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Boolean simplifier

Boolean simplifier

1.0 by sajith tiyenshan
(0 Reviews) August 11, 2025
Boolean simplifier Boolean simplifier Boolean simplifier Boolean simplifier Boolean simplifier Boolean simplifier

Latest Version

Version
1.0
Update
August 11, 2025
Developer
sajith tiyenshan
Categories
Education
Platforms
Android
Downloads
0
License
Free
Package Name
com.codeB.boolean
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More About Boolean simplifier

this is web view app of "https://www.boolean-algebra.com"
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 ≠ 0) 1b: $A=0$ (if A ≠ 1)
2a: $0∙0=0$ 2b: $0+0=0$
3a: $1∙1=1$ 3b: $1+1=1$
4a: $1∙0=0$ 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 $A∙B=B∙A$ $A+B=B+A$
Associative $A∙(B∙C)=(A∙B)∙C$ $A+(B+C)=(A+B)+C$
Distributive $A∙(B+C)=A∙B+A∙C$ $A+(B∙C)=(A+B)∙(A+C)$
THEOREMS that are defined in Boolean Algebra are the following:

1a: $A∙0=0$ 1b: $A+0=A$
2a: $A∙1=A$ 2b: $A+1=1$
3a: $A∙A=A$ 3b: $A+A=A$
4a: $A∙\overline{A}=0$ 4b: $A+\overline{A}=1$
5a: $\overline{\overline{A}}=A$ 5b: $A=\overline{\overline{A}}$
6a: $\overline{A∙B}=\overline{A}+\overline{B}$ 6b: $\overline{A+B}=\overline{A}∙\overline{B}$
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: $A∙(A+B)=A$ 7b: $A+A∙B=A$
8a: $(A+B)∙(A+\overline{B})=A$ 8b: $A∙B+A∙\overline{B}=A$
9a: $(A+\overline{B})∙B=A∙B$ 9b: $A∙\overline{B}+B=A+B$
10: $A⊕B=\overline{A}∙B+A∙\overline{B}$
11: $A⊙B=\overline{A}∙\overline{B}+A∙B$
⊕ = XOR, ⊙ = 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

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