Suppose you had some basic knowledge of Propositional Logic and Venn Diagrams.
Suppose you knew that there were three operators in Boolean Algebra (AND, OR, NOT) and two symbols for true and false. You know what AND, OR and NOT do, it’s just that you are used to seeing them with different symbols.
Assume that you know = is “equals” and you are aware of what parenthesis do. You would have this from Algebra and I can foresee someone getting to Propositional Logic without first going through Algebra.
With all that in mind, could you decode the following?
- (x+(x*y)=x
- (x*(x+y)=x
- x+(−x)=1
- x*(−x)=0
We know that AND and OR are binary operations and it appears that both ‘+’ and ‘-‘ are operating on both ‘x’ and ‘(-x)’. Seeing ‘-‘ up against x makes it look like its operating only on x so:
‘-‘ is NOT
Next, we ask, what is one of the first ideas learned when studying Logic and math that has true and false? A variable can’t be both true and false at the same time. The symbol AND has to go with either ‘+’ or ‘*’ and if it goes with ‘+’ then 1 has to be false or if it goes with ‘-‘ then 0 has to be false.
We could go either way. There is nothing else I can think of to prove one over the other. Sadly, this means we have to choose one way and work through it, and then come back and choose the other and work through it again. Well, I’ll do it one way and we’ll say doing it the other way is a homework assignment.
Who seems like a better candidate for ‘false’, 0 or 1? Let’s go with 0.
With this choice AND must be ‘*’ and that requires that OR must be ‘+’. Also, we have 1 being ‘true’.
Look at the first line: can we confirm that “x or (x and y)” will always be the value of x?
- If x is true, then we just need one side of the OR statement to be true; we get this from the ‘x’ on the left side of the OR.
- If x is false, then we need both sides of the OR statement to be false. We have that x is false so we are good on the left side of the OR. For the right side of the OR ‘x and y’ will go to false because x is false so we have the right side of the OR as well.
We’ve shown that the statement holds for both false x and true x without any need to a requirement on y, therefore we are finished.
Can you do similar work for “x and (x or y)”?
In the end we see it isn’t possible to find one unique solution (OR could be ‘*’ or it could be ‘+’). Perhaps that makes it feel like we didn’t win. However, after doing all that scrutiny, it will be very clear to you that distribution works both ways. This is in contrast to what we find in High School Algebra, where Multiplication distributes over Addition but Addition does not distribute over Multiplication.
Vectoreverything 