NO

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

Proof:
 String Reversal Processor:
  b(a(a(x1))) -> a(a(c(b(b(x1)))))
  c(a(x1)) -> a(a(x1))
  Unfolding Processor:
   loop length: 11
   terms:
    b(a(a(a(a(a(a(a(a(x29288)))))))))
    a(a(c(b(b(a(a(a(a(a(a(x29288)))))))))))
    a(a(c(b(a(a(c(b(b(a(a(a(a(x29288)))))))))))))
    a(a(c(b(a(a(c(b(a(a(c(b(b(a(a(x29288)))))))))))))))
    a(a(c(b(a(a(c(a(a(c(b(b(c(b(b(a(a(x29288)))))))))))))))))
    a(a(c(b(a(a(a(a(a(c(b(b(c(b(b(a(a(x29288)))))))))))))))))
    a(a(c(b(a(a(a(a(a(c(b(b(c(b(a(a(c(b(b(x29288)))))))))))))))))))
    a(a(c(b(a(a(a(a(a(c(b(b(c(a(a(c(b(b(c(b(b(x29288)))))))))))))))))))))
    a(a(c(b(a(a(a(a(a(c(b(b(a(a(a(c(b(b(c(b(b(x29288)))))))))))))))))))))
    a(a(c(b(a(a(a(a(a(c(b(a(a(c(b(b(a(c(b(b(c(b(b(x29288)))))))))))))))))))))))
    a(a(c(b(a(a(a(a(a(c(a(a(c(b(b(c(b(b(a(c(b(b(c(b(b(x29288)))))))))))))))))))))))))
   context: a(a(c([])))
   substitution:
    x29288 -> c(b(b(c(b(b(a(c(b(b(c(b(b(x29288)))))))))))))
   Qed