Question

Prove the first distributive law from Table 1 by showing that if A, B, and C are set, then A∪(B ∩ C) = (A∪B) ∩ (A∪C).

Short answer

Thus, we have to prove \(A \cup \left( {B \cap C} \right) = \left( {A \cup B} \right) \cap \left( {A \cup C} \right)\).

Step-by-step solution

Given

Given in the question , first distributive law

A∪(B ∩ C) = (A∪B) ∩ (A∪C).

Now we need to prove it .

explanation

here we written first derivative ,

\(A \cup \left( {B \cap C} \right) = \left( {A \cup B} \right) \cap \left( {A \cup C} \right)\)

We have to use symbol \( \vee = or\) , \( \wedge = and\)

Here we try to prove it taking LHS.

=\(A \cup \left( {B \cap C} \right)\)

Let ,

\( \Rightarrow m \in A \cup \left( {B \cap C} \right)\)

\( \Rightarrow m \in A \vee m \in \left( {B \cap C} \right)\)

\( \Rightarrow m \in A \vee \left( {m \in B \wedge m \in C} \right)\)( Definition of intersection )

\( \Rightarrow \left( {m \in A \vee m \in B} \right) \wedge \left( {m \in A \vee m \in C} \right)\)( by De- morgans law of logical question )

\( \Rightarrow m \in \left( {A \cup B} \right) \wedge m \in \left( {A \cup C} \right)\)

\( \Rightarrow m \in \left( {A \cup B} \right) \cap \left( {A \cup C} \right)\)

\( \Rightarrow m \in A \cup \left( {B \cap C} \right)\)= \(m \in \left( {A \cup B} \right) \cap \left( {A \cup C} \right)\)

\( \Rightarrow A \cup \left( {B \cap C} \right) \subseteq \left( {A \cup B} \right) \cap \left( {A \cup C} \right)\)___________(1)

Here we try to prove RHS

Let ,

\(n \in \left( {A \cup B} \right) \cap \left( {A \cup C} \right)\)