Prove.1. (AnB)\(CnD)=(A\C)n(B\D)

Question

2. (A∩B)\C=(A\C)∩(B\C)

Answer 1

Let C'=complement of C. A\C=A∩C'

1. (AnB)\(CnD)=(A\C)n(B\D)=A∩B∩C'∩D'

By DeMorgan's Law (C∩D)'=C'∪D'

(AnB)\(CnD)=(AnB)∩(C'∪D')=(A∩B∩C')∪(A∩B∩D') not even close to equal

 

 

 

 

2. (A∩B)\C=(A\C)∩(B\C)

(A∩B)\C=A∩B∩C'=A∩C'∩B∩C'=(A\C)∩(B\C)

Answer 2

1.  (AnB)\(CnD)=(A\C)n(B\D) 

 

A\C = A - A∩C

B\D = B - B∩D

(A - A∩C)∩(B - B∩D) = A∩B - C∪D = (A∩B)\(C∪D)   It seems they are not equal.

 

2. (A∩B)\C=(A\C)∩(B\C)

 

A\C = A - A∩C

B\C = B - B∩C

(A - A∩C)∩(B - B∩C) = (A∩B)\C