L1 intersection L2’ is equivalent to L1 – L2. So, if we see from the context of the Venn diagram, where we can see regular languages are a subset of CFL. L1 is RL and L2 is CFL, as every CFL is RL too, so doing RL – CFL = phi (null string) and phi is a regular language, so as it’s regular, hence it’s CFL too. So, why option (A) isn’t CFL?
In option (C), how L2 union L2’ gives sigma* ?
as from what I know is L2’ means apart from CFL (as CFL is removed, so RL is also removed). So sigma* won’t be able to consider sigma* but the twist is that those RL which are ignored earlier, that would be covered by CSL or TM. So, L2 union L2’ gives sigma*. As sigma* is RL, and thus it’s CFL too. According to me, all are CFL.
Please help me prove why option A isn’t CFL?