Boolean Algebra

Any equation must be within the <math> </math> tags. For Boolean alegbra the main issue is how to negate a term like:

[math] \overline{a}[/math] or [math] \overline{\overline{a}+b}[/math]

this can be done by adding the following around any term you wish to negate.:

<math> \overline{} </math>  

[math] \overline{a}[/math]

is

 <math> \overline{a} </math>

[math] \overline{\overline{a}+b}[/math]

is

 <math> \overline{\overline{a}+b} </math>.

Identities

AND Identities

[math] A.1 = A [/math]

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

[math] 0.A = 0 [/math]

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

[math] A.A = A[/math]

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

[math] A.\overline{A}=0 [/math]

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.

OR Identities

[math] 0+A = A [/math]

0 or A can be simplified as just A.

[math] 1+A = 1 [/math]

1 or A can be simplified as just A.

[math] A+A=A[/math]

A or A can be simplified as just A.

[math] \overline{A}+A=1[/math]

NOT A or A can be simplified as just 1.

Laws

Commutative Law

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

[math] A+B = B+A [/math]

or

[math] A.B = B.A [/math]

Associate Law

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

[math] A+(B+C) = B + (A+C) [/math]

[math] A+(B+C) = B + (A+C) [/math]

[math] A+(B+C) = C + (A+B) [/math]

So:

[math] A.(B.C) = B . (A.C) [/math]

[math] A.(B.C) = B . (A.C) [/math]

[math] A.(B.C) = C . (A.B) [/math]

Distributive Law

The distributive law is these two equations.

[math] A.(B+C) = A.B + A.C [/math]

[math] A+(B.C) = (A+B).(A+C) [/math]

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

[math] A.B + A.C = A.(B+C)[/math]

[math] A+B.A+C = A+(B.C) [/math]

Redundancy Law

Law 1 : [math] A + A \overline{B} = A + B [/math]

Proof :

[math]= A + A \overline{B} \\ = (A + \overline{A})(A + B) \\ = 1 . (A + B) \\ = A + B [/math]

Law 2: [math] A.(\overline{A} + B) = A.B[/math]

Proof :

[math]= A.(\overline{A} + B) \\ = A.\overline{A} + A.B \\ = 0 + A.B \\ = A.B [/math]

Identity Law

This is also in the identities section:

[math] A.A = A [/math]

[math] A+A = A [/math]

Negation Law

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

[math] \overline{ \overline{A} } = A [/math]

Equations

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

Example 2

[math] (\overline{A} + B) . (\overline{A} + C) [/math] [math](\overline{A} + A) = 1[/math] [math]= 1.(\overline{A} + C)[/math]

Example 3

[math](X + Y) . (X + \overline{Y})[/math]
Distributive: [math]X . (Y + \overline{Y})[/math]
Identity laws: [math]Y + \overline{Y} = 1 , X.1 = X[/math]
Therefore, the answer is X

Example 4

Expression Rule(s) Used C + BC Original Expression C + (B + C) DeMorgan's Law. (C + C) + B Commutative, Associative Laws. T + B Complement Law. T Identity Law.

Example 6

__ _ _

AB(A + B)(B + B) Original Expression

__ _

AB(A + B) Complement law, Identity law.

_   _  _

(A + B)(A + B) DeMorgan's Law

_ _

A + B.B Distributive law. This step uses the fact that or distributes over and.

_

A Complement, Identity.

Example 7

SIMPLIFY (A + C)A + AC + C

(A + C)A + AC + C Complement, Identity.

A((A + C) + C) + C Commutative, Distributive.

A(A + C) + C Associative, Idempotent.

AA + AC + C Distributive.

A + (A + T)C Idempotent, Identity, Distributive.

A + C Identity, twice.

Example 8

[math] Simplify (X+Y).(X+\overline{Y}) X.X + X.\overline{Y} + Y.X + Y.\overline{y} [/math]