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L1= Regular 

L2= DCFL

L2 – L1 = L2 ∩ L1’ = DCFL ∩ REG. = DCFL   , so  L3 = ∑* – DCFL = DCFL (closed under complementation)

Now ,

      i. L1 – L3 = L1 ∩ L3’ = REG. ∩ DCFL = DCFL  .

     ii. L3 – L2 = L3 ∩ L2’ = DCFL ∩ DCFL = DCFL may or may not.

    iii. L3 – L1 = L3 ∩ L1’ = DCFL ∩ REG. = DCFL  & closed under complementation.

    iv. L^-1(L3) = L^-1(DCFL) = DCFL  ( closed under inverse homomorphism )
   
 Only (iv) is TRUE.

 

 

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