MAYBE

Problem:
 R(2(x1)) -> 2(R(x1))
 R(3(x1)) -> 3(R(x1))
 R(1(x1)) -> L(3(x1))
 3(L(x1)) -> L(3(x1))
 2(L(x1)) -> L(2(x1))
 0(L(x1)) -> 2(R(x1))
 R(b(x1)) -> c(1(b(x1)))
 3(c(x1)) -> c(1(x1))
 2(c(1(x1))) -> c(0(R(1(x1))))
 2(c(0(x1))) -> c(0(0(x1)))

Proof:
 Open