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 8 from this chapter can be summarized as follows:
Inspector Craig arrived at the eighth asylum, and found out that
How did Craig prove that something was wrong with this asylum?
- (1) two residents either trust each other, or they don't;
- (2) every resident has another resident as his teacher;
- (3) residents are only willing to teach those who trust themselves;
- (4) for every resident x there is a resident, who trusts precisely those residents who have a teacher trusted by x;
- (5) there is a resident who trusts all patients, and patients only.
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 will also use the following notations:
xVy == "x trusts y" xLy == "x is a teacher of y"(Since both 'trusts' and 'teacher' starts with a 't', I used the initials of the two Dutch words: 'vertrouwt' en 'leraar'.)
We can model the given facts as follows: for any residents x and y,
(2) (Ea: aLx) (3) xLy => x[yVy] (4) (Ea: (Ab: aVb == (Ec| cLb: xVc))) (5) QVx == !D.xHere we've given the 'queer' resident from property (5) the name Q.
When we look at our goal, (Ea: S.a =/= D.a), and compare this to our givens, we see that only fact (5) deals with D. Similarly only fact (3) deals with S:
xLy => x[yVy] --- (3) == "(A)" xLy => (S.x == yVy)In both of these cases, S and D are directly related to an expression of the form ...V... Looking now at fact (4), we see that it contains two V relations. The first one can be connected to S by substituting b:=a; the second one can be connected to D by substituting x:=Q. Therefore our strategy is to rewrite and simplify (4).
However, in our goal S.a and D.a are close together, but in fact (4) the two V relations are separated. This can be fixed by applying the following law of predicate logic:
(Ex| p.x: q.x) == r => "(Ex: p.x)" (Ex| p.x: q.x == r)In other words: equivalence distributes (in one direction) over an existential quantification with a non-empty domain.
We now have enough to start rewriting (4):
(Ax: (Ea: (Ab: aVb == (Ec| cLb: xVc)))) --- (4) => "the above-mentioned law: (Ec: cLb) is given as fact (2)" (Ax: (Ea: (Ab: (Ec| cLb: aVb == xVc)))) => "substitute b:=a, x:=Q, as was suggested above" (Ea,c| cLa: aVa == QVc) == "fact (5), with x:=c" (Ea,c| cLa: aVa == !D.c) == "rewrite aVa using (A), to prepare for the use of fact (3)" (Ea,c| cLa: c[aVa] == S.c == !D.c) == "fact (3), with x:=c and y:=a" (Ea,c| cLa: S.c == !D.c) == "separate (Ea...)" (Ec| (Ea: cLa): S.c == !D.c) => "weaken range of (E...)" (Ec: S.c == !D.c)And that is what we set out to prove.
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