Surprising as this fact may be, it is not difficult to prove. Moreover, this fact holds even for normal reasoners of type 1.
8
Prove that if a normal reasoner of type 1 is afraid to believe in his own consistency, then he really shouldn’t believe in his own consistency.
Remarks. In the above problem, we see that for a normal reasoner of type 1, his belief in the proposition B~B⊥⊃B⊥ is self-fulfilling in the sense that his believing that proposition is a sufficient condition for the proposition being true. The theme of self-fulfilling beliefs will play a major role in the next few chapters.
9
A reasoner of type 4 is also a normal reasoner of type 1, hence according to the last problem, if he is afraid to believe in his own consistency, then he shouldn’t believe in his own consistency. This means that for a reasoner of type 4, the proposition B(B~B⊥⊃B⊥)⊃(B~B⊥⊃B⊥) is true. Prove that any reasoner of type 4 knows that this proposition is true (he knows that if he is afraid to believe in his own consistency, then he shouldn’t believe in it).
10
Suppose a reasoner of type 4 believes that he is afraid to believe in his own consistency. Does it follow that he really is afraid to believe in his own consistency?
SOLUTIONS
1. Suppose a reasoner believes that he is inconsistent. I see no reason why he is necessarily inconsistent, but he must be inaccurate for the following reasons.
A reasoner who believes he is inconsistent is either right or wrong in this belief. If he is wrong in this belief, then he is obviously inaccurate (he has the false belief that he is inconsistent). If he is right in this belief, then he really does believe the false proposition ⊥. In either case, he has at least one false belief.
2. If he were wrong, then he would be accurate, which is a contradiction.
3. Suppose he believes ~B⊥. Then he believes the logically equivalent proposition B⊥⊃⊥. Also he is regular (since he is of type 1*), and hence he will believe BB⊥⊃B⊥, hence he will believe the logically equivalent proposition ~B⊥⊃~BB⊥. Since he also believes ~B⊥, then he will believe ~BB⊥.
4. We know by Theorem 1 of Chapter 3 that for any proposition p, if a native of a knight-knave island says, “If I am a knight then p,” the native must be a knight and p must be true. Now, any reasoner—even if type 1—knows this as well as you and I, and so if the reasoner believes that the rules of the island hold, then if a native says to him, “If I am a knight, then p,” the reasoner will believe that the native is a knight and that p is true. In this particular problem, the native has said, “If I am a knight, then you will believe that I am a knave,” and so the reasoner will believe that the native is a knight and also that he (the reasoner) will believe that the native is a knave. And so the reasoner will believe k and also believe B~k (k is the proposition that the native is a knight). So far, we have used only the fact that the reasoner is of type 1. However, he is of type 4, and since he believes k, he will also believe Bk. Hence he will believe Bk and believe B~k, but he knows that (Bk&B~k)⊃B⊥ (as we showed in Chapter 11, this page). Therefore he will believe B⊥—i.e., he will believe that he is (or will be) inconsistent. He also believes that the native is a knight.
So far, we have not used the fact that the rules of the island really hold; our preceding argument used only the fact that the reasoner believes that the rules hold. Now, suppose the rules really do hold. Then the native really is a knight and the reasoner really will believe that the native is a knave (according to Theorem 1, Chapter 3). But since the reasoner also believes that the native is a knight, he will become inconsistent.
5. This time the reasoner believes k≡(k⊃~Bk), hence he believes k and believes ~Bk, which are logical consequences of k≡(k⊃~Bk). Since he believes k, he will believe Bk, and believing ~Bk, he will become inconsistent.
If the rules of the island really hold, then k≡(k⊃~Bk) is not only believed by the reasoner, but is actually true, hence k&~Bk (which is logically implied by it) is true, hence ~Bk is true, contrary to the fact that the reasoner does believe the native is a knight (and hence Bk, rather than ~Bk, is true). Thus the rules of the island don’t really hold.
Discussion. Let us look at the last two problems in the form of the student and his theology professor. Suppose the professor says: “God exists, but you will never believe that God exists.” If the student is of type 4 and believes the professor, he will have to believe himself inconsistent. If also the professor’s statement was true, then the student really will become inconsistent.
On the other hand, suppose the professor says: “God exists, but you will never believe that God exists.” Then if the student is of type 4—or even of type 3—and believes the professor, he will become inconsistent, and also, the professor’s statement is false.
6. The reasoner reasons: “Suppose he is a knave. Then his second statement is false, which means that I will be inconsistent, hence I’ll believe everything—in particular, that he’s a knave. But this will validate his first statement and make him a knight. Therefore it is contradictory to assume that he is a knave, hence he must be a knight. Since he is a knight, his first statement is true, hence I will believe he is a knave. But I now believe he is a knight, hence I will be inconsistent. This proves that I will be inconsistent. However, he’s a knight and said that I will always be consistent. Therefore I won’t ever be inconsistent.”
At this point the reasoner has come to the conclusion that he will become inconsistent and that he won’t become inconsistent, and so he is now inconsistent.
Since the reasoner has become inconsistent, the native’s second statement is false. Also, since the reasoner has become inconsistent, he will believe everything, including the fact that the native is a knave. This makes the native’s first statement true. Since the native has made one true statement and one false statement, then the rules of the island don’t really hold.
7. The reasoner believes the following two statements:
(1) k≡~Bk
(2) k≡(Bk⊃B⊥)
Since he believes (1), then, according to (b) of Theorem G#, Chapter 12 (this page), the reasoner will believe Bk⊃B⊥. And, since he believes (2), he will believe k. So, according to (a) of Theorem G#, Chapter 12, he will become inconsistent.
It further follows that the native’s first statement was false and his second statement true. Therefore the rules of the island don’t hold.
8. We assume that the reasoner is normal and of type 1 and that he believes B~B⊥⊃B⊥. We are to show that if he believes he is consistent, he will become inconsistent (and therefore that his fear is justified). So, suppose he ever does believe ~B⊥. Being normal, he will then believe B~B⊥. This, together with his belief in B~B⊥⊃B⊥, will cause him to believe B⊥ (because he is of type 1). And so he will believe both B⊥ and ~B⊥ (he will believe that he is inconsistent and that he is not inconsistent), which will make him inconsistent.
9. The reasoner reasons: “Suppose I become afraid to believe in my own consistency. This means that I’ll believe that I shouldn’t believe in my own consistency—i.e., I’ll believe B~B⊥⊃B⊥. Since I am of type 1 (being of type 4), then I will believe ~B⊥⊃~B~B⊥. Now suppose I should also believe that I am consistent—i.e., suppose I believe ~B⊥. Then, since I will believe ~B⊥⊃~B~B⊥, I’ll believe ~B~B⊥. But I’ll also believe B~B⊥ (since I’ll believe ~B⊥ and I am normal), and hence I’ll be inconsistent. Therefore, if I am afraid to believe in my own consistency, I really cannot believe in my own consistency without becoming inconsistent. Thus the proposition B(B~B⊥⊃B⊥)⊃(B~B⊥⊃B⊥) is true.”
10. We have just seen that the reasoner believes the proposition B(B~B⊥⊃B⊥)⊃(B~B⊥⊃B⊥). Therefore, if he believes B(B~B⊥⊃B⊥), he will believe B~B⊥⊃B⊥—which means that if he believes that he is afraid to believe in his own consistency, then he really will be afraid to believe in his own consistency.
• Part VI •
SELF-FULFILLING BELIEFS A
ND LÖB’S THEOREM
• 15 •
Self-Fulfilling Beliefs
THE PROBLEMS of this chapter are all related to Löb’s Theorem, a famous result that is important to the thrust of this book.
We now have a change of scenario: A reasoner of type 4 is thinking of visiting the Island of Knights and Knaves because he has heard a rumor that the sulfur baths and mineral waters there might cure his rheumatism. Before embarking, however, he discusses the situation with his family physician. He asks the doctor whether the “cure” really works. The doctor replies: “The cure is largely psychological; the belief that it works is self-fulfilling. If you believe that the cure will work, then it will work.”
The reasoner trusts his doctor implicitly, and so he goes to the island with the prior belief that if he believes the cure will work, then the cure will work. He takes the cure, which lasts only a day but which is not supposed to work for several weeks, if it works at all. The next day he starts worrying about the situation and thinks: “If only I can believe that the cure works, then it will work. But how do I know whether I will ever believe that it works? I have no rational evidence that the cure will work, nor do I have any evidence that I will ever believe that the cure will work. For all I know, I may never believe that the cure will work, and the cure might accordingly not work!”
A native of the island passes by and asks the reasoner why he looks so disconsolate. The reasoner explains the entire situation, then summarizes it by saying: “If I ever believe the cure will work, then it will, but will I ever believe the cure will work?” The native replies: “If you ever believe I’m a knight, then you will believe that the cure will work.”
At first, this did not seem particularly reassuring to the reasoner. He thinks: “What good does this do me? Even if what he says is true, this will only reduce the problem to whether I will ever believe that he is a knight. How do I know whether I will ever believe he is a knight? And even if I do, he may be a knave and his statement may be false, hence I may still not believe that the cure will work.” But then the reasoner thought some more about his problem, and after a while he heaved a sigh of relief. Why?
Well, as we will see, the amazing thing is that the reasoner will believe that the cure will work and, assuming that the doctor is right, the cure will work! I might remark that the rules of the island don’t have to really hold for the argument to go through; it is enough that the reasoner believes that they hold.
This problem is closely related to M. H. Löb’s important theorem, which we will examine later. But first, let’s consider a slightly simpler problem, one that comes even closer to Löb’s original argument. Suppose that the native, instead of saying the above, says: “If you ever believe that I’m a knight, then the cure will work.”
1. (After Löb)
Prove that under the above conditions, the reasoner will believe that the cure will work (and hence, if the doctor was right, then the cure will work).
Solution. It will be easiest to give the solution partly in words and partly in symbols. We let k be the proposition that the native is a knight, and we let C be the proposition that the cure will work. At the outset, the reasoner believes the proposition BC⊃C.
The reasoner reasons: “Suppose I ever believe that he is a knight. Then I’ll believe what he says—I’ll believe that Bk⊃C. Also, if I ever believe he’s a knight, I’ll believe that I believe he’s a knight—I’ll believe Bk. And so, if I ever believe he is a knight, I’ll believe both Bk and Bk⊃C, hence I’ll believe C. Thus, if I ever believe he is a knight, then I’ll believe that the cure works. But if I ever believe that the cure works, then the cure will work (as my doctor told me). And so, if I ever believe he’s a knight, then the cure will work. Well, that’s exactly what he said. He said that if I ever believe he’s a knight, then the cure will work, and he was right! Hence he is a knight.”
At this point the reasoner believes that the native is a knight, and since the reasoner is normal, he continues: “Now I believe he is a knight. I have already proved that if I believe he is a knight, then the cure will work, and since I do believe that he is a knight, the cure will work.”
At this point the reasoner believes that the cure will work. Then, assuming his doctor was right, the cure will work.
The solution to Problem I could have been established more quickly had we first proved the following lemma, which will have other applications as well.
Lemma 1. Given any proposition p, suppose a native of the island says to reasoner of type 4: “If you ever believe that I’m a knight, then p is true.” Then the reasoner will believe: “If I ever believe he is a knight, then I will believe p.” More generally, for any two propositions k and p, if a reasoner of type 4 believes the proposition k≡(Bk⊃p), or even believes the weaker proposition k⊃(Bk⊃p), then he will believe Bk⊃Bp.
Exercise 1. How is Lemma 1 proved? (This is the same problem found in Exercise 3, Chapter 11, this page.)
Exercise 2. How does the use of Lemma 1 facilitate the solution of Problem 1?
Answer to Exercise 1. Let’s first show this in the knight-knave version. We let k be the proposition that the native is a knight. The native has asserted the proposition Bk⊃p. The reasoner reasons: “If I ever believe that he’s a knight, then I’ll believe what he says—I’ll believe Bk⊃p. But if I believe he’s a knight, I’ll also believe Bk (I’ll believe that I believe he’s a knight). Once I believe Bk⊃p and I believe Bk, then I’ll believe p. And so, if I ever believe he’s a knight, then I’ll believe p.”
Of course the more general form can be proved in essentially the same manner, or alternatively as follows: Suppose a reasoner of type 4 believes k⊃(Bk⊃p), which he will certainly believe, if he believes the stronger proposition k≡(Bk⊃p). Then, according to Problem 2, Chapter 11, he will believe Bk⊃(BBk⊃Bp). He also believes Bk⊃BBk. Believing these two propositions, he will believe Bk⊃Bp, which is a logical consequence of them. (For any proposition X, Y, and Z, the proposition X⊃Z is a logical consequence of X⊃(Y⊃Z) and X⊃Y. In particular, this is so if X is the proposition Bk, Y is the proposition BBk, and Z is the proposition Bp.)
Answer to Exercise 2. The reasoner believes k≡(Bk⊃C)—because the native asserted Bk⊃C. Then, according to Lemma 1, the reasoner will believe Bk⊃BC. He also believes BC⊃C, hence he will believe Bk⊃C. Then he will believe k (since he believes both Bk⊃C and k≡(Bk⊃C)), hence he will believe Bk (he is normal). Now that he believes Bk and believes Bk⊃C, he will believe C.
The upshot of Problem 1 is the following theorem.
Theorem 1 (After Löb). For any proposition k and C, if a reasoner of type 4 believes BC⊃C and believes k≡(Bk⊃C), then he will believe C.
Theorem 1 yields the following curious result.
Exercise 3. Suppose a theology student is worried about such things as the existence of God and his own salvation. He asks his professor: “Does God exist?” And “Will I be saved?” The professor then makes the following statements:
(1) “If you believe that you will be saved, then you will be saved.”
(2) “If God exists and you believe that God exists, then you will be saved.”
(3) “If God doesn’t exist, then you will believe that God exists.”
(4) “You will be saved only if God exists.”
Assuming that the student is a reasoner of type 4 and that he believes his professor, prove: (a) The student will believe that he will be saved; (b) If the professor’s statements are true, then the student will be saved.
Solution. Let g be the proposition that God exists and let S be the proposition that the student will be saved. The student then believes the following four propositions:
(1) BS⊃S
(2) (g&Bg)⊃S
(3) ~g⊃Bg
(4) S⊃g
The proposition ~Bg⊃g is a logical consequence of (3). This proposition, together with (4), has as a logical consequence the proposition (~BgvS)⊃g. Also ~BgvS is logically equivalent to Bg
⊃S, hence (~BgvS)⊃g is logically equivalent to (Bg⊃S)⊃g. Also g⊃(Bg⊃S) is logically equivalent to (2), and g≡(Bg⊃S) is a logical consequence of (Bg⊃S)⊃g and g⊃(Bg⊃S). Therefore g≡(Bg⊃S) is a logical consequence of (1), (2), and (3). Since the student believes (1), (2), and (3), he will also believe g≡(Bg⊃S). Since by (1) he also believes BS⊃S, then by Theorem 1, he will believe S. Thus BS is true, and if the professor was right, BS⊃S is true, hence S is true, which means that the student will be saved.
Now let us return to Problem 1 and its proposition: “If you ever believe I’m a knight, then the cure will work.”
2
Suppose a reasoner of type 4 again believes BC⊃C, but this time the native says: “If you ever believe I’m a knight, then you will believe that the cure works.” Prove that the reasoner will again believe that the cure will work.
Solution. This time the native is asserting Bk⊃BC (instead of Bk⊃C). Then by Lemma 1, the reasoner will believe Bk⊃BBC (instead of Bk⊃BC). However, the reasoner believes BC⊃C, and since he is of type 4, he is regular, hence he will believe BBC⊃BC. Believing this and Bk⊃BBC, he will believe Bk⊃BC. He also believes k≡(Bk⊃BC), so he will believe k. Then he will believe Bk, and since he will believe Bk⊃BC, he will believe BC. But he also believes BC⊃C, hence he will believe C.
Forever Undecided Page 10
