Consider a $\Sigma = \{\binom{0}{0},\binom{1}{0},\binom{1}{1},\binom{0}{1}\}$ and a language $L$ over $\Sigma$ such that strings in the language have the "bottom row" of the string as the reverse of the "top row". e.g. $\binom{0}{1}\binom{0}{0}\binom{0}{0}\binom{1}{0} \in L$. How can I prove this language is regular, or rather prove it is not regular?
My available skillsets are automata and pumping lemma, I'm just not sure how to parse the question in an easier to approach manner. I thought maybe it would be equivalent to showing the set of palindromes are not regular. I could encode the "two rows" as a single "row" to make a palindrome:
$$ \binom{0}{1}\binom{0}{0}\binom{0}{0}\binom{1}{0} \equiv 00011000 \\ \binom{1}{1}\binom{1}{0}\binom{0}{1}\binom{1}{1} \equiv 11011011 \\ \binom{0}{0}\binom{1}{0}\binom{0}{1}\binom{0}{0} \equiv 01000010 \\ \dots $$
Note I've just appended the bottom "row" to the top "row" and found every string accepted by $L$ encodes to a palindrome using this method. The encoded language is just the language over $\{0,1\}$ where each string is a palindrome and of even length; I know this is not a regular language but would proving this be enough to prove the original language is not regular?
Any ideas?
Hint. Let $$ K = L \cap \binom{0}{0}^*\binom{1}{1}\binom{0}{0}^* = \left\{ \binom{0}{0}^n\binom{1}{1}\binom{0}{0}^n \mid n \geqslant 0 \right \} $$ If $L$ is regular, then $K$, which is the intersection of two regular languages, is regular. Up to a change of alphabet, it is the same langage than $S =\{0^n10^n \mid n \geqslant 0 \}$. To get a contradiction, it remains to show that $S$ is not regular...