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The class of Strictly Local Languages (in general) is closed under
• intersection but is not closed under
• union
• complement
• concatenation
• Kleene- and positive closure
Proof: For intersection, we can adapt the construction and proof for the SL2 case again to get closure under intersection for SLk. This is still not quite enough for SL in general, since one of the languages may be in SLi and the other in SLj for some i = j. Here we can use the hierarchy theorem to show that, supposing i < j, the SLi language is also in SLj . Then the adapted construction will establish that their intersection is in SL .
For non-closure under union (and consequently under complement) we can use the same counterexample as we did in the SL2 case:
To see that this is not in SLk for any k we can use the pair
which will yield abk-1 a under k-local suffix substitution closure.
For non-closure under concatenation we can use the counterexample
The two languages being concatenated are in SL2, hence in SLk for all k ≥ 2 but their concatenation is not in SLk for any k, as we showed in the example above.
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I want a proof for any NP complete problem
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