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ALJABAR BOOLE

ALJABAR BOOLE. Pertemuan ke-2 Oleh : Muh . Lukman Sifa , Ir. Aljabar Boole adalah suatu bentuk aljabar dimana variabel-variabel dan fungsi-fungsinya memiliki nilai 0 dan 1 .

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ALJABAR BOOLE

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  1. ALJABAR BOOLE Pertemuan ke-2 Oleh : Muh. LukmanSifa, Ir.

  2. Aljabar Boole adalahsuatubentukaljabardimanavariabel-variabeldanfungsi-fungsinyamemilikinilai 0 dan1. • Keluaran (output) darisatuataubeberapabuahkombinasigerbangdapatdinyatakandalamsuatuteoremaAljabar Boole. • Aljabar Boole dapatdigunakanuntukmenyederhanakanpersamaanlogika. ApakahAljabar Boole itu ?

  3. Postulat (Dalil) Boolean • Postulat 1 A + 0 = A ; A + 1 = 1 ; A . 0 = 0 ; A . 1 = A • Postulat 2 A + B = B + A ; A . B = B . A • Postulat 3 A + (B + C) = (A + B) + C; A . (B.C) = (A.B) . C • Postulat 4 A + (B . C) = (A + B). (A +C) ; (A . B) +C = (A + C) . (B + C) • Postulat 5 A = 0 ; atau A = 1 • Postulat 6 A + A = 1; A . A = 0; Hukum-hukumdanTeorema- teoremaAljabar Boole sebagaiberikut :

  4. T1: Rumuskomutatif a. A + B = B + A b. A.B = B.A • T2: Rumusasosiatif a. (A + B) + C = A + (B +C) b. (A.B).C = A. (B.C) • T3: Rumusdistributif a. A.(B +C) = AB + AC b. A+(B . C) = (A+B) . (A+C) • T4: Rumusidentif a. A + A = A b. A.A = A • T5: Rumusnegatif a. (A’) = A’ b. (A)” = A • T6: Rumus redundant a. A + A.B = A b. A.(A + B) = A • T7: Rumuseliminasi a. A + A’.B = A+B b. A.(A’ + B) = A.B • T8: Rumus Van De Morgan a. A + B = A . B b. A.B = A + B TeoremaAljabar Boolean

  5. 1. A.B = A + B 2. A + B = A.B Cobaandabuktikankeduateoremadiatasdengancara menurunkantabelkebenaran Teorema De Morgan :

  6. A.(A.B + B) = A.AB + A.B = A.B + A.B = A.B 2. AC + ABC = AC(1 + B) = AC 3. ABC + AB’C + ABC’ = AC(B + B’) + ABC’ = AC + ABC’= A(C + BC’) = A(C + B) = A(B + C) 4. (A + BC) = A (B + C) = A.B + A.C ContohsoalpenyelesaiandenganAljabar Boole :

  7. Selesai…. Terimakasih.

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