NO

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

Proof:
 String Reversal Processor:
  a(x1) -> x1
  b(a(x1)) -> a(b(b(c(a(a(x1))))))
  b(x1) -> x1
  c(c(x1)) -> x1
  Unfolding Processor:
   loop length: 7
   terms:
    b(a(c(a(x1210))))
    a(b(b(c(a(a(c(a(x1210))))))))
    a(b(b(c(a(c(a(x1210)))))))
    a(b(b(c(c(a(x1210))))))
    a(b(b(a(x1210))))
    a(b(a(b(b(c(a(a(x1210))))))))
    a(b(a(b(c(a(a(x1210)))))))
   context: a([])
   substitution:
    x1210 -> a(x1210)
   Qed