MAYBE

Problem:
 b(a(a(x1))) -> a(b(c(x1)))
 c(a(x1)) -> a(c(x1))
 b(c(a(x1))) -> a(b(c(x1)))
 c(b(x1)) -> d(x1)
 a(d(x1)) -> d(a(x1))
 d(x1) -> b(a(x1))
 L(a(a(x1))) -> L(a(b(c(x1))))
 c(R(x1)) -> c(b(R(x1)))

Proof:
 Open