9. Which of the following are true for all regular languages and all homomorphisms? (a) h (L1 ∪ L2)= h (L1) ∩ h (L2). (b) h (L1 ∩ L2)= h (L1) ∩ h (L2). (c) h (L1L2) = h (L1) h (L2).

h (L1 ∪ L2)= h (L1) ∩ h (L2).

This is true.

There will be w ∈  (L1 ∪ L2) for any s ∈ h (L1 ∪ L2) in such a way that s=h(w)

we can assume that w ∈ L1

So In this case h(w) ∈ L (S1). Hence s ∈ L(S1)

for any s ∈ h (L1) U h(L2)

We can assume that s ∈ L(S1)

There exists w ∈ L1 such that s= h(w)

In this case it is w ∈ L1 U L2 as well.

Hence , s ∈ h (L1 U L2)

Explanation

consider  = 0,1 and  = a,b and h(0) = a , h(1) = ab

(a) Consider L1 = 10,01 and L2 = 00,11

   Now L1 ∪ L2 = 00,01,10,11

   h (L1 ∪ L2) = h(00) , h(01) , h(10) , h(11) = h(0)h(0) , h(0)h(1) , h(1)h(0) , h(1)h(1)

  = aa, aab , aba , abab

Hence h (L1 ∪ L2) = aa, aab , aba , abab .

Here h (L1) = h(10) , h(01) = h(1)h(0) , h(0)h(1) = aba , aab

Hence h (L1) = aba , aab .

Here h (L2) = h(00) , h(11) = h(0)h(0), h(1)h(1) = aa, abab

Hence h(L2) = aa, abab.

Finally Hence , h (L1 ∪ L2)= h (L1) ∩ h (L2).

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this seems to be in the wrong category. but i'll take some free points my dude

posting this in wrong category will result in people seeing it who won't know the answer.