Philosophy Now Forum

A syllogism problem.

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


Is this valid or invalid and why? Please explain by using Venn Diagram also.

I thought it was valid, but something wrong when using Venn Diagram. I don't under stand, please help. Thank you.

It's invalid. For example, think of a category (eg. Food, or Animals), and any two distinct sub-categories (eg. Vegetables and Meat, or Birds and Insects).

All vegetables are food, and some food is meat; so some vegetables are meat.
All birds are animals, and some animals are insects, so some birds are insects.

(Sorry, I can't be arsed to find or draw a venn diagram.)

B(x) = x is mammal
A(x) = x is an Antelope.
C(x) = x is a Cat.

"All A are B"
(All antelope are mammals. Check.)

"Some B are C"
(Some mammals are cats. Check.)

"Therefore some A are C"
(...some antelope are cats??!)

Cecily wrote:A syllogism problem.

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


Is this valid or invalid and why? Please explain by using Venn Diagram also.

I thought it was valid, but something wrong when using Venn Diagram. I don't under stand, please help. Thank you.

I think the problem is this.If we plug in some actual terms the problem may become apparent.

For example:

All dogs are vertebrates

All vertebrates are animals

Therefore all dogs are animals.

This satisfies All A's are B's. All B's are C's All A's are C's. If the syllogism were of the universal affirmative type then it would be valid. The problem with your formulation is that allows the exclusion of some members of a group that should not be excluded. This can be done by using the word "some".

For example:

All dogs are vertebrates

Some vertebrates are animals

Therefore. Some dogs and animals.

Imagine a big circle with all Bs and a smaller circle of all As inside the circle of Bs.

If some Bs are Cs then the circle of Bs and the circle of Cs have an intersection.

Now:

The small circle of As can be either inside of the Bs and not touch the intersection.
Or it can be inside the intersection.
Or it can be partly in the intersection and partly in the circle of Bs.

(In the third case it will make an additional intersection with the other intersection.)

So we cannot say anything definite about Cs. Three options are available.

Is it possible to draw here ? I don´t see a pen or a pencil.

mickthinks wrote:It's invalid. For example, think of a category (eg. Food, or Animals), and any two distinct sub-categories (eg. Vegetables and Meat, or Birds and Insects).

All vegetables are food, and some food is meat; so some vegetables are meat.
All birds are animals, and some animals are insects, so some birds are insects.

(Sorry, I can't be arsed to find or draw a venn diagram.)

This is one of the three possible options.

A is a small circle inside B circle and it does not touch the intersection of B and C.

But we can find an example of A being inside the intersection of B and C.
And an example of A being partly in B and partly in the intersection between B and C.

Ginkgo wrote: This satisfies All A's are B's. All B's are C's All A's are C's. If the syllogism were of the universal affirmative type then it would be valid.

The universal affirmative type? You may be over-thinking this.

It is not the case that "some A are C" is necessarily false. It could be true. The problem with the thread is that the original poster may be asking if it is necessarily false, and if not it must be necessarily true. When the reality is the conclusion "..therefore some A are C" does not follow logically. This does not mean it is false, it is just that drawing that conclusion definitively would be a bad idea, since it may be false. (But it could be true in some situations!)

The three pictures on the right are three possible correct pictures of

all A are B and some B are C.

But what about the pictures on the left ?

What do you need them for ?


The syllogism propsed is not constructed correctly and that is why there is no conclusion.
I will try to figure out what the flaw is and offer a correct version.

If I remember correctly the matrix for this type of syllogism was:

All/some of Major term is Middle term.
All/some of Minor term is Middle term.

Therefore All/some of Major term is minor term.

duszek wrote:The three pictures on the right are three possible correct pictures of

all A are B and some B are C.

But what about the pictures on the left ?

What do you need them for ?

The english word "Some" has no analogy in formal logic.

1.
"Some" could be translated as a proper subset. Some mammals are cats. (Cats proper subset of mammals. Leftside case in the venn diagrams.)

2.
"Some" could be translated as an overlap. Some mammals are brown. (Brown things overlap mammals, but trees and rocks are also brown. Rightside case in the venn diagrams)

Kuznetzova wrote:The english word "Some" has no analogy in formal logic.
...

I thought it the ∃? As in ∃x - 'there is at least one object x'.

I will try to figure it out with a logician friend on another forum.

At the moment:

Look at the conversions:

Some mammals are cats. All cats are mammals.

Some cats are brown. Some brown things are cats.

Arising_uk wrote:

Kuznetzova wrote:The english word "Some" has no analogy in formal logic.
...

I thought it the ∃? As in ∃x - 'there is at least one object x'.

Okay Arising_uk that's very clever. I got into a very long (hours long) argument with someone outside this forum regarding how exactly to depict the english word "some" in formal notation. This happened more than once. The hottest contention in these arguments centers around the connotation of the word "some" to also entail "not all".

You will notice that "∃x such that..." does not specifically rule out every x also having this property. The contention comes in when we want it to be "some but not all". It is the "not all" connotation that makes this complicated for everyone.

There exists an integer N, that can be represented as a prime or a product of primes. We cannot conclude that therefore, "some" integers have a prime decomposition. Again, we cannot say that because the word "some" connotes that there exist integers without this property as well. The fact of the matter is illustrated most plainly here.

∃n in Z such that P(n) {A}
For all n in Z, P(n) {B}

In formal notation, {A} and {B} can both be true. There are situations where {A} can be true and {B} can be false. e.g. let the property P(x) = x is odd.

Kuznetzova wrote:Okay Arising_uk that's very clever. ...

Wasn't meant to be, Its just that in my limited knowledge of formal logic I just thought this was the analogy for "some".

I got into a very long (hours long) argument with someone outside this forum regarding how exactly to depict the english word "some" in formal notation. This happened more than once. The hottest contention in these arguments centers around the connotation of the word "some" to also entail "not all".

I can understand this interpretation and agree that Formal Logic does not capture all the nuances of a natural language. Just look at the discussions about the material conditional and "If...then..."

You will notice that "∃x such that..." does not specifically rule out every x also having this property. The contention comes in when we want it to be "some but not all". It is the "not all" connotation that makes this complicated for everyone.

Hmm... I think I get what you say but when would we say this? As if all have the property then ∃x is also true and if they don't ∃x is still true. Its why we use 'some' in the first place.

There exists an integer N, that can be represented as a prime or a product of primes. We cannot conclude that therefore, "some" integers have a prime decomposition. Again, we cannot say that because the word "some" connotes that there exist integers without this property as well. The fact of the matter is illustrated most plainly here.

∃n in Z such that P(n) {A}
For all n in Z, P(n) {B}

In formal notation, {A} and {B} can both be true. There are situations where {A} can be true and {B} can be false. e.g. let the property P(x) = x is odd.

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?

I have figured out this:

not everything that appears to be a syllogism is a syllogism.

A syllogism is an argument which leads to only one clear conclusion.
If something leads to several possible conclusions then it does not lead to any conclusion at all.

An example of what I mean by "something that appears to be a syllogism":

Some cats are white.
Some dogs are black.
Therefore some mice are ... ?

It appears to be a syllogism because its structure reminds one of a syllogism but it is not one.

Some logicians have studied syllogisms and have made tables of them.
We can learn from them.

One reliable and waterproof syllogism is Barbara.

Cecily wrote:A syllogism problem.

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

If A and B switched places in the first part of the question to where it were phrased this way :
All 'B' are 'A', some 'B' are 'C', so some 'A' are 'C'.

7hen, it would make sense to me.