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Using properties of sets prove the given statement.
For all sets A and B, A – (A – B) = A ∩ B
Using properties of sets prove the given statement.
For all sets A and B, A – (A – B) = A ∩ B
Using properties of sets prove the given statement.
For all sets A and B, A ∪ (B – A) = A ∪ B
Using properties of sets prove the given statement.
For all sets A and B, A ∪ (B – A) = A ∪ B
According to the question,
There are two sets A and B
To prove: A ∪ (B – A) = A ∪ B
L.H.S = A ∪ (B – A)
Since, A – B = A ∩ B’, we get,
= A ∪ (B ∩ A’)
Since, distributive property of set ⇒ (A ∪ B) ∩ (A ∪ C) = A ∪ (B ∩ C), we get,
= (A ∪ B) ∩ (A ∪ A’)
Since, A ∪ A’ = U, we get,
= (A ∪ B) ∩ U
= A ∪ B
= R.H.S
Hence Proved
Using properties of sets prove the given statement.
For all sets A and B, A – (A ∩ B) = A – B
Using properties of sets prove the given statement.
For all sets A and B, A – (A ∩ B) = A – B
According to the question,
There are two sets A and B
To prove: A – (A ∩ B) = A – B
L.H.S = A – (A ∩ B)
Since, A – B = A ∩ B’, we get,
= A ∩ (A ∩ B)’
= A ∩ (A ∩ B’)’
Since, (A ∩ B)’ = A’ ∪ B’, we get,
= A ∩ (A’ ∪ B’)
Since, Distributive property of set ⇒ (A ∩ B) ∪ (A ∩ C) = A ∩ (B ∪ C), we get,
= (A ∩ A’) ∪ (A ∩ B’)
Since, A ∩ A’ = Φ, we get,
= Φ ∪ (A ∩ B’)
= A ∩ B’
Since, A – B = A ∩ B’, we get,
= A – B
= R.H.S
Hence Proved
Using properties of sets prove the given statement.
For all sets A and B, (A ∪ B) – B = A – B
Using properties of sets prove the given statement.
For all sets A and B, (A ∪ B) – B = A – B
According to the question,
There are two sets A and B
To prove: (A ∪ B) – B = A – B
L.H.S = (A ∪ B) – B
Since, A – B = A ∩ B’, we get,
= (A ∪ B) ∩ B’
Since, Distributive property of set: (A ∩ B) ∪ (A ∩ C) = A ∩ (B ∪ C), we get,
= (A ∩ B’) ∪ (B ∩ B’)
Since, A ∩ A’ = Φ, we get,
= (A ∩ B’) ∪ Φ
= A ∩ B’
Since, A – B = A ∩ B’, we get,
= A – B
= R.H.S
Hence Proved
According to the question,
There are two sets A and B
To prove: A – (A – B) = A ∩ B
L.H.S = A – (A – B)
Since, A – B = A ∩ B’, we get,
= A – (A ∩ B’)
= A ∩ (A ∩ B’)’
Since, (A ∩ B)’ = A’ ∪ B’, we get,
= A ∩ [A’ ∪ (B’)’]
Since, (B’)’ = B, we get,
= A ∩ (A’ ∪ B)
Since, distributive property of set ⇒ (A ∩ B) ∪ (A ∩ C) = A ∩ (B ∪ C), we get,
= (A ∩ A’) ∪ (A ∩ B)
Since, A ∩ A’ = Φ, we get,
= Φ ∪ (A ∩ B)
= A ∩ B
= R.H.S
Hence Proved