#### Answer

given below

#### Work Step by Step

(a)
To find the set \[A\cup \left( B'\cap C' \right)\],
First perform the operation inside parenthesis:
Now, compute \[B'\cap C'\]
The complement of set Bcontaining all the elements of U which are not in B.
So, set \[B'=\left\{ \text{c,d,e,f} \right\}\]
Similarly, set \[C'=\left\{ \text{a,c,e,f} \right\}\]
So, the intersection of both the sets containing their common elements.
Therefore,
\[\begin{align}
& B'\cap C'=\left\{ \text{c,d,e,f} \right\}\cap \left\{ \text{a,c,e,f} \right\} \\
& =\left\{ \text{c,e,f} \right\}
\end{align}\].
Now, the union of sets A and\[B'\cap C'\]containing elements either of the set A or \[B'\cap C'\]or both. So,
\[\begin{align}
& A\cup \left( B'\cap C' \right)=\left\{ \text{c} \right\}\cup \left\{ \text{c,e,f} \right\} \\
& =\left\{ \text{c,e,f} \right\}
\end{align}\]
Now, to find the set \[\left( A\cup B' \right)\cap \left( A\cup C' \right)\]
First perform operation inside parenthesis:
So,
\[\begin{align}
& A\cup B'=\left\{ \text{c} \right\}\cup \left\{ \text{c,d,e,f} \right\} \\
& =\left\{ \text{c,d,e,f} \right\} \\
& A\cup C'=\left\{ \text{c} \right\}\cup \left\{ \text{a,c,e,f} \right\} \\
& =\left\{ \text{a,c,e,f} \right\}
\end{align}\]
Their intersection is:
\[\begin{align}
& \left( A\cup B' \right)\cap \left( A\cup C' \right)=\left\{ \text{c,d,e,f} \right\}\cap \left\{ \text{a,c,e,f} \right\} \\
& =\left\{ \text{c,e,f} \right\}
\end{align}\]
The set \[A\cup \left( B'\cap C' \right)=\left\{ \text{c,e,f} \right\}\]and\[\left( A\cup B' \right)\cap \left( A\cup C' \right)=\left\{ \text{c,e,f} \right\}\].
(b)
To find the set \[A\cup \left( B'\cap C' \right)\],
First perform the operation inside parenthesis:
Now, compute \[B'\cap C'\]
The complement of set Bcontaining all the elements of U which are not in B.
So, set \[B'=\left\{ \text{1,4,5,8} \right\}\]
Similarly, set \[C'=\left\{ \text{1,2,3,5} \right\}\]
So the intersection of both the sets containing their common elements.
Therefore,
\[\begin{align}
& B'\cap C'=\left\{ 1,4,5,8 \right\}\cap \left\{ 1,2,3,5 \right\} \\
& =\left\{ 1,5 \right\}
\end{align}\].
Now the union of sets A and\[B'\cap C'\]containing elements either of the set A or \[B'\cap C'\]or both. So,
\[\begin{align}
& A\cup \left( B'\cap C' \right)=\left\{ 1,3,7,8 \right\}\cup \left\{ \text{5} \right\} \\
& =\left\{ \text{1,3,5,7,8} \right\}
\end{align}\]
Now to find the set \[\left( A\cup B' \right)\cap \left( A\cup C' \right)\]
First perform operation inside parenthesis:
So,
\[\begin{align}
& A\cup B'=\left\{ 1,3,7,8 \right\}\cup \left\{ \text{1,4,5,8} \right\} \\
& =\left\{ \text{1,3,4,5,7,8} \right\} \\
& A\cup C'=\left\{ 1,3,7,8 \right\}\cup \left\{ 1,2,3,5 \right\} \\
& =\left\{ \text{1,2,3,5,7,8} \right\}
\end{align}\]
Their intersection is:
\[\begin{align}
& \left( A\cup B' \right)\cap \left( A\cup C' \right)=\left\{ \text{1,3,4,5,7,8} \right\}\cap \left\{ \text{1,2,3,5,7,8} \right\} \\
& =\left\{ \text{1,3,5,7,8} \right\}
\end{align}\]
The set \[A\cup \left( B'\cap C' \right)=\left\{ \text{1,3,5,7,8} \right\}\]and\[\left( A\cup B' \right)\cap \left( A\cup C' \right)=\left\{ \text{1,3,5,7,8} \right\}\]
(c)
Inductive reasoning is the process of arriving at a general conclusion based on observations of specific examples.
So, we consider examples given in part (a)and part (b).
In part (a), observe that set\[A\cup \left( B'\cap C' \right)=\left\{ \text{c,e,f} \right\}\]and set\[\left( A\cup B' \right)\cap \left( A\cup C' \right)=\left\{ \text{c,e,f} \right\}\]
So, they are equal.
Also, in part (b), set \[A\cup \left( B'\cap C' \right)=\left\{ \text{1,3,5,7,8} \right\}\]and\[\left( A\cup B' \right)\cap \left( A\cup C' \right)=\left\{ \text{1,3,5,7,8} \right\}\].
So they are equal.
Generalize the concept from the above two observations that for any set A, B and CIntersection is Distributive over Union of sets. So,
\[A\cup \left( B'\cap C' \right)=\left( A\cup B' \right)\cap \left( A\cup C' \right)\].
For any set A, B and C:
\[A\cup \left( B'\cap C' \right)=\left( A\cup B' \right)\cap \left( A\cup C' \right)\]
(d)
Deductive reasoning is the process of proving a specific conclusion from one or more general statements. A conclusion that is proved to be true by deductive reasoning is called a theorem.
Now to prove \[A\cup \left( B'\cap C' \right)=\left( A\cup B' \right)\cap \left( A\cup C' \right)\]
Consider a Venn diagram of set A, B, C and universal set U.
To find the region which represent the set \[A\cup \left( B'\cap C' \right)\]in the above Venn diagram,
First perform the operation inside parenthesis:
Now, find the region of set \[B'\cap C'\].
The complement of set Bcontaining all the elements of U which are not in B.
Regions II, III, V and VI together represents set B.
Then removing these regions from universal set U (containing all the regions), the left part is complement of set B.
So, regions I, IV, VII and VIII represents the set\[B'\].
Similar way,
Regions IV, V, VI, and VII together represent set C.
Then removing these regions from universal set U (containing all the regions), the left part is complement of set C.
So, regions I, II, III and VIII represents the set\[C'\].
Then, intersection\[B'\cap C'\] represented by the common regions of both the sets, which are I and VIII.
Now, set A is represented by the regions I, II, IV and V.
Then,\[A\cup \left( B'\cap C' \right)\]is represented by union of the regions of the set A and set\[B'\cap C'\].
So, regions I, II, IV, V and VIII represent the set\[A\cup \left( B'\cap C' \right)\].
Now, to find the set\[\left( A\cup B' \right)\cap \left( A\cup C' \right)\].
First perform operation inside parenthesis:
So, set\[A\cup B'\]is represented by I, II, IV, V, VII and VIII and set\[A\cup C'\]is represented by
I, II, III, IV, V and VIII.
Then, their intersection \[\left( A\cup B' \right)\cap \left( A\cup C' \right)\] is represented by I, II, IV, V and VIII.
Since, set\[A\cup \left( B'\cap C' \right)\]is represented by regions I, II, IV, V and VIII and also the set\[\left( A\cup B' \right)\cap \left( A\cup C' \right)\]represented by regions I, II, IV, V and VIII.
So, they are equal for any sets A, B and C.
i.e.\[A\cup \left( B'\cap C' \right)=\left( A\cup B' \right)\cap \left( A\cup C' \right)\]
\[A\cup \left( B'\cap C' \right)=\left( A\cup B' \right)\cap \left( A\cup C' \right)\]for any set A, B and C. Hence, it is a theorem.