GameLoop एमुलेटर के साथ पीसी पर Boolean simplifier डाउनलोड करें

पीसी पर Boolean simplifier

Boolean simplifier, डेवलपर CodeBeex से आ रहा है, अतीत में Android सिस्टर्म पर चल रहा है।

अब, आप Boolean simplifier को पीसी पर GameLoop के साथ आसानी से खेल सकते हैं।

इसे GameLoop लाइब्रेरी या खोज परिणामों में डाउनलोड करें। अब और गलत समय पर बैटरी या निराशाजनक कॉलों पर ध्यान नहीं देना चाहिए।

बस मुफ्त में बड़ी स्क्रीन पर Boolean simplifier पीसी का आनंद लें！

Boolean simplifier परिचय

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