Boolean Algebra

Boolean Algebra Precedence

the order of precedence for boolean algebra is:

1. Brackets
2. Not
3. And
4. Or

Boolean Identities

Using AND

$A.1 = A$

This equation means that the output is determined by the value of A. So if A =0, The output is 0, and vice versa.

$0.A = 0$

Because there is a 0 in this equation, the output of this will always be 0 regardless of the value of A.

$A.A = A$

The output is determined by A alone in this equation. This can be simplified to just "A".

$A.\overline{A}=0$

Here the output will be 0, regardless of A's value. A would have to be 1 and 0 for the output to be 1. This means we can simplify this to just 0.

Using OR

$0+A = A$

0 or A can be simplified as just A.

$1+A = 1$

1 or A can be simplified as just 1.

$A+A=A$

A or A can be simplified as just A.

$\overline{A}+A=1$

NOT A or A can be simplified as just 1.

Boolean Laws

Commutative Law

The Commutative Law is where equations are the same no matter what way around the letters are written. For example

$A+B = B+A$

or

$A.B = B.A$

Associate Law

If all of the symbols are the same it doesn't matter which order the equation is evaluated.

$A+(B+C) = B + (A+C)$

$A+(B+C) = B + (A+C)$

$A+(B+C) = C + (A+B)$

So:

$A.(B.C) = B . (A.C)$

$A.(B.C) = B . (A.C)$

$A.(B.C) = C . (A.B)$

Distributive Law

The distributive law is these two equations.

$A.(B+C) = A.B + A.C$

$A+(B.C) = (A+B).(A+C)$

This is essentially factorising or expanding the brackets, but you can also:

$A.B + A.C = A.(B+C)$

$A+B.A+C = A+(B.C)$

Redundancy Law

Law 1 :

$A + \overline{A} B = A + B$

Proof :

$= A + \overline{A} B \\ = (A + \overline{A})(A + B) \\ = 1 . (A + B) \\ = A + B$

Law 2:

$A.(\overline{A} + B) = A.B$

Proof :

$= A.(\overline{A} + B) \\ = A.\overline{A} + A.B \\ = 0 + A.B \\ = A.B$

Law 3:

$A.(A + B) = A$

Proof using distributive law:

$A.(A + B) = (0+A) . (A + B)$

So: $A + (0 . B)$

So: $A + 0 = A$

Law 4:

$A+(A . B) = A$

Proof using distributive law:

$A+(A . B) = (1 . A) + (A . B)$

So: $A . (1 + B)$

So: $A . 1 = A$

Identity Law

This is also in the identities section:

$A.A = A$

$A+A = A$

Negation Law

Just like in any other logic negating a negative is a positive so:

$\overline{ \overline{A} } = A$

Solving Boolean Equations

Solving equations is a matter of applying the laws of boolean algrebra, followed by any of the identities you can find:

Example 1

$𝐶+(𝐶.𝐷)$

---

Take out the common factor C:

$(C.D)+(C.1)=C.(D+1)$,

We know that $1+A=1$,

Therefore, $C.1$,

Use identity $A.1=A$,

Answer = $C$

Example 2

𝐴.(𝐶+𝐴) Use Distributive Law ->(A.C)+(A.A) Use Identity A.A=A ->(A.C)+A ->A

Example 3

$𝐵.(𝐴+\overline{𝐵})$

Example 4

$𝑋.(\overline{𝑋}+𝑌)$

Example 5

$𝑋.(X+\overline{Y})$

Example 6

$(𝐴.𝐴)+(𝐴.1)+(𝐵.\overline{𝐵})$

Example 7

$𝐴.(𝐶+𝐵+\overline{𝐴})$

Example 8

$𝐷.𝐸+𝐸.\overline{𝐷}$

Example 9

$𝐴.(𝐴+𝐴).(𝐵+\overline{𝐴})$

Example 10

$𝐴.𝐷.\overline{𝐵}+𝐴.𝐷.𝐵$

Example 11

$𝐶.𝐷.(𝐷.𝐵+𝐶)$

Example 12

$\overline{𝐵}.(𝐷+𝐵)$

Example 13

$(\overline {A}+\overline {B}).B$

Example 14

$\overline{B} + (A.B)$

Example 15

$X+(\overline{X}.\overline{Y})$

Example 16

$(\overline{A} . A) + (\overline{A} . C)$

= A

Because the centre symbol is an OR gate and at least one of the outputs of the and gates is A therefore by redundancy rule it is A.

Example 17

$(\overline{A} + A) . (\overline{A} + C)$

Example 18

$X.(\overline{X}+Y)$

Example 19

$(X + Y) . (X + \overline{Y})$
Distributive:
$X . (Y + \overline{Y})$
Identity laws:
$Y + \overline{Y} = 1$

$X.1 = X$

Alternative

$X.X + X.\overline{Y} + Y.X + Y.\overline{Y}$ Expanding the brackets

$X + X.\overline{Y} + Y.X + 0$ Use of $X.X = X$ and $Y.\overline{Y} = 0$

$X + X(\overline{Y}+Y)$ Taking X out of the brackets

$X + X(1)$ Use of $Y + \overline{Y} = 1$

$X(1)$

$X$