don't know how to verify one way or the other — DavidJohnson

Drawing a Venn diagram is always the right thing to do.

D
There are no Sees that are not Bees. Because all Sees are Bees.
Therefore nothing that is not a Bee is a See. In other words, if it's not a Bee, it is impossible that it is a See.
There are Ayes that are not Bees. That is, at least one Aye is a Bee. But there could or could not be more Ayes that are not Bees.
So if it's an Aye, and it exists inside Bees, it has a chance of being a See, but not necessarily.
But since there are no Sees outside of Bee, therefore Ayes outside of Bee are certainly not Sees.
Hence, no Ayes that are outside of all Bees are a See... precisely what D says.

---------------------

For the above, I used the concepts of "Aye" "Bee" and "See" as individuals contained necessarily or potentially or impossibly inside of all that are Ayes, Bees and Sees. Furthermore, Ayes and Bees and Sees I used as all of them, or I used them as more than zero.

Why not A? Seas and Bees are equivalent at least in their aspect of all seas being bees. If all seas are Bees and if some ayes are bees as well than some ayes must also be seas since they are equivalent to bees.

Why not A? — Tobias

Some mugs(Ayes) are copper(Bees).
All pennies(Seas) are copper(Bees).

[Pennies] and [Copper] are equivalent at least in their aspect of all [Pennies] being [copper]. If all [pennies] are [Copper] and if some [mugs] are [copper] as well than some [mugs] must also be [pennies] since they are equivalent to [copper]. — Tobias

Nope. Can't drink mead from a penny.

Thanks InPizotl, makes a lot of sense. It depends on the Ayes Bees and Seas being defined as substances or properties, right. Thanks!

It depends on the Ayes Bees and Seas being defined as substances or properties, right. — Tobias

Not quite. It's just that the logical statement
"All seas are bees",
on it's own doesn't allow the conclusion that the two things are equivalent. It doesn't explain their relation any further than "All seas are bees" - we don't know whether:
"All seas are bees and all bees are seas." or "All seas are bees but not all bees are seas."

Yes, it is toying with the ambivalence of 'is'... 'is' of identity and the 'is' of predication. I always sucked at formal logic ;)

Ok, thanks! And if it ends up being that no Ayes are Seas then D is still a valid conclusion like you said. That's the part which was confusing me. I was thinking, "D is right, but is it still valid if it ends up being that no Ayes are Seas?"

Alternatively, to solve this sort of problem using the smallest number of brain cells and the least amount of graphics, restate the given premises formally, using a single variable x of universal type U

i) ∃x:U , A(x) ∧ B(x)
ii) ∀x:U, S(x) ⇒ B(x), which is equivalent to ∀x:U, ¬B(x) ⇒ ¬S(x) which immediately gives D.

Then state the theorems

A) ∃x:U , A(x) ∧ S(x) (which can instantly be seen to not be derivable from i and ii)
B) ∃x:U , A(x) ∧ S(x) (same as A)
C) ∀x:U , S(x) ⇒ ¬A(x) (instantly recognizable as not derivable)

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No sweat. I am pleased I could help. And you have a number of alternative "right" answers following mine. "Gimme the wisdom to choose the right answer (when I don't know which one is right.)"

Some Ayes are Bees
All Seas are Bees — DavidJohnson

The only conclusion that seems reasonable to state is some kind of categorical relationship between Ayes and Seas (Bees being the middle term).

There are 4 possibilities:

1. Some Ayes are Seas (true, but a premise)
2. Some Ayes are not Seas (undecidable)
3. All Ayes are Seas (undecidable)
4. No Ayea are Seas (false)

That's all from me. :smile:

Thank you for the visual! That helped a lot. I was trying to do it in my head. :lol:

If the proposition, "No Ayes are Seas" were added, would D still be valid? I don't think "No Ayes that aren't Bees are Seas" necessarily implies that there exist Ayes that are Seas, just that Ayes can't possibly be Seas if they aren't also Bees, so maybe D is still valid regardless. If adding that extra proposition invalidates D then I think E is definitively the correct answer since it's not necessarily true that any Ayes that are Seas even exist.

do it in my head — DavidJohnson

Some can! Of all the people in the world, these are the ones that arouse my envy. My usual reaction/response is to use my finger and draw on the ground. I think this is some kind of genetic memory. Our genes "remember" a time when we used sand to draw pictures on, like Archimedes! Now there was a guy who could do things in his head! :smile:

Nōlī turbāre circulōs meōs!

In modern times, universals are always interpreted as conditionals. Just translate in your head like this:

“All F’s are G” means “If anything is F, then it’s G”
“No F’s are G” means “If anything is F, then it’s not G”

The upshot of the translation is that the bits on the right are still true, even when there’s nothing that’s F.

You may also have to sit a while with this understanding of conditionals (known as “material implication”) until it feels natural.

D looks tempting, but depends on the implicit added proposition that if some As are Bs, then some As are not Bs — tim wood

That is not needed to see that (D) is the correct answer.

It is quite simple, without even need for Venn diagrams or symbolic logic.

(D) can be couched in two equivalent ways:

"If a thing is an Aye and not a Bee, then it is not a Sea".

"There is no thing that is an Aye and not a Bee and a Sea".

And either way you couch (D) it is entailed by the premise:

All Seas are Bees, or couched equivalently:

"If a thing is a Sea then it is a Bee."

If the proposition, "No Ayes are Seas" were added, would D still be valid? — DavidJohnson

The question is not the validity of (D) but the validity of inferring (D) from the premises.

(D) is validly inferred from the premises. Adding an additional premise cannot vitiate an otherwise valid argument. This is the monotonic property of basic logic.

Some Ayes are Bees
All Seas are Bees

.

1. Some Ayes are Seas (true, but a premise)
2. Some Ayes are not Seas (undecidable)
3. All Ayes are Seas (undecidable)
4. No Ayes are Seas (false) — Agent Smith

1. Is not a premise and it is not entailed by the premises.

4. Is not made false by the premises.

If the proposition, "No Ayes are Seas" were added, would D still be a valid conclusion? In the two propositions given, we can't be sure that there are any Ayes that are Seas, but I think D is valid regardless since, even if no Ayes are Seas, it's still true that no Ayes that aren't Bees aren't Seas.

Ok, thanks a million! That really helped clarify what the problem was asking. :grin:

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Why would D only follow iff there was an Aye that isn't a Bee? Even if all Ayes are Bees and all Bees are Seas, wouldn't "No Ayes that are not Bees are Seas" still follow?

Can't believe this thread is still going on. :roll:

Some Ayes are Bees
All Seas are Bees — DavidJohnson

What about bees swimming in all seas? Aye!

Lol, me too. The general consensus is that the answer is "D" but Tim Wood is making an argument for answer "E", and as a non-logistician I'm waiting to see where this goes. :rofl:

as a non-logistician I'm waiting to see where this goes. — DavidJohnson

If this forum taught me something, it is to pick your bottles well. Not to win, but in order not to get caught in the quagmire of non-understanding.

So I shalt remain silent.

And picking the right battles is good too, but that's not as important.

the middle term is not distributed, which means no valid conclusion can be drawn. — tim wood

Whether the conclusion is or not inferred according to syllogistic forms, it is inferred validly.

This follows iff there is an A that is not a B. — tim wood

All Seas are Bees

entails

No Ayes that are not Bees are Seas

whether or not there is an Aye that is not a Bee. Ayes don't even have anything to do with it. And there is no need for an existence assumption.

You've got something stuck in your head that is not the case.

Run it through your mind, or write it in symbols [where 'U' and 'E' are the universal and existential quantifiers]:

Ux(Sx -> Bx)

entails

Ux((Ax & ~Bx) -> ~Sx)

or couched equivalently:

Ux(Sx -> Bx)

entails

~Ex(Ax & ~Bx & Sx)

It's clear as day:

If there were a thing that is both not a Bee and is a Sea, then that would contradict that all Seas are Bees. And that is the case whether there is or is not anything that is an Aye or a Bee or Sea or any combination.

Okay everyone, I contacted a logic professor from UNC who formerly worked at Yale and his response was that, in modern logic systems the answer is "D" and that only in outdated, Aristotelian categories it's "E", but that those traditional systems haven't been used since the late 19th century.

Conclusion: the answer is "D" and Tim Wood is a vampire no younger than 122 years old.