Question: Suppose a Looking-Glass logician believes that all gryphons have wings. Does it follow that there are any gryphons?
1. Understanding the problem
To quote Hamlet, the difficulty of this question is largely due to the ambiguity of the "words, words, words" (II. 2-192). The question is presented in an over-the-top, verbose way which is meant to confuse the reader. Thus, in order to more clearly see the logical flow of the statements, we should convert the given information into symbolic form.
1. A Looking-Glass logician is completely honest. He will claim those and only those statements which he actually believes.
2. Whenever a Looking-Glass logician claims a statement to be true, then he also claims that he doesn’t believe the statement.
3. Given any true statement, the Looking-Glass logician always claims that he believes the statement.
4. If a Looking-Glass logician believes something, then he cannot also believe its opposite.
5. Given any statement, a Looking-Glass logician either believes the statement or he believes its opposite.
Let x be a statement. Let T(x) mean that a statement is true. Let B(x) mean that he (the Looking-Glass logician) believes a statement. Let C(x) be that he claims a statement.
The conditions can therefore be translated into:
1. B(x) => C(x) ^ C(x) => B(x). That is, C(x) iff B(x)
2. C(T(x)) => ¬B(x)
3. T(x) => C(B(T(x))). Note: it is not T(x) => C(B(x)), even though this would be a literal translation -- since we previously assumed that the statement was true. Hence, x => T(x).
4. B(x) ∨exclusive B(¬x). Where ∨ is the exclusive-or (only one of the predicates is true at any given time)
5. x => B(x) ∨exclusive B(¬x).
From a quick observation of the data so far, I immediately see that conditions 4 and 5 seem to be closely linked. Condition 4 tells us when the predicates are true, and condition 5 finishes the "incomplete" thought of condition 4 and puts it in a formalized implication statement. We also note that there are nuances of meaning between believing a statement and believing a statement to be true (they might be completely two different things). Also, thinking back to nested functions in calculus and programming, things like C(B(T(x))) can be thought of as a composition of C(u) where u = B(w) and w = T(x). This might help us later on if we wanted to break things down into smaller components.
From a quick observation of the data so far, I immediately see that conditions 4 and 5 seem to be closely linked. Condition 4 tells us when the predicates are true, and condition 5 finishes the "incomplete" thought of condition 4 and puts it in a formalized implication statement. We also note that there are nuances of meaning between believing a statement and believing a statement to be true (they might be completely two different things). Also, thinking back to nested functions in calculus and programming, things like C(B(T(x))) can be thought of as a composition of C(u) where u = B(w) and w = T(x). This might help us later on if we wanted to break things down into smaller components.
Now that we've translated the conditions, we should also translate the original question into symbolic form. Let x be the statement that "all gryphons have wings." Then what we are given is that: B(x) (from "Suppose a Looking-Glass logician believes that all gryphons have wings"), and we have to find out whether ∃ gryphons (which is the actual question -- "Does it follow that there are any gryphons?")
Hence, we are trying to solve: does B(x) => ∃ gryphons? We should be able to solve this, using the conditions given in the question. After all, everything that is given to us is perfectly reasonable within the context of the problem.
2. Devising a Plan
I am going to start looking for a link between the thing we are trying to solve and the conditions given. Since we are only given B(x), it seems like there are going to be many intermediate steps between B(x) and ∃ gryphons (at this point we do not know how to get there). Given that there are 5 conditions in total, and Humpty Dumpty seems to want us to use most of them (what a tricky guy), we should be able to form theorems, or at least, corollaries from the given information. (Step 1) This will help us get from a... to a1... to a2... to b!
Given that the problem is presented in a story that is meant for all audiences (under 80), it seems unlikely that the author meant for the question to be more complicated than what it is. Thus, I am assuming that the author is assuming that we have no mathematical knowledge and skills (forget about calculus, linear algebra, statistics!) other than logic. The scope of the problem should be limited in this sense -- and devising a plan should be made easier by knowing that we are not expected to know anything other than the conditions given. Based on what we learned in class, about logical statements and their negations, I am sure that plugging in B(x) into certain statements, negating them, taking the contrapositives, and generally tweaking things around will eventually generate something meaningful. (Step 2)
But what is meaningful in this case? Well, since we want to find a connection between the conditions and whether there exists gryphons, we have to find out whether Looking-Glass logicians believe the right things -- that is, whether they believe things that are true or false. If we know this, then the validity of the phrase, "all gryphons have wings" will tell us whether or not there exists gryphons. There are two cases:
a) He believes true things. Then all gryphons have wings is TRUE. Since we are trying to prove the existential, that ∃ a gryphon... knowing ∀gryphons => the gryphon has wings does not tell us that there exists a gryphon!
b) He believes false things. Then all gryphons have wings is FALSE <=> "some gryphons have wings" <=> ∃ gryphons ^ the gryphon does not have wings. Then this implies that there exists a gryphon (a wingless gryphon is still a gryphon nonetheless!) (Step 3)
But what is meaningful in this case? Well, since we want to find a connection between the conditions and whether there exists gryphons, we have to find out whether Looking-Glass logicians believe the right things -- that is, whether they believe things that are true or false. If we know this, then the validity of the phrase, "all gryphons have wings" will tell us whether or not there exists gryphons. There are two cases:
a) He believes true things. Then all gryphons have wings is TRUE. Since we are trying to prove the existential, that ∃ a gryphon... knowing ∀gryphons => the gryphon has wings does not tell us that there exists a gryphon!
b) He believes false things. Then all gryphons have wings is FALSE <=> "some gryphons have wings" <=> ∃ gryphons ^ the gryphon does not have wings. Then this implies that there exists a gryphon (a wingless gryphon is still a gryphon nonetheless!) (Step 3)
After manipulating the conditions to try to deduce the answer, we should go back and make sure that it logically makes sense. Since one thing leads to another... it would be quite horrendous if it turns out that we built our assumptions on falsity. (Step 4)
Hence, this is the approach that I will be taking in solving this problem:
1. Make theorems or corollaries from the given conditions.
2. Plug in the assumptions from the question into the theorems, corollaries, and conditions and try to find a meaningful chain of statements that will eventually generate what I am looking for.
3. After finding out whether the Looking-Glass logician believes true statements or false statements, I will be able to deduce whether there exists at least one gryphon.
3. After finding out whether the Looking-Glass logician believes true statements or false statements, I will be able to deduce whether there exists at least one gryphon.
4. Go back and check the logic of the assumptions that I have made. If there is a single link that is off, I have to fix it (or risk having my entire solution be wrong!)
3. Carrying Out the Plan
3. Carrying Out the Plan
1. B(x) => C(x) ^ C(x) => B(x). That is, C(x) iff B(x)
2. C(T(x)) => ¬B(x)
3. T(x) => C(B(T(x))). Note: it is not T(x) => C(B(x)), even though this would be a literal translation -- since we previously assumed that the statement was true. Hence, x => T(x).
4. B(x) ∨exclusive B(¬x). Where ∨ is the exclusive-or (only one of the predicates is true at any given time)
5. x => B(x) ∨exclusive B(¬x).
It seems that combining condition 1 with conditions 2 and 3 yields particularly interesting results:
Corollary I: Whenever a Looking-Glass logician believes something, then he believes that he doesn't believe it. i.e, B(x) => B(¬B(x))
Proof:
If a Looking-Glass logician believes something, then by definition B(x) implies C(x) (from condition 1). Hence, if he believes something, he will claim it. And since he claims that something, we can use condition 2. If the Looking-Glass logician claims T(x) is true, then ¬B(x) follows. That is, if he claims a statement to be true, he will also not believe it. Using C(x) => B(x) (the second predicate of condition 1), we plug in the result from condition 2, C(T(x)) => ¬B(x) into condition 1 such that x = ¬B(x). Then C(¬B(x)) => B(¬B(x)). Hence, whenever a Looking-Glass logician believes something, then he also believes that he doesn't believe it. (i.e. he is delusional or inaccurate)
Corollary II: Given any true statement, a Looking-Glass logician believes that he does believe the statement. i.e, T(x) => B(B(T(x)))
Proof:
This case is slightly different -- whereas for Corollary I, we claimed that if he claims that a statement is true (it might not actually be true, since he is only claiming that it is), then he also does not believe it. But now, we are saying that if the statement were true, he believes that he believes it. From condition 3: T(x) => C(B(T(x))). Then from the second predicate of condition 1, C(x) => B(x) we have C(B(T(x))), where x = B(T(x)) implying B(B(T(x)). Thus given any true statement, a Looking-Glass logician believes that he does believe the statement.
As you can see, combining condition 1 and condition 2 gives us Corollary I and combining conditions 1 and 3 give us Corollary II. Now, there appears to be a contradiction! By the second corollary, given a true statement, a Looking-Glass logician believes that he does believe the statement. But the first corollary says that whenever a Looking-Glass logician believes something, then he believes that he doesn't believe it. So how can he believe that he does believe it AND believe that he doesn't believe it at the same time? (This seems illogical. As well, condition 4 says that he can only believe a statement or its opposite -- not both at the same time -- so what are we to do?) Remembering that we first assumed that T(x) was true (from Corollary II), this must mean that T(x) cannot be so. Thus, T(x) has to be false! Hence, Looking-Glass logicians only believe false statements.
Since they only believe false statements, then this implies that Looking-Glass logicians must believe the false statement that "all gryphons have wings." The negation of this would be clearer if this phrase was converted to symbols. Let the phrase be denoted ∀G(y) => W(y) where G(y) means that y is a gryphon and W(y) mean that y has wings. Then the negation would be ∃G(y) ^ ¬W(y).
Therefore, the existence of a wingless gryphon still proves that there exists at least one gryphon! The answer to Suppose a Looking-Glass logician believes that all gryphons have wings. Does it follow that there are any gryphons? is true. From the assumption that a Looking-Glass logician believes that all gryphons have wings, it necessarily follows that there are indeed gryphons.
4. Looking Back
It seems that combining condition 1 with conditions 2 and 3 yields particularly interesting results:
Corollary I: Whenever a Looking-Glass logician believes something, then he believes that he doesn't believe it. i.e, B(x) => B(¬B(x))
Proof:
If a Looking-Glass logician believes something, then by definition B(x) implies C(x) (from condition 1). Hence, if he believes something, he will claim it. And since he claims that something, we can use condition 2. If the Looking-Glass logician claims T(x) is true, then ¬B(x) follows. That is, if he claims a statement to be true, he will also not believe it. Using C(x) => B(x) (the second predicate of condition 1), we plug in the result from condition 2, C(T(x)) => ¬B(x) into condition 1 such that x = ¬B(x). Then C(¬B(x)) => B(¬B(x)). Hence, whenever a Looking-Glass logician believes something, then he also believes that he doesn't believe it. (i.e. he is delusional or inaccurate)
Corollary II: Given any true statement, a Looking-Glass logician believes that he does believe the statement. i.e, T(x) => B(B(T(x)))
Proof:
This case is slightly different -- whereas for Corollary I, we claimed that if he claims that a statement is true (it might not actually be true, since he is only claiming that it is), then he also does not believe it. But now, we are saying that if the statement were true, he believes that he believes it. From condition 3: T(x) => C(B(T(x))). Then from the second predicate of condition 1, C(x) => B(x) we have C(B(T(x))), where x = B(T(x)) implying B(B(T(x)). Thus given any true statement, a Looking-Glass logician believes that he does believe the statement.
As you can see, combining condition 1 and condition 2 gives us Corollary I and combining conditions 1 and 3 give us Corollary II. Now, there appears to be a contradiction! By the second corollary, given a true statement, a Looking-Glass logician believes that he does believe the statement. But the first corollary says that whenever a Looking-Glass logician believes something, then he believes that he doesn't believe it. So how can he believe that he does believe it AND believe that he doesn't believe it at the same time? (This seems illogical. As well, condition 4 says that he can only believe a statement or its opposite -- not both at the same time -- so what are we to do?) Remembering that we first assumed that T(x) was true (from Corollary II), this must mean that T(x) cannot be so. Thus, T(x) has to be false! Hence, Looking-Glass logicians only believe false statements.
Since they only believe false statements, then this implies that Looking-Glass logicians must believe the false statement that "all gryphons have wings." The negation of this would be clearer if this phrase was converted to symbols. Let the phrase be denoted ∀G(y) => W(y) where G(y) means that y is a gryphon and W(y) mean that y has wings. Then the negation would be ∃G(y) ^ ¬W(y).
Therefore, the existence of a wingless gryphon still proves that there exists at least one gryphon! The answer to Suppose a Looking-Glass logician believes that all gryphons have wings. Does it follow that there are any gryphons? is true. From the assumption that a Looking-Glass logician believes that all gryphons have wings, it necessarily follows that there are indeed gryphons.
4. Looking Back
Checking the answers, everything seems to fit together. We started the problem with a motivating question and a whole lot of fragmented conditions. After combining several of the conditions into more relevant corollaries, we saw that Looking-Glass logicians always believe false statements. This indirect method of accumulating information resulted in us finding out whether or not Looking-Glass logicians believe TRUE or FALSE statements. For the case where the statement is true, that "all gryphons have wings," we would not be able to conclude whether there exists a gryphon, since "for all" does not mean the same thing as "there exists" (Quite the opposite, actually). The only way for us to prove the implication in the question would be to arrive at a false statement -- that "all gryphons have wing" is false, since this will introduce ∃ (the negation of ∀ is ∃). Seeing that our conclusion matched our predicted hypothesis in 2. Devising a Plan, we can conclude that we have found the solution!
Also, just to double check and reassure ourselves, we used all of the conditions. Conditions 1, 2, and 3 were combined profitably to make important corollaries. We also used condition 4 (B(x) ∨exclusive B(¬x)), which is more or less exactly the same as condition 5; x => B(x) ∨exclusive B(¬x). Since we can treat condition 4 and 5 as having the same effect in this case, this means that we have used conditions 1, 2, 3, 4, and 5! This is definitely a good sign that we solved the problem that way it was meant to be solved. (Problems of mathematical or logical nature often want you to use all of the information that is given). Based on our reasoning as outlined throughout this problem-solving episode, it seems like we have arrived at a reasonable conclusion (and just in the nick of time -- thirty minutes before the SLOG deadline!)
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Since I have less than half an hour remaining until I have to submit this post, I will just quickly share my final thoughts on CSC165. Honestly, it was quite a fun experience. Initially, I was afraid that a class that concentrated solely on mathematical expression and symbols was going to be at least a bit dry. However, this was not the case. The course content was actually interesting and the stellar combination of lectures, tutorials, and (well purposed) assignments helped me see things in a more defined way. Specifically, I chose to do a logical puzzle for my problem solving episode because in the past, these questions always intimidated me. I would be the type of person who would trip up over the "exact" meaning of words, and get lost while trying to decipher what things meant. Now that I am able to parse words into symbols, and see the logical structure of things hiding behind words (not that words aren't concise.. but symbols are often better for making the logic of an argument explicit), I feel more confident as a student. The tools and concepts learned from this course will definitely help me move on to bigger and better things. And for this, I am thankful.
Cheers to a great class.
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