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Let U be the universal set and let A, B and C be subsets of U. By using the properties of U, ∩ and ~(complement), Theorem 4.5.1 and Corollary 4.5.2, how can I simplify the following?

(i) (C ∩ U) ∪ ~C
(ii) ~(A ∩ U) U ~A  
(iii) ~[~(C U 0) U C]
(iv)  (A ∩ B) ∩ ~A
Asaf Karagila
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Julfikar
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1 Answers1

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$X^c$ denotes the complement set of $X$

  • (i)

(i) $= C \cup C^c = U$

I used that $C\subset U$ because $U$ is the universal set, that implies that $C\cap U = C$.

  • (ii)

(ii) $= A^c \cup A^c = A^c$

I used that $A\subset U \implies A\cap U = A$. You can also use De Morgan's law:

(ii) $= (A^c\cup U^c) \cup A^c = (A^c\cup \emptyset) \cup A^c = A^c \cup A^c = A^c$

  • (iii)

(iii) $= (C^c \cup C)^c = U^c = \emptyset$

The first stepp is because $\emptyset \subset C$ implies that $C\cup \emptyset = C$.

Another possibility is using De Morgan's law $(M\cup N)^c = M^c\cap N^c$ and $(M^c)^c = M$:

(iii) $= ((C\cup \emptyset)^c)^c \cap C^c = (C\cup \emptyset)\cap C^c = C\cap C^c = \emptyset$

  • (iv)

(iv) $= (B\cap A)\cap A^c = B\cap (A\cap A^c) = B\cap \emptyset = \emptyset$

Here using the commutative and associative properties of the intersection.

Abelois
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