TITLE: On logic vs information theory
QUESTION [5 upvotes]: If the statements

All crows are black

and

All non black things are non crows

are equal, then why is the former so much easier to communicate by giving examples? What implications does this have on information theory?

REPLY [2 votes]: Let $\mathcal{U}$ be the finite universal set of all things under the Sun. Let $\mathcal{B}$ be the set of all black things. Let $\mathcal{C}$ be the set of all crows. Since there are non-black things and black things that are not crows, we have
$$\mathcal{C} \subset \mathcal{B} \subset \mathcal{U}$$
Suppose that a friend living far away is thinking about a thing. As this friend has many interests and thinks about many things, we are conservative and assume that the PMF of the thing being thought of is uniform over $\mathcal{U}$. Hence, the measure of our uncertainty regarding our friend's thought is $\log_2 |\mathcal{U}|$ bits. If our friend sends us the message

I am thinking about a crow.

then our uncertainty has been reduced to $\log_2 |\mathcal{C}|$ bits, i.e., our friend's message contained
$$\log_2 |\mathcal{U}| - \log_2 |\mathcal{C}| = \log_2 \left(\frac{|\mathcal{U}|}{|\mathcal{C}|}\right) > 0$$
bits of information. However, if our friend sends us the message

I am thinking about a non-black thing.

then our uncertainty has been reduced to $\log_2 (|\mathcal{U}|-|\mathcal{B}|)$ bits, i.e., our friend's message contained
$$\log_2 |\mathcal{U}| - \log_2 (|\mathcal{U}|-|\mathcal{B}|) = \log_2 \left(\frac{|\mathcal{U}|}{|\mathcal{U}|-|\mathcal{B}|}\right) > 0$$
bits of information. If there are more non-black things than crows, which is a most reasonable assumption to make, then $|\mathcal{U}|-|\mathcal{B}| > |\mathcal{C}|$ and, thus,
$$\log_2 \left(\frac{|\mathcal{U}|}{|\mathcal{C}|}\right) > \log_2 \left(\frac{|\mathcal{U}|}{|\mathcal{U}|-|\mathcal{B}|}\right)$$
i.e., the former message ("crow") contains more information than the latter one ("non-black").