Download Boolean simplifier on PC With GameLoop Emulator

Boolean simplifier on PC

Boolean simplifier, coming from the developer CodeBeex, is running on Android systerm in the past.

Now, You can play Boolean simplifier on PC with GameLoop smoothly.

Download it in the GameLoop library or search results. No more eyeing the battery or frustrating calls at the wrong time any more.

Just enjoy Boolean simplifier PC on the large screen for free！

Boolean simplifier Introduction

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