MAYBE

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

Proof:
 Open