(~ABC)+(A ~ B ~ C) + (AB ~ C) + (ABC)
I am trying to solve this Formula:
(~ABC)+(A~B~C)+(AB~C)+(ABC)
(~ABC)+(A~B~C)+AB
(~ABC)+A(B+~B~C)
But I don't know how to get out of this last part. I know the end result has to be a~c + bc. But I don't know how to get into it.
2 answers
Let's start with this:
(~ABC)+(A~B~C)+(AB~C)+(ABC)
Let's reorder the expressions:
(~ABC)+(ABC)+(A~B~C)+(AB~C)
Let's put BC and A~C in evidence:
BC(~A+A)+(A~C)(~B+B)
Every expression in the form X+~X is true. Logo:
BC+(A~C)
Note that its original expression (~ABC)+(A~B~C)+(AB~C)+(ABC) has an interesting property: it says exactly what are the four rows of the truth table in which the expression is true, since each subexpression in parentheses has all three variables A, B and C exactly once each.
Victor's answer is very good, but this solution I will give you is a direct continuation of yours.
Basic properties of Boolean algebra used.
A ( B + C) = AB + AC (Distributiva AND)
A + ( BC) = (A+B)(A+C) (Distributiva OR) /*Cuidado com essa propriedade,
ela não existe na aritmética
convencional*/
AB = BA (Comutativa AND)
A ^ 1 = A (Elemento Neutro AND)
A + B = B + B (Comutativa OR)
A + 1 = 1 (Elemento Máximo)
A + ~A = 1 (Involução)
All these properties can be proved using a truth table.
Step by step:
(~ABC) + (A~B~C) + (AB~C) + (ABC)
(~ABC) + (A~B~C) + AB
(~ABC) + A(B+ ~B~C) (Seu Checkpoint)
To get out of this point it is necessary to develop B +~B ~ C, I will develop it separately in (1).
(~ABC) + A(B+ ~B~C) /* Por (1) */
(~ABC) + A(B + ~C) /* Distributiva AND */
(~ABC) + (AB) + (A~C) /* Comutativa AND, irei colocar o B no inicio*/
(B~AC) + (BA) + (A~C) /* Distributiva AND, "isolar" o B*/
B(~AC + A) + (A~C)
To get out of this step you need to develop ~AC + A, I will develop it separately in (2).
B(~AC + A) + (A~C) /* Por (2) */
B(A + C) + (A~C) /* Distributiva AND*/
(BA) + (BC) + (A~C)
Notice that we are 1 term from the answer:
(BA) + (BC) + (A~C) /* Colocaremos um elemento neutro no primeiro termo */
(BA) (C + ~C) + (BC) + (A~C) /* Distributiva AND*/
Você pode notar que se desenvolvermos (C + ~C) temos 1 que é neutro no AND.
(BAC) + (BA~C) + (BC) + (A~C) /* Comutativa AND, deixarei em ordem alfabética*/
(ABC) + (AB~C) + (BC) + (A~C) /* Comutativa OR, juntarei os termos semelhantes */
(ABC) + (BC) + (AB~C) + (A~C) /* Distributiva AND, "isolar" os termos*/
BC(A + 1) + (A~C)(B + 1) /* Elemento Máximo OR */
BC + A~C //
Justification
(1) B + ~ B ~ C
B + ~B~C /* Distributiva OR */
(B + ~B) (B + ~C) /* Elemento Máximo */
1 (B + ~C) /* Elemento Neutro */
B + ~C
That is, B + ~ B~C = B + ~C. You can confirm that B + ~B~C B + ~C using a true table.
(2) ~AC + A
~AC + A /* Comutativa OR */
A + ~AC /* Distributiva OR */
(A + ~A) (A + C) /* Elemento Máximo */
1 (A + C) /* Elemento Neutro AND */
A + C
I.e. ~AC + A = A + C.
If you compare (1) with (2) you will see that we essentially justify the same thing.
Remark I believe that at doing Boolean algebra exercises you can end up in two ways: (1) having an initial balcony, as Victor did, that quickly solves the problem or (2) doing it randomly and needing to use several properties to achieve the answer;)