Boolean simplifier

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Mayelana nalolu hlelo lokusebenza

lolu uhlelo lokusebenza lokubuka iwebhu lwe-"https://www.boolean-algebra.com"
I-Boolean Postulate, Properties, kanye namathiyori
I-postulate elandelayo, izakhiwo, kanye nethiyori kuvumelekile ku-Boolean Algebra futhi kusetshenziselwa ukwenza lula izinkulumo ezinengqondo noma imisebenzi:

AMA-POSTULATE angamaqiniso asobala.

1a: $A=1$ (uma A ≠ 0) 1b: $A=0$ (uma 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: $\overline{0}=1$
IZIMPILO ezisebenzayo ku-Boolean Algebra ziyafana nalezo eziku-algebra evamile

I-$A∙B=B∙A$ $A+B=B+A$
I-Associative $A∙(B∙C)=(A∙B)∙C$A+(B+C)=(A+B)+C$
Ukusabalalisa $A∙(B+C)=A∙B+A∙C$ $A+(B∙C)=(A+B)∙(A+C)$
IZIHLOKO ezichazwe ku-Boolean Algebra yile elandelayo:

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: $\overline{A+B}=\overline{A}∙\overline{B}$
Ngokusebenzisa i-Boolean postulates, izakhiwo kanye/noma ithiyori singenza lula izisho ze-Boolean eziyinkimbinkimbi futhi sakhe idayagramu yebhulokhi elinengqondo elincane (isifunda esingabizi kakhulu).

Isibonelo, ukwenza lula i-$AB(A+C)$ sine:

$AB(A+C)$ umthetho wokusabalalisa
=$ABA+ABC$ umthetho oqongelelekayo
=$AAB+ABC$ theorem 3a
=$AB+ABC$ umthetho wokusabalalisa
=$AB(1+C)$ theorem 2b
=$AB1$ theorem 2a
=$AB$
Nakuba lokhu okungenhla yikho konke okudingayo ukuze wenze lula i-Boolean equation. Ungasebenzisa isandiso sethiyori/imithetho ukwenza kube lula ukwenza lula. Okulandelayo kuzonciphisa inani lezinyathelo ezidingekayo ukuze kube lula kodwa kuzoba nzima kakhulu ukuzibona.

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: $A⊕B=\phezu kwe-inthanethi{A}∙B+A∙\phezu kwe-inthanethi{B}$
11: $A⊙B=\ngaphezulu{A}∙\ngaphezulu{B}+A∙B$
⊕ = XOR, ⊙ = XNOR
Manje sisebenzisa le theory/mithetho emisha singenza lula inkulumo yangaphambili kanje.

Ukuze senze i-$AB(A+C)$ ibe lula sine:

$AB(A+C)$ umthetho wokusabalalisa
=$ABA+ABC$ umthetho oqongelelekayo
=$AAB+ABC$ theorem 3a
=$AB+ABC$ theorem 7b

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