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 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.
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]
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]
[math] \overline {AB} (\overline {A}+B) \lt \math\gt
Complement law, Identity law
\lt math\gt (\overline {A}+\overline {B})(\overline {A}+B) \lt \math\gt
DeMorgan's law
\lt math\gt \overline {A}+/overline {B}B \lt \math\gt D
istributive law.
\lt math\gt \overline {A} \lt \math\gt
==Example 11==
==Example 12==
==Example 13==
\lt math\gt \overline{\overline{A} + \overline{(B.A)}}[/math]
First simplify using De Morgam's Law
[math] \overline{\overline{A} + \overline{B} + \overline{A}} [/math]
This can be simplified to
[math] \overline{\overline{A} + \overline{B}}
Then use De Morgans Law Again
\lt math\gt \overline{\overline{A}} + \overline{\overline{B}} [/math]
Finaly 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
[math]\overline{A}(A + B) + (B + AA)(A + \overline{B})[/math]
[math]\overline{A}A + \overline{A}B + (B + A)A + (B + A)\overline{B}[/math]
[math]\overline{A}B + (B + A)A + (B + A)\overline{B}[/math]
[math]\overline{A}B + BA + AA + B\overline{B} + A\overline{B}[/math]
[math]\overline{A}B + BA + A + A\overline{B}[/math]
[math]\overline{A}B + AB + AT + A\overline{B}[/math]
[math]\overline{A}B + A(B + T + \overline{B})[/math]
[math]\overline{A}B + A[/math]
[math]A + \overline{A}B[/math]
[math](A + \overline{A})(A + B)[/math]
[math]A + B[/math]
Example 16
