MAYBE

Problem:
 f(x,a(b(y))) -> f(a(b(x)),y)
 f(a(x),y) -> f(x,a(y))
 f(b(x),y) -> f(x,b(y))

Proof:
 DP Processor:
  DPs:
   f#(x,a(b(y))) -> f#(a(b(x)),y)
   f#(a(x),y) -> f#(x,a(y))
   f#(b(x),y) -> f#(x,b(y))
  TRS:
   f(x,a(b(y))) -> f(a(b(x)),y)
   f(a(x),y) -> f(x,a(y))
   f(b(x),y) -> f(x,b(y))
  Usable Rule Processor:
   DPs:
    f#(x,a(b(y))) -> f#(a(b(x)),y)
    f#(a(x),y) -> f#(x,a(y))
    f#(b(x),y) -> f#(x,b(y))
   TRS:
    
   Open