Fundamental Hardware Elements of Computers: Boolean identities

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