Boolean Algebra - A Form of Logic

Let's talk about logic. For Boolean algebra, it is logic in the form of algebra. True and false are represented by 1 and 0. The “or” and “and” correspond to addition and multiplication. In short, it takes logic and its operations and laws and transforms it to a more familiar language, algebra. I think Boolean algebra makes logic easier.

In this post, I will go over the symbols, properties, and laws of Boolean algebra. I will compare it to symbolic logic and regular algebra. I will briefly discuss how implication (IF-THEN) can be converted to Boolean algebra. Then I will show using Boolean algebra that four common arguments are valid arguments.

Symbolic Logic	Boolean Algebra
True False Statement x X’ X˄Y X˅Y X→Y	1 0 X X’ or (1-x) XY or X∙Y X+Y N/A

Properties and Laws of Boolean algebra

1.  Idempotent laws

a.       X+X=X

b.      XX=X

2. Bound laws

a.       X+1=1

b.      X0=0

3. Absorption laws

a.       X+XY=X

b.      X(X+Y)=X

4. Involution laws

a.       (X’)’=X

5. 0 and 1 laws

a.       0’=1

b.      1’=0

6. De Morgan's laws for Boolean algebra

a.       (X+Y)’=X’Y’

b.      (XY)’=X’+Y’

The laws and properties for addition and multiplication can be used as well including commutative, associative, and distributive properties.

1. Commutative property

a.       X+Y=Y+X

b.      XY=YX

2. Associative property

a.       (X+Y)+Z=X+(Y+Z)

b.      (XY)Z=X(YZ)

3. Distributive property

a.       X(Y+Z)=XY+XZ

However, unlike regular algebra, you can do addition over multiplication as well.

X+YZ=(X+Y)(X+Z)

Boolean algebra is the language of logic, so you have both the distribution of conjunction (and) and disjunction (or).

This also means that X implies Y can be converted to Boolean algebra. From logic, we know the tautology, implication to disjunction.

(X→Y)↔(X’˅Y)

So X→Y would be X’+Y or (1-X)+Y in Boolean algebra. This will be useful for the next section.

Four Common Valid Arguments

If p, then q P Thus, q

If p, then q Not q Thus, not p

If p, then q If q, then r Thus, if p, then q

P or Q Not P Thus, Q

Valid argument can found using the laws and properties of Boolean algebra to reach the conclusion. Valid arguments assumes the premises are true. This also can be used. Remember, an argument is valid when both the premises and the conclusion are true. With the assumption that the premises are true, if the conclusion is false or unknown, then the argument is invalid.

First argument

If p, then q p

(1-p)+q using implication to disjunction p=1 since assumption of true premises

(1-p)+q

(1-1)+q

0+q

Thus, q

An argument is a large conditional statement where the “if” is a line of premises connected by “AND”.

IF p1p2p3…,THEN conclusion.

Another way

((1-p)+q)p

p(1-p)+pq

0+pq

pq

1q since p=1

Thus, q

Second argument

If p, then q Not q

(1-p)+q using implication to disjunction 1-q=1 since assumption of true premises

1-q=1

q=0

1-p+q

1-p+0

1-p

Thus, not p.

Third argument

If p, then q If q, then r

(1-p)+q or p’+q using implication to disjunction (1-q)+r or q’+r using implication to disjunction

(p’+q)(q’+r)

p’q’+p’r+qq’+qr

p’q’+p’r+0+qr

p’q’+p’r+qr

p’q’+p’r+p’r+qr

p’(q’+r)+r(p’+q)

p’1+r1

Thus, p’+r

Fourth argument

P or Q Not P

p+q 1-p

1-p=1

p=0

p+q

0+q

Thus, q.

For more on logic, check out my youtube channel. If you have any questions or blog ideas, you can e-mail me at jdmathguy@gmail.com