From 596b888c75b6bc2750bf4b6b67c3360bb85234e0 Mon Sep 17 00:00:00 2001
From: tomit4 <mosssap@gmail.com>
Date: Fri, 24 Jul 2026 15:39:35 -0700
Subject: [PATCH] :construction: Mid of 6.4

---
 chapter_6/exercises.md     | 125 +++++++++++++++++++++++++++++++++++++
 chapter_6/notes.md         |   2 +-
 chapter_6/test_yourself.md |   6 ++
 3 files changed, 132 insertions(+), 1 deletion(-)

diff --git a/chapter_6/exercises.md b/chapter_6/exercises.md
index e182cc5..7e12add 100644
--- a/chapter_6/exercises.md
+++ b/chapter_6/exercises.md
@@ -4221,6 +4221,16 @@ $$ = a \cdot a $$
 
 __ (e) __
 
+a. by the identity law for $\cdot$
+
+b. by the complement law for $+$
+
+c. by the distributive law for $+$ over $\cdot$
+
+d. by the complement law for $\cdot$
+
+e. by the identity law for $+$
+
 2. _Universal bound law for $+$:_ For every $a$ in $B$, $a + 1 = 1$.
 
 **Proof:**
@@ -4243,6 +4253,12 @@ $$ = 1 $$
 
 __ \(c\) __
 
+a. by the complement law for $+$
+
+b. by the associative law for $+$
+
+c. by the complement law for $+$
+
 3. _Absorption law for $\cdot$ over $+$:_ For all $a$ and $b$ in $B$,
    $(a + b) \cdot a = a$.
 
@@ -4280,24 +4296,133 @@ $$  = a $$
 
 __ (f) __
 
+a. by the commutative law for $\cdot$
+
+b. by the distributive law of $\cdot$ over $+$
+
+c. because $1$ is an identity for $\cdot$
+
+d. by the distributive law of $\cdot$ over $+$
+
+e. by the commutative law for $+$
+
+f. because $1$ is an identity for $\cdot$
+
 In 4-10 assume that $B$ is a Boolean algebra with operations $+$ and $\cdot$.
 Prove each statement using only the axioms for a Boolean algebra and statements
 proved in the text or in lower-numbered exercises.
 
 4. _Universal bound for $0$:_ For every $a$ in $B$, $a \cdot 0 = 0$.
 
+**Proof:**
+
+$$ a \cdot 0 = a \cdot (a \cdot \overline{a}) $$
+
+by the complement law for $\cdot$
+
+$$ = (a \cdot a) \cdot \overline{a} $$
+
+by the associative law for $\cdot$
+
+$$ = a \cdot \overline{a} $$
+
+by exercise 1
+
+$$ = 0 $$
+
+by the complement law for $\cdot$
+
 5. _Complements of $0$ and $1$:_
 
 a. $\overline{0} = 1$
 
+**Proof:**
+
+$$ 0 = 0 \cdot 1 $$
+
+because $1$ is an identity for $\cdot$, and
+
+$$ 0 + 1 = 1 + 0 $$
+
+because $+$ is commutative and $0$ is an identity for $+$.
+
+Since $0 = 0 \cdot 1$ and $0 + 1 = 1 + 0$, $1 = \overline{0}$ by the uniqueness
+of the complement laws.
+
 b. $\overline{1} = 0$
 
+$$ 1 = 1 + 0 $$
+
+$$ 1 = 1 + \overline{1} $$
+
+by the complement law for $+$
+
+$$ 0 = \overline{1} $$
+
+by the uniquness of $0$ law.
+
 6. _Uniqueness of $0$:_ There is only one element of $B$ that is an identity for
    $+$.
 
+**Proof:**
+
+Suppose $0$ and $0'$ are elements of $B$ both of which are identities for $+$.
+Then both $0$ and $0'$ satisfy the identity, complement, and universal bound
+laws.
+
+_[We will show that $0 = 0'$.]_
+
+By the identity law for $+$, for every $a \in B$,
+
+$$ a + 0 = a(*) \quad \text{ and } \quad a + 0' = a(**) $$
+
+It follows that
+
+$$ 0' = 0' + 0 $$
+
+by (*) with $a = 0'$
+
+$$ = 0 + 0' $$
+
+by the commutative law for $+$
+
+$$ = 0 $$
+
+by (**) with $a = 0$.
+
+_[This is what was to be shown.]_
+
 7. _Uniqueness of $1$:_ There is only one element of $B$ that 8s an identity for
    $\cdot$.
 
+**Proof:**
+
+Suppose $1$ and $1'$ are elements of $B$ both of which are identities for
+$\cdot$. Then both $1$ and $1'$ satisfy the identity, complement, and universal
+bound laws.
+
+_[We will show that $1 = 1'$.]_
+
+By the identity law for $\cdot$, for every $a \in B$,
+
+$$ a \cdot 1 = a(*) \quad \text{ and } \quad a \cdot 1' = a(**) $$
+
+It follows that
+
+$$ 1' = 1' \cdot 1 $$
+
+by (*) with $a = 1'$
+
+$$ = 1 \cdot 1' $$
+
+by the commutative law for $\cdot$
+
+$$ = 1 $$
+
+by (**) with $a = 1$.
+
+_[This is what was to be shown.]_
+
 8. _De Morgan's law for $\cdot$:_ For all $a$ and $b$ in $B$,
    $\overline{a \cdot b} = \overline{a} + \overline{b}$. (_Hint:_ Prove that
    $(a \cdot b) + (\overline{a} + \overline{b}) = 1$ and that
diff --git a/chapter_6/notes.md b/chapter_6/notes.md
index 2b11fce..bcc1d57 100644
--- a/chapter_6/notes.md
+++ b/chapter_6/notes.md
@@ -558,7 +558,7 @@ $$ \text{(a) } a + 1 = 1 \quad \text{ and } \quad \text{(b) } a \cdot 0 = 0 $$
 
 6. _De Morgan's Laws:_ For all $a$ and $b \in B$,
 
-$$ \text{(a) } \overline{a + b} = \oveline{a} \cdot \overline{b} \quad \text{ and } \quad \text{(b) } \overline{a \cdot b} = \overline{a} + \overline{b}  $$
+$$ \text{(a) } \overline{a + b} = \overline{a} \cdot \overline{b} \quad \text{ and } \quad \text{(b) } \overline{a \cdot b} = \overline{a} + \overline{b}  $$
 
 7. _Absorption Laws:_ For all $a$ and $b \in B$,
 
diff --git a/chapter_6/test_yourself.md b/chapter_6/test_yourself.md
index 5a8bf1f..19225ba 100644
--- a/chapter_6/test_yourself.md
+++ b/chapter_6/test_yourself.md
@@ -127,10 +127,16 @@ Page 445
    $\mathbf{t}$ corresponds to _____, a contradiction $\mathbf{c}$ corresponds
    to _____, and the negation operation, denoted $\neg$, corresponds to _____.
 
+$\cup$;$\cap$,$U$,$\emptyset$,$^c$
+
 2. The operations of $+$ and $\cdot$ in a Boolean algebra are generalizations of
    the operations of _____ and _____ in the set of all statement forms in a
    given finite number of variables and the operations of _____ and _____ in the
    set of all subsets of a given set.
 
+$\vee$;$\wedge$;$\cup$;$\cap$
+
 3. Russell showed that the following proposed "set definition" could not
    actually define a set: _____.
+
+the set of all sets that are not elements of themselves