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Suppose that the Bursar is telling the truth. Then, given our premisses, we can deduce in turn that the Chaplain is telling the truth, that the Dean is not, that the Warden is, and finally that the Chaplain is not. But this constitutes a contradiction. So our original supposition must have been false, and the Bursar is lying.
Some of you wished to extract a formal pattern, so as to display your reasoning text-book style. Good for you. There are many virtues of such an approach. Here is how it might then go.
Let ‘B' stand for ‘The Bursar is telling the truth', ‘C' for ‘The Chaplain is telling the truth', and so on. We can extract four premises, numbered for convenience:
[1] If B, then C
[2] It is either false that C or false that D
[3] Either D or W
[4] If W, then it is not true that C
Suppose B. Then C, from [1]. So not D, from [2]. Hence W, by [3], and so not C, by [4]. This is a contradiction, so our supposition must have been false. Therefore the Bursar is lying .
Two points about the notation. First notice how the introduction of a notation and the numbering of premises aids brevity and ease of exposition. And second, notice that we now have a solution to any puzzle with the same formal structure. For instance:
Yet one more State Occasion, and as always, it's a minefield of competing interests. “If Brian goes, we'll have to take Camilla as well,” mused Brenda, “but we can't have both Camilla and Di.” “If we drop Di, we'll have to have Wills,” interposed Phil. “Someone has to represent that side.” “But if Wills goes, Camilla won't” countered Brenda. “It's all too confusing. I can't even work out whether we can take Brian.” Can you help her?
Back to the proof. How did I know where to begin? Where did I get my proof idea ? To the logician's eye, this is an obvious candidate for a Reductio proof. The full name is Reductio ad Absurdum , often abbreviated to RAA , and the proof technique is to assume the opposite of what you want to prove, and deduce a contradiction from that assumption. Whereupon the assumption must have been false, because it leads to an impossible state of affairs, and the required result now follows. Logicians use RAA as the most likely proof technique when they can't see anything more direct. When your imagination fails you, try a reductio proof.
Some of you used a different proof method (labelled Constructive Dilemma by mediaeval logicians), arguing as follows:-
From [3], either D or W. If D, then not C, from [2]. And if W, then not C from [4]. So either way, not C.
And now it follows from [1] that not B.
Both statements by Berkeley , and the first by Hume, are irrelevant. That leaves us with
[1] Locke: “ Berkeley did it”
[2] Locke: “Hume's next statement will be true”
[3] Hume: “Locke has made two false statements”
And now we reason: Suppose that [2] is true. Then so is [3], whereupon [2] is false after all. So [2] is false. So Hume's next statement, [3] is false, and Locke must have made at least one true statement. This cannot be [2], which we know to be false, so it must be [1]. So the culprit is Berkeley .
An important part of general problem-solving technique is to sift out the relevant . How did I know what to drop? To the experienced solver it's obvious, but let me spell out the thought process in detail. The problem turns on whether or not [1] is true, for no other statement says anything about who did it. So [1] is relevant. The first statements from Berkeley and Hume have no logical links with anything else in the puzzle – no other statement even mentions them – so they can be ditched: they cannot play a role in any argument. The same applies to Berkeley 's second statement. Nothing else mentions it, so it cannot feature in any deduction.
How did I know where to begin? What gave me the idea for the proof? Answer: experience. Philosophers and Logicians have been discussing such oddities for more than 2,000 years. Epimedes the Cretan is remembered by us for his remark that “All Cretans are liars”. If this is taken to mean that every statement made by a Cretan is false, then, Epimedes being a Cretan, it cannot be true. For if it were true, then being a statement made by a Cretan, it would be false. Therefore it is false.
At least, with Epimedes, we could show that his statement was false. But there are also statements which can be neither true nor false, yielding genuine paradox. For instance
[X] This statement is false
If [X] is true, then [X] is false. And if [X] is false, then it is true. So it can't be either.
One response might be to banish all propositions which refer to themselves. But the paradox only returns. Compare these two:
The proposition expressed by the next sentence is false.
The poposition expressed by the last sentence is true.
Taken together they produce the same paradox. But each on its own is quite harmless.
To pursue this puzzle further, you could try some reading:-
Douglas R. Hofstadter Godel, Escher, Bach: an Eternal Golden Braid Penguin, 1980. I suggest you begin with the introduction, and then follow up whatever you like. Or perhaps:-
Raymond Smullyan What is the Name of this Book? Prentice-Hall, 1978.
Ernest Nagel & James R. Newman Godel's Proof , New York University Press, 1958
The most anyone can shake is 10 hands, and if anyone did shake 10, no-one shook 0. If we leave out the Warden of Keble for the moment, the other ten all shook different numbers of hands. So their total of handshakes must be either
0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 or 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55
One of these numbers is duplicated by the Warden of Keble. It must be an odd number, since the grand total of handshakes has to be even – each handshake counts twice. So it cannot be twice the number shaken by the Master of Pembroke. Q.E.D.
Difficult, but only because of the number of proof-ideas required. You need to realize that if anyone shakes 10, no-one shakes 0, a point easily missed. And also that handshakes count twice. And that on top of the logic of odds and evens, plus an application of the pigeonhole principle. (The pigeonhole principle was dicovered and named by Jobsworth, the porter, reminding us that no life is without value. In its most basic form the principle tells us that if you have, say, fifteen letters to be distributed among fourteen pigeonholes, then at least one pigeonhole must receive at least two letters).
So if you solved this one unaided, you did extremely well.
And notice again how the solution covers not just this particular puzzle, but any one of a large class. For instance, you now know the solution to this one:
Ofra, David and I play for the SCR cricket team, and on the last Saturday of the season there was a flurry of e-mail activity amongst the chosen side. All e-mails were reciprocated, and only Ofra and David sent sent the same number. Ofra reckons that she sent twice as many as me. Can you prove her wrong?
-oOo-