MAYBE

Problem:
 b(b(x1)) -> c(c(c(c(x1))))
 c(x1) -> x1
 b(c(b(x1))) -> b(b(b(x1)))

Proof:
 DP Processor:
  DPs:
   b#(b(x1)) -> c#(x1)
   b#(b(x1)) -> c#(c(x1))
   b#(b(x1)) -> c#(c(c(x1)))
   b#(b(x1)) -> c#(c(c(c(x1))))
   b#(c(b(x1))) -> b#(b(x1))
   b#(c(b(x1))) -> b#(b(b(x1)))
  TRS:
   b(b(x1)) -> c(c(c(c(x1))))
   c(x1) -> x1
   b(c(b(x1))) -> b(b(b(x1)))
  Open