Boolean Algebra

Boolean Algebra Precedence

the order of precedence for boolean algebra is:

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

Boolean Identities

TRC Video

https://www.youtube.com/watch?v=ym73-rgnrOQ

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

TRC Video

https://www.youtube.com/watch?v=Cdqj4XDsUVY

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 remove the common factor:

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

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

You can also remove the common factor if you only have 1 term on one side:

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

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

if the symbol inside the brackets is a '+' you can add '+0' or if the symbol inside the brackets is '.' you can add '.1'. Doing this will not change the nature of the brackets because 'A' is the same as 'A+0' and is the same as 'A.1'.

Redundancy Law

Law 1 :

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

Proof :

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

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Law 2:

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

Proof :

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

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Law 3:

$A.(A + B) = A$

Proof using distributive law:

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

So: $A + (0 . B)$

So: $A + 0 = A$

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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:

TRC Video

https://www.youtube.com/watch?v=N1r1D__NMGg

Example 1

$𝐶+(𝐶.𝐷)$

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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 = $C$

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Example 2

A.(C+A)

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|Use Distributive Law|

->(A.C)+(A.A)

|Use Identity| A.A=A

->(A.C)+A

|This is the same as writing (could straight apply redundancy rule here)|

->(A.C)+(A.1)

|Take out the common factor|

->A.(C+1)

|Use Identity| C+1 = 1

->A

Example 3

$𝐵.(𝐴+\overline{𝐵})$

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B.(A+ NOT B) REDUNDANCY (A + NOT B) REDUNDANCY

ANSWER = NOT B

Example 4

$𝑋.(\overline{𝑋}+𝑌)$

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$𝑋.\overline{𝑋} = 0$

$0+𝑌 = 𝑌$

Example 5

$𝑋.(X+\overline{Y})$

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$(0+𝑋).(X+\overline{Y})$

$𝑋+(0.\overline{Y})$

$𝑋+(0)$

$𝑋$

Example 8

$𝐷.𝐸+𝐸.\overline{𝐷}$

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D.E+E.D Distributivetive Law D.(E+D) Redundancy Law D

Example 13

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

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Expand the brackets: $(\overline {A} . B) + (\overline{B} . B)$

Not B AND B = 0: $\overline {A}.B) + (0)$

Something OR 0 is Something: $\overline {A}.B$

Example 14

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

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(B) + (A.B) Distributive Law. (B + A) . (B + B) Not B cancels out. B + A . 1 = B+A

Example 19

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

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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 $Y + \overline{Y} = 1$

$X(1)$

$X$