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Intuitively, closure of SL2 under intersection is reasonably easy to see, particularly if one considers the Myhill graphs of the automata. Any path through both graphs will be a path through the intersection of the graphs (by which we mean the graph resulting by taking the intersection of the vertex sets and the intersection of the edge sets).
For the union, on the other hand, the corresponding construction won't work. An automaton built from the union of the two automata will still recognize all of the strings in L1 and all of the strings in L2, but it is likely to also recognize strings made up of adjacent pairs from L1 combined with adjacent pairs from L2 that aren't in either language. And, in fact, we can use Suffx Substitution Closure to show that there are languages that are the union of two SL2 languages that are not, themselves, SL2.
One of the first issues to resolve, when exploring any mechanism for defining languages is the question of how to go about constructing instances of the mechanism which define part
Construct a PDA that accepts { x#y | x, y in {a, b}* such that x ? y and xi = yi for some i, 1 = i = min(|x|, |y|) }. For your PDA to work correctly it will need to be non-determin
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The fact that SL 2 is closed under intersection but not under union implies that it is not closed under complement since, by DeMorgan's Theorem L 1 ∩ L 2 = We know that
All that distinguishes the de?nition of the class of Regular languages from that of the class of Star-Free languages is that the former is closed under Kleene closure while the lat
Kleene called this the Synthesis theorem because his (and your) proof gives an effective procedure for synthesizing an automaton that recognizes the language denoted by any given r
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In Exercise 9 you showed that the recognition problem and universal recognition problem for SL2 are decidable. We can use the structure of Myhill graphs to show that other problems
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