Question

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

Short answer

Thus, it is true \(A \cup \left( {B \cup C} \right) = \left( {A \cup B} \right) \cup C\).

Step-by-step solution

Given

Given in the question A, B, and C are set. now we proof here,

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

explanation

Here we solve the LHS part .

Let \(x \in A \cup \left( {B \cup C} \right)\).

\( \Rightarrow x \in A\)or \(x \in B\)or\(x \in C\).

= (\(x \in A\) or \(x \in B\)) or \(x \in C\).

\( \Rightarrow x \in \left( {A \cup B} \right)\)or \(x \in C\)

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

\(\therefore A \cup \left( {B \cup C} \right) \subseteq \left( {A \cup B} \right) \cup C\)___________(i)

Now , here we solve RHS part

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

\( \Rightarrow x \in \left( {A \cup B} \right)\)or \(x \in C\).

\( \Rightarrow x \in A\)or \(x \in B\)or\(x \in C\)

\( \Rightarrow x \in A\)or ( \(x \in B\) or \(x \in C\))

\( \Rightarrow x \in A\)or \(x \in \left( {B \cup C} \right)\).

\( \Rightarrow x \in A\)or \(x \in \left( {B \cup C} \right)\)

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

\(\left( {A \cup B} \right) \cup C \subseteq A \cup \left( {B \cup C} \right)\)____________-(ii)

Frome equation (i) and (ii)

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

Hence ,

LHS = RHS

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