This is a app that can simplify Boolean algebra using law and Kmaps
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| 名前 | Boolean simplifier |
|---|---|
| バージョン | 1.0 |
| アップデート | 2022年03月16日 |
| サイズ | 5 MB |
| カテゴリー | 教育 |
| インストール | 5千+ |
| 開発元 | sajith tiyenshan |
| Android OS | Android 5.1+ |
| Google Play ID | com.codeB.boolean |
Boolean simplifier · スクリーンショット
Boolean simplifier · 説明
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
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
