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Prove the second De Morgan law from the Table 1 by showing that if \(A\) and \(B\) are sets, then \(\overline {A \cup B} = \overline A \cap \overline B \)

(a) by showing each side is a subset of the other side.

(b) using a membership table.

Short Answer

Expert verified

(a) it is proved that \(\overline {A \cup B} = \overline A \cap \overline B \).

(b) it is proved that \(\overline {A \cup B} = \overline A \cap \overline B \) using a membership table.

Step by step solution

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01

(a) Showing each side is a subset of the other side.

Solve the left hand side of set \(\overline {A \cup B} \) as follows:

\(\begin{aligned}LHS &= \overline {A \cup B} \\ &= \left\{ {x|x \notin A \cup B} \right\}\;{\rm{by}}\;{\rm{definition}}\;{\rm{of}}\;{\rm{compliment}}\\ &= \left\{ {x|\neg \left( {x \in A \cup B} \right)} \right\}\;{\rm{by}}\;{\rm{definition}}\;{\rm{of}}\;{\rm{does}}\;{\rm{not}}\;{\rm{belong}}\;{\rm{symbol}}\\ &= \left\{ {x|\neg \left( {x \in A \vee x \in B} \right)} \right\}\;{\rm{by}}\;{\rm{definition}}\;{\rm{of}}\;{\rm{union}}\end{aligned}\)

\(\begin{aligned}LHS &= \left\{ {x|\neg \left( {x \in A} \right) \wedge \neg \left( {x \in B} \right)} \right\}\;{\rm{by}}\;{\rm{De Morgan's law (for logical equivalence)}}\\ &= \left\{ {x|x \notin A \wedge x \notin B} \right\}\;{\rm{by}}\;{\rm{definition}}\;{\rm{of}}\;{\rm{does}}\;{\rm{not}}\;{\rm{belong}}\;{\rm{symbol}}\\ &= \left\{ {x|x \in \overline A \wedge x \in \overline B } \right\}\;{\rm{by}}\;{\rm{definition}}\;{\rm{of}}\;{\rm{complement}}\end{aligned}\)

\(\begin{aligned}LHS &= \left\{ {x|x \in \overline A \cap \overline B} \right\}\;{\rm{by}}\;{\rm{definition}}\;{\rm{of}}\;{\rm{intersection}}\\ &= \overline A \cap \overline B \;{\rm{by}}\,{\rm{set}}\;{\rm{builder}}\;{\rm{notation}}\\ &= RHS\end{aligned}\)

Hence, it is proved that \(\overline {A \cup B} = \overline A \cap \overline B \).

02

(b) Using membership table \(\overline {A \cup B}  = \overline A  \cap \overline B \).

Draw the table to solve the set as follows:

\(A\)

\(B\)

\(\overline A \)

\(\overline B \)

\(A \cup B\)

\(\overline {A \cup B} \)

\(\overline A \cap \overline B \)

1

1

0

0

1

0

0

1

0

0

1

1

0

0

0

1

1

0

1

0

0

0

0

1

1

0

1

1

Hence, it is proved that \(\overline {A \cup B} = \overline A \cap \overline B \).

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