All A are B, Some B are C, therefore some A are C?

[206UE]

My reply hasn't even shown up yet, and I already feel stupid. Changing the place of A and B doesn't seem to have any effect at all. I don't know what I was thinking.

[206UE]

I must have been thinking : just because all A are B doesn't intend that All B are A.

[Ginkgo]

Just because all A are B doesn't intend that All B are A.

This would probably be right in terms of the contradictory ie., Some A is not B

The inverse need not be the same as the original proposition. I think this is how it works.

[duszek]

I see a huge difference between all A are B and all B are A.

All birds are animals.
All animals are birds.


All birds are animals.
Some birds are canaries.
Therefore some animals are canaries.

All animals are birds.
Some birds are canaries.
Therefore some animals are canaries.



You have fixed the syllogism, 206UE.

[duszek]

I think that

All A are B and all B are A

is only true in the case of synonyms, when A and B are two different names for exactly the same thing.

All napkins are serviettes and all serviettes are napkins.

If we assume that exact synonyms really exist.

[Kuznetzova]

Logic as applied to Maths is not my forte, but I'm not sure what the problem is here or I don't understand what you are trying to demonstrate?

Conversational english (to remove some of your confusion, formalists call this "natural language").

In conversational english there are these three things that are all unequal.

1. "all"
2. "some"
3. "none"

You know this because you yourself are an english speaker. If you assert that ∃x s.t. is directly translatable into english as: "there are some x such that", you would be completely wrong with that assertion.

The word "some" in english does not mean "all". In conversational english, "some" excludes "'all". You know this plainly.

However, in formal logic, "∃x s.t." does not exclude the for-all case whatsoever. Let me prove this to you. Pick a property that is true of all integers. Call this property P. I can safely assert ∃x s.t. P(x). Think about it. Yes I can assert this. In fact, I can assert it with logical impunity. I know there is one such x, because this property is true of all of them. I can safely select one of them and then assert the existence of it. Voila.

And of course, you have can "∃x s.t." and it not be that case that it is true for all x. Define property V(x) as "x is evenly divisible by 17". I can assert ∃x s.t V(x), instantiate x with 34 and be on my merry way. In no shape or form does V(x) hold for all x.

So I have shown you the following.

1. ∃x in a case which it is also true for all x.
2. ∃x in a case which it is also false for particular x's.

Go back to the list I gave above. 1. All. 2. Some. 3. None.

With complete clarity, you should see now that the conversational word "some" is not translatable to "∃x". They have two different meanings.

[Kuznetzova]

Logic as applied to Maths is not my forte, but I'm not sure what the problem is here or I don't understand what you are trying to demonstrate?

The conversational word "some" is not translatable to "∃x". They have two different meanings.

Do not argue with me further. Read above.

[duszek]

I see a huge difference between all A are B and all B are A.

All birds are animals.
All animals are birds.


All birds are animals.
Some birds are canaries.
Therefore some animals are canaries.

All animals are birds.
Some birds are canaries.
Therefore some animals are canaries.



You have fixed the syllogism, 206UE.

I was wrong, the syllogism is not fixed because "some" is still ambiguous.
So no clear-cut conclusion can be drawn from the two premises.


Some birds are canaries.

(All canaries are birds, canaries are a subsection of birds.)

Some birds are fast runners.

(Not all fast runners are birds, emus are fast running birds, but cheetahs are fast running cats. Running birds are a cross-section between birds and runners.)

Before we resolve this ambiguity of the word "some" the syllogism does not lead to a clear-cut conclusion and remains defective.

[Impenitent]

I didn't know the British women's track team competed in syllogisms...

-Imp

[duszek]

British or continental, man or woman, we all use syllogisms all the time when we speak, without noticing it.

It is good to make a reasoning explicit and to examine it.

It helps to create a clear mind and to avoid misunderstandings, misery and chaos.

Team work and cooperation instead of competition.

Please contribute.

[Arising_uk]

Conversational english (to remove some of your confusion, formalists call this "natural language").

In conversational english there are these three things that are all unequal.

1. "all"
2. "some"
3. "none"

No need for this as I understand what you are saying.

My question was because I do not understand Maths so i was confused how you could say that "{A} and {B} can both be true" as, as you say, Vx makes ∃x redundant?

You know this because you yourself are an english speaker. If you assert that ∃x s.t. is directly translatable into english as: "there are some x such that", you would be completely wrong with that assertion.

I do not assert such a thing as the problems of logical translation of equivalent natural language terms is well understood. All I ever said was that your assertion that there is no such logical formalism was wrong. I also said that the translation is "There is at least one x such that ...". not "there are some x such that ..." as this would be assuming the term under formalization?

I'm confused, in a formal way, what you mean by "∃x s.t."? What does this mean? I.e. what is "s.t." in formal logical notation? As I have it as "∃x(p(X)). i.e. there is at least one object such that it has the property p".

The word "some" in english does not mean "all". In conversational english, "some" excludes "'all". You know this plainly.

Which is why I asked you why in your example of the set Z you said both could be true?

However, in formal logic, "∃x s.t." does not exclude the for-all case whatsoever. Let me prove this to you. Pick a property that is true of all integers. Call this property P. I can safely assert ∃x s.t. P(x). Think about it. Yes I can assert this. In fact, I can assert it with logical impunity. I know there is one such x, because this property is true of all of them. I can safely select one of them and then assert the existence of it. Voila.

But this is exactly what I was pointing out to you!? That some does not dis-include there being an all and that that despite what people think about the natural language expression that this applies in its case as well. I may well say that some things are X and I may not know that all those things are X and if I do find out that all things are X it does not make my some statement false. It just makes it redundant if I assert it in future.

Still dont understand what the "s.t." stands for?

And of course, you have can "∃x s.t." and it not be that case that it is true for all x. Define property V(x) as "x is evenly divisible by 17". I can assert ∃x s.t V(x), instantiate x with 34 and be on my merry way. In no shape or form does V(x) hold for all x.

Okay, I think I get what you are trying to say but I have a couple of issues with your formalization. If you're going to say "For all x, x is evenly divisible by 17" the notation is Vx(P(x)) where P stand for "evenly divisible by 17" and it will only apply to the set where the numbers are all divisible by 17. Any other interpretation and it will be false. You can assert ∃x(P(x)) and i agree that there is nowhere in formal logic that from an ∃x you can deduce a Vx but there is nothing to say that it it not the case that Vx.

So I have shown you the following.

1. ∃x in a case which it is also true for all x.
2. ∃x in a case which it is also false for particular x's.

Sorry, I'm really slow in Maths so can you show me where again? As I get where an all x means at least one x but still don't get where at least one x is supposed to mean that if another x is found that does not meet the criteria that the former is false?

Go back to the list I gave above. 1. All. 2. Some. 3. None.

With complete clarity, you should see now that the conversational word "some" is not translatable to "∃x". They have two different meanings.

Never said it was, just said that your statement that there is no formal notation for "some" was untrue. That it doesn't capture all the nuances of a natural language is why we don't talk in logical notation and has been the issue of philosophical logic for yonks.

[Arising_uk]

The conversational word "some" is not translatable to "∃x". They have two different meanings.

Do not argue with me further. Read above.

See above and what a ridiculous request.

[duszek]

Perhaps the problem is not the word "some" but the predication that is used.

Predication means - more or less - the quality that we attach to a subject we are talking about.

The dog is big.

the dog = subject
is big = predication

Perhaps the only reasonable predications are universals.

(Some birds are yellow.)

A subsection of the subject not being a reasonable predication.

(Some birds are canaries.)


In that case

Some A are B

could only mean

a cross-section between A and B.

[duszek]

A friendly man helped me to solve the problem:

the middle term (the term that occurs in both premises, in our case B) has to be taken in at least one premise universally.

If all A are B then some B are A.

B is not taken universally here.

"Some B are C."

B is not taken universally either.

That is why nothing can follow from these two premises.

Somone suggested: let´s say "All B are A".

Which is by no means the same as saying "All A are B.".

So we have:

All B are A.
Some B are C.

B is taken universally in the first premise.

My friend changed the second premise into: Some C are B. (in order to get the so-called first figure, I suppose).

So now we have:

All B are A.
Some C are B.
Conclusion: Some C are A.