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Re: Negation of a relational quantified term?

I gave you the way to prove the desired conclusion. I added the owned to dogs and the owned to beagles and made the beagles proposition general.

Not all dogs or beagles have owners, of course.
If all you have is that some (owned) beagles are (owned) dogs, then the conclusion would be particular too, i.e. some owners of beagles are kind.

Re: Negation of a relational quantified term?

I should add, in case it is not obvious, that you cannot always get a conclusion from given premises. E.g. if your minor premise is really only that some dogs are beagles, you cannot be sure that the dogs that are beagles are owned, or that the beagles that are dogs are owned, and so you have no conclusion at all about owners of beagles.

As for my previous particular conclusion, viz. some owners of beagles are kind - come to think of it, that could still be every owner of beagles is kind, provided we understand that only beagles that are known to be dogs and owned are intended, since we are given that all owners of dogs are kind.

Your concern here and before is evidently with terms within terms. As shown here and above, the way to deal with these is to draw out the implicit propositions. But also to accept that in some cases, the desired conclusion is not possible.

Re: Negation of a relational quantified term?

Avi, this is how I would tackle it, but the correctness of the proof depends on "a dog" (undistributed) becoming "any dog" (distributed).

1. Every owner of a dog is a kind person. ("a dog" is undistributed)
2. Some dogs are beagles.
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3. Every unkind person is a non[owner of a dog], by (1), contraposition.

Now the predicate "non[owner of a dog]" when expanded out, so to speak, becomes

"a person who doesn't own any dog", where "dog" is now distributed. So (3) becomes

3. Every unkind person is a person who doesn't own any dog.

Now since we are talking about all dogs, we can substitute what is predicated of "some dogs" in premise 2, ie, beagles.

Thus :

4. Every unkind person is a person who doesn't own any beagle, by (2) and (3).

Finally, we can obvert (4) to get

5. No unkind person owns a beagle.

Where in (5) "doesn't own" has been contradicted to get "owns" and "any beagle" has been changed back from distributed to undistributed.

Re: Negation of a relational quantified term?

You mention the process of substitution in your book "Future Logic". It seems to me that all mediate inference is based on it. The rules of the traditional syllogism are really just a special case of the more general principle of substitution.

Re: Negation of a relational quantified term?

Contraposition of All dog-owners are kind persons All not(kind person) are non(dog-owners). It is not dog that is distributed in the original subject, but dog-owners. The negative term non(dog-owner) means not any dog-owner. A dog-owner in either case is an owner of some dog (dog undistributed).

Given your ONLY two premises, all dog-owners are kind and some dogs are beagles, you cannot draw the desired conclusion by substitution or any other means, for reasons already explained.

Non(dog-owner) cannot be changed to non(beagle-owner), because according to your statement, only some beagles are dogs (i.e. you refused all beagles are dogs). You cannot be sure that the beagles you have in mind are the dog variety, and therefore you cannot place the term beagle in lieu of the term dog.

Re: Negation of a relational quantified term?

Avi Sion
Contraposition of All dog-owners are kind persons All not(kind person) are non(dog-owners). It is not dog that is distributed in the original subject, but dog-owners. The negative term non(dog-owner) means not any dog-owner. A dog-owner in either case is an owner of some dog (dog undistributed).

Given your ONLY two premises, all dog-owners are kind and some dogs are beagles, you cannot draw the desired conclusion by substitution or any other means, for reasons already explained.

Non(dog-owner) cannot be changed to non(beagle-owner), because according to your statement, only some beagles are dogs (i.e. you refused all beagles are dogs). You cannot be sure that the beagles you have in mind are the dog variety, and therefore you cannot place the term beagle in lieu of the term dog.
Given your ONLY two premises, all dog-owners are kind and some dogs are beagles, you cannot draw the desired conclusion by substitution or any other means, for reasons already explained.

Avi, with respect, you CAN draw the conclusion I derived, but not using standard syllogistic, because it doesn't allow for complex terms (including relational ones). I'm a little surprised that you're taking this line because in your book you mention both complex terms and substitution, and in fact you've also demonstrated in another thread on this forum that such inferences are possible.

Here:

http://pub12.bravenet.com/forum/static/show.php?usernum=1014493548&frmid=6526&msgid=900296&cmd=show

Also, the conclusion can be derived using modern predicate logic fairly easily, and since you deny that modern logic is no way superior to the traditional logic (and I agree, by the way), it follows that the conclusion can also be drawn from those two premises ALONE using term logic.

Non(dog-owner) cannot be changed to non(beagle-owner), because according to your statement, only some beagles are dogs (i.e. you refused all beagles are dogs). You cannot be sure that the beagles you have in mind are the dog variety, and therefore you cannot place the term beagle in lieu of the term dog.

I was objecting to your inference that "all beagles are dogs" can be drawn FORMALLY from "some dogs are beagles". You don't need me to tell you that All B is D cannot be inferred from Some D is B.

The philosopher Hanoch Ben-Yami has written an interesting book called "Logic & Natural Language", which you can download from here :

http://publications.ceu.edu/publications/ben-yami/2004/15107

There is also the work by Fred Sommers and George Englebretsen. They developed a "new syllogistic" which is capable of anything modern predicate logic can do, and more. It uses a simple algebra of plus and minus signs, but there's no reason why it can't be expressed using the standard notation of traditional logic (and in fact that's what I'm working on, because I know most people are put off by mathematics).

Re: Negation of a relational quantified term?

Joe, sorry to disappoint you. But so far you have not managed to convince me. I'm not taking any line - I'm open-minded. But so far you have not shown me a convincing proof. Try formulating your thesis without dogs and beagles and without owners. Just symbols A, B, C, etc. for the terms, but still in ordinary language for the rest of it. You will, I wager, see for yourself that it can't be done.

One more comment I can make is that you do not realize that the negation of a complex term of the sort you have put forward is broader than you conceive. For example, the negation of [a creature that likes some dogs] is not [a creature that dislikes all dogs] - for it includes all other things in the universe. Or to put it differently, or more narrowly, not to like is not to dislike, because there is a third alternative, viz. indifference. Similarly with your other terms.