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The fact that SL2 is closed under intersection but not under union implies that it is not closed under complement since, by DeMorgan's Theorem
L1 ∩ L2 =
We know that the intersection of SL2 languages is also SL2. If the complement of SL2 languages was also necessarily SL2, then would be SL2 contradicting the fact that there are SL2 languages whose union are not SL2.
Lemma The class of strictly 2-local languages is not closed under complement .
A common approach in solving problems is to transform them to different problems, solve the new ones, and derive the solutions for the original problems from those for the new ones
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We'll close our consideration of regular languages by looking at whether (certain) problems about regular languages are algorithmically decidable.
Lemma 1 A string w ∈ Σ* is accepted by an LTk automaton iff w is the concatenation of the symbols labeling the edges of a path through the LTk transition graph of A from h?, ∅i to
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We got the class LT by taking the class SL and closing it under Boolean operations. We have observed that LT ⊆ Recog, so certainly any Boolean combination of LT languages will also
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