Three questioning ladies

Chapter 5, Problem 10 from [The Lady or the Tiger]

Here is a [calculational solution] to this problem.

Chapter 5 of "The Lady or the Tiger" is about two groups of people: those who can only ask questions that should be answered with "Yes" (type A), and those who can only ask questions that should be answered with "No" (type B).

Problem 10 from this chapter is as follows (as found somewhere on the Internet):

"I met three sisters: Alice, Betty and Cynthia. Alice asked Betty: 'Are you the type, who could ask Cynthia, if she is the type to ask you, if you both belong to different groups?' You can derive the type of one of the sisters! Who is what?"

Representation

For all the problems from this chapter, we can use the following notations. x stands for an arbitrary person, and P.x for an arbitrary statement involving (at least) x.

  A.x == "x is of type A"
  B.x == "x is of type B"
  x[P.x] == "x (is of the type that) can/could ask P.x"
  x<P.x> == "x asks P.x"

With these notations, we can put down the following axioms:

  (A1) A.x =/= B.x
  (A2) x[P.x] == A.x == P.x
  (A3) x<P.x> => x[P.x]

Solution

We will call the three persons from Problem 10 a, b, and c. Converting the given fact into a formula, we are given that

  a< b[c[A.b =/= A.c]] >

Reasoning from this using our axioms, we get

   a< b[c[A.b =/= A.c]] >
=>   "(A3) with x:=a --the only thing we know about x<P.x>"
   a[ b[c[A.b =/= A.c]] ]
==   "(A2) with x:=a --the only thing we know about x[P.x]"
   A.a == b[c[A.b =/= A.c]]
==   "(A2) with x:=b --the only thing we know about x[P.x]"
   A.a == A.b == c[A.b =/= A.c]
==   "(A2) with x:=c --the only thing we know about x[P.x]"
   A.a == A.b == A.c == A.b =/= A.c
==   "logical manipulation on == and =/=; see subproof" 
 
         A.b == A.c == A.b =/= A.c
      ==   "== and =/= are symmetric and associative"
         (A.b == A.b) == (A.c =/= A.c)
      ==   "logic: meaning of == and =/="
         true == false
      ==   "logic"
         false
 
   A.a == false
==   "(A1) with x:=a"
   B.a

So the only thing that we can derive from the given fact is that Alice is of type B.

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Changes to calc/tlott-5-10.html:

Wed Nov 10 13:27:59 MET 2004  Marnix Klooster
  * Make clearer that this computation is the only thing we can do.

Sat Oct 30 12:38:39 MEST 2004  Marnix Klooster
  * Move from CVS to darcs: do not show the CVS date anymore.

Thu Feb  7 19:43:35 MET 2002  marnix
  * Updated all files to use the m4 template

Wed Sep  5 21:20:00 MEST 2001  marnix
  * Made explicit that last changed date is in GMT

Wed Sep  5 21:08:51 MEST 2001  marnix
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Wed Sep  5 10:10:47 MEST 2001  marnix
  * Added date last changed

Wed Sep  5 08:50:24 MEST 2001  marnix
  * Initial revision