# Can anyone help me rationalize De Morgan's Law in terms of Venn Diagram?

The logic is suppose to be very simple, but only in terms of truth table. When the same thing is visualized through Venn Diagram. The logic seems self contradicting.

This is the Venn Diagram expressions of $A and B$ and $A or B$:

De Morgan's Law states that not(A and B) is equivalent to not A or not B, and vice versa. However, based on the above expressions, these diagrams do not look logical at all:

• Could you be a bit more precise about how you think "these diagrams do not look logical"? Its very unclear what your complaint about them is. Which contradiction do you think you see? – Henning Makholm Mar 14 '15 at 1:25
• I thought for the not A and notB case, only the intersection will be white, following from the logic in A and B diagram. – most venerable sir Mar 14 '15 at 1:27
• not-A is everything outside the circle named A. Likewise for not-B. Thus intersection of not-A and not-B is everywhere that is both outside of A and also outside of B. Thus it is the area in gray. The area in white is either A or B. Hence $\operatorname{not}(A)\operatorname{ and }\operatorname{not}(B) = \operatorname{not}\Big(A\operatorname{ or }B\Big)$ , – Graham Kemp Mar 14 '15 at 2:10

Let us use an explicit example. Let A be the event that Adam shows up at the party. Let B be the event that Bill shows up at the party. Suppose that Adam and Bill hate eachother and if are together will start a fight. You have two choices, either to make sure to invite only one or the other, or to invite neither at all.

The phrase "not A and not B" (in symbols: $\sim\!\! A * \sim\!\! B$) can be reworded in more common English as "Neither A nor B is true." In our example, this means that Adam is not at the party, and also Bill is not at the party. So, in this case, we choose not to invite Adam and we choose not to invite Bill. We don't want to be friends with them anymore since they are the type to start a fight. Adam Fred and Charlie being at the party but Bill is stuck at home is not okay. We didn't want Adam there at all.

The phrase "not A and B" (in symbols: $\sim\!\!(A*B)$) can be reworded in more common English as "It is not true that A and B are simultaneously true." In our example, this is the case where we allow ourselves to invite exactly one of Adam or Bill to the party. With them separate, they shouldn't cause any trouble. It is okay for us to have Adam Fred and Charlie at the party with Bill stuck at home. Since Adam is without Bill it is okay.

If your confusion stems from which image to pair with which, the $\sim\!\!(A * B)$ is the opposite shading of the $(A * B)$ image.

The $(\sim\!\! A) * (\sim\!\!B)$ is the opposite shading of the $(A+B)$ image.

• I really like you second analogy and the phrase that it's not true that both are simultaneously true. Hence I can also think of ~(A+B) as it is not true that either one of them is true. So both are false. This is same as ~A*~B, because this one means A is not true and B is also not true. – most venerable sir Mar 14 '15 at 14:48
• But in my diagrams, A and B is when both are satisfied. This is why I like to think that notA and notB, is when both are satisfied simultaneously. Hence it is only the white intersected region. – most venerable sir Mar 14 '15 at 14:53
• Try drawing the diagram one step at a time. Complete things in parenthesis first. AND keeps only the intersection (what is common to both). OR keeps the union (what is in one or more of the sets being OR'd). NOT inverts the shading. – JMoravitz Mar 14 '15 at 15:04
• You mean draw A and B, A or B first. – most venerable sir Mar 14 '15 at 15:07