True/False Logic Question (may, or may not, involve Bonsai Planets)

I ran across this question a day or so ago but didn’t see the answer and now it’s bugging the sanity outta me.

If some groks are graks and some graks are griks then some groks are griks. T/F***

If pressed, I’d answer ‘false’ but, the why of it escapes me. Changing the nonsense words to real world examples seemed the way to solve this. However, the following examples appear to conflict.

A: If some dogs are poodles and some poodles are bald then some dogs are bald.
B: If some adults are girls and some girls are infants then some adults are infants.
C: If some singers are dancers and some dancers are actors then some singers are actors.
D: If some planets are dwarfs and some dwarfs are bonsais then some planets are bonsais.***

My head hurts. In my estimation A is obviously true. B is obviously false. C depends on unknown variables. And D just makes me say 'Fut the wuck?"

Does anyone here have an explanation to back up the correct answer? I’m off in search of aspirin…

I am disappointed; I thought this thread would be about the ancient Japanese practice of cultivating tiny planets :frowning:

Clearly the answer is: Undetermined.

If some A are B, and some B are C,
this gives us no information as to whether some A are C.

Look at it this way: the intent of the original question is to ask “Is it the case that, no matter what groks, graks, and griks are, if some groks are graks and some graks are griks, then there must be some groks that are griks?”.

Your examples B and C show that this this needn’t be the case for all definitions. Therefore, the answer is False, because the question is specifically, albeit implicitly, asking whether it’s the case for all definitions, not just some. The fact that it works for example A is not a problem.

You view A as true only because you possess knowledge outside of the statement. If you evaluated it according to its structure:

set x [m, n, o, p] overlaps with set y [o, p, q, r]. Although you know that set y also overlaps with set z, you will see that the structure does not guarantee that set x overlaps with set z, as set z could be [q, r, s, t], which would not overlap with x, even though it does overlap set y.

Of course, it could be the case that z overlaps with x, as it may be defined as [m, q, s, v].

Or, as other posters have said, the relationship between “dogs and baldness” or “planets and bonsais” is undetermined via the logic and your statement of “if, then” is not logical, regardless of whether the statements are true.

I think, in questions like these, you’re supposed to say whether or not it’s necessarily true—whether the conclusion necessarily follows from the premises.

In this case, it does not. (So, False).

One way to see this is to come up with a counterexample. Like, “If some dogs are black animals and some black animals are cats then some dogs are cats.”

Another way is with a Venn Diagram. “Some groks are graks” would have circle O overlapping with circle A, and “some graks are griks” would have circle A overlapping with circle I; but in such a case, circle O may or may not overlap with circle I.

Because all poodles are* dogs*, then a characteristic of some poodles is a characteristic of some dogs. But there’s no information to determine if griks and graks have that kind of relationship. Example C above shows a case where it isn’t true. Not all dancers are singers. The word are doesn’t convey the nature of the relationship, it’s too general purpose.

No. The key word here is “some”. Not ALL dogs are poodles, just some are. Not ALL poodles are bald, just some are. The status of the ones excluded is not determined.

So it could be that some dogs are NOT poodles. Some of those might NOT be bald. Therefore “Some dogs are bald” cannot be determined and is not true.

In other words, the two groups might not overlap. Or they might – we just don’t know from the premise.

Venn diagrams are very useful in this type of thing, showing that all of the statements are undetermined. Even the one about poodles. You think it’s valid because you know that all poodles are dogs. But “poodle” has another meaning (a servile person) that has nothing to do with dogs. So there are poodles that are not dogs, and some of them might be the bald ones.

I’m not following you here. It’s not relevant that some dogs are not poodles. Even if there are Great Danes or whatever, that doesn’t tell us about bald dogs. It also doesn’t matter if some Great Danes are not bald- they all, some, or none could be bald and it wouldn’t change anything so far as we know.

“Some dogs are bald” is unknown because it’s possible that the bald poodles are not dogs, not for any reason dealing with non-poodle or non-bald dogs. If you postulate that all poodles are, in fact, dogs, then the conclusion
“Some dogs are bald” becomes true, as somewhere there’s a bald poodle dog.

Not according to your premise. You are stating a conclusion that does not follow the premise. (It may very well be true, but not in a logic sense).

That’s all you need. You figured it out for yourself. In these sorts of questions, “true” means “necessarily true”, and if you can think of counterexamples within the confines of the conditions, then you’ve disproven the claim.

Pardon me, I’m going to the E for existential because I don’t feel like looking up the symbol code

E x,y,z (x=y ^ y=z) -> (x=z)
(There is a dancer such that there is a singer that is the same person, and there is an actor such that it is the same person as both the dancer and the singer, therefore there is a dancer that is an actor).

Is what it’s supposed to look like it’s saying, it’s not. It’s actually this statement

E x,y,z,a (x=y ^ a=z) -> (x=z)
(There is a dancer such that there is a singer that is the same, incidentally, there is also an actor who is the same as some dancer, therefore there is also a dancer who is an actor).

The missing bit of information is “and y=a”, otherwise the answer is unknowable.

The correct statement then, is

“If some X are Y and some of the Y that are also X are Z, then some X are Z.”

This is why the poodle example seems to work, since all poodles are dogs, then necessarily all Y are X, so therefore some X are Z.

The poodle statement infers

forAll y, E x (y=x)
(For every poodle, there exists a dog such that the poodle is that dog).

The statement itself, absent any other information, is bad and wrong, despite it possibly being true in some cases.

Edit: We’re also assuming that in our set relations, “=” is transitive, which probably happens more often in logic and set theory than anywhere else to not be true. But we’ll take it as a given here.

Sorry, missed something (added in red):

ETA: And we can quibble about my phrasing of the FOL, that the implication should be “If there is a dancer… then…” etc, but I think what I said gets the point across fine.

It’s can be true, but not necessarily so, that:

Some sheep are small (groks), and some sheep are black (graks), and some sheep are shorn (griks), then some small sheep are shorn.

It’s false that:

Some sheep are tall (groks), some sheep are clean (graks), some sheep are short (griks), then some tall sheep are short sheep.

Conclusion is “False” to the grok/grak/grik proposition.

The crucial thing about A is that all poodles are dogs. That’s why some dogs are bald. Change “poodles” to “animals” and you will see.

The answer they want is based on “Is it necessarily true”. Since it fails in one or more examples, it is not necessarily true. Of course, if I had posed the question, I would have included that word.

It’s not a logical question, because it cannot be answered true or false. According to the premise, some groks may be griks. Maybe dosen’t fly.

I believe in this case “true” is supposed to reflect “tautology” (reducible using syllogistic identities to “True”) vs not a tautology.