NO

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

Proof:
 String Reversal Processor:
  a(x1) -> x1
  b(a(x1)) -> a(b(c(a(a(b(x1))))))
  b(x1) -> x1
  c(c(x1)) -> x1
  Unfolding Processor:
   loop length: 7
   terms:
    b(a(a(x14973)))
    a(b(c(a(a(b(a(x14973)))))))
    a(b(c(a(b(a(x14973))))))
    a(b(c(b(a(x14973)))))
    a(b(c(a(b(c(a(a(b(x14973)))))))))
    a(b(c(b(c(a(a(b(x14973))))))))
    a(b(c(c(a(a(b(x14973)))))))
   context: a([])
   substitution:
    x14973 -> b(x14973)
   Qed