3
I’m 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 that the end result has to be a~c + bc
. But I don’t know how to get to him.
3
I’m 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 that the end result has to be a~c + bc
. But I don’t know how to get to him.
7
Let’s start with that:
(~ABC)+(A~B~C)+(AB~C)+(ABC)
Let’s reorder the expressions:
(~ABC)+(ABC)+(A~B~C)+(AB~C)
Let’s put the BC
and the A~C
in evidence:
BC(~A+A)+(A~C)(~B+B)
Every expression in form X+~X
is true. Soon:
BC+(A~C)
Note that its original expression (~ABC)+(A~B~C)+(AB~C)+(ABC)
has an interesting property: It tells exactly which are the four lines 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.
1
Victor’s answer is very good, but this solution I’ll give you is a direct continuation of his.
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.
Footsteps:
(~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)
Note 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 //
Justifications
(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 truth 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
That is, ~AC + A = A + C.
If you compare (1) to (2) you will see that we essentially justify the same thing.
Observing I believe that when doing Boolean algebra exercises you can end up in two ways: (1) Having an initial draw, as did Victor, that quickly solves the problem or (2) Doing randomly and needing to use several properties to achieve the answer ;)
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Thank you for the answer. The formula I took from a truth table. That’s why it has this characteristic of saying the 4 true expressions.
– Wincenty Bertoni Lech