mcilroy-regex

o my regular expression crossword puzzle solver. For that, I'm converting the hint regexps to DFAs, that match strings of some fixed length, and concatenating and intersecting them, until a single regexp is yielded, which should be a string literal, if the puzzle has a single solution. For backreferences, it's more tricky, but I plan on rewriting backreferences to the captured expression, where the lengths of both match, then either executing it with a stack like a pushdown automata or constructing a set of constraints on the characters by index. As an aside: In my proof of the pumping lemma[3], I got stuck on the case for intersection and I'd love insight. Regular languages are closed under intersection, so the pumping lemma should hold for my implementation. I need to prove that if s =~ re1 and s =~ re2 can be pumped, then so can s =~ And re1 re2. s is is split into different substrings for re1 and re2, s = s11 ++ s12 ++ s13 = s21 ++ s22 ++ s23, then repeated an arbitrary number of times, (forall n, s11 ++ repeat s12 n ++ s13 =~ re1) and (forall n, s21 ++ repeat s22 n ++ s23 =~ re2). My intuition is that s11 = s21, s12 = s22, and s13 = s23, because they both match for the intersection, but I'm not convinced of that and haven't been able to formulate a proof for that. 0: https://github.com/thaliaarchi/recross-coq 1: https://github.com/thaliaarchi/re1-rust 2: https://github.com/arnoldrobbins/mcilroy-regex 3: https://github.com/thaliaarchi/recross-coq/blob/main/theorie...