So how would the argument of "Most people who know how to play the violin are not exceptional. All violinists at the New York Philharmonic know how to play the violin. So, some violinists at the New York Philharmonic are not exceptional." work compared to the original argument?
Here's some examples I made for myself, hope this provides some clarity!
Trap 4: Swapping "most" and "all" arrows
Remember: a valid conclusion about intersecting sets can only be made if the most arrow appears before the all arrow.
Ex.
All wrestlers in the WMMA know how to fight. Most people who know how to fight suck at it. Therefore, some fighters in the WMMA suck at it
WMMA -> F --m-> Suck@F
____
WMMA <-s-> Suck@F
The above is NOT a valid conclusion. The first set is not indicative of any other set, even if their qualities (fighting) overlap. So, that quality cannot be used to draw valid conclusions .
Ex.
Most WMMA women know how to fight. All women who know how to fight are strong. Therefore, most WMMA women are strong.
WMMA -m-> F -> S
____
WMMA -m-> S
The above conclusion IS valid
Remember, it's not "all most" chocolate is good, its "most all" chocolate is good
I don't want to "almost" ace the test, i want to get "most-all" questions correct
I'm understanding this but I think that it would be far easier to learn if you would use acronyms that match what is being discussed in the arguments rather than always using A, B, C.
If most comes before all, it is valid; if most comes after all, it is invalid.
This makes sense because if you say
Premise: All lawyers are smart people. Most smart people enjoy coffee.
Conclusion: Some lawyers enjoy coffee.
In lawgic:
lawyer --> smart ‑m→ enjoy coffee
From this, you cannot infer that some lawyers enjoy coffee because we don't know how big the set of lawyers and smart people is. It can be that there are 300 lawyers, all of whom are smart, but an overall total of 1,000 smart people - so there is still 700 smart people who could enjoy coffee and 300 who happen to be the lawyers who don't enjoy it. This makes the statement that most smart people enjoy coffee still stand but not the inference that some lawyers enjoy coffee. So, it can be the case that no lawyer at all enjoys coffee. This invalid argument boils down to the idea that we are unsure how big the set for the first two sets is.
An example of this argument in a valid way would be
Premise: Most lawyers are smart people. All smart people enjoy coffee.
Conclusion: Some lawyers enjoy coffee.
In lawgic:
lawyer ‑m→smart→enjoy coffee
From this, you can infer that some lawyers enjoy coffee. This is because we know that over 51% of lawyers are smart and ALL smart people enjoy coffee; therefore, it must be the case that SOME (at least one or all) lawyers do enjoy coffee because most of them (over half of them) fall in the set of being smart, all of whom enjoy coffee.
if the all arrow shows up first in the chain and then you see the most arrow, there are no valid conclusions to be drawn via the chain.
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110 comments
Why not videos this is frustrating
#FEEDBACK We need the videos back
So how would the argument of "Most people who know how to play the violin are not exceptional. All violinists at the New York Philharmonic know how to play the violin. So, some violinists at the New York Philharmonic are not exceptional." work compared to the original argument?
Can this general idea of A->B -m-> C; A<-s->C be valid if set sizes are within reasonable quantities?
where are the videos I dont like reading 😠
Does anyone else draw Venn Diagrams for these problems? It makes so much more sense when you can just visualize the buckets.
This explanation uses a flawed argument itself, the ecological fallacy.
A->B-m->C
-----
A <-s-> C
There could be a chance that all none of the A that are B were scooped into the C bucket when most B became C. So we cannot conclude that Some A are C
I am really confused by the "let's review" portion. The "Most before all" argument DOES say that:
A -m-> B --> C
Therefore, A -m-> C
But in the "let's review" part of this lesson it says that:
A -m-> B --> C yields no valid conclusions via the chain.
Can someone explain this please?
Example
All non-caffeinated teas are relaxing. Most teas help with sleep. Therefore, most relaxing teas help with sleep.
I understand the logic behind this stim. My issue is with the lawgic. If I did it myself, I would've done:
Premise: NYP → PV ‑m→ /EG
Conclusion: NYP ←s→ /EG
Though, in the lesson, 'C' was not negated. I'm not sure if i'm not getting it or i've been reading too long today. #help
Here is a rule of thumb that could help to understand this concept.
Most --> All chains can work.
All --> most chains don't preserve logic.
Hope this helps!
Has anybody found an actual LSAT question where this flaw is done, so we can see how the answers would look?
Here's some examples I made for myself, hope this provides some clarity!
Trap 4: Swapping "most" and "all" arrows
Remember: a valid conclusion about intersecting sets can only be made if the most arrow appears before the all arrow.
Ex.
All wrestlers in the WMMA know how to fight. Most people who know how to fight suck at it. Therefore, some fighters in the WMMA suck at it
WMMA -> F --m-> Suck@F
____
WMMA <-s-> Suck@F
The above is NOT a valid conclusion. The first set is not indicative of any other set, even if their qualities (fighting) overlap. So, that quality cannot be used to draw valid conclusions .
Ex.
Most WMMA women know how to fight. All women who know how to fight are strong. Therefore, most WMMA women are strong.
WMMA -m-> F -> S
____
WMMA -m-> S
The above conclusion IS valid
Remember, it's not "all most" chocolate is good, its "most all" chocolate is good
I don't want to "almost" ace the test, i want to get "most-all" questions correct
I am actually crying. I have no idea what this is saying
Would you switch the "All →" to a "most ‑m→" and then you would have the "Two Mosts split" argument? Is this possible?
this argument flaw does not lend itself well to his style of logic.
Where are the videos? I prefer the videos to just reading.
#feedback please include a video for this lesson. It was so confusing
I'm understanding this but I think that it would be far easier to learn if you would use acronyms that match what is being discussed in the arguments rather than always using A, B, C.
I don't get this :/
Need Videos here so confusing to read it
If most comes before all, it is valid; if most comes after all, it is invalid.
This makes sense because if you say
Premise: All lawyers are smart people. Most smart people enjoy coffee.
Conclusion: Some lawyers enjoy coffee.
In lawgic:
lawyer --> smart ‑m→ enjoy coffee
From this, you cannot infer that some lawyers enjoy coffee because we don't know how big the set of lawyers and smart people is. It can be that there are 300 lawyers, all of whom are smart, but an overall total of 1,000 smart people - so there is still 700 smart people who could enjoy coffee and 300 who happen to be the lawyers who don't enjoy it. This makes the statement that most smart people enjoy coffee still stand but not the inference that some lawyers enjoy coffee. So, it can be the case that no lawyer at all enjoys coffee. This invalid argument boils down to the idea that we are unsure how big the set for the first two sets is.
An example of this argument in a valid way would be
Premise: Most lawyers are smart people. All smart people enjoy coffee.
Conclusion: Some lawyers enjoy coffee.
In lawgic:
lawyer ‑m→smart→enjoy coffee
From this, you can infer that some lawyers enjoy coffee. This is because we know that over 51% of lawyers are smart and ALL smart people enjoy coffee; therefore, it must be the case that SOME (at least one or all) lawyers do enjoy coffee because most of them (over half of them) fall in the set of being smart, all of whom enjoy coffee.
Hopefully this breaks it down well enough.
A —→ B
B --m→ C
therefore: A--m-->C
correct?
if the all arrow shows up first in the chain and then you see the most arrow, there are no valid conclusions to be drawn via the chain.