NO

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

Proof:
 String Reversal Processor:
  a(x1) -> x1
  b(a(x1)) -> a(c(b(x1)))
  c(a(c(x1))) -> a(a(b(x1)))
  Unfolding Processor:
   loop length: 12
   terms:
    b(a(a(a(a(a(a(x162197)))))))
    a(c(b(a(a(a(a(a(x162197))))))))
    a(c(a(c(b(a(a(a(a(x162197)))))))))
    a(a(a(b(b(a(a(a(a(x162197)))))))))
    a(a(a(b(a(c(b(a(a(a(x162197))))))))))
    a(a(a(b(a(c(a(c(b(a(a(x162197)))))))))))
    a(a(a(b(a(a(a(b(b(a(a(x162197)))))))))))
    a(a(a(b(a(a(a(b(a(c(b(a(x162197))))))))))))
    a(a(a(b(a(a(a(a(c(b(c(b(a(x162197)))))))))))))
    a(a(a(b(a(a(a(a(c(b(c(a(c(b(x162197))))))))))))))
    a(a(a(b(a(a(a(a(c(b(a(a(b(b(x162197))))))))))))))
    a(a(a(b(a(a(a(a(c(a(c(b(a(b(b(x162197)))))))))))))))
   context: a(a(a([])))
   substitution:
    x162197 -> b(b(a(b(b(x162197)))))
   Qed