Prove that A-(B⋃C) = (A-B)⋂(A-C)

Question

Please let me know if this proof is right, I think it is but I still want confirmation

A - (B⋃C)
= A ⋂ (B⋃C)'           [X-Y = X⋂Y']
= A ⋂ (B'⋂C')          [(X⋃Y)' = (X'⋂Y') - De-Morgan's Law]
= (A⋂A) ⋂ (B'⋂C')      [A = (A⋂A) - Idempotent Law]
= A ⋂ A ⋂ B' ⋂ C'
= A ⋂ B' ⋂ A ⋂ C'      [Commutative Law]
= (A⋂B') ⋂ (A⋂C')
= (A-B) ⋂ (A-C)        [X⋂Y' = X-Y]

Hence, Proved

Let me know what you think

Thanks...

Answer 1

Yes, your proof is correct (Good Work!) but note it is not complete in terms of justification; you need to note that fact that $\cap$ is associative (it may seem obvious to you that the unjustified steps will be understood, but when you're new to proofs, it's best to not cut corners):

A - (B⋃C)
= A ⋂ (B⋃C)'                                         [X-Y = X⋂Y']
= A ⋂ (B'⋂C')                                        [(X⋃Y)' = (X'⋂Y') - DeMorgan's Law]
= (A⋂A) ⋂ (B'⋂C')                                    [A = (A⋂A) - Idempotent Law]
= A ⋂ A ⋂ B' ⋂ C'                                    [Associative Law]
= A ⋂ B' ⋂ A ⋂ C'                                    [Commutative Law]
= (A⋂B') ⋂ (A⋂C')                                    [Associative Law]
= (A-B) ⋂ (A-C)                                       [X⋂Y' = X-Y]

Answer 2

Your proof is fine, although I would advise that you use MathJax in future.