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CSE322
Comparison of deterministic and
nondeterministic versions
&
closure properties
Lecture #34
Union Closure Properties
Lemma: Let A1 and A2 be two CF languages,
then the union A1∪A2 is context free as well.
Proof: Assume that the two grammars are
G1=(V1,Σ,R1,S1) and G2=(V2,Σ,R2,S2).
Construct a third grammar G3=(V3,Σ,R3,S3) by:
V3 = V1 ∪V2 ∪ { S3 } (new start variable) with
R3 = R1 ∪ R2 ∪ { S3 → S1 | S2 }.
It follows that L(G3) = L(G1) ∪ L(G2).
Intersection & Complement?
Let again A1 and A2 be two CF languages.
One can prove that, in general,
the intersection A1 ∩ A2 ,
and
the complement Ā1= Σ*  A1
are not context free languages.
One proves this with specific counter examples
of languages (see homework).

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A890743837 22750 24_2018_comparison of deterministic and non-deterministic version

  • 1. CSE322 Comparison of deterministic and nondeterministic versions & closure properties Lecture #34
  • 2. Union Closure Properties Lemma: Let A1 and A2 be two CF languages, then the union A1∪A2 is context free as well. Proof: Assume that the two grammars are G1=(V1,Σ,R1,S1) and G2=(V2,Σ,R2,S2). Construct a third grammar G3=(V3,Σ,R3,S3) by: V3 = V1 ∪V2 ∪ { S3 } (new start variable) with R3 = R1 ∪ R2 ∪ { S3 → S1 | S2 }. It follows that L(G3) = L(G1) ∪ L(G2).
  • 3. Intersection & Complement? Let again A1 and A2 be two CF languages. One can prove that, in general, the intersection A1 ∩ A2 , and the complement Ā1= Σ* A1 are not context free languages. One proves this with specific counter examples of languages (see homework).