MAYBE

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

Proof:
 Open