0 votes 0 votes If $A$ and $B$ are languages, define $A \diamond B = \{xy \mid x \in A\: \text{and}\: y \in B \;\text{and} \mid x \mid = \mid y \mid \}$. Show that if $A$ and $B$ are regular languages, then $A \diamond B$ is a CFL. Theory of Computation michael-sipser theory-of-computation regular-language proof + – admin asked Oct 12, 2019 • edited Oct 12, 2019 by Lakshman Bhaiya admin 291 views answer comment Share Follow See 1 comment See all 1 1 comment reply Sanandan commented Oct 3, 2020 reply Follow Share Clearly there is a comparison between x and y , so its a cfl. 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes Suppose, A = { 0^n | n > 0 } and B = { 1^n | n > 0 } By def. of a and b having any no of 0s and 1s repectively both are regular languages BUT, A ∘ B = { 0^n.1^n | n >0 } which is a renowned cfl . same applies for → B∘A akash_chauhan answered Jul 18, 2022 akash_chauhan comment Share Follow See all 0 reply Please log in or register to add a comment.