A---->B<-----C Forum

[mbmoss]

All A are B and all C are B

written A--->B<---C

I found somewhere that the inference that can be made is that

Not A Some Not C

which you would say as: some things that are not A are not C

is this true?

if so why?

Thanks

[HorseThief]

You can't make that inference. All you know is that A -->B and C --> B.

It could be the case that A is a subset of B, while C = B. If, additionally, B is everything (the entire space), then A divides B into 2 parts: the A part and the not-A part. Everything in the not-A part is in B, which is also C.

Sure, you could construct a case where what you say is true, but from a logic standpoint, we can only say that sometimes the statement is true.

[Christine (MLSAT)]

That's not a valid inference. Now, it would be valid to say something like "if there are any Not-B's in the world, then some Not-A's are Not-C's".

You can rewrite your original two statements as one 'or' statement, since either of the triggers will work to get the result:
If A or C --> B

The contrapositive of this is:
~B --> ~A and ~C

So, if there is a ~B in the world, it also must have the characteristics of being both ~A and ~C. But there's no guarantee there ARE any ~Bs in the universe. What if everything is a B?

If everything is a B, then you have no idea if some ~As are also ~Cs. But if you know that at least one ~B exists, then you can say that some ~As are ~Cs.

Does that help?

[Pneumonia]

I'm sensing some lackluster prep materials here. That diagram is weird. Maybe consider something else?

[redcrosse]

That's not a valid inference. Now, it would be valid to say something like "if there are any Not-B's in the world, then some Not-A's are Not-C's".

You can rewrite your original two statements as one 'or' statement, since either of the triggers will work to get the result:
If A or C --> B

The contrapositive of this is:
~B --> ~A and ~C

So, if there is a ~B in the world, it also must have the characteristics of being both ~A and ~C. But there's no guarantee there ARE any ~Bs in the universe. What if everything is a B?

If everything is a B, then you have no idea if some ~As are also ~Cs. But if you know that at least one ~B exists, then you can say that some ~As are ~Cs.

Does that help?

Is it not true that given D<---E--->F, we can infer D some F? (For reference, see PowerScore LR Bible, pp. 328-329)

The pattern that is provided in the subject of this post is merely the contrapositive of this pattern. This can be seen if we stipulate that D=~A, E=~B, and F=~C, or ~A<---~B--->~C, of which A--->B<---C is the contrapositive (arrows reversed and all terms negated).

So, if D some F is a valid inference, then why is not ~A some ~C a valid inference? And why is it necessary to condition the inference on the existence of ~B?

[Pneumonia]

That's not a valid inference. Now, it would be valid to say something like "if there are any Not-B's in the world, then some Not-A's are Not-C's".

You can rewrite your original two statements as one 'or' statement, since either of the triggers will work to get the result:
If A or C --> B

The contrapositive of this is:
~B --> ~A and ~C

So, if there is a ~B in the world, it also must have the characteristics of being both ~A and ~C. But there's no guarantee there ARE any ~Bs in the universe. What if everything is a B?

If everything is a B, then you have no idea if some ~As are also ~Cs. But if you know that at least one ~B exists, then you can say that some ~As are ~Cs.

Does that help?

Is it not true that given D<---E--->F, we can infer D some F? (For reference, see PowerScore LR Bible, pp. 328-329)

The pattern that is provided in the subject of this post is merely the contrapositive of this pattern. This can be seen if we stipulate that D=~A, E=~B, and F=~C, or ~A<---~B--->~C, of which A--->B<---C is the contrapositive (arrows reversed and all terms negated).

So, if D some F is a valid inference, then why is not ~A some ~C a valid inference? And why is it necessary to condition the inference on the existence of ~B?

Both are valid IF the sufficient condition is met. Some D's are F's is only valid if it is true that E. The same holds true for your point.

[foggynotion]

That's not a valid inference. Now, it would be valid to say something like "if there are any Not-B's in the world, then some Not-A's are Not-C's".

You can rewrite your original two statements as one 'or' statement, since either of the triggers will work to get the result:
If A or C --> B

The contrapositive of this is:
~B --> ~A and ~C

So, if there is a ~B in the world, it also must have the characteristics of being both ~A and ~C. But there's no guarantee there ARE any ~Bs in the universe. What if everything is a B?

If everything is a B, then you have no idea if some ~As are also ~Cs. But if you know that at least one ~B exists, then you can say that some ~As are ~Cs.

Does that help?

Is it not true that given D<---E--->F, we can infer D some F? (For reference, see PowerScore LR Bible, pp. 328-329)

The pattern that is provided in the subject of this post is merely the contrapositive of this pattern. This can be seen if we stipulate that D=~A, E=~B, and F=~C, or ~A<---~B--->~C, of which A--->B<---C is the contrapositive (arrows reversed and all terms negated).

So, if D some F is a valid inference, then why is not ~A some ~C a valid inference? And why is it necessary to condition the inference on the existence of ~B?

Both are valid IF the sufficient condition is met. Some D's are F's is only valid if it is true that E. The same holds true for your point.

Pneumonia and MSLsat Christine are saying what they are saying because, in modern logic, when you have statements like "all E's are D" and "all E's are F", there is no promise that there are, in fact, any E's--only that IF there are, then they are D (and F). But when there are statements like "some D's are F", "some" implies that there is, in fact, at least one D is existence (and an F too, since this is the same as saying "some F's are D"). So "all" doesn't guarantee the existence of anything, only that IF it exists...here's what's true of it. In symbolic logic, you'd symbolize "all E's are D" in a way that would translate to read "for all x, if x is an E then x is a D". But a the statement "some D's are F" would be symbolized in a way that translates to "there exists an x such that x is both D and F" (I hope this makes sense). So I believe that's where Pneumonia and Christine are coming from.

BUT...on most questions on the test that involve statements like these, it is often included in the information somewhere that (using the E F D example) there is at least one E. For instance, there's a question about parrots that can learn to speak (I don't remember the test number off the top of my head). They say something like, "all parrots can learn to speak a few words and phrases", then they tell you about specific subgroups of parrots and what's true of them ("those parrots native to Australia are sweet-tempered", etc.). The inference they wanted you to make was something like "some sweet-tempered parrots can learn to speak", or whatever it actually was, but that's the idea. So basically in this question, because they told you that parrots exits, there's no issue anymore--you can safely draw your "some" conclusion using the "all" statement without having to worry about whether or not there are any parrots, because they told you there were. I hope this helps.

[redcrosse]

I hope this helps.

Surely, it does. Thanks for the clarity.

[Christine (MLSAT)]

That's not a valid inference. Now, it would be valid to say something like "if there are any Not-B's in the world, then some Not-A's are Not-C's".

You can rewrite your original two statements as one 'or' statement, since either of the triggers will work to get the result:
If A or C --> B

The contrapositive of this is:
~B --> ~A and ~C

So, if there is a ~B in the world, it also must have the characteristics of being both ~A and ~C. But there's no guarantee there ARE any ~Bs in the universe. What if everything is a B?

If everything is a B, then you have no idea if some ~As are also ~Cs. But if you know that at least one ~B exists, then you can say that some ~As are ~Cs.

Does that help?

Is it not true that given D<---E--->F, we can infer D some F? (For reference, see PowerScore LR Bible, pp. 328-329)

The pattern that is provided in the subject of this post is merely the contrapositive of this pattern. This can be seen if we stipulate that D=~A, E=~B, and F=~C, or ~A<---~B--->~C, of which A--->B<---C is the contrapositive (arrows reversed and all terms negated).

So, if D some F is a valid inference, then why is not ~A some ~C a valid inference? And why is it necessary to condition the inference on the existence of ~B?

In that example, we can only infer that some D are F if we know that E exist, just as pneumonia and foggynotion indicate.

Let's take an example:

All tap-dancing elephants are gray and can travel backwards in time.

From this, can we conclude that some gray things can travel backwards in time? If there are tap-dancing elephants in the world, then sure, there would have to be some gray things that travel backwards in time - all those tap-dancing elephants! But what if there aren't any? What if there are ZERO tap-dancing elephants in existence? If that were true, then we don't know anything.

Now, foggynotion has a great point that often in questions where this kind of thing comes up, the stimulus will tell you that the trigger condition exists - and if it does, then you don't have to worry about it. The question about parrots is from PT43-S2-Q22, and the second sentence of the stimulus confirms the existence of parrots ("some of those native to...").

Also, if you have a statement like "all apples are fruits", assuming that there could be no apples in the universe would surely violate the prohibition on "mak[ing] assumptions that are by commonsense standards implausible...", so making inferences based on the existence of apples would taken for granted.

But if the trigger element is not something you know to exist (either from the stimulus or because it's implausible that it doesn't exist), then you can't assume that it does.