Tugas 5 [Nabila Putri Shalehah] Aljabar Boolean

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 Laws & Rules of Boolean Algebra

  • Commutative law of addition

           Commutative law of addition, A+B = B+A
          Urutan dimana dua variabel ORed tidak membuat perbedaan 

  • Commutative law of Multiplicatiom

           Commutative law of Multiplication, AB = BA
           Urutan dimana dua variabel ANDed tidak membuat perbedaan
  • Associative law of multiplication
           Asosiatif- Hukum ini memungkinkan penghapusan tanda kurung dari ekspresi dan                   penglompokkan ulang variabel.
           A(BC) = (AB)C  (Hukum Asosiasi AND)          
            Urutan dimana variabel ANDed tidak membuat perbedaan
  • Distributive Law

           Hukum ini mengizinkan penggandaan atau pemfaktoran dari ekspresi 
            A(B+C) = AB + AC
           (A+B)(C+D) = AC + AD + BC + BD
         
               Boolean Rules
1). A + 0 = A
  • In math if  you add 0 you have changed nothing
  • In Boolean Algebra ORing with 0 changes nothing
2). A + 1 = 1
  • ORing woth 1 must give a 1 since if anya input is 1 an OR will give a 1
3). A • 0 = 1
  •  In math if 0 is multiplied with anything you get 0. if you byAND anything with 0 you get 0
4). A • 1 = A
  • ANDing anything with 1 will yield anything
5). A + A = A 
  • ORing woth itself will guve the same result 
6). A + Ā = 1
  • Either A or Ā must be 1 so A + Ā = 1
7). A • A = A 
  • ANDing with itself will give the same result
8). A • Ā = 0
9). A = Ā
  • If you not something twice you are back to the beginning 
10). A + AB = A
11). A + ĀB = A + B
  • If A is 1 the output is 1, If A is 0 the output is B proof :
A + ĀB = (A + AB) + ĀB    RULE 10 
= (AA +AB) + ĀB               RULE 7 
= AA + AB + AĀ +ĀB        RULE 8
= (A + Ā)(A + B)                FACTORING 
= 1·(A + B)                         RULE 6
= A + B                               RULE 4

12). (A + B) (A + C) = A + BC
Proof:

(A + B)(A +C) = AA + AC +AB +BC       DISTRIBUTIVE LAW 
= A + AC + AB + BC                                 RULE 7 
= A(1 + C) +AB + BC                                FACTORING
= A.1 + AB + BC                                        RULE 2
= A(1 + B) + BC                                         FACTORING 
= A.1 + BC                                                  RULE 2 
= A + BC                                                     RULE 4




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