Why does + represent OR in boolean algebra for circuit design?

Question

I'm currently learning the basics of Boolean algebra in my Digital Circuits course. One thing that I didn't get much clarity on is why we use the operator + to represent OR rather than AND. Literally, if I have 2 cookies in one hand AND 3 cookies in the other, I have 5 cookies. Further, summation is compared to ORing and multiplication to ANDing in discussions on Product of Sums and Sum of Products forms of Boolean expression. Is there a mathematical reason that ORing is equivalent to summation of terms and ANDing to multiplication, or is it simply convention?

Answer 1

Hint

How many are the cookies that are in your right hand or in your left hand ?

Answer 2

Because False is the $0$ in a boolean system.
True is represented as $1$ and False is represented as $0$.
True OR False is True $\iff 1+0=1$
True AND False is False $\iff 1\times 0=0$

Answer 3

With the correspondence $\text{true}\leftrightarrow1,\text{false}\leftrightarrow0$, there is a lot of similarity between the boolean operations and usual arithmetic. Namely

$$a\land b\leftrightarrow a\cdot b$$ and $$a\lor b\leftrightarrow a+ b$$ with the exception that $1+1=1$.

With these definitions, we have the ordinary properties

$$a\cdot b=b\cdot a,\\a+b=b+a,\\0\cdot a=a\cdot 0=0,\\1\cdot a=a\cdot 1=a,\\a+0=0+a=a,\\a\cdot(b+c)=a\cdot b+a\cdot c$$

and a few new ones

$$a+1=1+a=1,\\a+b\cdot c=(a+b)\cdot(a+c),\\a\cdot a=a,\\a+a=a$$

which have their logical equivalents.

In this notation we keep the precedence of $\cdot$ over $+$, which introduces a little of asymmetry. Though not completely equivalent to usual arithmetic, this notation is very compact and is compatible with our habits of expression manipulation, which is quite convenient.

Answer 4

It is just a convention that follows from a choice to use $0$ to represent false and $1$ to represent true (so that multiplication corresponds exactly to AND and addition corresponds to OR, at least when the operands are not both true). That choice goes back to Boole himself, who was very likely trying to give an account of the algebra of sets (rather than propositions) that looked like the algebra of numbers. See this nice artucke by Stanley Burris for an interesting discussion of Boole's thinking.

The conventional choice of $0$ for false and $1$ for true is intuitive if you think of a proposition as denoting the set of situations in which the proposition holds. The convention can be counter-intuitive if you think about the propositions as assertions: $A \le B$ means $A$ is logically stronger than $B$. However, a choice had to be made and $0$ for false and $1$ for true is the one that is generally adopted (although not in recursion theory).