Difference between revisions of "Boolean Algebra"

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(Example 11)
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  <nowiki> <math> \overline{\overline{a}+b} </math></nowiki>.
 
  <nowiki> <math> \overline{\overline{a}+b} </math></nowiki>.
  
=Identities=
+
=Boolean Identities=
==AND Identities==
+
===Using AND==
  
 
<math> A.1 = A </math>
 
<math> A.1 = A </math>
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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.
 
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==
+
===Using OR===
 
<math> 0+A = A </math>
 
<math> 0+A = A </math>
  
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NOT A or A can be simplified as just 1.
 
NOT A or A can be simplified as just 1.
  
=Laws=
+
=Boolean Laws=
 
==Commutative Law==
 
==Commutative Law==
 
The Commutative Law is where equations are the same no matter what way around the letters are written. For example
 
The Commutative Law is where equations are the same no matter what way around the letters are written. For example
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<math> \overline{ \overline{A} } = A </math>
 
<math> \overline{ \overline{A} } = A </math>
  
=Equations=
+
=Boolean Equations=
 
Solving equations is a matter of applying the laws of boolean algrebra, followed by any of the identities you can find:
 
Solving equations is a matter of applying the laws of boolean algrebra, followed by any of the identities you can find:
  

Revision as of 13:44, 29 June 2018

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>.

Boolean Identities

=Using AND

[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.

Using OR

[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 1.

[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.

Boolean 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 + \overline{A} B = A + B [/math]

Proof :

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

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

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

[math] \overline{\overline{A}} = A [/math] Use Negation Law

[math] \overline{(B+C)} = (\overline{B} . \overline{C}) [/math] Use De Morgan's Law

[math] A . (\overline{B} . \overline{C}) = A . \overline{B} . \overline{C} [/math] Use Associate Law

Example 2

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

OR Identity

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

AND Identity

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

Simplify

[math]= (\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[/math]

[math] X.1 = X[/math]

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 5

Simplify: AB(A + B)(B + B):
Expression Rule(s) Used
AB(A + B)(B + B) Original Expression
AB(A + B) Complement law, Identity law.
(A + B)(A + B) DeMorgan's Law
A + BB Distributive law. This step uses the fact that or distributes over and. It can look a bit strange since addition does not distribute over multiplication
A Complement, Identity

Example 6

[math] \overline{A.B}(\overline{A} + B)(\overline{B} + B) [/math] Original Expression

[math] \overline{AB}(\overline{A} + B) [/math] Complement law, Identity law.

[math] (\overline{A} + \overline{B})(\overline{A} + B) [/math] DeMorgan's Law

[math] \overline{A} + \overline{B}.B [/math] Distributive law. This step uses the fact that or distributes over and.

[math] \overline{A} [/math] Complement, Identity.

Example 7

SIMPLIFY [math](A + C)A + AC + C[/math]


[math](A + C)A + AC + C[/math] Complement, Identity.


[math]A((A + C) + C) + C[/math] Commutative, Distributive.


[math]A(A + C) + C[/math] Associative, Idempotent.


[math]AA + AC + C[/math] Distributive.


[math]A + (A + T)C[/math] Idempotent, Identity, Distributive.


[math]A + C[/math] Identity, twice.

Example 8

Simplify

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

solution

[math] X.X + X.\overline{Y} + Y.X + Y.\overline{Y} [/math] Expanding the brackets

[math] X + X.\overline{Y} + Y.X + 0 [/math] Use of [math] X.X = X [/math] and [math] Y.\overline{Y} = 0 [/math]

[math] X + X(\overline{Y}+Y) [/math] Taking X out of the brackets

[math] X + X(1) [/math] Use of [math] Y + \overline{Y} = 1 [/math]

[math] X(1) [/math]

[math] X [/math]

Example 9

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

[math] X + (Y.\overline{Y}) [/math] after distributive law applied

[math] (Y.\overline{Y}) = 0 [/math]

[math] X + 0 [/math]

[math] X [/math]

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

Example 10

Simplify:

[math] \overline {AB} (\overline {A}+B)(\overline {B}+B) [/math]

Complement law, Identity law

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

DeMorgan's law

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

Distributive law

[math] \overline {A}+\overline {B}B [/math]

Complement, Identity

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

Example 11

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

Example 12

Example 13

Simplify

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

First, using De Morgan’s Law simplify it to

[math] \overline{\overline{A} + \overline{B} + \overline{A}} [/math]

This can then be simplified to

[math] \overline{\overline{A} + \overline{B}} [/math]

Then using De Morgan’s Law again, you get

[math] \overline{\overline{A}} + \overline{\overline{B}} [/math]

Then there are double negatives, so you end up with

[math] A + B [/math]

Example 14

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

[math] (X.\overline{X})+(X.Y) [/math] using the Distributive Law

[math] 0 + (X.Y) [/math] using the identity [math]A.\overline{A}=0[/math]

[math] X.Y [/math] using the identity [math]A+0=A[/math]

Fully simplified the equation

Example 15

Simplify the following: [math] \overline{ \overline{(A.A)}. \overline{(B.B)}} [/math]

Using De Morgan's law we can put the equation into another form:

1. Change All operations,

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

2. Negate letters:

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

3. Negate the whole expression:

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

Using the identity law, we can then simplify this expression to:

[math] A+B [/math]

Example 16

For this example the Boolean Algebra itself is shown above while the description/the rules that have been used are listed below it:

[math]\overline{A}(A + B) + (B + AA)(A + \overline{B})[/math]

Original expression

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

Idempotent (AA to A), then Distributive, used twice.

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

Complement, then Identity. (Strictly speaking, we also used the Commutative Law for each of these applications.)

[math]\overline{A}B + BA + AA + B\overline{B} + A\overline{B}[/math]

Distributive, two places.

[math]\overline{A}B + BA + A + A\overline{B}[/math]

Idempotent (for the A's), then Complement and Identity to remove BB.

[math]\overline{A}B + AB + AT + A\overline{B}[/math]

Commutative, Identity; setting up for the next step.

[math]\overline{A}B + A(B + T + \overline{B})[/math]

Distributive.

[math]\overline{A}B + A[/math]

Identity, twice (depending how you count it).

[math]A + \overline{A}B[/math]

Commutative.

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

Distributive.

[math]A + B[/math]

Complement, Identity.

Example 16