Unduh Boolean simplifier di PC Dengan Emulator GameLoop

Boolean simplifier di PC

Boolean simplifier, berasal dari pengembang CodeBeex, berjalan di sistem Android di masa lalu.

Sekarang, Anda dapat memainkan Boolean simplifier di PC dengan GameLoop dengan lancar.

Unduh di perpustakaan GameLoop atau hasil pencarian. Tidak ada lagi memperhatikan baterai atau panggilan frustasi pada waktu yang salah lagi.

Nikmati saja Boolean simplifier PC di layar besar secara gratis！

Boolean simplifier Pengantar

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