# A-level Computing/AQA/Paper 2/Fundamentals of computer systems/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:
| 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$ 000 111 |
| $A.{\overline {A}}=0$ | 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}}$ 010 100 |
| $1+A=1$ | 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$$1+A$ 101 111 |
| $0+A=A$ | 1+2=4 OR It is raining. So it doesn't matter about the 1+2=4 statement, the only thing that will make the statement true or not is whether it's raining | $0$$A$$0+A$ 000 011 |
| $A+A=A$ | It is raining OR It is raining is the equivalent of saying It is raining | $A$$A$$A+A$ 000 111 |
| $A+{\overline {A}}=1$ | It is raining OR It isn't raining is always true | $A$${\overline {A}}$$A+{\overline {A}}$ 011 101 |
| $0.A=0$ | 1+2=4 AND It is raining. It is impossible to make 1+2=4 so this equation so this equation is always false | $0$$A$$0.A$ 000 010 |
| $1.A=A$ | 2+2=4 AND It is raining. This statement relies totally on whether it is raining or not, so we can ignore the 2+2=4 part | $1$$A$$1.A$ 100 111 |
| $A+B=B+A$ | It is raining OR I have a cold, is the same as saying: I have a cold OR It is raining | $A$$B$$A+B$$B+A$ 0000 0111 1011 1111 |
| $A.B=B.A$ | 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$$B.A$ 0000 0100 1000 1111 |
| $A+(A.B)=A$ | It is raining OR (It is raining AND I have a cold). If It is raining then both sides of the equation are true. Or if It is not raining then both sides are false. Therefore everything relies on A and we can replace the whole thing with A. Alternatively we could play with the boolean algebra equation: $A+(A.B)=(1.A)+(A.B)$ Using the identity rule $1.A=A$ $(1.A)+(A.B)=A.(1+B)$ Take out the A, common to both sides of the equation $A.(1+B)=A.1$ Using the identity rule $1+B=1$ $A.1=A$ Using the identity rule $1.A=A$ | $A$$B$$A.B$$A+(A.B)$ 0000 0100 1001 1111 |
| $A.(A+B)=A$ | 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)$ 0000 0110 1011 1111 |
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 $(A.{\overline {B}})+(A.B)=A$