Monday, September 13, 2004


For discussion of the Body-Minus (Tib and Tibbles/Pezdelaneous and Pezminus/The statue guy/the cat with many tails/whatever)

(Admin note: What follows is the content of the first comment, which was accidentally deleted.)

Here is a recap of what has happened so far:

Steve: Is anyone writing about the Body-minus paradox? What do you think is the weakest premise and why? I separated the paradox into 4 premises and a conclusion. I found the premise "Both Body-minus and Tibbles exist after the accident" to be the weakest premise. I gave the following reasoning:Tibbles, in the sense used in premise (1) [the premise that states that both Tibbles and Body-minus exist before the accident], no longer exists. In premise (1), the word “Tibbles” means “The set of molecules composing the cat named ‘Tibbles.’” In premise (3) it is clear that “Tibbles” no longer refers to this same set of molecules. Instead, the sense of premise (3) is that “Tibbles” means “The set of all molecules composing Tibbles, except for the molecules composing the tail,” which is exactly how Body-minus would have to be defined in premise (1). Hence, the argument uses the word “Tibbles” to refer to two distinct objects, while assuming that the meaning of the word does not change.

Then steve said: Just a follow up....Strictly speaking, by my reasoning, that premise is OK. The one that fails is "Body-minus and Tibbles are identical after the accident." This may be true, but the author must explicity make clear what is meant by "Tibbles." By my interpretation, this premise is false.

Then myself (howeman, Brendan whatever) said: (First of all, I assume that our conclusion is that a paradox exists?) The point about Tibbles (from my understanding), is that the cat continues to exist as a cat, even without his tail. Thus, Tibbles and body-minus are identical.Personally, I dont' think body-minus is a paradox at all. 1. If you can't have 2 objects occupy the same space at teh same time, then the initial set up of the problem is the flaw, becuase body-minus is in teh same space as tibbles. 2. What this really is is an exercise in names, just like the statue. If I call an object Bob, and someone else calls it Sally, then I guess we must have a paradox because two identical objects in the same space. The statue is a little more entertaining, but it's still really all semantics. I suppose the Theseus is also the same in the end, but I enjoy it more and I think it poses a more interesting problem.

Then steve replyed: Howeman, 1. If you can't have 2 objects occupy the same space at teh same time, then the initial set up of the problem is the flaw, becuase body-minus is in teh same space as tibbles. That's really a brilliant observation... I agree that there's no paradox. However, I wasn't able to put the reason why into such clear terms. It appears to be merely a matter of semantics. Actually, I had at first wanted to attack the first premise, that Body-minus exists before the accident, because just giving a name to something doesn't necessarily make it exist. In fact, you are correct: Body-minus isn't so much distinct from Tibbles, as it is a part of Tibbles. P.S. thanks for the compliment steve

Then Chris added: This is a nice discussion of the Tib-Tibbles case. Hopefully we can address some of these issues in class. (There were some other things said about making this thread, so I did)


Blogger Howeman said...

This comment has been removed by a blog administrator.

September 13, 2004 at 7:39 PM  
Blogger Chris Tillman said...

How does the argument go? Does anyone have a formulation other than the one presented in class? (There is more than one acceptable formulation of practically any argument.) Which premise is most suspect, and why? Or should we accept the conclusion? If so, what should we make of its weird consequences?

September 13, 2004 at 9:06 PM  
Blogger Steve said...

I think this paradox suffers from the fact that we are dealing with life. Suppose I have a collection of 100 1-pound distinguishable weights (let's say they are numbered 1-100). I give this collection of weights a name, say A. Now I mention to you that within this collection, I can conceive of a collection consisting just of the weights numbered 1-99. I give this collection of weights a name also, say B. It is evident that the collection of weights A is distinct from the collection of weights B. Now I remove from A the weight numbered 100, and I call the resulting collection A'. It is clear that A' is equal to B and NOT to A. We have in this case no tendency to believe that A'=A as we do in the case of Tibbles. We believe that Tibbles remains Tibbles even after the accident, because we believe that Tibbles has some inner, intangible quality of whatever sort. By cutting of Tibbles' tail, we change the physical body of Tibbles, but we do not change the internal character of Tibbles. Since we believe the inner part to be more essential, we do not believe that Tibbles no longer exists. I have thought to myself, "How much of Tibbles can we cut off and still have Tibbles?" What if I told you that Tibbles lost its head, but through some miracle of science the remainder of Tibbles was still alive (in the normal sense)? Would the Tibbles without the head be Tibbles? What if the head remained living? Would the head then be Tibbles? I think the problem is that we view the tail of Tibbles to be such an inessential component to Tibbles that it can survive without its tail.

September 14, 2004 at 11:28 AM  
Blogger Howeman said...

Okay, I've been thinking about it, and while I don't necessarily think Quine's opinion is wrong, I don't think it is necessary. My qualm is that say I showed you pezminus. Unless I told you the whole story, you would be hard pressed to see two distinct objects that look exactly the same.

The argument again (so you don't have to find your notes)

1. Pezdelaneous has an arm at T and pezminus does not
2. If (1), then pezdelaneous and pezminus are distinct
3. So, Pezminus and pezdelaneous are distinct.
4. At T', pezdelaneous and pezminus are coincident
5. Therefore (3) and (4)
6. If (5), then there are distinct coincident objects
7. So, there are distinct coiincident objects

Therefore, given the proof discussed in class, they are obviously distinct at time T because one has an arm and the other doesn't. At T', ignoring the temperospatial argument, they are coincident because they are made up of all the same pieces, and there is no longer a distinction between the two. So it would appear that I'm just agreeing a paradox exists. However, let's re-write (5) "So, (3) and (4)"

5. So, Pezminus and Pezdelaneous are distinct, and pezminus and pezdelaneous are coincident.

and 6 becomes

6. If Pezminus and Pezdelaneous are distinct, and pezminus and pezdelaneous are coincident, then there are distinct coincident objects.

Sounds logical. But, what's missing from (5) is the time at which they occured. So, (5) should really be written:

5. So, Pezminus and Pezdelaneous are distinct AT T, and pezminus and pezdelaneous are coincident AT T'.

and then (6) becomes

6. If Pezminus and Pezdelaneous are distinct at T, and pezminus and pezdelaneous are coincident at T', then we have coincident distinct objects at the same time

So what we are really doing is taking two circumstances at two different points in time, and say they are occurring together, which really isn't true.

1. If Brendan is in CAS 105, then he is in Lattimore 431
2. At time T, Brendan is in CAS 105
3. Therefore, Brendan is in Lattimore 431.
4. At T', Brendan is eating lunch in Danforth
5. Therefore, Brendan is in Danforth
6. (3) and (5)
7. If Brendan is in Lattimore and Danforth, then he is in 2 places at once
8. So, Brendan is in two places at once.

I fail to see how the treatment of time in this argument is any different from the treatment in Tib and Tibbles

So, therefore my qualm with this argument is in what would seem like the obvious premise (6)


P.S. Sorry for the length of that, I hope some of you got through it!

September 17, 2004 at 9:00 AM  
Blogger Chris Tillman said...

Is the suggestion that P and P-minus are only distinct at T, and not distinct simpliciter?

This is an excellent suggestion. But let's follow it a bit further. We considered an argument for the conclusion that P and P-minus are distinct period, not just distinct at T. Our justification of it was given in the explanaion of premise (2), which says if P has an arm at T and P-minus does not, then P and P-minus are distinct. If we really ought to replace the claim that P and P-minus are distinct with the weaker claim that P and P-minus are distinct at T, then something must be wrong with the explanaion of (2).

Here's a recap of that explanation:
Suppose P has an arm at T, and P-minus does not. (We're just assuming the 'if'-part here.) We want to show that P and P-minus are distinct (the 'then'-part). Since P has an arm at T, P has the following property:

having an arm at T.

P-minus lacks that property. Recall Leibniz's Law:

If x = y, then, for absolutely any property F, x has F just in case y has F.

Leibniz's Law implies the following:

If there is some F that x has but y lacks, then x is not identical to y.

But now let x = Pezdelaneous and y = Pezminus and F = having an arm at T.

Pezdelaneous has that property, whereas Pezminus does not. So, by Leibniz's Law, they are distinct. So (2) is true.

If it is true that P and P-minus are distinct at T but not distinct simpliciter, then something in this argument for (2) must go wrong.

So the challenge is to determine what goes wrong and why. (I am not trying at all to suggest that this challenge cannot be met.)

I want to emphasize that this is a good idea. While most would accept that P and P-minus are distinct at T (I can't think of any who wouldn't), some philosophers do reject that they are distinct simpliciter. It is instructive to try to determine what in the above argument such philosophers reject, and how plausible their rejection of (2) really is.

September 19, 2004 at 7:22 PM  
Blogger Howeman said...

But to use that definition of Leibenz's law, you are assuming that objects have a 4-D path.

Say we have a line on a ship working perfectly at T. Then it snaps. Now it does not work.

Does the fact that the part was working 10 minutes ago matter? No because the line is snapped and that's what is here and now.

Thus is the case with your P and P-minus scenario. You use Leibenz's law to say the line was working 10 minutes ago, so it is working now; the two objects were distinct before, so they are distinct now.

If you reject that objects have temporal parts--- no, not reject that objects have temporal parts, merely reject that their characteristics carry through time, then you cannot use Leibenz's law in such a way. Thus, my argument that they are distinct all the way up until T', where they coincide, and the two points never overlap.

September 23, 2004 at 9:49 PM  

Post a Comment

<< Home