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Boolean Postulate, Properties, and Theorems
The following postulate, properties, and theorems are valid in Boolean Algebra and are used in simplification of logical expressions or functions:
POSTULATES are self - evident truths.
1a: $A=1$ (if A ā 0) 1b: $A=0$ (if A ā 1)
2a: $0ā0=0$ 2b: $0+0=0$
3a: $1ā1=1$ 3b: $1+1=1$
4a: $1ā0=0$ 4b: $1+0=1$
5a: $\overline{1}=0$ 5b: $\overline{0}=1$
PROPERTIES that are valid in Boolean Algebra are similar to the ones in ordinary algebra
Commutative $AāB=BāA$ $A+B=B+A$
Associative $Aā(BāC)=(AāB)āC$ $A+(B+C)=(A+B)+C$
Distributive $Aā(B+C)=AāB+AāC$ $A+(BāC)=(A+B)ā(A+C)$
THEOREMS that are defined in Boolean Algebra are the following:
1a: $Aā0=0$ 1b: $A+0=A$
2a: $Aā1=A$ 2b: $A+1=1$
3a: $AāA=A$ 3b: $A+A=A$
4a: $Aā\overline{A}=0$ 4b: $A+\overline{A}=1$
5a: $\overline{\overline{A}}=A$ 5b: $A=\overline{\overline{A}}$
6a: $\overline{AāB}=\overline{A}+\overline{B}$ 6b: $\overline{A+B}=\overline{A}ā\overline{B}$
By applying Boolean postulates, properties and/or theorems we can simplify complex Boolean expressions and build a smaller logic block diagram (less expensive circuit).
For example, to simplify $AB(A+C)$ we have:
$AB(A+C)$ distributive law
=$ABA+ABC$ cumulative law
=$AAB+ABC$ theorem 3a
=$AB+ABC$ distributive law
=$AB(1+C)$ theorem 2b
=$AB1$ theorem 2a
=$AB$
Although the above is all you need to simplify a Boolean equation. You can use an extension of the theorems/laws to make it easier to simplify. The following will reduce the amount of steps required to simplify but will be more difficult to identify.
7a: $Aā(A+B)=A$ 7b: $A+AāB=A$
8a: $(A+B)ā(A+\overline{B})=A$ 8b: $AāB+Aā\overline{B}=A$
9a: $(A+\overline{B})āB=AāB$ 9b: $Aā\overline{B}+B=A+B$
10: $AāB=\overline{A}āB+Aā\overline{B}$
11: $AāB=\overline{A}ā\overline{B}+AāB$
ā = XOR, ā = XNOR
Now using these new theorems/laws we can simplify the previous expression like this.
To simplify $AB(A+C)$ we have:
$AB(A+C)$ distributive law
=$ABA+ABC$ cumulative law
=$AAB+ABC$ theorem 3a
=$AB+ABC$ theorem 7b