Editing Common knowledge (logic) (section)

==== Proof ====
The solution can be seen with an inductive argument. If ''k''&nbsp;=&nbsp;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''&nbsp;=&nbsp;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 (<math>C_G[\forall x\! \in\! G ( \neg K_{x}Bl_{x})]</math>). 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''&nbsp;>&nbsp;1; and also that the other blue-eyed person does not think that everyone except themself are not blue-eyed <math>\neg K_{a}[\forall x\! \in\! (G-a) (\neg Bl_{x})]</math>, so ''another'' blue-eyed person <math>\exists x\! \in\! (G-a) (Bl_{x})</math>), will leave on the second dawn. Inductively, it can be reasoned that no one will leave at the first ''k''&nbsp;−&nbsp;1 dawns if and only if there are at least ''k'' blue-eyed people. Those with blue eyes, seeing ''k''&nbsp;−&nbsp;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''&nbsp;>&nbsp;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 (logic)|mutual knowledge]].

For ''k''&nbsp;=&nbsp;2, it is merely "first-order" knowledge (<math>E_G[\exists x\! \in\! G ( Bl_{x})]</math>). 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''&nbsp;=&nbsp;3, it is "second order" knowledge (<math>E_GE_G[\exists x\! \in\! G ( Bl_{x})]=E_G^2[\exists x\! \in\! G ( Bl_{x})]</math>). 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''&nbsp;>&nbsp;1, it is "(''k''&nbsp;&minus;&nbsp;1)th order" knowledge (<math>E_G^{k-1} [\exists x\! \in\! G ( Bl_{x})]</math>). 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''&nbsp;&minus;&nbsp;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:
# <math>E_G^{i} [|G ( Bl_{x})| \ge j]</math> is free (i.e. known prior to the outsider's statement) [[iff]] <math>i + j \le k</math>.
# <math>E_G^{i} [|G ( Bl_{x})| \ge j]</math>, with a passing day where no one leaves, implies the next day <math>E_G^{i - 1} [|G ( Bl_{x})| \ge j + 1]</math>.
# <math>E_G^{i} [|G ( Bl_{x})| \ge j]</math> for <math>j \ge k</math> is thus reached iff it is reached for <math>i + j > k</math>.
# The outsider gives <math>E_G^{i} [|G ( Bl_{x})| \ge j]</math> for <math>i = \infty, j = 1</math>.