29. For all sets A ,B and C , (A-c) n (B-C) N(A-B) = Ø Proof (by contradicton) Let A,B, and c be any sets. Suppose not . That is Suppose E (A-c) n(B-C) n (A -B) + Ø . Then is an elenment x in there...

Can you please check my prove and please let me know if i did it correctly in the answer. Thank you! Directions: Use element argument to prove number 29. Assume that all sets are subsets of a universal set of U.

Extracted text: 29. For all sets A ,B and C , (A-c) n (B-C) N(A-B) = Ø Proof (by contradicton) Let A,B, and c be any sets. Suppose not . That is Suppose E (A-c) n(B-C) n (A -B) + Ø . Then is an elenment x in there (A-C) n (B-C) n (A-B).. In particular Z t'LB-C) nCA-B) by definiton of intersectun. X E (B-c) -> 'atB and x¢C and x¢ B X E (A-B) Ž XE. A 1 E B. and x¢ B Thus , XE (B-C) n(A-B) > in partiaular (B-C) n(A-B)= Ø Therefore (A-C) n d = d because intersectuon of an empty set an empty set is There fore the assumption (x). Buesn sI SO) (A-c)n (B-C) (A-B)=Ø