Is the Algebraic Manipulation of Set Theory Expressions Accurate?

[Natasha1]

Can anyone just check if I got it right please?
And if so could you just explain the theorems that come with each line? Many many thanks in advance

(A-B) n (B-A) = (AuB’) n (BuA’)
= (Au(BuA’)) u(B’n (BuA’))
= ((AnB) u (AnA’)) u ((B’nB) u (B’nA’))
= (AnB) u Ø u Ø u (B’nA’)
= (AnB) u (B’nA’)
= (AnB) u (BuA)’
= (AnB) - (BuA)
= (AnB) - (AuB)

 

[EnumaElish]

If A = {1,2,3,4} and B = {3,4,5,6}, what is A-B? What is B-A? What is their intersection?

 

[Architeuthis Dux]

Yes, this is correct. The theorem being used here is the distributive law, which states that (A n B) u (A n C) = A n (B u C). This means that the intersection of two sets can be distributed over the union of two other sets. In this case, we are distributing (A-B) n (B-A) over (AuB') n (BuA').

The first line simply applies the distributive law.

The second line uses the fact that A n B = A n (B u A') = A n (AuB').

The third line applies the distributive law again, this time distributing B' over (BuA').

The fourth line uses the fact that A n A' = Ø (empty set) and B n B' = Ø.

The fifth line simplifies the expression by removing the empty sets.

The sixth line uses the fact that (A u B)' = A' n B'.

The seventh line uses the fact that (A-B) = A - (A n B) and (B-A) = B - (A n B).

The final line simplifies the expression by removing the common term (AuB).