Question

In Exercise 5-10 assume that \(A\) is a subset of some underlying universal set \(U\).

Prove the complement laws in Table 1 by showing that

(a) \(A \cup \overline A = U\)

(b) \(A \cap \overline A = \phi \)

Short answer

(a) it is proved that \(A \cup \overline A = U\).

(b) it is proved that \(A \cap \overline A = \phi \).

Step-by-step solution

Showing that \(A \cup \overline A  = U\).

Let the sets be \(U = \left\{ {1,2,3,4,5,6} \right\}\) and \(A = \left\{ {1,2,3} \right\}\)

Now the set is written as:

\(\overline A = \left\{ {4,5,6} \right\}\)

Thus, write as follows:

\(\begin{aligned}LHS &= A \cup \overline A \\ &= \left\{ {1,2,3} \right\} \cup \left\{ {4,5,6} \right\}\\ &= \left\{ {1,2,3,4,5,6} \right\}\\ &= U\end{aligned}\)

\(LHS = RHS\)

Hence, it is proved that \(A \cup \overline A = U\).

Showing that \(A \cap \overline A  = \phi \).

Let the sets be \(U = \left\{ {1,2,3,4,5,6} \right\}\) and \(A = \left\{ {1,2,3} \right\}\)

Now the set is written as:

\(\overline A = \left\{ {4,5,6} \right\}\)

Thus, write as follows:

\(\begin{aligned}LHS &= A \cap \overline A \\ &= \left\{ {1,2,3} \right\} \cap \left\{ {4,5,6} \right\}\\ &= \phi \\ &= RHS\end{aligned}\)

Hence, it is proved that \(A \cap \overline A = \phi \).