You meet three blonde women, A, B, and C. Each of them is either a knight or a liar. Exactly one of them is Elle Woods.

A:I am Elle Woods.
B:I am Elle Woods.
C:At least two of us are liars.

What are A, B, and C, and who is Elle Woods?

The discussion between Ady Tzidon and Steve Herman raises the interesting question "When does an argument have a sufficient logical transparency?"
Ady Tzidon delivers a very short and speed argument, holding several steps in his head, because he thinks the problem is too easy. Good old Aristoteles would have said that Ady Tzidon uses some "enthymemes", in that he omits several premises (not worth to mention).

We believe that in

Henry is a brother of Joan.
So Joan is a sister of Henry.

the truth of the premise succeeds in guaranteeing the conclusion. But this is only the case, if we supply mentally:

If Henry is a brother of Joan and if Joan is a female, then Joan is a sister of Henry.

Note that "Joan" is also a Spanish male first name (Joan Coromines, Joan Miro ...).

The whole argument would have to be:

If Henry is a brother of Joan and if Joan is a female, then Joan is a sister of Henry.
Henry is a brother of Joan.
Joan is a female.
So Joan is a sister of Henry.

Therefore it's understandable that we often leave out some things, as Ady Tzidon did. On the other side, it's understandable that Steve Herman finds that too many things have been left out.
About 60 years ago, a proof of a thesis in set theory by the Harvard philosopher W.V.O. Quine was done by Ernst Specker, professor of mathematical logic at the ETH in Zurich, Switzerland, and this proof was 3 pages long. Although this proof was convincing for everyone in the field, Quine engaged a student to present this proof in a way that it met the requirements of a formal proof. The result was a proof of 70 pages, and even experts in the field were hardly able to understand it.

Now, here is this other extreme to Ady Tzidon's solution, namely a formal proof of Math Man's problem in the good old style.

The premises of the problem are in the steps 1 to 10.
Our first goal is to show that either A or B (or both) are liars. This is done in the conclusion step 11.
The second goal is to show a contradiction between that step 11 and the assumption that C is also a liar.
This is done in a subproof (step 12 until 21). As a result, we see that C must be a knight.
Thanks to this and to the original premises, we can draw further conclusions, arriving at the result that C is Elle Woods.
The trick of this proof is to find the key relation (contradiction) between step 11 and step 19. By transforming step 19 in step 20, we can see that this is nothing but the negation of step 11.

Of course, there are other and better ways possible.
And, of course, a formal proof is not better than an informal proof.
Between the two extremes of Ady Tzidon's "fast brain" argument and the formal argument below, we find the ideal of logical transparency in Charlie's argument.

Note that in the 6 tautologous conclusions, several steps have been made together. So, in a strict sense, the proof would have about 40 steps instead of only those 27!


 1. A -> (a = e)
 2. (a = e) -> A
 3. B -> (b = e)
 4. (b = e) -> B
 5. C -> [(~A & ~B) v (~A & ~C) v (~B & ~C) v (~A & ~B & ~C)]
 6. [(~A & ~B) v (~A & ~C) v (~B & ~C) v (~A & ~B & ~C)] -> C
 7. (a = e) v (b = e) v (c = e)
 8. ~[(a = e) & (b = e)]
 9. ~[(a = e) & (c = e)]
10. ~[(b = e) & (c = e)]
___

11. ~A v ~B                                                 Tautol.conclus. 1,3,8

  12. ~C                                                        SUBPROOF

  13. ~[(~A & ~B) v (~A & ~C) v (~B & ~C) v (~A & ~B & ~C)]

                                                                   Tautol.conclus. 12,6

  14. ~(~A & ~ B) & ~(~A & ~ C) & ~(~B & ~C) & ~(~A & ~B &  ~C)                                   
                                                                    Equiv.transform 13

  15. (A v B) & (A v C) & (B v C) & (A v B v C)                       

                                                                    Equiv.transform 14

  16. (~A -> B) & (~A -> C) & (~B -> C) & [~(A v B) -> C]

                                                                    Equiv.transform 15

  17. (~A -> B) & (~A -> C) & (~B -> C)        &-Elim. 16
  18. (~A -> C)    & (~B -> C)                       &-Elim. 17
  19. A & B                                                   Tautol.conclus. 12,18
  20. ~(~A v ~B)                                          Equiv.transform 19
  21. Contradiction                                        Contra.intro 11,20

22. ~~C                                   ~ Intro 12 - 21
23. C                                       Equiv.transform 22
24. ~A & ~B                            Tautol.conclus. 23,5
25. ~(a = e) & ~(b = e)           Tautol.conclus. 24,2,4
26. c = e                                 Tautol.conclus. 25,7
27. C & ~A & ~B & (c = e)        &-Intro 23,24,26

q.e.d.

Edited on December 20, 2016, 8:40 am

Edited on December 20, 2016, 8:43 am

Edited on December 20, 2016, 8:53 am

Edited on December 20, 2016, 10:59 am

Posted by ollie on 2016-12-20 08:34:01