MAYBE

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

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