I understand the implied logic of the formal argument all A are B and most A are C; therefore, some B are C. Could we stretch this to say it is a valid inference that most B are C?
I think the answer would be no, but I would like to understand this better.
Think of this example. All really cool guys own a red sports car. Most really cool guys are tennis players. Therefore, most red sports car owners are tennis players.
The reaons I believe this is not valid is because All A are B only tells me that the subset A is completely consumed by the superset B. I can not make a "most" statement about superset B because I do not know how large that superset is.
This seems obviously valid to me but I can easily imagine a world where the conclusion is untrue. If this were the question on the LSAT, would we have to say the conclusion follows from the premises even if it is very easy to imagine a world where that is not Actually true?
Another way to get the same valid conclusion, albeit longer, is this:
Since most A's are C's (A‑m→C), you can flip that and make it a some (C←s→A). In other words some C's are A's because some (1-100%) includes most (51-100%) within it. Then, substitute B for A according to the first premise (A→B) in (C←s→A), and you'll get (C←s→B), which is just the flipping of the valid conclusion (B←s→C) given the some property.
I think if we change the “most” into “some,” it works as well. Please let me know if I got it wrong!
A→B
A‑s→C
—————
B←s→C
My line of thinking is this: the first line indicates that A is a subset of B, and thus, since something in that subset is C, something in the superset(B) must be C as well.
It’s like:
All Italians find carbonara gross.
Some Italians are humans (the original meme: https://x.com/radio_netas/status/1405322285771280390)
————————
Some humans find carbonara gross.
Methinks the form in this lesson (using most instead of some) is just an iteration with an even stronger premise . Perhaps it’s one of the forms that they want us to figure out ourselves?
think of this way if you know that all (everyone) of this thing likes/is another thing and most of those same people (everyone) also likes/is something else U CAN INFER THAT some people that are/like the first thing also are/like the 2nd thing.
Since "all" also implies "some" can we translate line 1 to be A ←s→ B?
And "most" also implies "some" so can we translate line 2 to A ←s→ C?
0
Topics
PT Questions
Select Preptest
You've discovered a premium feature!
Subscribe to unlock everything that 7Sage has to offer.
Hold on there, stranger! You need a free account for that.
We love that you want to get going. Just create a free account below—it only takes a minute—and then you can continue!
Hold on there, stranger! You need a free account for that.
We love that you came here to read all the amazing posts from our 300,000+ members. They all have accounts too! Just create a free account below—it only takes a minute—and then you’re free to discuss anything!
Hold on there, stranger! You need a free account for that.
We love that you want to give us feedback! Just create a free account below—it only takes a minute—and then you’re free to vote on this!
Subscribers can learn all the LSAT secrets.
Happens all the time: now that you've had a taste of the lessons, you just can't stop -- and you don't have to! Click the button.
53 comments
Starting to get the hang of it a little more
A medical study found that all people who are obese have hypertension. Most of the obese people in this study had sleep apnea.
Therefore, some people with hypertension had sleep apnea.
#Help
Why couldn't the conclusion be "Most B's are C's?"
I'm trying to visualize it with the buckets.
Here is a question #feedback
I understand the implied logic of the formal argument all A are B and most A are C; therefore, some B are C. Could we stretch this to say it is a valid inference that most B are C?
I think the answer would be no, but I would like to understand this better.
Think of this example. All really cool guys own a red sports car. Most really cool guys are tennis players. Therefore, most red sports car owners are tennis players.
The reaons I believe this is not valid is because All A are B only tells me that the subset A is completely consumed by the superset B. I can not make a "most" statement about superset B because I do not know how large that superset is.
This was SOO helpful to know that u can see the all statement as a most statement in order to combine the 2 most statements into a some statement.
So is this where we get into valid vs True?
All X wings have Hyper drives
X->H
Most X wings have battle scars
X‑m→BS
Therefore, some hyperdrives have battle scars
H←s→BS
This seems obviously valid to me but I can easily imagine a world where the conclusion is untrue. If this were the question on the LSAT, would we have to say the conclusion follows from the premises even if it is very easy to imagine a world where that is not Actually true?
So this is interesting. Could we not say that for the above argument (listed below):
A → B
A —m→ C
B ←s→ C
could we not say instead that MOST B's are C's?
ACB
ACB
ACB
AB
AB
Another way to get the same valid conclusion, albeit longer, is this:
Since most A's are C's (A‑m→C), you can flip that and make it a some (C←s→A). In other words some C's are A's because some (1-100%) includes most (51-100%) within it. Then, substitute B for A according to the first premise (A→B) in (C←s→A), and you'll get (C←s→B), which is just the flipping of the valid conclusion (B←s→C) given the some property.
All french people like music
Most music in France is electronic
Therefore, some french people like electronic music.
All girls who like purses buy Chanel
Most girls who like purses have trouble saving money
Therefore, some girls who buy Chanel have trouble saving money
All dogs like bacon
Most dogs play fetch
Some animals that like bacon play fetch.
1. All people with anxiety feel a sense of impending doom
2. Most people with anxiety have trouble sleeping
Therefore, some people who feel a sense of impending doom also have trouble sleeping
A → D
A ‑m→ TS
_
D ←s→ TS
All water has flavor (A → B)
Most water has minerals in it (A ‑m→ C)
Therefore, some flavor comes from minerals (B ←s→ C)
Negated:
Some water doesn’t have flavor (A ←s→ /B)
Most water don’t have minerals in it (A ‑m→ /C)
Therefore, all/no flavor comes from minerals (B → C)
#help Is this argument valid because "most" also implies "some"?
Is it valid to conclude that all implies some as well?
I think if we change the “most” into “some,” it works as well. Please let me know if I got it wrong!
A→B
A‑s→C
—————
B←s→C
My line of thinking is this: the first line indicates that A is a subset of B, and thus, since something in that subset is C, something in the superset(B) must be C as well.
It’s like:
All Italians find carbonara gross.
Some Italians are humans (the original meme: https://x.com/radio_netas/status/1405322285771280390)
————————
Some humans find carbonara gross.
Methinks the form in this lesson (using most instead of some) is just an iteration with an even stronger premise . Perhaps it’s one of the forms that they want us to figure out ourselves?
here's an example
all cats drink milk,
most cats eat watermelon.
some animals that drink milk also eat watermelon.
LAW GIC
C -> DM
C -> EW
DM EW
think of this way if you know that all (everyone) of this thing likes/is another thing and most of those same people (everyone) also likes/is something else U CAN INFER THAT some people that are/like the first thing also are/like the 2nd thing.
hope this helps!!!!
So if all pets are nice, and most pets are dogs, then that means some dogs (the ones that are pets) have to be nice.
pets → nice
pets ‑m→ dogs
dogs ←s→ nice
This one is confusing
#question:
is this used for logic games or for LR too? if so how
#feedback Would have been great to have an example for this one...
Pretty sure we can also infer that B‑m→C
#help Just a hypothetical here:
Since "all" also implies "some" can we translate line 1 to be A ←s→ B?
And "most" also implies "some" so can we translate line 2 to A ←s→ C?