Regular languages: words from A that do not contain any word from B as a substring

Question

How to prove, by construction and/or regular closures, that if $A$, $B$ are regular then the language $L$ defined by

$$L=\{w\mid w\in A,\ \not\exists y\in B\ :\ w=xyz\text{ for some }x,z∈Σ^∗\}$$

is also regular?

Answer 1

Your language is $A$ intersected with $$\{w\mid \not\exists y\in B\ :\ w=xyz\text{ for some }x,z∈Σ^∗\},$$ i.e., the complement of $$\{w\mid\exists y\in B\ :\ w=xyz\text{ for some }x,z∈Σ^∗\}$$ which is $$\overline{\Sigma^* B \Sigma^*},$$ so $$L=A\cap \overline{\Sigma^* B \Sigma^*}$$ which is regular by the usual closure properties of regular languages.