structural:summarization, information:classification
Proof

The solution can be seen with an inductive argument. If k = 1 (that is, there is exactly one blue-eyed person), the person will recognize that they alone have blue eyes (by seeing only green eyes in the others) and leave at the first dawn. If k = 2, no one will leave at the first dawn, and the inaction (and the implied lack of knowledge for every agent) is observed by everyone, which then becomes common knowledge as well (C G {\displaystyle C_{G}}). The two blue-eyed people, seeing only one person with blue eyes, and that no one left on the first dawn (and thus that k > 1; and also that the other blue-eyed person does not think that everyone except themself are not blue-eyed ¬ K a {\displaystyle \neg K_{a}}, so another blue-eyed person ∃ x ∈ (G − a) (B l x) {\displaystyle \exists x\!\in \!(G-a)(Bl_{x})}), will leave on the second dawn. Inductively, it can be reasoned that no one will leave at the first k − 1 dawns if and only if there are at least k blue-eyed people. Those with blue eyes, seeing k − 1 blue-eyed people among the others and knowing there must be at least k, will reason that they must have blue eyes and leave.

For k > 1, the outsider is only telling the island citizens what they already know: that there are blue-eyed people among them. However, before this fact is announced, the fact is not common knowledge, but instead mutual knowledge.

For k = 2, it is merely "first-order" knowledge (E G {\displaystyle E_{G}}). Each blue-eyed person knows that there is someone with blue eyes, but each blue eyed person does not know that the other blue-eyed person has this same knowledge.

For k = 3, it is "second order" knowledge (E G E G = E G 2 {\displaystyle E_{G}E_{G}=E_{G}^{2}}). Each blue-eyed person knows that a second blue-eyed person knows that a third person has blue eyes, but no one knows that there is a third blue-eyed person with that knowledge, until the outsider makes their statement.

In general: For k > 1, it is "(k − 1)th order" knowledge (E G k − 1 {\displaystyle E_{G}^{k-1}}). Each blue-eyed person knows that a second blue-eyed person knows that a third blue-eyed person knows that.... (repeat for a total of k − 1 levels) a k th person has blue eyes, but no one knows that there is a " k th" blue-eyed person with that knowledge, until the outsider makes his statement. The notion of common knowledge therefore has a palpable effect. Knowing that everyone knows does make a difference. When the outsider's public announcement (a fact already known to all, unless k=1 then the one person with blue eyes would not know until the announcement) becomes common knowledge, the blue-eyed people on this island eventually deduce their status, and leave.

In particular: