MAYBE

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

Proof:
 Open