor:
  DPs:
   g#(x) -> g#(x)
   g#(x) -> f#(g(x))
  TRS:
   f(g(a())) -> a()
   f(f(x)) -> b()
   g(x) -> f(g(x))
  TDG Processor:
   DPs:
    g#(x) -> g#(x)
    g#(x) -> f#(g(x))
   TRS:
    f(g(a())) -> a()
    f(f(x)) -> b()
    g(x) -> f(g(x))
   graph:
    g#(x) -> g#(x) -> g#(x) -> f#(g(x))
    g#(x) -> g#(x) -> g#(x) -> g#(x)
   SCC Processor:
    #sccs: 1
    #rules: 1
    #arcs: 2/4
    DPs:
     g#(x) -> g#(x)
    TRS:
     f(g(a())) -> a()
     f(f(x)) -> b()
     g(x) -> f(g(x))
    Open