From c902719b0734b30c9e7a2ccfc7dd7b49cda34ccd Mon Sep 17 00:00:00 2001
From: tomit4 <mosssap@gmail.com>
Date: Tue, 21 Jul 2026 17:25:46 -0700
Subject: [PATCH] :construction: Omitted last few questions of 6.2

---
 chapter_6/exercises.md | 14 ++++++++++++++
 1 file changed, 14 insertions(+)

diff --git a/chapter_6/exercises.md b/chapter_6/exercises.md
index 45f20f2..cc254f0 100644
--- a/chapter_6/exercises.md
+++ b/chapter_6/exercises.md
@@ -2665,29 +2665,43 @@ Prove each statement in 39-44.
 
 a. $(A - B) \cup (B - A) \cup (A \cap B) = A \cup B$
 
+Omitted.
+
 b. The sets $(A - B)$, $(B - A)$, and $(A \cap B)$ are mutually disjoint.
 
+Omitted.
+
 40. For every positive integer $n$, if $A$ and $B_1, B_2, B_3, \dots$ are any
     sets, then
 
 $$ A \cap \left(\bigcup_{i = 1}^{n}B_i\right) = \bigcup_{i = 1}^{n}(A \cap B_i) $$
 
+Omitted.
+
 41. For every positive integer $n$, if $A_1, A_2, A_3, \dots$ and $B$ are any
     sets, then
 
 $$ \bigcap_{i = 1}^{n}(A_i - B) = \left(\bigcup_{i = 1}^{n}A_i\right) - B $$
 
+Omitted.
+
 42. For every positive integer $n$, if $A_1, A_2, A_3, \dots$ and $B$ are any
     sets, then
 
 $$ \bigcap_{i = 1}^{n}(A_i - B) = \left(\bigcap_{i = 1}^{n}A_i\right) - B $$
 
+Omitted.
+
 43. For every positive integer $n$, if $A$ and $B_1, B_2, B_3, \dots$ are any
     sets, then
 
 $$ \bigcup_{i = 1}^{n}(A \times B_i) = A \times \left(\bigcup_{i = 1}^{n}B_i\right) $$
 
+Omitted.
+
 44. For every positive integer $n$, if $A$ and $B_1, B_2, B_3, \dots$ are any
     sets, then
 
 $$ \bigcap_{i = 1}^{n}(A \times B_i) = A \times \left(\bigcap_{i = 1}^{n}B_i\right) $$
+
+Omitted.