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Exercise 3.3 · Q7

Q.Fill in the blanks.

(i) A′∩ϕ=A' \cap \phi = ___
(ii) A∩ϕ′=A \cap \phi' = ___
(iii) A∪U′=A \cup U' = ___
(iv) A∩U′=A \cap U' = ___
(v) A∩ϕ′=A \cap \phi' = ___
(vi) U′∩ϕ=U' \cap \phi = ___
(vii) U′∩ϕ′=U' \cap \phi' = ___
(viii) U′∪ϕ=U' \cup \phi = ___
(ix) U′∪ϕ′=U' \cup \phi' = ___
(x) U∪ϕ′=U \cup \phi' = ___
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Since U′=ϕU' = \phi (complement of the universal set is empty) and ϕ′=U\phi' = U (complement of the empty set is the universal set), substitute and apply the identity laws A∩U=AA\cap U = A, A∪ϕ=AA\cup \phi = A, A∩ϕ=ϕA\cap\phi=\phi, A∪U=UA\cup U = U.

Identity laws: A∪ϕ=AA\cup\phi = A, A∩U=AA\cap U = A, A∩ϕ=ϕA\cap\phi = \phi, A∪U=UA\cup U = U. Complement of universal/empty set: U′=ϕU' = \phi, ϕ′=U\phi' = U.

  1. (i) A′∩ϕA'\cap\phi: intersection with the empty set is always empty. =ϕ= \phi.

  2. (ii) A∩ϕ′A\cap\phi': since ϕ′=U\phi' = U, this is A∩U=AA\cap U = A.

  3. (iii) A∪U′A\cup U': since U′=ϕU' = \phi, this is A∪ϕ=AA\cup\phi = A.

  4. (iv) A∩U′A\cap U': since U′=ϕU' = \phi, this is A∩ϕ=ϕA\cap\phi = \phi.

  5. (v) A∩ϕ′A\cap\phi': same as step 2 — ϕ′=U\phi'=U, so A∩U=AA\cap U = A.

  6. (vi) U′∩ϕU'\cap\phi: U′=ϕU'=\phi, so this is ϕ∩ϕ=ϕ\phi\cap\phi = \phi. …

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