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.