seems like the bucket is helping people, but you can also do circles here again, to build off previous visuals. it’s not perfect, but i do better with static visuals vs the animated buckets.
most A are B = most of A circle intersects B circle.
all B are C = B is a subset of C, so the C circle completely surrounds B circle.
just imagine the C circle meets exactly at the B circle edge that’s inside A circle.
if all of B is inside C, then the A intersection is the same for B and C. most A are in B so most A are in C by default.
If most of A consists of B and all of B consists of C, then B and C are now inseparable and one in the same. So naturally, most of A must be C. I am looking at B and C interchangeably and one in the same.
Most hippos can sing. All hippos that can sing fart glitter. therefore most hippos fart glitter . Hippos---m-->sing HIPPOS-->Fart G connect it : hippos--m->sing-->fart glitter conclusion hippos--m> fart g
We don't know which electric cars are red. The Teslas might all be among the minority that aren't red. It's possible, but not necessary — and logic only allows conclusions that are guaranteed by the premises.
To use numbers: say you have 1,000 Teslas. So you know you have 1,000 (at least) electric cars (from the first premise).
We don’t know how many electric cars exist in total, but clearly, it’s more than 1,000 (since not all electric cars are Teslas). So let’s suppose there are 5,000 electric cars total.
You know (based on the 5,000 number) that you have at least 2,501 electric cars that are Red, and at most 2,499 electric cars that are NOT red. And, you also know you have 1,000 electric cars (i.e. Teslas). But how do you know that this 1,000 Teslas (electric cars) are either among the 2,501 red ones or the 2,499 non-red ones? The two premises don't guarantee that Teslas belong in the Red category, which is why you cannot say that some Teslas are red (what if they are actually among the 2,499 that are not red? It's possible: you have 2,499 non-red cars, which include the 1,000 Teslas).
It should be "most students in Professor Slughorn's class can brew potions masterfully . . ." The Slug Club is Professor Slughorn's club and only exists when he is the potions professor at Hogwarts.
for ''most'' or conditional relations, if A it follows that B. So negating that means if A is true, B does not follow. But then why Isnt it A -> /B. If A follows then not the opposite of B follows is what it is saying. Does this mean Negating with a conjunction means that A and B can never be true? e.g A &
Let's see what an expert says, but I don't think that this is valid. "Most" does infer "some", but I don't think you can just deduce, from "most" a "some" statement.
its intersting when you use the all quantifier because if you say All b's are c's, youre equating the two. if some of a's are b's and all c's are b's, you can just combine the two some a's are c's because you meshed together b and c as two equal things.
For this argument: Most bars that serve wine also serve fancy cocktails. All bars that serve fancy cocktails play loud music. Therefore, most bars that serve wine play loud music.
Can the conclusion also be some restaurants that service wine play loud music?
I'm experimenting with Formal Argument #5, is this valid?
Most bars that serve wine also serve fancy cocktails. All bars that serve wine play loud music. Therefore, most bars that serve serve fancy cocktails play loud music.
There might be ways to trick someone here that people might want to watch out for. Let's take the first example:
Most students in Prof. Snape's class can brew potions masterfully. All students who can masterfully brew potions are invited to join the Slug Club. Therefore, most students in Prof. Snape's class are invited to join the Slug Club.
This works, is valid. But, if we just remove the modifier to "in professor Snape's Class" we change the domain, superset/subset relationship. whatever. We get:
Most students in Prof. Snape's class can brew potions masterfully. All students who can masterfully brew potions are invited to join the Slug Club. Therefore, most students are invited to join the Slug Club.
Respectfully Snape was never a potions professor when Slughorn was a professor. Thus, because I am a big fat nerd, the first argument is factually invalid.
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47 comments
The buckets saved my life.
Thank God they sterilized the letter moving around like that was disturbing (though I still appreciated the effort to animate them).
the WHAT club???
These videos make so much more sense than the set prior lol.
@kmartz I keep waiting for them to just start not making sense it seems too good to be true lol
@kmartz RIGHT
Is there a way to get a summary of all 5 formal arguments?
So you can basically think about the "all" statement passing through whatever quantifier came first to C, whether it is some or most.
seems like the bucket is helping people, but you can also do circles here again, to build off previous visuals. it’s not perfect, but i do better with static visuals vs the animated buckets.
most A are B = most of A circle intersects B circle.
all B are C = B is a subset of C, so the C circle completely surrounds B circle.
just imagine the C circle meets exactly at the B circle edge that’s inside A circle.
if all of B is inside C, then the A intersection is the same for B and C. most A are in B so most A are in C by default.
I just wanted to say that the visual of the buckets is extremely helpful. It made the logic behind the argument so clear and I really appreciated it!
If most of A consists of B and all of B consists of C, then B and C are now inseparable and one in the same. So naturally, most of A must be C. I am looking at B and C interchangeably and one in the same.
Is my reasoning correct?
Most hippos can sing. All hippos that can sing fart glitter. therefore most hippos fart glitter . Hippos---m-->sing HIPPOS-->Fart G connect it : hippos--m->sing-->fart glitter conclusion hippos--m> fart g
#feedback
it would be helpful to see an argument that is NOT valid. So we could see an example of how an invalid argument would be fed to us on the LSAT.
@EmilyMacaluso An example would be:
All Teslas are electric cars.
Most electric cars are red.
____________
Some Teslas are red --> WRONG conclusion.
We don't know which electric cars are red. The Teslas might all be among the minority that aren't red. It's possible, but not necessary — and logic only allows conclusions that are guaranteed by the premises.
To use numbers: say you have 1,000 Teslas. So you know you have 1,000 (at least) electric cars (from the first premise).
We don’t know how many electric cars exist in total, but clearly, it’s more than 1,000 (since not all electric cars are Teslas). So let’s suppose there are 5,000 electric cars total.
You know (based on the 5,000 number) that you have at least 2,501 electric cars that are Red, and at most 2,499 electric cars that are NOT red. And, you also know you have 1,000 electric cars (i.e. Teslas). But how do you know that this 1,000 Teslas (electric cars) are either among the 2,501 red ones or the 2,499 non-red ones? The two premises don't guarantee that Teslas belong in the Red category, which is why you cannot say that some Teslas are red (what if they are actually among the 2,499 that are not red? It's possible: you have 2,499 non-red cars, which include the 1,000 Teslas).
It should be "most students in Professor Slughorn's class can brew potions masterfully . . ." The Slug Club is Professor Slughorn's club and only exists when he is the potions professor at Hogwarts.
for ''most'' or conditional relations, if A it follows that B. So negating that means if A is true, B does not follow. But then why Isnt it A -> /B. If A follows then not the opposite of B follows is what it is saying. Does this mean Negating with a conjunction means that A and B can never be true? e.g A &
If
A most B
B -> C
Then obvi: A most C
Question; is it also valid then to conclude
A some C
C some A
Thanks
Let's see what an expert says, but I don't think that this is valid. "Most" does infer "some", but I don't think you can just deduce, from "most" a "some" statement.
its intersting when you use the all quantifier because if you say All b's are c's, youre equating the two. if some of a's are b's and all c's are b's, you can just combine the two some a's are c's because you meshed together b and c as two equal things.
By extension then, if all As are Bs and all As are Cs , is the relationship between B and C that some Bs are Cs (and vice versa) ?
You can't get some B's are C's from all As are Bs and all As are Cs. Where do Bs come in?
#feedback I really like the bucket analogy, I would like to see more of it!
am i the only one that understood the lawgic and the minute he pulled out the buckets visual it made it confusing
Nope, I felt the same way. The logic checked out for me without the bucket visualization. This video could have been cut in half.
#feedback venn diagrams (as used in past lessons) express these relationships more clearly than this convoluted 3-D bucket analogy.
Agreed! I always use them
If it is the same as some is it valid to say most A are B. All B are C. Therefore most A are C?
Yes, you are correct.
For this argument: Most bars that serve wine also serve fancy cocktails. All bars that serve fancy cocktails play loud music. Therefore, most bars that serve wine play loud music.
Can the conclusion also be some restaurants that service wine play loud music?
[ Serve Wine Loud Music]
Looks right to me. In 'Quantifier Inferences', the order is:
All -> Most -> Some, which makes sense.
All is more than 50%, and some can go from 1% to 100%.
Thus, yeah.
W ‑m→ LM
gets at the same point as
W ←s→ LM
I'm experimenting with Formal Argument #5, is this valid?
Most bars that serve wine also serve fancy cocktails. All bars that serve wine play loud music. Therefore, most bars that serve serve fancy cocktails play loud music.
SW--m-->SFC
SW---->PLM
therefore,
SFC---m-->PLM
There might be ways to trick someone here that people might want to watch out for. Let's take the first example:
Most students in Prof. Snape's class can brew potions masterfully. All students who can masterfully brew potions are invited to join the Slug Club. Therefore, most students in Prof. Snape's class are invited to join the Slug Club.
This works, is valid. But, if we just remove the modifier to "in professor Snape's Class" we change the domain, superset/subset relationship. whatever. We get:
Most students in Prof. Snape's class can brew potions masterfully. All students who can masterfully brew potions are invited to join the Slug Club. Therefore, most students are invited to join the Slug Club.
This would be invalid.
Invalid because it's generalizing the students right? versus when it says in Prof. Snapes class you're specifically talking about his/her students..
Very true!
that is correct. subtle wording change, but a totally different conclusion. great point in calling out!
#feedback
The diagrams offered for the other formal arguments were beneficial. Please make one for this formal argument.
Respectfully Snape was never a potions professor when Slughorn was a professor. Thus, because I am a big fat nerd, the first argument is factually invalid.