But because "some" is weaker than "most", wouldn't that actually help an argument in this case? If "most" violinists aren't exceptionally good at playing the violin, I'm struggling to see how changing "most" to "some" would be invalid just because the "most" argument is invalid.
I understood it by thinking of the Formal Arguments #4 and #6. They all have to have some or most before all because we needs As in the other three two buckets.
A <--s--> B --> C
A <--s--> C
Some As are Cs. Because there is some As in the C bucket.
Not like the invalid argument:
A -->B <--s--> C
A <-s-> C
How do I have some As if I put all of them B. It does not make a correct conclusion.
I found it helpful to draw out the sets in a way where it makes sense why the invalid form doesn't work. Hope this helps someone who's struggling to understand!
It's really helpful for me to think of this and the previous flaw with the buckets scenario. Think of it, if you only take some of the A's and put them into B, but then you take that entire bucket and put it into C, then of course there will be A's in there. But if you take all the A's into B (remember, we don't know how many A's that is, it could be literally 2), then take only some of that entire bucket into C, it's less likely that there will be A's in C. Hope this helped!!
@Gisellednava rjon27 comment really helped me understand. I think it possible that there are humans that have tails but there is no 100% chance there that some mammals that have tails happened to be human. The "some" could all be different mammals that are not humans.
Here's my attempt at an invalid and valid argument. Someone correct me if I'm going about it wrong!
Invalid: All peaches have pits. Some pits are rotten. Therefore, some peaches are rotten.
The reason why this is invalid is because we don't know the peach is rotten. It could be because the air had weird chemicals in it, or maybe a bug was hiding in the peach. Simply knowing that some peach pits are rotten isn't enough to know that the peach itself is rotten.
Valid: Some peaches have pits. All pits are rotten. Therefore, some peaches are rotten.
This is valid because we know ALL of B is true. If all pits are rotten and we know some peaches have pits, we at least know some of the peaches are rotten.
What I have found that trips me up is thinking 'all A are B'. To me, it makes it sound like they are interchangeable ideas, when really it just means all of A is in the B group. It does not guarantee that if some (or most) of B is part of the C group, that that part of B has any A in it at all.
My understanding of this is that all A have B, but just because they have B doesn't mean that A will take on C. Because it can't be confirmed that just because you have B you also have C. The key indicator is "some". A is sufficient for B but does not guarantee that B brings about C.
All Options at Marys dessert shop is frozen yogurt, some frozen yogurt is mint flavored; therefore some of the Options at Marys shop is mint.
this in incorrect because the statement about mint is about all frozen yogurt, it doesn't tell us anything about if frozen yogurt at Marys dessert shop. It is just as likely she only serves vanilla and chocolate. we cannot disprove or prove she has mint, just that it is possible.
Remember: a valid conclusion about intersecting sets can only be made if the some arrow appears before the all arrow.
Ex.
Artists in the master class are painters. Some people who call themselves painters suck at art. Therefore, some artists in the master class suck at art.
Art's -> Pnt's <-s-> Suck@it
____
Art's <-s-> Suck@it
Thus is NOT valid
Again, just because the first set shares a characteristic with the second (painters), does not mean you can equate the two.
Ex.
Some artists in the master class are painters. All people who paint suck at art. Therefore, some artists in the master class suck at art.
Art's <-s-> Pnt's -> Suck@it
____
Art's <-s-> Suck@it
Here, we know 100% of the subset (painters) ALL share the quality we are discussing (sucking at it). So, we can easily conclude that the portion of artists in the class that are painters definitely suck at it.
im confused on the quantifiers. If we say all A's are B's, and we equate themselves logically, can they be interchangeable? A B and All B's are C's (or A = B), we cans say A B/C (because theyre interchangeable .
Why cant it work with some? All A's are B's (or A = B) and B C, therefore A/B C.
Is it all just contextual basing it on what sets youre talking about?
We are not trying to equate A and B... when we say all A's are B's we are definitely not saying A = B, we are saying that if you are A then you are B. This is distinct from saying A=B because A=B also means B=A, and we do not want to misconclude that B →A (that would be the oldest mistake in the book!). A and B are not interchangable, A is still just sufficient for B.
It cannot work with 'some' when some comes after 'all' for the following reason:
Consider that I own 5 marbles, and all of the marbles I own (A) are red marbles (B).
So thus far we can conclude A → B.
Now, some red marbles that exist also have blue stripes (C).
B ←s→ C
Does that mean that some of the marbles I own have blue stripes? Not necessarily. It's not IMPOSSIBLE that I could own one with blue stripes... there's just no way we can conclude that. Just because there are red marbles with blue stripes that exist, does not mean that I own them. therefore A → B ←s→ C warrants no valid conclusions.
Conversely, A ←s→B → C, does warrant a valid conclusion. Let's use the same example:
Consider that I own 5 marbles, and some of the marbles I own (A) are red marbles (B). We conclude that A ←s→ B.
If all red marbles (B) have blue stripes (C), that means B → C.
If I own some red marbles (A), and every single red marble has blue stripes(C), that means some of my red marbles have blue stripes, too.
I think it helps to think about it like order of operations.
If you dump some of you A's into your B's
then dump all of your B bucket into your C's, then you will definitely have some A's in your C's.
But if you dump all of your A's into your B's
and then dump some of your B's into your C's, then you won't necessarily have any A's in your C's, you could, but its not guaranteed. If it is not guaranteed then it is not valid.
I think this reasoning also work for the most rule as well.
By stronger they mean because ‑m→ refers to most and ←s→ refers to some. It is much stronger to say that most people agree with me on a certain topic than that some people agree with me on a certain topic (some = at least one, it is not a lot). So most is much stronger. In this case, they are referring to the idea that if we say that "most people do x" that is equivalent to saying that "some people do x" because most also encompasses some. Hope this makes sense!
To me, it looks as if you just repeated one of the statements as a conclusion. It'll be valid because it's part of the initial set of claims... it's not really something you'd conclude from the previous claims.
All CHICKENS are BIRDS. Some BIRDS are GREEN.
C: Chickens
B: Birds
G: Green
Written out, it'd look like this.
C → B ←s→ G
You effectively stated that this would be valid if the conclusion was B ←s→ G
You're not wrong, but at the same time, you just kind of repeated the second claim: Some BIRDS are GREEN.
As for your second point, I think you're right.
If we take the same sentence and before and got C ←s→ G, that wouldn't be right.
All CHICKENS are BIRDS. Some BIRDS are GREEN.
Therefore some CHICKENS are GREEN? That wouldn't be the case.
Hopefully I'm right about this and getting my point across.
So in the previous lesson they said that A←s→B→C leads to a valid inference of A←s→C. This in plain English would be if some of A are B and all of B is C, then some of A are C. (For example, if some fish in the fish tank are red and all red fish are venomous then we can make the inference that some fish in the fish tank are venomous).
However, in this lesson they introduce the formal argument A←s→B←C (or in other words C→B←s→A). In this case, you would be saying that all of C is B and some of B are A. For this type of formal argument you CANNOT make an inference. When you first see this argument, you would want to make the inference A←s→C but this cannot be an inference made because it could be true but it could also be false.
Here's how that works. Imagine a circle diagram where B is a big circle and within B there is a smaller circle called C. Because the only thing that this lawgic statement tells us is that all of C is in B, the size of C can be large or small. So C could take up a big spot within B or it could be a small dot within the circle "B." However, the lawgic statement goes onto say that some of B are A. In this case, could we guarantee that at least some of C (in other words, at least one of C) will be a part of A? No. Because what if the "some of B" that is a A is a part of the circle that C does not cover? If within the circle "B," circle C and A does not overlap, it would mean that some of C is NOT A and this is a possible scenario.
For example, let's replace A with fish in the fish tank, B with red and C with venomous fish. We can make the exact same formal argument as above (A←s→B←C) by saying "all venomous fish are red and some red fish are in the fish tank." (in fish tank ←s→ red fish ← venomous) Here we can't say that some fish in the fish tank are venomous. This example argument only states that the only possible fish in the world that can be venomous are red (venomous → red). Does that mean all red fish are venomous(red → venomous)? No, that would be a mistaken reversal. The owner of the fish tank could have gotten very lucky or only picked red fish that are not venomous. In that case, are any fish in the fish tank venomous? No! Because the owner only picked the non-venomous red fish which is possible under this lawgic statement. So if the only fish in the world that are venomous are red fish and the fish tank has red fish, it does not necessarily mean that the fish tank has venomous fish.
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!
Hold on there, you need to slow down.
We love that you want post in our discussion forum! Just come back in a bit to post again!
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.
63 comments
But because "some" is weaker than "most", wouldn't that actually help an argument in this case? If "most" violinists aren't exceptionally good at playing the violin, I'm struggling to see how changing "most" to "some" would be invalid just because the "most" argument is invalid.
Add a video for these flaw explanations please.
I understood it by thinking of the Formal Arguments #4 and #6. They all have to have some or most before all because we needs As in the other three two buckets.
A <--s--> B --> C
A <--s--> C
Some As are Cs. Because there is some As in the C bucket.
Not like the invalid argument:
A -->B <--s--> C
A <-s-> C
How do I have some As if I put all of them B. It does not make a correct conclusion.
DRAW OR THINK OF VENN DIAGRAMS! It helps a lot :D
I found it helpful to draw out the sets in a way where it makes sense why the invalid form doesn't work. Hope this helps someone who's struggling to understand!
It's really helpful for me to think of this and the previous flaw with the buckets scenario. Think of it, if you only take some of the A's and put them into B, but then you take that entire bucket and put it into C, then of course there will be A's in there. But if you take all the A's into B (remember, we don't know how many A's that is, it could be literally 2), then take only some of that entire bucket into C, it's less likely that there will be A's in C. Hope this helped!!
@SheridanMcGadden thank you so much. This helped me understand it way better.
A--->B
B<s>C
A<s>c
this is invalid. but if there is at least 1 B in the set of C
then shouldnt there be at least 1 A in the set of C. Since all A's are B's?
Here's my attempt at explaining:
All of A are B: all humans are mammals (A --> B)
Some B are C: some mammals have tails (B <s> C)
Therefore, some humans have tails (A <s> C)
We cannot conclude that some humans have tails just because all humans are mammals. I tried to use an example that was obvious, I hope this helps!
who else is confused :p
@Gisellednava rjon27 comment really helped me understand. I think it possible that there are humans that have tails but there is no 100% chance there that some mammals that have tails happened to be human. The "some" could all be different mammals that are not humans.
Here's my attempt at an invalid and valid argument. Someone correct me if I'm going about it wrong!
Invalid: All peaches have pits. Some pits are rotten. Therefore, some peaches are rotten.
The reason why this is invalid is because we don't know the peach is rotten. It could be because the air had weird chemicals in it, or maybe a bug was hiding in the peach. Simply knowing that some peach pits are rotten isn't enough to know that the peach itself is rotten.
Valid: Some peaches have pits. All pits are rotten. Therefore, some peaches are rotten.
This is valid because we know ALL of B is true. If all pits are rotten and we know some peaches have pits, we at least know some of the peaches are rotten.
@kyorofan20 This was helpful thank you!
What I have found that trips me up is thinking 'all A are B'. To me, it makes it sound like they are interchangeable ideas, when really it just means all of A is in the B group. It does not guarantee that if some (or most) of B is part of the C group, that that part of B has any A in it at all.
My understanding of this is that all A have B, but just because they have B doesn't mean that A will take on C. Because it can't be confirmed that just because you have B you also have C. The key indicator is "some". A is sufficient for B but does not guarantee that B brings about C.
why is there no video for the important subjects :'(
Ok here is my attempt at an invalid argument.
All Options at Marys dessert shop is frozen yogurt, some frozen yogurt is mint flavored; therefore some of the Options at Marys shop is mint.
this in incorrect because the statement about mint is about all frozen yogurt, it doesn't tell us anything about if frozen yogurt at Marys dessert shop. It is just as likely she only serves vanilla and chocolate. we cannot disprove or prove she has mint, just that it is possible.
Here's an example I made for myself if it helps:
Trap 5: Swapping "some" and "all" arrows:
Remember: a valid conclusion about intersecting sets can only be made if the some arrow appears before the all arrow.
Ex.
Artists in the master class are painters. Some people who call themselves painters suck at art. Therefore, some artists in the master class suck at art.
Art's -> Pnt's <-s-> Suck@it
____
Art's <-s-> Suck@it
Thus is NOT valid
Again, just because the first set shares a characteristic with the second (painters), does not mean you can equate the two.
Ex.
Some artists in the master class are painters. All people who paint suck at art. Therefore, some artists in the master class suck at art.
Art's <-s-> Pnt's -> Suck@it
____
Art's <-s-> Suck@it
Here, we know 100% of the subset (painters) ALL share the quality we are discussing (sucking at it). So, we can easily conclude that the portion of artists in the class that are painters definitely suck at it.
im confused on the quantifiers. If we say all A's are B's, and we equate themselves logically, can they be interchangeable? A B and All B's are C's (or A = B), we cans say A B/C (because theyre interchangeable .
Why cant it work with some? All A's are B's (or A = B) and B C, therefore A/B C.
Is it all just contextual basing it on what sets youre talking about?
None of it is contextual!
We are not trying to equate A and B... when we say all A's are B's we are definitely not saying A = B, we are saying that if you are A then you are B. This is distinct from saying A=B because A=B also means B=A, and we do not want to misconclude that B →A (that would be the oldest mistake in the book!). A and B are not interchangable, A is still just sufficient for B.
It cannot work with 'some' when some comes after 'all' for the following reason:
Consider that I own 5 marbles, and all of the marbles I own (A) are red marbles (B).
So thus far we can conclude A → B.
Now, some red marbles that exist also have blue stripes (C).
B ←s→ C
Does that mean that some of the marbles I own have blue stripes? Not necessarily. It's not IMPOSSIBLE that I could own one with blue stripes... there's just no way we can conclude that. Just because there are red marbles with blue stripes that exist, does not mean that I own them. therefore A → B ←s→ C warrants no valid conclusions.
Conversely, A ←s→B → C, does warrant a valid conclusion. Let's use the same example:
Consider that I own 5 marbles, and some of the marbles I own (A) are red marbles (B). We conclude that A ←s→ B.
If all red marbles (B) have blue stripes (C), that means B → C.
If I own some red marbles (A), and every single red marble has blue stripes(C), that means some of my red marbles have blue stripes, too.
A ←s→B → C
So what about the following argument? Is this valid?
A→B
A ←s→ C
B←s→ C
Yes, this is valid.
An easier way to understand this in Lawgic would be to flip the initial some claim (as it is bidirectional) to:
C ←s→ A
and chain it with the other claim A → B to make:
C ←s→ A → B
From which we know C ←s→ B would be a valid conclusion, as it is now some before all. Flip that to B ←s→ C to confirm that your conclusion was valid.
I think it helps to think about it like order of operations.
If you dump some of you A's into your B's
then dump all of your B bucket into your C's, then you will definitely have some A's in your C's.
But if you dump all of your A's into your B's
and then dump some of your B's into your C's, then you won't necessarily have any A's in your C's, you could, but its not guaranteed. If it is not guaranteed then it is not valid.
I think this reasoning also work for the most rule as well.
Hands down the best explanation I've found on here. Way better than what is in the lessons. Thank you !!
Very helpful, thank you!
@A.Belle you are a BEAST. 7Sage should update this with your formula.
#YouRock
@A.Belle great explanation. I like this more than the "scoop" example 7sage has been using.
I found that drawing the circles for the subsets and supersets make it much easier to understand.
Same!
im going to need a video explanation on this one chief im starting to dive myself insane
Same! I'm so lost with these last 3 logic flaws.
So basically, the smaller amount needs to come before the larger amount in order for an argument to be true
Why B —m→ C is stronger than B ←s→ C? And what does it mean it is stronger?
By stronger they mean because ‑m→ refers to most and ←s→ refers to some. It is much stronger to say that most people agree with me on a certain topic than that some people agree with me on a certain topic (some = at least one, it is not a lot). So most is much stronger. In this case, they are referring to the idea that if we say that "most people do x" that is equivalent to saying that "some people do x" because most also encompasses some. Hope this makes sense!
Thank you for the great explanation. I appreciate it!
Okay so pretty much in order for this to be valid is by this:
A→B←s→C
----------------
B←s→C
And this would be valid because A is not affecting the conclusion??
Invalid:
A→B←s→C
-------------------
A←s→C
X-
This could also be valid if it said Some before All obviously.....
Yes and No?
To me, it looks as if you just repeated one of the statements as a conclusion. It'll be valid because it's part of the initial set of claims... it's not really something you'd conclude from the previous claims.
All CHICKENS are BIRDS. Some BIRDS are GREEN.
C: Chickens
B: Birds
G: Green
Written out, it'd look like this.
C → B ←s→ G
You effectively stated that this would be valid if the conclusion was B ←s→ G
You're not wrong, but at the same time, you just kind of repeated the second claim: Some BIRDS are GREEN.
As for your second point, I think you're right.
If we take the same sentence and before and got C ←s→ G, that wouldn't be right.
All CHICKENS are BIRDS. Some BIRDS are GREEN.
Therefore some CHICKENS are GREEN? That wouldn't be the case.
Hopefully I'm right about this and getting my point across.
Can someone please explain this in English and not Lawgic?
So in the previous lesson they said that A←s→B→C leads to a valid inference of A←s→C. This in plain English would be if some of A are B and all of B is C, then some of A are C. (For example, if some fish in the fish tank are red and all red fish are venomous then we can make the inference that some fish in the fish tank are venomous).
However, in this lesson they introduce the formal argument A←s→B←C (or in other words C→B←s→A). In this case, you would be saying that all of C is B and some of B are A. For this type of formal argument you CANNOT make an inference. When you first see this argument, you would want to make the inference A←s→C but this cannot be an inference made because it could be true but it could also be false.
Here's how that works. Imagine a circle diagram where B is a big circle and within B there is a smaller circle called C. Because the only thing that this lawgic statement tells us is that all of C is in B, the size of C can be large or small. So C could take up a big spot within B or it could be a small dot within the circle "B." However, the lawgic statement goes onto say that some of B are A. In this case, could we guarantee that at least some of C (in other words, at least one of C) will be a part of A? No. Because what if the "some of B" that is a A is a part of the circle that C does not cover? If within the circle "B," circle C and A does not overlap, it would mean that some of C is NOT A and this is a possible scenario.
For example, let's replace A with fish in the fish tank, B with red and C with venomous fish. We can make the exact same formal argument as above (A←s→B←C) by saying "all venomous fish are red and some red fish are in the fish tank." (in fish tank ←s→ red fish ← venomous) Here we can't say that some fish in the fish tank are venomous. This example argument only states that the only possible fish in the world that can be venomous are red (venomous → red). Does that mean all red fish are venomous(red → venomous)? No, that would be a mistaken reversal. The owner of the fish tank could have gotten very lucky or only picked red fish that are not venomous. In that case, are any fish in the fish tank venomous? No! Because the owner only picked the non-venomous red fish which is possible under this lawgic statement. So if the only fish in the world that are venomous are red fish and the fish tank has red fish, it does not necessarily mean that the fish tank has venomous fish.
Hope that makes sense.
Awesome, thank you so much
So in this and the last one, to put it very simply, if there is an arrow before the ←s→ or ‑m→ it is invalid? The ←s→ and ‑m→ must be first.
Exactly!