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
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 &
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.
most butterflies are magical, all magical creatures eat pineapple, therefore most butterflies eat pineapple :)
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43 comments
These videos make so much more sense than the set prior lol.
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.
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
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) ?
#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
#feedback venn diagrams (as used in past lessons) express these relationships more clearly than this convoluted 3-D bucket analogy.
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?
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]
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.
#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.
"ALL" and MOST" can also be explained as All is a subset of Most.
Most books are paperback. All paperback books smell good. Therefore, most books smell good.
B ‑m→ P → SG
B ‑m→ SG
most butterflies are magical, all magical creatures eat pineapple, therefore most butterflies eat pineapple :)