---
tags: logic,
---

# The Algebra of Boole

In the following "Boole's book" refers to George Boole, [The Mathematical Analysis of Logic](https://www.gutenberg.org/files/36884/36884-pdf.pdf), 1847.

I don't follow Boole's book in detail, but it seems to me that the following is fairly true to the original while at the same time written using modern mathematics.

**Operations:** Two constants $0$ and $1$, one unary operation $-$, two binary operations $+$, $\cdot$.

**Equations:** $(0,1,+,\cdot)$ satisfy the equations of a [ring](https://en.wikipedia.org/wiki/Ring_(mathematics)#Definition). In more detail, $(0,+)$ is an [abelian group](https://en.wikipedia.org/wiki/Abelian_group) with inverse $-$, while $(1,\cdot)$ satisfy the equations of a [monoid](https://en.wikipedia.org/wiki/Monoid) and $\cdot$ distributes over $+$. [^commutative] Moreover, the ring is a [Boolean ring](https://en.wikipedia.org/wiki/Boolean_ring), that is, multiplication is idempotent, or in symbolic notation $x\cdot x = x$ for all $x$. [^boole]

**Remark on Notation:** Negation $1-x$ is introduced on page 20 of Boole's book. It is not immediately clear whether Boole thinks of this as a unary operation or a binary operation. In modern algebra, we take $-$ to be a unary operation with the property that $x+(-x)=0$. This is part of $(0,+)$ being an abelian group. $x-y$ is treated as an abbreviation of $x+(-y)$.

An interesting exercise is to show that in a Boolean ring we have $1+1=0$, which is radically different from what we are used to with numbers. But before proving this it may be good to get some of the basic properties out of the way.

**Exercise:** The following is true in all rings:
- Neutral elements are unique:
- $a+x=x$ for all $x$ implies $a=0$
- $a\cdot x=x$ for all $x$ implies $a=1$.
- Additive inverses are unique: $x+y=0$ and $x+z=0$ implies $y=z$.
- $-(-x)=x$.
- $-0=0$.
- $0\cdot x = 0$.
- $(-1)\cdot x = -x$.
- If $0=1$, then the ring has only one element.

**Exercise:** The following is true in all Boolean rings:
- $1+1=0$.
- $x=-x$.
- The ring is commutative, that is, $x\cdot y=y\cdot x$. [^hint1]

The last one is (slightly) more difficult. In fact, Boole had this as an axiom, but as it turns out, it already follows from the other axioms.

It is clear from reading Boole's book that he was thinking in terms of Venn diagrams (page 19 and following):

**Exercise:** Explain the following correspondences of Boole (some are verbatim from his book, others I changed a little). The left-hand column consists of the building blocks of Aristotle's syllogisms.

|Aristotle|Boole|
|:---:|:---:|
| All Xs are Ys | $x=xy$ or also $x(1-y)=0$
| No Xs are Ys | $xy=0$
| Some Xs are Ys | $xy\not= 0$
| Some Xs are not Ys | $x(1-y)\not 0$

**Exercise:** Show that there is only one Boolean ring with 2 elements.

**Exercise:** Show that there is no Boolean ring with 3 elements.

**Exercise:** How many Boolean rings are there with 4 elements?

**Exercise:** For which $n$ is there a Boolean ring with $n$ elements?

## Further Reading

A starting point for [the historical development](https://www.maa.org/press/periodicals/convergence/origins-of-boolean-algebra-in-the-logic-of-classes-george-boole-john-venn-and-c-s-peirce).

[^commutative]: Boolean rings are always commutative, see the exercise below.

[^boole]: Here is a snippet from Boole's book (page 16):

are sufficient for the basis of a Calculus.

[^hint1]: This is not straightforward. The reason is that there are idempotent monoids that are not commutative, see eg [here](https://www.cmi.ac.in/~kumar/words/lecture01a.pdf). Hence the proof needs to make use of addition of the ring.