Showing Equivalence using Set Properties 2

In this class, We discuss Showing Equivalence using Set Properties 2.

The reader should have prior knowledge of set theory basics. Click Here

Example 1:

A ∪ (B – A) = A ∪ B

Solution:

A ∪ (B – A)

A ∪ (B ∩ A’)

(A ∪ B) ∩ (A ∪ A’) from distributive law

(A ∪ B) ∩ U

(A ∪ B)

Example 2:

A – (B ∪ C) = (A – B) ∩ (A – C)

Solution:

Take LHS A – (B ∪ C)

A ∩ (B ∪ C)’

A ∩ (B’ ∩ C’)

A ∩ B’ ∩ C’

Take RHS (A – B) ∩ (A – C)

(A ∩ B’) ∩ (A ∩ C’)

A ∩ B’ ∩ C’

Example 3:

A ∪(B – C) = (A ∪ B) – (C -A)

Solution:

A ∪ (B – C)

A ∪ ( B ∩ C’)

(A ∪ B) ∩ (A ∪ C’) from distributive law

(A ∪ B) ∩ ((A’)’ ∪ C’)

(A ∪ B) ∩ (A’ ∩ C)’ from de morgans law

(A ∪ B) ∩ (C ∩ A’)’

(A ∪ B) ∩ (C – A)’

(A ∪ B) – (C – A)