Boolean quotient algebra
Quotient Algebras via Congruences
Let $A$ be a Boolean algebra^{}. A congruence^{} on $A$ is an equivalence relation^{} $Q$ on $A$ such that $Q$ respects the Boolean operations:

•
if $aQb$ and $cQd$, then $(a\vee c)Q(b\vee d)$

•
if $aQb$, then ${a}^{\prime}Q{b}^{\prime}$
By de Morgan’s laws, we also have $aQb$ and $cQd$ implying $(a\wedge c)Q(b\wedge d)$.
When $a$ is congruent to $b$, we usually write $a\equiv b\phantom{\rule{veryverythickmathspace}{0ex}}(modQ)$.
Let $B$ be the set of congruence classes: $B=A/Q$, and write $[a]Q$, or simply $[a]$ for the congruence class containing the element $a\in A$. Define on $B$ the following operations^{}:

•
$[a]\vee [b]:=[a\vee b]$

•
${[a]}^{\prime}:=[{a}^{\prime}]$
Because $Q$ respects join and complementation, it is clear that these are welldefined operations on $B$. Furthermore, we may define $[a]\wedge [b]:={({[a]}^{\prime}\vee {[b]}^{\prime})}^{\prime}={([{a}^{\prime}]\vee [{b}^{\prime}])}^{\prime}={[{a}^{\prime}\vee {b}^{\prime}]}^{\prime}=[{({a}^{\prime}\vee {b}^{\prime})}^{\prime}]=[a\wedge b]$. It is also easy to see that $[1]$ and $[0]$ are the top and bottom elements of $B$. Finally, it is straightforward to verify that $B$ is a Boolean algebra. The algebra $B$ is called the Boolean quotient algebra of $A$ via the congruence $Q$.
Quotient Algebras via Ideals and Filters
It is also possible to define quotient algebras via Boolean ideals and Boolean filters. Let $A$ be a Boolean algebra and $I$ an ideal of $A$. Define binary relation^{} $\sim $ on $A$ as follows:
$$a\sim b\mathit{\hspace{1em}\hspace{1em}}\text{if and only if}\mathit{\hspace{1em}\hspace{1em}}a\mathrm{\Delta}b\in I,$$ 
where $\mathrm{\Delta}$ is the symmetric difference operator on $A$. Then

1.
$\sim $ is an equivalence on $A$, because

–
$a\mathrm{\Delta}a=0\in I$, so $\sim $ is reflexive^{}

–
$b\mathrm{\Delta}a=a\mathrm{\Delta}b$, so $\sim $ is symmetric^{}, and

–
if $a\sim b$ and $b\sim c$, then $a\sim c$; to see this, note that $(ab)\vee (bc)=((ab)\vee b)\wedge ((ab)\vee {c}^{\prime})=(a\vee b)\wedge ((ab)\vee {c}^{\prime})$. Since the LHS (and hence the RHS) is in $I$, and that $a\le a\vee b$ and ${c}^{\prime}\le (ab)\vee {c}^{\prime}$, RHS $\ge a\wedge {c}^{\prime}=ac\in I$ too. Similarly $ca\in I$ so that $a\sim c$.

–

2.
$\sim $ respects $\vee $ and ${}^{\prime}$, because

–
if $a\sim b$ and $c\sim d$, then $(a\vee c)(b\vee d)=(a\vee c)\wedge {(b\vee d)}^{\prime}=(a\vee c)\wedge ({b}^{\prime}\wedge {d}^{\prime})=(a\wedge ({b}^{\prime}\wedge {d}^{\prime}))\vee (c\wedge ({b}^{\prime}\wedge {d}^{\prime}))\le (a\wedge {b}^{\prime})\vee (c\wedge {d}^{\prime})\in I$, so that $(a\vee c)(b\vee d)\in I$ as well. That $(b\vee d)(a\vee c)\in I$ is proved similarly. Hence $(a\vee c)\sim (b\vee d)$.

–
${a}^{\prime}\mathrm{\Delta}{b}^{\prime}=a\mathrm{\Delta}b$, so $\sim $ preserves ${}^{\prime}$.

–
Thus, $\sim $ is a congruence on $A$. The quotient algebra $A/\sim $ is called the quotient algebra of $A$ via the ideal $I$, and is often denoted by $A/I$.
From this congruence $\sim $, one can recapture the ideal: $I=[0]$.
Dually, one can obtain a quotient algebra from a Boolean filter. Specifically, if $F$ is a filter of a Boolean algebra $A$, define $\sim $ on $A$ as follows:
$$a\sim b\mathit{\hspace{1em}\hspace{1em}}\text{if and only if}\mathit{\hspace{1em}\hspace{1em}}a\leftrightarrow b\in F,$$ 
where $\leftrightarrow $ is the biconditional^{} operator on $A$. Then it is easy to show that $\sim $ too is a congruence on $A$, so that one forms the quotient algebra of $A$ via the filter $F$, denoted by $A/F$. Of course, an easier approach to this is to realize that $F$ is a filter of $A$ iff ${F}^{\prime}:=\{{a}^{\prime}\mid a\in F\}$ is an ideal of $A$, and the process of forming $A/{F}^{\prime}$ turns out to be identical to $A/F$.
From $\sim $, the filter $F$ can be recovered: $F=[1]$.
In fact, given a congruence $Q$, the congruence class $[0]Q$ is a Boolean ideal and the congruence class $[1]Q$ is a Boolean filter, and that the quotient algebras derived from $Q,[0]Q$ and $[1]Q$ are all the same:
$$A/Q=A/[0]Q=A/[1]Q.$$ 
Title  Boolean quotient algebra 

Canonical name  BooleanQuotientAlgebra 
Date of creation  20130322 17:59:09 
Last modified on  20130322 17:59:09 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  8 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 06E05 
Classification  msc 03G05 
Classification  msc 06B20 
Classification  msc 03G10 