NO

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

Proof:
 String Reversal Processor:
  a(b(a(b(c(x1))))) -> c(b(c(b(a(b(a(b(a(x1)))))))))
  Unfolding Processor:
   loop length: 4
   terms:
    a(b(a(b(c(b(c(b(a(b(c(b(c(x180)))))))))))))
    c(b(c(b(a(b(a(b(a(b(c(b(a(b(c(b(c(x180)))))))))))))))))
    c(b(c(b(a(b(c(b(c(b(a(b(a(b(a(b(a(b(c(b(c(x180)))))))))))))))))))))
    c(b(c(b(a(b(c(b(c(b(a(b(a(b(c(b(c(b(a(b(a(b(a(b(c(x180)))))))))))))))))))))))))
   context: c(b(c(b(a(b(c(b(c(b([]))))))))))
   substitution:
    x180 -> b(a(b(a(b(a(x180))))))
   Qed