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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
Give the Myhill graph of your automaton. (You may use a single node to represent the entire set of symbols of the English alphabet, another to represent the entire set of decima
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Both L 1 and L 2 are SL 2 . (You should verify this by thinking about what the automata look like.) We claim that L 1 ∪ L 2 ∈ SL 2 . To see this, suppose, by way of con
Theorem The class of recognizable languages is closed under Boolean operations. The construction of the proof of Lemma 3 gives us a DFA that keeps track of whether or not a give
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We will specify a computation of one of these automata by specifying the pair of the symbols that are in the window and the remainder of the string to the right of the window at ea
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