Is it correct to say that this Boolean simplification is correct?

Question

We own the following truth table:

Soon after the minterms your boolean expression is the following:

A'.B'.C' + A'.B'.C + A'.B.C + A.B.C' + A.B.C

According to this site, it is said that the simplifying of his expression is given by:

y = A'.B' + A'.C + A.B

But with my own boolean simplification I managed to arrive at the following result:

S = A'.B'.C' + A'.B'.C + A'.B.C + A.B.C' + A.B.C
S = A’.B’(C’ + C) + A’.B.C + AB(C’ + C)
S = A’.B’ + A’.B.C + AB
S = A’.B’ + B(A’.C + A)
S = A’.B’ + B(C.1)
S = A’.B’ + BC

Checking in the following site, for the two answers I did not get the same output from the truth table.. I am not a doctor to be able to prove and affirm that it is correct and this would be the best solution, but what conclusion I must draw from this?

• a) That my simplification is correct, even different from the other?
• b) That my simplification is better than the other, by using fewer doors?

Answer 1

You have an error in your solution by pointing out that:

S = A’.B’ + B(A’.C + A)
S = A’.B’ + B(C.1)

You are implying that A'. C + A = 1 = (A' + A)

A   C   C.(A'+A)
0   0      0
0   1      1
1   0      0
1   1      1

A   C   A'.C+A 
0   0     0
0   1     1
1   0     1
1   1     1

It is worth noting that A'. B' + A'.B.C + A.B is also a solution. But somehow, A'. B' + A'. C + A.B is also a solution.

Answer 2

Answer your immediate questions: a)--- is my simplification correct, even different from the other? Its simplification is wrong from line 4 (line that starts its implication A'.C + A = C . 1). Being different from the other doesn’t necessarily mean being wrong.

b)--- my simplification is better than the other, by using fewer doors? If your answer was correct, yes and your biggest improvement would be by not using a three-input OR port but an OR port with two inputs.

Footsteps:

It includes some steps you omitted, as well as '.' points in places like AB, so as to stay, A.B, so as not to leave anything implicit. The comments next to a line concern what I had to do on that line to get to the next line.

Obs. I recommend not using the '.' point to represent the logical AND operation as well as the "+" for the OR, this overloads these operators when you are doing arithmetic. I recommend using the " " for the AND and "v" for the OR. Not to confuse it I did following its way ;)

1. S = A'.B'.C' + A'.B'.C + A'.B.C + A.B.C' + A.B.C  /* distributiva AND */

2. S = A’.B’.(C’ + C) + A’.B.C + A.B.(C’ + C)        /* elemento máximo OR */

3. S = A’.B’.1  + A’.B.C + A.B.1  (OMITIDO)          /* elemento neutro AND */

4. S = A’.B’ + A’.B.C + A.B                          /* associativa AND */ 

5. S = A’.B’ + B.A’.C + B.A   (OMITIDO)              /* distributiva AND */ 

6. S = A’.B’ + B(A’.C + A)                           /* por (1) */

S = A’.B’ + B(C.1)                                   /* ERRADO */
S = A’.B’ + BC                                       /* ERRADO */

7. S = A’.B’ + B.(A + C)                             /* distributiva AND */ 

8. S = A’.B’ + B.A  + B.C  /* e isso é equivalente a resposta do site por (2) */

Justifications

(1) Test: A'. C + A = A + C

A’.C + A                   /* associativa OR  */
A + A’.C                   /* distributiva OR     */
(A + A') . (A + C)         /* elemento máximo OR  */
1 . (A + C)                /* elemento neutro AND */
A + C

Therefore, A'. C + A = A + C

Proof by truth table:

(To follow the proof by the truth table recommend opening the image in a separate tab)

As there are 2 variables, A and C, 2 2 combinations = 4 combinations.

Passo 1, 
        Lado esquerdo,
             A Coluna(Col.) 3  recebe NEGAÇÃO da Col. 1.
             Col. 5 recebe Col.2;
             Col. 7 recebe Col.1 

        Lado direito,
             Col. 9 recebe Col. 1;
             Col. 11 recebe Col. 2


Passo 2,
        Lado esquerdo: Col. 4 recebe AND entre a Col. 3 e 5.

        Lado direito: Col. 10 recebe OR entre a Col. 9 e 11.

Passo 3, 
        Lado esquerdo: Col. 6 recebe OR entre a Col. 4 e 7.

Passo 4,(sinal de igual, nesse caso, representa o operador de equivalência)
        Pela definição de equivalência, as colunas 6 e 10 deveriam ser iguais,
        fato que acontece. Portanto, está provada a equivalência entre as funções lógicas. 

(2) Proof: A'. B' + B.A + B.C = A'. B' + A'. C + A.B

Proof by truth table:

3 variables A, B, C. 2 3 combinations = 8 combinations

Passo 1, 
        Lado esquerdo,
             Col. 4 recebe a NEGAÇÃO da Col. 1
             Col. 6 recebe a NEGAÇÃO da Col. 2
             Col. 8 e 12 recebe Col. 2
             Col. 10 recebe Col. 1
             Col. 14 recebe Col. 3

        Lado direito,
             Col. 16 e Col. 20 recebem a NEGAÇÃO da Col. 1
             Col. 18 recebe a NEGAÇÃO da Col. 2
             Col. 22 recebe Col. 3
             Col. 24 recebe Col. 1
             Col. 26 recebe Col. 2 

 Passo 2, 
         Lado esquerdo,
              Col. 5 recebe AND entre Col. 4 e 6;
              Col. 9 recebe AND entre Col. 8 e 10;
              Col. 13 recebe AND entre Col. 12 e 14;

         Lado direito,
              Col. 17 recebe AND entre Col. 16 e 18;
              Col. 21 recebe AND entre Col. 20 e 22;
              Col. 25 recebe AND entre Col. 24 e 26;

 Passo 3, (Lembre-se que o OR é associativo)
         Lado esquerdo, 
              Col. 7 recebe OR entre Col. 5 e 9;

         Lado direito, 
              Col. 19 recebe OR entre Col. 17 e 21;

 Passo 4, 
         Lado esquerdo, 
              Col. 11 recebe OR entre Col. 7 e Col. 13

         Lado direito, 
              Col. 23 recebe OR entre Col. 19 e 25;

 Passo 5,
         Pela definição de equivalência 11 deveria ser igual a 23, 
         fato que acontece. Portanto, está provada a equivalência 
         entre essas duas funções lógicas.  

Complimentary

(1) About Gabriel’s reply, he was wrong to say that you implied that A'. C + A = 1, you didn’t do it, which you did imply that A'. C + A = C.

(2) The reason why there are several solutions is that in fact there is not a single correct solution. But make no mistake, as was quoted by William A'. B' + A'.B.C + A.B is also a solution but it is not a "correct solution", in my opinion, because it can still be simplified.

For this problem there are two correct solutions. The karnaugh map justifies.

First:

Second:

Be it f the function represented by these Karnaugh maps then : (First Image) f = A'. B' + A'. C + A.B or (Second Image) f = A'. B' + B.C + A.B , respectively the answer you found on the site and the answer we found.