Question

Please #NoSpamming.

Prove : (A ⋂ B' ⋂ C') = A - (A ⋂ B) - (A ⋂ C) + (A ⋂ B ⋂ C)

Please Solve Step by Step..

Answer 1

Answer:

We can use the Venn Diagram.

Step-by-step explanation:

*It looks like you forgot to put n(A) in front of each set...

*Sorry for late answering... I'll put attachments to every explanation.

First n(A) - n(A ⋂ B)

= n(A - B) [Venn Diagram] ... *Refer to the 1st attachment.

= n(A ∩ B')

I've shown n(A ∩ B') - n(A ∩ C) in the 2nd attachment.

Then lastly add n(A ∩ B ∩ C)

= (A ∩ B' ∩ C') [Venn Diagram] ... *Refer to the 3rd attachment.

Answer 2

Answer:

It is easily visualised by Venn Diagram. But since the other answer has given you with approach of Venn diagram, I will provide my answer in terms of Set theory.

Given: ( A ⋂ B' ⋂ C' ) = A - ( A ⋂ B ) - ( A ⋂ C ) + ( A ⋂ B ⋂ C )

Considering LHS we get:

⇒ (A ⋂ B' ⋂ C')

Let us solve ( A ∩ B' ) first.

For any general form ( X ∩ Y' ) the equivalent form is: X - ( X ∩ Y )

Hence based on the above sentence we get:

→ A - ( A ∩ B ) = ( A ∩ B' )   ...Eqn. (1)

Now substituting the equivalent form in place of ( A ∩B' ) in equation 1, we get:

→ [ A - ( A ∩ B ) ] ∩ C' = ( A ∩ B' ∩ C' )

Assuming [ A - ( A ∩ B ) ] as X we get:

→ X ∩ C'

This looks similar to the general form. Hence we get the equivalent form to be:

→ X - ( X ∩ C )

Substituting back thee old values for X. Hence we get:

→ [ A - ( A ∩ B ) ] - ( [ A - ( A ∩ B ) ] ∩ C )

→ A - ( A ∩ B ) - [ ( A ∩ C ) - ( A ∩ B ∩ C ) ]  (Using distributive property)

→ A - ( A ∩ B ) - ( A ∩ C ) + ( A ∩ B ∩ C )  (After multiplying - sign inside)

Hence proved!