NO

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

Proof:
 String Reversal Processor:
  a(x1) -> x1
  b(a(x1)) -> x1
  b(a(x1)) -> a(a(c(b(x1))))
  c(c(x1)) -> b(x1)
  Unfolding Processor:
   loop length: 10
   terms:
    b(a(a(a(a(c(a(a(x73580))))))))
    a(a(c(b(a(a(a(c(a(a(x73580))))))))))
    a(a(c(a(a(c(b(a(a(c(a(a(x73580))))))))))))
    a(a(c(a(c(b(a(a(c(a(a(x73580)))))))))))
    a(a(c(c(b(a(a(c(a(a(x73580))))))))))
    a(a(b(b(a(a(c(a(a(x73580)))))))))
    a(a(b(a(a(c(b(a(c(a(a(x73580)))))))))))
    a(a(b(a(a(c(c(a(a(x73580)))))))))
    a(a(b(a(a(b(a(a(x73580))))))))
    a(a(b(a(a(a(a(c(b(a(x73580))))))))))
   context: a(a([]))
   substitution:
    x73580 -> c(b(x73580))
   Qed