Answer:
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).
