NO

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

Proof:
 String Reversal Processor:
  a(x1) -> x1
  b(a(x1)) -> a(a(b(c(b(x1)))))
  b(x1) -> a(x1)
  c(c(x1)) -> x1
  Unfolding Processor:
   loop length: 9
   terms:
    b(a(a(a(x92236))))
    a(a(b(c(b(a(a(x92236)))))))
    a(a(b(c(a(a(b(c(b(a(x92236))))))))))
    a(a(b(c(a(b(c(b(a(x92236)))))))))
    a(a(b(c(b(c(b(a(x92236))))))))
    a(a(b(c(a(c(b(a(x92236))))))))
    a(a(b(c(c(b(a(x92236)))))))
    a(a(b(b(a(x92236)))))
    a(a(b(a(a(b(c(b(x92236))))))))
   context: a(a([]))
   substitution:
    x92236 -> c(b(x92236))
   Qed