NO

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

Proof:
 Unfolding Processor:
  loop length: 10
  terms:
   a(b(c(b(c(c(x140853))))))
   b(b(c(b(c(c(x140853))))))
   c(c(a(a(a(b(c(c(x140853))))))))
   c(c(a(b(a(b(c(c(x140853))))))))
   c(c(a(b(b(b(c(c(x140853))))))))
   c(c(a(b(c(c(a(a(a(c(x140853))))))))))
   c(c(a(b(c(a(a(a(c(x140853)))))))))
   c(c(a(b(c(b(a(a(c(x140853)))))))))
   c(c(a(b(c(b(b(a(c(x140853)))))))))
   c(c(a(b(c(b(b(b(c(x140853)))))))))
  context: c(c([]))
  substitution:
   x140853 -> a(a(a(x140853)))
  Qed