Fundamental Hardware Elements of Computers: Boolean identities

 ← Simplifying boolean equations Boolean identities De Morgan's Laws →

Sometimes a very complex set of gates can be simplified to save on cost and make faster circuits. A quick way to do that is through boolean identities. Boolean identities are quick rules that allow you to simplify boolean expressions. For all situations described below:

A = It is raining upon the British Museum right now (or any other statement that can be true or false)
B = I have a cold (or any other statement that can be true or false)

Identity Explanation Truth Table
$A.A = A$ It is raining AND It is raining is the same as saying It is raining
$A$ $A$ $A.A$
0 0 0
1 1 1
$A.\overline{A}=0$ It is raining AND It isn't raining is impossible at the same time so the statement is always false
$A$ $\overline{A}$ $A.\overline{A}$
0 1 0
1 0 0
$1+A=1$ 2+2=4 OR It is raining. So it doesn't matter whether it's raining or not as 2+2=4 and it is impossible to make the equation false
1 $A$ $1+A$
1 0 1
1 1 1
$0+A=A$ 1+2=4 OR It is raining. So it doesn't matter about the 1+2=4 statement, the only thing that will make the statement true or not is whether it's raining
$0$ $A$ $0+A$
0 0 0
0 1 1
$A+A=A$ It is raining OR It is raining is the equivalent of saying It is raining
$A$ $A$ $A+A$
0 0 0
1 1 1
$A+\overline{A}=1$ It is raining OR It isn't raining is always true
$A$ $\overline{A}$ $A+\overline{A}$
0 1 1
1 0 1
$0.A=0$ 1+2=4 AND It is raining. It is impossible to make 1+2=4 so this equation so this equation is always false
$0$ $A$ $0.A$
0 0 0
0 1 0
$1.A=A$ 2+2=4 AND It is raining. This statement relies totally on whether it is raining or not, so we can ignore the 2+2=4 part
$1$ $A$ $1.A$
1 0 0
1 1 1
$A+B=B+A$ It is raining OR I have a cold, is the same as saying: I have a cold OR It is raining
$A$ $B$ $A+B$ $B+A$
0 0 0 0
0 1 1 1
1 0 1 1
1 1 1 1
$A.B=B.A$ It is raining AND I have a cold, is the same as saying: I have a cold AND It is raining
$A$ $B$ $A.B$ $B.A$
0 0 0 0
0 1 0 0
1 0 0 0
1 1 1 1
$A+(A.B)=A$ It is raining OR (It is raining AND I have a cold). If It is raining then both sides of the equation are true. Or if It is not raining then both sides are false. Therefore everything relies on A and we can replace the whole thing with A. Alternatively we could play with the boolean algebra equation:

$A+(A.B)=(1.A)+(A.B)$ Using the identity rule $1.A=A$
$(1.A)+(A.B)=A.(1+B)$ Take out the A, common to both sides of the equation
$A.(1+B)=A.1$ Using the identity rule $1+B=1$
$A.1=A$ Using the identity rule $1.A=A$

$A$ $B$ $A.B$ $A+(A.B)$
0 0 0 0
0 1 0 0
1 0 0 1
1 1 1 1
$A.(A+B)=A$ It is raining AND (It is raining OR I have a cold). If It is raining then both sides of the equation are true. Or if It is not raining then both sides are false. Therefore everything relies on A and we can replace the whole thing with A. Alternatively we could play with the boolean algebra equation:

$A.(A+B)=(0+A).(A+B)$ Using the identity rule $0+A=A$
$(0+A).(A+B)=A+(0.B)$ Take out the A, common to both sides of the equation
$A+(0.B)=A+0$ Using the identity rule $0.B=0$
$A+0=A$ Using the identity rule $0+A=A$

$A$ $B$ ${A+B}$ $A.(A+B)$
0 0 0 0
0 1 1 0
1 0 1 1
1 1 1 1

Examples of manipulating and simplifying simple Boolean expressions.

 Example: Simplifying boolean expressions Let's try to simplify the following: $A+B+B$  Using the rule $B + B = B$ $A+B+B = A+B$  Trying a slightly more complicated example: $(A.0)+B$  dealing with the bracket first $(0)+B$ as $0.A = 0$ $B$ as $0+B = B$ $(A.0)+B = B$ 
 Exercise: Simplifying boolean expressions $A+0$ Answer : $A$ $A.0$ Answer : $0$ $E+1$ Answer : $1$ $A+A+B+B+C$ Answer : $A+B+C$ $(A.B)+(A.B)$ Answer : $A.B$ $A.A.B.B.C$ Answer : $A.B.C$ $(A+\overline{A}).B$ Answer : $(A+\overline{A}).B$ applying the identity $A+\overline{A} = 1$ $(1).B$ applying the identity $1.B = B$ $B$

Sometimes we'll have to use a combination of boolean identities and 'multiplying' out the equations. This isn't always simple, so be prepared to write truth tables to check your answers:

 Example: Simplifying boolean expressions $(A.B) + A$  Where can we go from here, let's take a look at some identities $(A.B) + (A.1)$ using the identity A = A.1 $A.(B+1)$ taking the common denominator from both sides $A.1$ as B+1 = 1 $A$ Now for something that requires some 'multiplication' $(\overline{A}.B) + A$ $(\overline{A}+A).(B+A)$multiply it out $1.(B+A)$cancel out the left hand side as $(\overline{A}+A)=1$ $B+A$using the identity $1.Q = Q$
 Exercise: Simplifying boolean expressions $(A.\overline{B}) + B$ Answer : $(A.\overline{B}) + B$ $(A+B).(\overline{B}+B)$ multiplying out $(A+B).(1)$ $(A+B)$  $(A+B).\overline{A}$ Answer : This takes some 'multiplying' out: $(A+B).\overline{A}$ $(A.\overline{A})+(\overline{A}.B)$ $0+(\overline{A}.B)$ $B.\overline{A}$  $B.(A+A.B)$ Answer : This takes some 'multiplying' out: $B.(A+(A.B))$ treat the brackets first and the AND inside the brackets first $(B.A)+(B.A.B)$ multiply it out $(B.A)+(A.B)$ as $B.A.B = A.B$ $A.B$ as $(B.A)=(A.B)$  $(A+B).(A+A)$ Answer : $(A+B).(A+A)$ $(A+B).A$ as $A = A+A$ $(A+B).(A+0)$ as $A = A+0$ $A+(B.0)$ take A out as the common denominator $A$ as $(B.0) = 0$  $(A.\overline{B})+\overline{A}$ Answer : This takes some 'multiplying' out: $(A.\overline{B})+\overline{A}$ $(A+\overline{A}).(\overline{B}+\overline{A})$ $1.(\overline{B}+\overline{A})$ $\overline{B}+\overline{A}$  $(A.B)+\overline{A}$ Answer : This takes some 'multiplying' out: $(A.B)+\overline{A}$ $(A+\overline{A}).(B+\overline{A})$ multiplied out $(1).(B+\overline{A})$ as $(A+\overline{A}) = 1$ $B+\overline{A}$ as $1.Q = Q$  $(A.\overline{B})+(A.B)$ Answer : Take the common factor, $A$ from both sides: $A.(\overline{B}+B)$ As $\overline{B}+B = 1$ Then $A.(\overline{B}+B) = A . 1$ As $A . 1 = A$ Then $(A.\overline{B})+(A.B) = A$