Raymond Raymond Smullyan "As the name of this book?" - 15

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From the paradox of truth

Paradoxes

252. The paradox of Protagoras.
One of the most ancient paradoxes tells about a teacher of Greek law Protagoras, who took to the poor students, but very capable young man and agreed to teach him for free, provided that when he finished the course and win his first trial, then pay a certain amount of Protagoras. The student accepted the terms of Protagoras, but, having completed his education, did not appear in court. After some time Protagoras filed for his student to court, demanding payment of the sum promised by him. Here are the readings gave Protagoras and his student at the trial.
Trainee. If I win this process is, by definition, I will not have to pay anything to Protagoras. If I lose this process, by the same token I do not win their first trial, and by agreement, I have to pay Protagoras only after winning his first trial. Therefore, I will win this lawsuit or lose, I do not care to pay do not have to.
Protagoras. If my former student lose this lawsuit, then by definition he will have to pay the appropriate amount to me (after all, it is for the payment of the amount owed to me and I initiated the process). If my former student will win this lawsuit, by the same token, he won his first trial, and by agreement will have to pay my debt. Consequently, he will win this lawsuit or lose, but he would have to pay anyway.
Who is right: Protagoras or his apprentice?

Note. I'm not sure I know the right answer to the problem. As the very first puzzle (on whether I was or was not fooled), Protagoras paradox is the prototype of a series of paradoxes. Best known to me making this paradox offered a lawyer, which I outlined the problem arises here. He said: "The court shall render a decision in favor of the student, ie student will not have to pay Protagoras, as at the beginning of the process the pupil is not yet won his first trial. When the judgment is over, the student by agreement will have to Protagoras what -That money. Therefore, Protagoras must go back to court and sue the student a second case. this time, the court will have to decide in favor of Protagoras, since the beginning of the second process, the pupil has already won its first trial. "

253. The paradox of the liar.
The so-called "paradox of the liar," or Epimenides paradox is actually a whole family of the founder of the paradoxes of a particular type, known as the paradox of the liar (sounds like a tautology, is not it?). In its original version, the paradox of the story about a certain Cretan named Epimenides expressed by the statement "All Cretans are liars."
No paradox here yet. In any case, the statement of Epimenides paradox no more than a statement that a certain inhabitant of the island of knights and liars statement expresses "all the inhabitants of this island are liars." From such a statement it follows that, first, a liar and saying that, secondly, there is at least one knight on the island. Similarly, the original version of the paradox of Epimenides, we conclude only that Epimenides is a liar, and that at least one Cretan says
only the truth. No paradox here, as you can see, no.
Now, if Epimenides was the only Cretan, the paradox would really originated. In this case, the only inhabitant of the island of knights and liars would argue that all the inhabitants of the island are liars (that is, in the long run would argue that he is a liar, and it is impossible).
The improved version of the liar paradox refers to the person uttering the statement "I am lying." Lying or not?
The next version of the improved version will be called the paradox of the liar in the future. Consider the statement:
This statement is false.
True or false? If it is false, it is true. If it is true, it is false. The solution of the paradox of the liar, we will discuss a little later.

254. Paradox Jourdain.
The next version of the liar paradox was first proposed in 1913 by English mathematician P.E.B. Jourdain. Sometimes it is called "the paradox of Jourdain with the card." Imagine a card on one side of which is written:
(1) The statement on the other side of this card is true.
Turning the card to the other side, you see the words:
(2) The statement on the other side of this card is false.
The paradox is this. If the first statement is true, the second statement is true (as in the first statement said that the second statement is true). Consequently, the first statement is false (as it says in the second assertion, that the first statement is false). If the first statement is false, the second statement is false. Consequently, the first statement is not false, but true. Thus, the first assertion is true if and only if it is false, and this is impossible.

255. Another option.
In another embodiment, the paradox of the liar on the card following three assertions are written:
(1) This statement contains five words.
(2) This statement contains eight words.
(3) Exactly one statement on this card is true.
Assertion (1) is certainly true, and (2) is obviously false. The problem arises in connection with the approval (3). If the statement (3) true, then the card - two true statements, namely the statement (3) and approval (1), contrary to what is said in the statement (3). Consequently, the assertion (3) must be false. On the other hand, if the statement (3) is false, then the statement (1) - the only true statement on the card, which means that (3) must be true! Thus, the statement (3) is true if and only if it is false.

Note. Where is the error in the reasoning in all these paradoxes? This question is very thin and rather controversial. Some (mostly philosophers, mathematicians and not) feel completely unacceptable any statement that contains a reference to itself. By counting the number of words contained in it, you will see that it is true.

The statement "this statement contains six words" false, however its meaning is clear, and the true value is established without difficulty: it says that the number of words contained in it is six, whereas they are only five. There is no doubt about the meaning of the statements in both instances there is no considered by us. Let us now consider the following statement:
This statement is true.
It does not lead to any paradoxes. No contradiction arises, regardless of whether we suppose it is true, and we assume it to be false. However, this statement does not make sense for the following reasons.
Whenever there is a need to establish what is the truth of a statement, we begin to find out what it means to the statement itself. For example, let the X - approval "twice two is four." Before I can understand the meaning of the truth of statements X, I need to figure out what each of the members of the X words and what is the meaning of the statement X. In this case, I know the meaning of each word in the X, and I clear meaning of the statement X: it says that two plus two equals four. Because I know that two plus two really equal four, then I know that X must be true. But I could not know that X is true, if it did not know that twice two is four. Moreover, I could not know what the truth is X, if I did not, which means that the statement "twice two is four." This example clearly shows me that a true statement "X is true" depends on what is meant by the statement X. If X is arranged so that its value depends on the truth of the statement "X is true", we find ourselves in a trap, because we walk in a circle .

It is arranged and outwardly innocuous statement "this statement is true." Before I can understand the meaning of the truth of this assertion, I need to understand the meaning of the statement itself. What did it say? B It reported only that it is true, but I still do not know what it means for this statement to be true. I do not know what is the truth of this statement (not to mention the fact that I do not know whether it is true or false), until I know what it means and to find out what it means, I can not as long as I do not know, which means that it is true. Thus our statement does not contain any information. Such statements are referred to as not well-founded.

liar paradox (and its variants) is based on the unsubstantiated claims. (Unjustified, I call for the sake of brevity are not well-founded allegations.) Bed and task 253 ( "The paradox of the liar") is not justified by the statement "this statement is false." In task 254 ( "Paradox Jourdain") is not substantiated allegations on both sides of the card. In task 255 ( "Another option"), two statements are reasonable, and the third is not justified.
Note, incidentally, that we can now say more as to whether, where a mistake in their reasoning contender for the hand Portions of the N-th (see. Sec. 5 of the caskets Portions). All of her ancestors on his mother's side used only well-founded allegations, and Portia N-I, wanting to make fun of his ardent admirer, skilfully used unfounded allegations. The same error occurs in a number of evidence given at the beginning of the previous chapter.

256. What do you think?
Let's go back to our good old friend Bellini and Cellini from the history of the caskets Portia. These two great masters not only produced the box, but engraved on their covers various inscriptions. Cellini engraved caskets on their false allegations, and Bellini graced the cover boxes of his work as true statements. Suppose that, in addition to Bellini and Cellini, in those days no one engraved inscriptions on the covers of the boxes (their sons were engaged in the manufacture of boxes, but did not know how to engrave).
You met box, the lid of which is engraved:
This inscription engraved Cellini
Whose autograph? If the inscription left Cellini, it would mean that he engraved a true statement, and we would have a contradiction. If the inscription left by Bellini, it would mean that he engraved a false statement, and again we have a contradiction. Who left an inscription?
You can not answer the question, referring to the fact that the statement "this inscription engraved Cellini" is not justified. It is quite justified. It tells us a historical fact, namely, that the inscription was engraved Cellini. If the recording was indeed made Cellini hand, it is true. If it made another master, then it is false. What's the deal here?
The difficulty arose from the fact that I have supplied you conflicting information. If you really got into the hands of a casket with the inscription "This inscription engraved Cellini" on the cover, it would mean that either Cellini in ancient times sometimes engraved not only false, but true statements (despite the fact that I will talk about it ), or at least that there was once some other master, sometimes engraved on the lids of boxes false statements (again, contrary to the information that you have received from me). Consequently, we have not a genuine paradox, but a kind of fraudulent trick.

By the way, you still have not managed to find out the name of that book?

257. drown or hang?
This puzzle is known widely enough. Someone has committed a crime punishable by death. At trial, he is given the last word. He should say one statement. If it proves to be true, the offender will drown. If it is false, the offender will be hanged. Which statement it must make to bring torturers totally confused?

258. The paradox of the barber.
Here is another well-known paradox. In a small town barber shaves everyone who does not shave himself, and does not shave any of those who shaves himself. Does the barber shave himself? If the barber shaves himself, then he thereby violates the rule, since shaves one of those who shaves himself. If the barber does not shave himself, he again violates the rule, because they do not shave one of those who does not shave himself. What barber?

259. What do you say?
One of the islands knights and liars sparsely populated: it lived only two native A and B. They made the following statement:
A: B - liar.
B: A - knight.
Who is A: knight or liar? A What about B?

Problem Solving 257, 258, 259.

257. The offender has to say: "I'll be hanged."

258. Nothing: the existence of a barber is logically impossible.

259. In response to questions you problem should be stated that the author was lying again! This situation can not be me. In fact, this problem is not that other, as Jourdain paradox slightly "disguised" form (see. The task 254).
If A were a knight, then B would actually knight. Therefore, A is not actually knight. If A was a liar, then B was in fact not a liar, and a knight. Thus, his statement would be true and A would have been a knight. Consequently, A can be neither a knight nor a liar, as in this and in another case, we arrive at a contradiction.

PARADOX OF TRUTH TO

Someone has defined a paradox as the truth, delivered on its head. Indeed, in many paradoxes contained ideas that after minor modifications lead to important discoveries. The following three tasks can serve as convincing proof of this principle.

260. Where a catch in this story?
Once the inspector Craig visited a certain community and talked with one of its members - Maksnurdom sociologist, who said the following:
- Community members have organized several clubs. Each community member may be a member of more than one club. Each club is named after one of the members of the community. No two clubs are not named in honor of the same member of the community, and the name of each member of the community has a certain club. A member of the community does not have to be a member of the club that bears his name. Anyone who is a member of the club that bears his name, we call nominabelnym. Anyone who is not a member of the club that bears his name, we call nenominabelnym. The most amazing thing in our community - that's what all nenominabelnye its members belong to the same club. (Apparently, the problem of missing condition that nominabelnye can not go to the club nenominabelnyh)
Inspector Craig thought for a moment and suddenly realized that Maksnurd not very strong in logic: in his stories ends do not converge with the ends. Why?

Decision. B reality, this problem is not that other, as the barber paradox in a new guise.
Let us assume that the story told by Maksnurdom would correspond to the truth. The club, which unites all the members of the community nenominabelnyh, named in honor of a member of the community, for example, in honor of Jack. We call this club just for the sake of brevity, the club Jack. Jack himself may be either nominabeliym or nenominabelnym. In this and in another case, we arrive at a contradiction.
Suppose that Jack nominabelen. Then Jack is Jack Club. But Jack comprised members of the club can only nenominabelnye members of the community, and we have a contradiction. On the other hand, if Jack nenominabelen, he is a member of the club nenominabelnyh members of the community. So, Jack Jack is a member of the club, which brings together all members of the community nenominabelnyh. But then Jack should be nominabelnym member of the community. Therefore, we in this case leads to a contradiction.

261. Are there any secret agent in the community?
Once the inspector Craig visited another community, where he met his old friend the sociologist Maksnaffa. Craig knew Maksnaffa from college (both studied at Oxford) as the person who owns the logic perfectly. Maksnaff told Craig about his community as follows:
- As in other communities, we organized at clubs. The name of each member of the community is exactly one club, each club is named in honor of a member of the community. Each member of our community, join a club, can either openly declare it, or keep its membership secret. Anyone who has not said publicly about his membership in a club that bears his name, we call suspicious. Anyone about whom we know that he is secretly a member of the club that bears his name, we call the undercover agent. Our community has a very curious feature: all suspect are members of a club. (Again, I missed the condition that the club suspect can not enter non-suspect)
Inspector Craig after a moment's reflection realized that unlike previous history professor Maksnaffa report does not contain the slightest contradiction. Moreover, it was found out an interesting fact: a purely logical way it was possible to determine whether there is in the community of secret agents.
So, whether in the community of secret agents?

Decision. Club suspicious named after some of the members of the community, for example, in honor of John. We call this club in the future the club John.
John himself is a member of the club, John, or is not a member. Let us assume that he is not a member. Then John can not be suspicious (suspicious because every member of the community is a member of the club, John). This means that John has publicly declared his membership in the club of John. Therefore, if John is not a member of the club, John, then John loudly announces its membership in the club, John, and we arrive at a contradiction. So, John must be a member of the club, John. And since each club member John is suspicious, then John should be suspicious. So John did not loudly announced its membership in the club of John and at the same time, John is a member of the club. Therefore, John the secret agent, or, to put it simply, fat!
Note that if you use a solution of 260, then this problem can be solved easily. Indeed, if the community was not secret agents, the suspect would not differ from nenominabelnyh, so a lot of suspicious would possess all the properties of the set nenominabelnyh members of the community. This means that all members of the community nenominabelnye have been members of the club. But in problem 260 we have proved that all nenominabelnye community members may not be members of a club. Therefore, the assumption that the community is not secret agents, leads to a contradiction. Hence, in the community is sure to be a secret agent (although we do not know who he is).
On these two evidence clearly shows the difference between the so-called "constructive" and "non-constructive" proof. The second proof is not constructive: we come to the conclusion that the community can not be a secret agent, but the evidence should not be who the undercover agents. In contrast, the first proof is constructive: it allows you to determine who the secret agent (member named John community), which is named in honor of the club suspicious.

262. The problem of the universe.
One member of each set of the universe inhabitants consist in its special club. Registrar of the universe would like to give each club the name of one of the inhabitants, so that no two clubs have not been named in honor of the same inhabitant of the universe, and every inhabitant was a club; named after him.
If the number of the inhabitants of this universe was finite, the registrar failed to carry out his grand plan, as the club would be more than the inhabitants of the universe, such as if the entire universe was only 5 inhabitants, the number of clubs has reached to 32 ( one club would be an empty set). If in the universe would be 6 inhabitants, the number of clubs has reached to 64, and in the universe with the inhabitants of the n number of clubs have was 2 ^ n. But to the universe of which we are now speaking, the number of inhabitants was infinite, so the recorder hoping for a favorable outcome of his venture. Over billions of years, he worked day after day trying to carry out his plan, but any attempt always fails. Why is this: not enough the best one scheme or impracticable in principle venture?

Decision. Failures associated with intentions impracticable in principle registrar. This remarkable mathematical fact opened mathematician Georg Cantor. Assume that the Registrar could assign names to all clubs inhabitants of the universe with all the rules (no two clubs have not named the name of the same inhabitant of the universe, and each inhabitant has a club named after him). We call nenominabelnym inhabitant of the universe, unless he is a member of the club, named in his honor. All the inhabitants of the universe nenominabilnye form a well-defined set, and we know that the members of each set of the inhabitants of the universe consists in its special club. Consequently, there should be a club nenominabelnyh inhabitants of the universe, it is impossible for the reasons stated in the problem 260 (this club is to be named in honor of one of the inhabitants of the universe, which may not be any nominabelnym nor nenominabelnym as both a contradiction ).

263. The problem of the recorded sets.
You see the same problem in a new garb. Some of the concepts introduced here will be needed in the next chapter.
One mathematician kept "book set". On every page describes some set of numbers (a set of numbers, we understand the subset of positive integers 1, 2, 3, ..., n, ...). Any set that is mentioned on any page of the book, called discounted set. Pages of the book are numbered in order of positive integers. What set whose description no one "sets Books" page.

Decision. Let n - any positive integer. We call extraordinary number n if n belongs to that described for n-th page and an ordinary, if not belong to that described for n-th page.
Many ordinary numbers can not be described on any "Books sets" page. Indeed, if it was listed on the k-th page, the number k might not be any extraordinary or ordinary, as in this and in the other case we would have a contradiction.