[Math] how to make a truth table from an boolean expression

boolean-algebra

I am trying to make a truth table from an SOP boolean algebra expression. I understand AND, OR, NOT truth tables. I just don't understand these types of tables and their outputs.

This is the expression:
$$A'BD' + BCD + ABC' + AB'D = A'BD' + BCD + ABC' + AB'D + BC'D' + A'BC + ABD.$$

I can use either side whichever is easier. Just let me know which side.

Would $A'$ be a $1$ and the others be a zero? I am also not sure how they get the output?

I understand the outputs of a AND, OR truth tables.

But I can't figure out these outputs. Would this be considered an OR table since the expression is $+$?

Would I just construct $A$, $B$, and $D$ with nots = 1 or zero?
Then, how do I determine the output?

 -----------------------
 A  | B  | D  |  output
-----------------------
| 1 | 0  | 1  |   1?    |  A'BD'
------------------------
| 0 | 0  | 0  |   0?    |  BCD
------------------------
| 0 | 0  | 1  |   1?    |  ABC'
-------------------------

something like that.

What I am trying to achieve is how the below expression is true using theorems.

$$A'BD' + BCD + ABC' + AB'D = A'BD' + BCD + ABC' + AB'D + BC'D' + A'BC + ABD$$

Best Answer

A, B, C, and D are boolean variables, meaning that each takes the value "true" or "false". More complex expressions have value "true" or "false" depending on the values of these variables, so for example A'BD' is true if A is false, B is true, and D is false, and C is either true or false.

To say that F = G, where F and G are complex expressions, means that, no matter what values the boolean variables have, the value of F is the same as the value of G (that is, F and G are either both true or both false). So, for example, we have

     AB + AC = A(B+C)

because if A is true and either B or C is true, then both sides are true, and in any other case both sides are false---there's no way to assign values to A, B, C such that the two side come out differently.

As previous posters have mentioned, you can always prove (or disprove!) an equality by going through all possible assignments of "true" and "false" to the variables. The goal seems to be to prove "by theorems", that is, using operations previously proven true. As you say, we can either manipulate one side of the equation until it has the form of the other side, or we could manipulate both sides and get them into a common form.

In this case, the first thing to notice is that the right-hand side (RHS) is a copy of the LHS with some extra stuff tacked onto the end. For this reason it's simplest to manipulate just the RHS to get rid of the surplussage! We can start with this basic theorem:

  if Q is true whenever P is true, then Q = Q + P

We can prove this theorem by systematically considering all possibilities for P and Q. Or look at it this way: If Q is true, then both sides of the equation are true. And if Q is false, then P must be false (since, by assumption, if P were true Q would be true) hence both sides are false.

Given this theorem we prove:

  XY + X'Z = XY + X'Z + YZ

Proof: Let P be YZ and let Q be XY+X'Z. Suppose P is true, that is, Y and Z are both true. But if Y and Z are both true, then XY+X'Z must be true. (Reason: If X is true then, since Y is true, XY is true. And if X is false then, since Z is true, X'Z is true. So in either case XY+X'Z is true.) So we've shown that Q is true whenever P is true, hence by the previous theorem Q = Q + P, which in this case is what we wanted to prove.

Now we can prove

 A'D' + AC' = A'D' + AC' + C'D'

This is just the same as the previous theorem, putting A for X, D' for Y, and C' for Z.

This last theorem implies

 B(A'D' + AC') = B(A'D' + AC' + C'D')

which is the same as

 A'BD' + ABC' = A'BD' + ABC' + BC'D'

Given this, we can take the RHS of the original and substitute A'BD' + ABC' for A'BD' + ABC' + BC'D', which is to say, we can drop the term BC'D'.

With steps similar to the above we can prove these two theorems:

 A'BD' + BCD = A'BD' + BCD + A'BC
 BCD + ABC' = BCD + ABC' + ABD

which permit us to drop the final two terms of the RHS of the original, completing the proof.

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