Boolean Algebra Precedence
the order of precedence for boolean algebra is:
- Brackets
- Not
- And
- Or
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]
Law 3:
[math] A.(A + B) = A[/math]
Proof using distributive law:
[math]
A.(A + B) = (0+A) . (A + B)
[/math]
So:
[math]
A + (0 . B)
[/math]
So:
[math]
A + 0 = A
[/math]
Law 4:
[math] A+(A . B) = A[/math]
Proof using distributive law:
[math]
A+(A . B) = (1 . A) + (A . B)
[/math]
So:
[math]
A . (1 + B)
[/math]
So:
[math]
A . 1 = A
[/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]
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
[math]
πΆ+(πΆ.π·)
[/math]
Take out the common factor C:
[math](C.D)+(C.1)=C.(D+1)[/math],
We know that [math]1+A=1[/math],
Therefore, [math]C.1[/math],
Use identity [math]A.1=A[/math],
Answer = [math]C[/math]
Example 2
A.(C+A)
|Use Distributive Law|
->(A.C)+(A.A)
|Use Identity|
A.A=A
->(A.C)+A
->A
Example 3
[math]
π΅.(π΄+\overline{π΅})
[/math]
B.(A+ NOT B)
REDUNDANCY
(A + NOT B)
REDUNDANCY
ANSWER = NOT B
Example 4
[math]
π.(\overline{π}+π)
[/math]
[math]
π.\overline{π} = 0
[/math]
[math]
0+π = π
\lt \math\gt
===Example 5===
\lt math\gt
π.(X+\overline{Y})
[/math]
[math]
(0+π).(X+\overline{Y})
[/math]
[math]
π+(0.\overline{Y})
[/math]
[math]
π+(0)
[/math]
[math]
π
[/math]
Example 6
[math]
(π΄.π΄)+(π΄.1)+(π΅.\overline{π΅})
[/math]
Example 7
[math]
π΄.(πΆ+π΅+\overline{π΄})
[/math]
B.C
Example 8
[math]
π·.πΈ+πΈ.\overline{π·}
[/math]
D.E+E.D
Distributivetive Law
D.(E+D)
Redundancy Law
D
Example 9
[math]
π΄.(π΄+π΄).(π΅+\overline{π΄})
[/math]
Example 10
[math]
π΄.π·.\overline{π΅}+π΄.π·.π΅
[/math]
Example 11
[math]
πΆ.π·.(π·.π΅+πΆ)
[/math]
Example 12
[math]
\overline{π΅}.(π·+π΅)
[/math]
Example 13
[math]
(\overline {A}+\overline {B}).B
[/math]
[math]
(\overline {A}+\overline{B}). (B+1)
[/math]
[math]
\overline {A}+(b.1)
[/math]
[math]
\overline {A}+B
[/math]
Example 14
[math]
\overline{B} + (A.B)
[/math]
Example 15
[math] X+(\overline{X}.\overline{Y}) [/math]
Example 16
[math] (\overline{A} . A) + (\overline{A} . C) [/math]
= NOT A
Because the centre symbol is an OR gate and neither of the outputs of the and gates is A therefore by redundancy rule it is NOT A.
Example 17
[math] (\overline{A} + A) . (\overline{A} + C) [/math]
Example 18
[math] X.(\overline{X}+Y) [/math]
Example 19
[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]
Alternative
[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]
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