◀ ▲ ▶Branches / Logic / Definition: Boolean Algebra

George Bool, a poor shoemaker's son, never attended secondary school, college, or university. He was a gifted autodidact and at the age of 34, in 1849, he was appointed Professor of Mathematics at Queen's College, Cork, where he published his book entitled An Investigation of the Laws of Thought. In this book, he introduced a new type of algebra, later known as Boolean Algebra, which is a kind of ordinary numerical algebra studying the rules of equations involving variables such as $a,b,x,y,...$, in which the values of all variables are limited to $0$ and $1$. The Boolean Algebra was dormant until 1939, when Shannon discovered that it was appropriate language to describe switching circuits, laying the foundations for modern computers.

We will now define Boolean algebra formally and show how it can be used to study propositional logic.

Definition: Boolean Algebra

Any set \((B,\sqcap,\sqcup,1,0)\) with the operations

• product $"\sqcap":B\times B\to B$, $(x,y)\to x\sqcap y$ for all $x,y\in B$,
• sum $"\sqcup":B\times B\to B$, $(x,y)\to x\sqcup y$ for all $x,y\in B$,
• and the designated elements $1,0\in B$

is called a Boolean algebra if and only if:

1. $\sqcap$ and $\sqcup$ are associative in \(B\), i.e. for all \(x,y,z\in B\) $$\begin{array}{c}x\sqcap(y\sqcap z)=(x\sqcap y)\sqcap z,\\x\sqcup (y\sqcup z)=(x\sqcup y)\sqcup z,\end{array}$$.
2. $\sqcap$ and $\sqcup$ are commutative in \(B\), i.e. for all \(x,y \in B\) $$\begin{array}{c}x\sqcap y=y \sqcap x,\\x\sqcup y=y\sqcup x,\end{array}$$.
3. "\(\sqcap\)" and "\(\sqcup\)" are distributive over each other in \(B\), i.e. for all \(x,y,z\in B\) $$\begin{array}{c}x\sqcap(y\sqcup z)=(x\sqcap y)\sqcup (x\sqcap z),\\x\sqcup (y\sqcap z)=(x\sqcup y)\sqcap (x\sqcup z),\end{array}$$.
4. There is an element $0\in B$ with $0\sqcap x=0$ and $x\sqcup 0=x$ for all $x\in B$; $0$ is called the smallest element of $B,$.
5. There is an element $1\in B$ with $1\sqcap x=x$ and $x\sqcup 1=1$ for all $x\in B$; $1$ is called the greatest element of $B,$.
6. There is an element $y\in B$ with $x\sqcap y=0$ and $x\sqcup y=1$ for all $x\in B$; $y$ is called the complement element of $B.$.

Table of Contents

1. Lemma: Boolean Algebra of Propositional Logic

Mentioned in:

Lemmas: 1
Proofs: 2 3 4 5 6

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References

Bibliography

1. Govers, Timothy: "The Princeton Companion to Mathematics", Princeton University Press, 2008,