Fdtxjr

I am having a difficult time comprehending De Morgan's second law:
(A n B)' = A' u B' or ¬
(
P
and
Q
)
⇔
(
¬
P
)
or
(
¬
Q
)

I am having a difficult time grasping this law because through my own (flawed) interpretation of it, I have come up with examples in which the law doesn't hold. For example, according to the law, the following statement is true:

(I do not like shrimp and ice cream is to say that I do not like shrimp or
I do not like ice cream.)

The intuition is that if I don't like shrimp and ice cream, then I must dislike at least one of those things. However, I actually love shrimp and I love ice cream, but the combination/intersection of those foods is something I find quite gross.

So where am I going wrong here? I just can't escape/get around the notion that negating the intersection of two sets does not necessarily imply negating one set or the other. I even tried some computations in set theory to see if both statements calculated the same result and they didn't. For example:

Let the universal set, U, be all the letters of the English alphabet.

Let the following sets be created:

Set A = a, b, c, d

Set B = b, c, d, e

Let us apply the formulas which constitute De Morgan's second law:

(A n B) = b, c, d

(A n B)' = x: x is the set of all English letters except b, c, d

A' = x: x is the set of all English letters except, a, b, c, d

B' = x: x is the set of all English letters except b, c, d, e

A' u B' = x: x is the set of all English letters except a, b, c, d, e

Conclusion: (A n B)' = x: x is the set of all English letters except b, c, d and A' u B' = x: x is the set of all English letters except a, b, c, d, e

(A n B)' does not equal A' u B'

I feel so frustrated that I cannot figure out where my reasoning is going wrong. Any help is extremely appreciated. Please break down my thinking as far as you want to and show me where there is a gap in my logic and where I fail to proceed correctly. Thank you!

Best,

AJ

You're making a mistake when you work out what $A' cup B'$ is.

You don't union the "excepts", but rather what is leftover. $A'$ is the set of letters from $e$ to $z$. $B'$ is the set of letters $a$ plus $f$ to $z$. So the union is $a$ plus the letters from $e$ to $z$. That is very different from "all letters except a,b,d,c,e.$

You stuck a "union" inside the word "except", but "except" is a "not" and DeMorgan is about dealing with "not's", so you can't be so cavalier here.

Edit: For your ice cream and shrimp example. You're using "and" in two different ways. Is it "I like ice cream and I like shrimp" or is it "I like shrimp in my ice cream"?

You're misinterpreting "I do not like shrimp and ice cream" due to how informal language can be. "I do not like having shrimp with ice cream" would be the correct statement that you are referring to, and it's not $Awedge B$, it is one statement. Now let statement $A$ be "I like ice cream", and statement $B$ be " I like shrimp", then $A wedge B$ is "I like ice cream and I like shrimp", the negation of this is that you do not like both, but you may like one of them or neither, thus it fits $barA vee barB$.
As for your second example, you are once again making a mistake. $overlineAwedge B$ is indeed all letters except $b,c,d$, but $barAvee barB$ is not all letters except $a,b,c,d,e$, that would be $overlineAvee B$.

To elaborate on the "ice cream and shrimp" part: let's break it down into the the different logical components. Set:
beginalign*
P = text"I like ice cream" &&
Q = text"I like shrimp"
endalign*
Now, the conjunction of these to statements, $P wedge Q$, is usually read as: "I like ice cream and I like shrimp", and in English, this can sometimes be shortened to "I like ice cream and shrimp".

However, saying "I like ice cream and shrimp" can have a second meaning: "I like ice cream combined with shrimp". This is important, so I'll reiterate: in English, the sentence "I like ice cream and shrimp" can be interpreted in two ways:

1. I like ice cream and I like shrimp


2. I like ice cream combined with shrimp

The first one is the correct interpretation of $P wedge Q$. To convince yourself on this, consider that $P wedge Q$ is the statement which is true if and only if $P$ and $Q$ are both true. As you remarked, you like shrimp and you like ice cream, but you don't like shrimp and ice cream mixed together. So if the second statement represented $P wedge Q$, we'd have that $P$ is true and $Q$ is true, but $P wedge Q$ is false. This is nonsense!

So, to interpret $neg (P wedge Q)$, instead of saying "I don't like shrimp and ice cream", it might be clearer to say "It is not true that I like shrimp and ice cream", or better yet, "It is not true that (I like shrimp and I like ice cream)". Then you should clearly see that this is equivalent to the statement that you either dislike shrimp or dislike ice cream.