Chapter 3 of "The Lady or the Tiger" is about asylums that contain sane and crazy people: the sane always speak the truth, and the crazy always lie. Every one of them is either a doctor or a patient. Something is wrong with an asylum if it has either a crazy doctor or a sane patient.

Problem 10 from this chapter can be summarized as follows:

Inspector Craig encounters an asylum where several committees have been formed. It is given thatHow did Craig prove that something was wrong with this asylum?

- (1) all patients form a committee;
- (2) all doctors form a committee;
- (3) everyone has exactly one best friend and one worst enemy within the asylum;
- (4a) the best friends of all persons that form a committee, also form a committee;
- (4b) the worst enemies of all persons that form a committee, also form a committee;
- (5) for any two committees 1 and 2, there is a person P whose best friend believes that P is in committee 1, and whose worst enemy believes that P is in committee 2.

D.x == "x is a doctor" S.x == "x is sane" x[p.x] == "x believes p.x"With these notations, we can put down the following axiom:

(A) x[p.x] == S.x == p.xFor this specific problem we additionally write k or l for a set of people, and

Comm.k == "the people in set k form a committee" Fr.x = "the best friend of x" En.x = "the worst enemy of x"The last two are allowed because of (3) above.

We can model the given facts as follows, for every k and l:

(1) Comm.{x| !D.x} (2) Comm.{x| D.x} (4a) Comm.{x| k.(Fr.x)} <= Comm.k (4b) Comm.{x| l.(En.x)} <= Comm.l (5) (Ex: Fr.x[k.x] /\ En.x[l.x]) <= Comm.k /\ Comm.lIt is easily seen that from (5) follow two other statements that deal with just one committee: (5a) for each committee there is a person whose best friend believes he is in it; and similarly: (5b) for each committee there is a person whose worst enemy believes he is in it. Formally:

(5a) (Ex: Fr.x[k.x]) <= Comm.k (5b) (Ex: En.x[l.x]) <= Comm.lIt turns out that we only need (1), (4a), and (5a) to solve this problem. We calculate as follows:

Comm.{x| !D.x} --- given (1) => "(4a) with k:={x| !D.x}" Comm.{x| !D.(Fr.x)} => "(5a) with k:={x| !D.(Fr.x)}" (Ex: Fr.x[!D.(Fr.x)]) == "(A)" (Ex: S.(Fr.x) == !D.(Fr.x)) == "give Fr.x a name, working towards our goal" (Ex,y| y = Fr.x: S.y == !D.y) == "move (E...)" (Ey| (Ex: y = Fr.x): S.y == !D.y) => "weaken range of (E...)" (Ey: S.y == !D.y)which is our goal.

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