I'm still stuck and afraid but at least not as frustrated as the beggining I feel that I've learned some things, still a little afraid of taking the LSAT drills or prep questions but I feel a little better, maybe because it feels like I've learned some stuff
I’m amazed at how much I picked up in the first week of curriculum. I’m glad I took the LSAT the hard way first because now this instruction feels like a massive cheat code!
@DeborahAdel I spent lots of time just taking sections and small chunks directly from law hub. I later learned are 4 test within law hub that provide great feedback but I found them too late.
@JeremyK I think it's a sufficient and necessary thing. While all A's are B's, not all B's are A's. The A's that are C might represent only a small portion of all B's, so we can't make a must be true conclusion from that
@JeremyK in the lesson before this one they gave a great visualizer to explain this.
B and C have some overlap and the only thing we can infer is that that overlap is more than 0 so we cannot say Most Bs are C. The instructor actually said in the formal argument 6 lesson regarding this with the buckets example doesn’t work as well. This is the example they gave
if we have 5 A’s and most (more than half) A’s are B and C then we know it must be true that at least 3 A’s are B and same goes for C so that leaves one A to look like this
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.
@ImanMozaffarian. Exactly, we don't know anything about B except that it has some A.
A = students, B = people, C = smart
A -> B, all students are people
A -m-> C, most students are smart
It'd be pretty outlandish to claim most people are smart from these two statements. While it COULD be true that most people are smart, these two statements ALLOW for the possibility that everyone else who isn't a student is not smart, therefore we can't make B -m-> C
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?
The questions on the LSAT are concerned about if the argument is valid or logical. Truth (in relation to the real world) is irrelevant.
So, yes, on the LSAT you do need to say the conclusion follows from the premises even if it is very easy to imagine a world (or even the real world ) where the argument is not actually true.
A is only sufficient to be a B. It is not necessary for B to be A. The oldest trick in book problem. (A -> B /= B -> A)
So, if the second premise tells us, most A are C. It is not saying most B are C, because our premises only tell us that A is sufficient for being a B. There could be other ways to become a B, not promising us the majority. In english take this example;
All dogs go to heaven. Most dogs have long tails. Therefore, some things in heaven have long tails.
A = Dogs
B = Heaven
C = Long Tails
Dogs -> Heaven.
Dogs -m> Long Tails.
Heaven ←s→ Long Tails.
It would be insane to say that majority of things in heaven have long tails (Heaven -m> Long Tails) or at least we have no way of knowing that based solely on our given premises.
If it was a numbers game say the total number of creatures in heaven is 100. And there are 10 dogs in all of existence.
We know all 10 of those dogs are in heaven. (A->B) And we know at least 6 of those dogs have long tails (A -m> C).
Out of our population of heaven we have at least 6 creatures that have long tails, because the majority of our dogs did. (B←s→C)
There are 90 others creatures in heaven, they could also have long tails but we don’t know that. We can only logically conclude that at least 6 of those creatures have long tails. We couldn’t guarantee 51 or more creatures have long tails. (B-m>C) Because we only know about the 6 long-tailed dogs.
Not to over explain / beat a dead horse. Just trying to strengthen my own understanding of the concept.
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.
1. "French people" describing a population and "France" describing a geographic location are not interchangeable. Just because most music found in France is electronic, it does not speak to the music preferences of French people.
2. Your argument structure is not valid.
Your argument structure:
F->M
M-Most->E
------------
F some E
To make this valid it would need to be
F->M
F-Most->E
-----------
M some E
Translated, that would be:
All French people like music
Most French people like electronic
Therefore, some people who enjoy music, enjoy electronic
That conclusion could be true, but it isn't a logical conclusion from the argument you provided. A more appropriate one would be some dogs that like bacon play fetch.
the only thing that i would say to be careful about in this example is assumming the connection between flavor and minerals. we know water is sufficient for flavor and water is sufficient for minerals. we do not know the relationship between flavor and minerals. the only assumption that could be made is "some drinks that have flavor have minerals as well." i make this sort of mistake when doing LR and the test loves to exploit it so just thought i'd let you know!
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?
Technically what G. W. Roriksson asked is really just formal argument #4, but if it helps, I actually found a question where JY does exactly this when mapping out the stimulus, check it out:
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.
I noticed ur comments on the last couple posts and they got me inspired to force myself to make my own examples which has really helped me see if I ACTUALLY understand a concept
This is a great example, but I did want to raise one point because it did stop me in my tracks and make me reread the lesson a few times. I can't explain it, but the wording tripped me up about the relationship between the "all" arrow and the "most" arrow.
In your example, be careful not to swap the "most" arrow sign for the "all" arrow sign. Just because all implies most does not mean you can swap them the other way around. Most does not imply all.
Not trying to be a stickler. Just speaking up in case anyone did a double take like me. This example was helpful!
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62 comments
The example I created was this: All Bugs are Black (in this world I am creating here.)
Most Bugs are Crawly.
(So if I have 5 bugs, all of them must be Black. At least 3 of them are Crawly.)
Thus, there are some Black things that are Crawly.
Bugs (arrow) Black.
Bugs (most arrow) are Crawly.
Black [things] (some arrow) are Crawly.
I'm sorry but I am just not understanding the diagramming, and it's so frustrating.
Do you think I could flip B <s> C to be C<s> B so that some of the C's that are on A are also B (at least 1?). I don't know if that make sense:
A C B
A C
A C
A
A
I'm still stuck and afraid but at least not as frustrated as the beggining I feel that I've learned some things, still a little afraid of taking the LSAT drills or prep questions but I feel a little better, maybe because it feels like I've learned some stuff
@LauraBolivar the questions can be intimidating! I find them super helpful though.
Think of it this way:
All questions make us do better on the exam.
Most questions can feel scary.
Thus, some things that can make us do better on the exam feel scary.
I’m amazed at how much I picked up in the first week of curriculum. I’m glad I took the LSAT the hard way first because now this instruction feels like a massive cheat code!
@mav73 what was the hard way for you? purely out of curiosity.
@DeborahAdel taking the test without mastering the foundations (I assume).
@DeborahAdel I spent lots of time just taking sections and small chunks directly from law hub. I later learned are 4 test within law hub that provide great feedback but I found them too late.
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.
@CeciliaBurton1 thank you for the example
@CeciliaBurton1 Great example, thank you.
#Help
Why couldn't the conclusion be "Most B's are C's?"
I'm trying to visualize it with the buckets.
@JeremyK I think it's a sufficient and necessary thing. While all A's are B's, not all B's are A's. The A's that are C might represent only a small portion of all B's, so we can't make a must be true conclusion from that
@JeremyK in the lesson before this one they gave a great visualizer to explain this.
B and C have some overlap and the only thing we can infer is that that overlap is more than 0 so we cannot say Most Bs are C. The instructor actually said in the formal argument 6 lesson regarding this with the buckets example doesn’t work as well. This is the example they gave
if we have 5 A’s and most (more than half) A’s are B and C then we know it must be true that at least 3 A’s are B and same goes for C so that leaves one A to look like this
A B
A B
A B C some B are C
A C
A C
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.
@ImanMozaffarian. Exactly, we don't know anything about B except that it has some A.
A = students, B = people, C = smart
A -> B, all students are people
A -m-> C, most students are smart
It'd be pretty outlandish to claim most people are smart from these two statements. While it COULD be true that most people are smart, these two statements ALLOW for the possibility that everyone else who isn't a student is not smart, therefore we can't make B -m-> C
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?
The questions on the LSAT are concerned about if the argument is valid or logical. Truth (in relation to the real world) is irrelevant.
So, yes, on the LSAT you do need to say the conclusion follows from the premises even if it is very easy to imagine a world (or even the real world ) where the argument 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
i was wondering the same thing
A is only sufficient to be a B. It is not necessary for B to be A. The oldest trick in book problem. (A -> B /= B -> A)
So, if the second premise tells us, most A are C. It is not saying most B are C, because our premises only tell us that A is sufficient for being a B. There could be other ways to become a B, not promising us the majority. In english take this example;
All dogs go to heaven. Most dogs have long tails. Therefore, some things in heaven have long tails.
A = Dogs
B = Heaven
C = Long Tails
Dogs -> Heaven.
Dogs -m> Long Tails.
Heaven ←s→ Long Tails.
It would be insane to say that majority of things in heaven have long tails (Heaven -m> Long Tails) or at least we have no way of knowing that based solely on our given premises.
If it was a numbers game say the total number of creatures in heaven is 100. And there are 10 dogs in all of existence.
We know all 10 of those dogs are in heaven. (A->B) And we know at least 6 of those dogs have long tails (A -m> C).
Out of our population of heaven we have at least 6 creatures that have long tails, because the majority of our dogs did. (B←s→C)
There are 90 others creatures in heaven, they could also have long tails but we don’t know that. We can only logically conclude that at least 6 of those creatures have long tails. We couldn’t guarantee 51 or more creatures have long tails. (B-m>C) Because we only know about the 6 long-tailed dogs.
Not to over explain / beat a dead horse. Just trying to strengthen my own understanding of the concept.
thank youuu @mitsubaatlaw
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.
There's a few problems with your argument.
1. "French people" describing a population and "France" describing a geographic location are not interchangeable. Just because most music found in France is electronic, it does not speak to the music preferences of French people.
2. Your argument structure is not valid.
Your argument structure:
F->M
M-Most->E
------------
F some E
To make this valid it would need to be
F->M
F-Most->E
-----------
M some E
Translated, that would be:
All French people like music
Most French people like electronic
Therefore, some people who enjoy music, enjoy electronic
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
it’s me I’m the girl
All dogs like bacon
Most dogs play fetch
Some animals that like bacon play fetch.
That conclusion could be true, but it isn't a logical conclusion from the argument you provided. A more appropriate one would be some dogs 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)
the only thing that i would say to be careful about in this example is assumming the connection between flavor and minerals. we know water is sufficient for flavor and water is sufficient for minerals. we do not know the relationship between flavor and minerals. the only assumption that could be made is "some drinks that have flavor have minerals as well." i make this sort of mistake when doing LR and the test loves to exploit it so just thought i'd let you know!
#help Is this argument valid because "most" also implies "some"?
That would be a valid way of visualizing the argument
Is it valid to conclude that all implies some as well?
Yes, All is 100%
Some is at least one.
If its true that all grass is green, then it is valid to conclude that some grass is green.
This is also true for most.
Yes, all negates to some
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?
I'm also wondering this!
Hey how's it going
Technically what G. W. Roriksson asked is really just formal argument #4, but if it helps, I actually found a question where JY does exactly this when mapping out the stimulus, check it out:
https://7sage.com/lsat_explanations/lsat-116-section-3-question-21/
I recommend you start from "map" (0:49 seconds) until the end of answer choice B. He also "Negates All", so don't confuse that with this topic.
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!!!!
I noticed ur comments on the last couple posts and they got me inspired to force myself to make my own examples which has really helped me see if I ACTUALLY understand a concept
This is a great example, but I did want to raise one point because it did stop me in my tracks and make me reread the lesson a few times. I can't explain it, but the wording tripped me up about the relationship between the "all" arrow and the "most" arrow.
In your example, be careful not to swap the "most" arrow sign for the "all" arrow sign. Just because all implies most does not mean you can swap them the other way around. Most does not imply all.
Not trying to be a stickler. Just speaking up in case anyone did a double take like me. This example was helpful!
Where did she do this?
Oh, in the lawgic portion. Okay, yes I agree. Didn't see that. I kept looking at the text and couldn't see what you were referring to.
C->EW means all cats eat watermelon.
Very different than C m->W which means most cats eat watermelon.
I wasn’t trying to be a jerk, but her lawgic translation was off. Most is not interchangeable with all.
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