MAYBE

Problem:
 f(tt(),x) -> f(isDouble(x),s(s(x)))
 isDouble(s(s(x))) -> isDouble(x)
 isDouble(0()) -> tt()

Proof:
 DP Processor:
  DPs:
   f#(tt(),x) -> isDouble#(x)
   f#(tt(),x) -> f#(isDouble(x),s(s(x)))
   isDouble#(s(s(x))) -> isDouble#(x)
  TRS:
   f(tt(),x) -> f(isDouble(x),s(s(x)))
   isDouble(s(s(x))) -> isDouble(x)
   isDouble(0()) -> tt()
  Open