lecture 38

6
1 Lecture 38 • Showing CFL’s not closed under set intersection and set complement

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Lecture 38. Showing CFL’s not closed under set intersection and set complement. Nonclosure Properties for CFL’s. CFL’s not closed under set intersection. How do we prove that CFL’s are not closed under set intersection? State closure property as IF-THEN statement - PowerPoint PPT Presentation

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Page 1: Lecture 38

1

Lecture 38

• Showing CFL’s not closed under set intersection and set complement

Page 2: Lecture 38

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Nonclosure Properties for CFL’s

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CFL’s not closed under set intersection

• How do we prove that CFL’s are not closed under set intersection?– State closure property as IF-THEN statement

• If L1 and L2 are CFL’s, then L1 intersect L2 is a CFL

– Proof is by counterexample• Find 2 CFL’s L1 and L2 such that L1 intersect L2 is

NOT a CFL

Page 4: Lecture 38

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Counterexample

• What is a possible L1 intersect L2?

– What non-CFL languages do we know?

• What could L1 and L2 be?

– L1 =

– L2 =

– How can we prove that L1 and L2 are context-free?

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CFL’s not closed under complement

• How can we prove that CFL’s are not closed under complement?– We could do the same thing, find a

counterexample– Another way

• Use fact that any language class which is closed under union and complement must also be closed under intersection

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Language class hierarchy

All languages over alphabet

RE

REG

HH

EqualCFLREC

Equal-3