YES

Problem:
 a(b(x1)) -> C(x1)
 b(c(x1)) -> A(x1)
 c(a(x1)) -> B(x1)
 A(C(x1)) -> b(x1)
 C(B(x1)) -> a(x1)
 B(A(x1)) -> c(x1)
 a(a(a(a(x1)))) -> A(A(A(x1)))
 A(A(A(A(x1)))) -> a(a(a(x1)))
 b(b(b(b(x1)))) -> B(B(B(x1)))
 B(B(B(B(x1)))) -> b(b(b(x1)))
 c(c(c(c(x1)))) -> C(C(C(x1)))
 C(C(C(C(x1)))) -> c(c(c(x1)))
 B(a(a(a(x1)))) -> c(A(A(A(x1))))
 A(A(A(b(x1)))) -> a(a(a(C(x1))))
 C(b(b(b(x1)))) -> a(B(B(B(x1))))
 B(B(B(c(x1)))) -> b(b(b(A(x1))))
 A(c(c(c(x1)))) -> b(C(C(C(x1))))
 C(C(C(a(x1)))) -> c(c(c(B(x1))))
 a(A(x1)) -> x1
 A(a(x1)) -> x1
 b(B(x1)) -> x1
 B(b(x1)) -> x1
 c(C(x1)) -> x1
 C(c(x1)) -> x1

Proof:
 KBO Processor:
  weight function:
   w0 = 1
   w(B) = w(A) = w(c) = w(C) = w(a) = w(b) = 1
  precedence:
   A > C > a > B > c > b
  problem:
   
  Qed