The fact that I could not understand a single letter here one week ago and now I can read it out loud without even checking my notes speaks volumes, even create quick examples on my mind that are valid when translated to lawgic. I'm gonna destroy this exam and you will also do it <3
@dancingqueen138 translating each question and each of the possible answer to lawgic. It truly gets to the point in which u start mapping things out on your mind as you read. When u get to that point u truly feel that u are becoming faster without necessarily loosing accuracy. Which I think is the goal.
@lauren0683 Hello! I understood the formal argument #6 with the "floating" letters example. Most As are Bs. Most As are Cs. Therefore, it would be valid to say that some Bs are Cs (Remember the "range" that is covered by the word "some"- The range could be from at least one, which would be the minimum, to all of the members of the set).
Assume that the population of As is 5. If we know that most As are Bs, then it would mean that at least three of the As are Bs. If we also say that most As are Cs, then it would also mean that at least three of the As are Bs. This would cause an overlap of an A to have both a B and a C.
for "Some Before All" and "Most Before All" does it have to be before? can't it be an all statement, and then a some/most statement and it still works??
If all implies most and most implies some, cant we just convert all these most arrows to some arrows making it easier for us to conclude that some A are E or whatever else we want to conclude?: A --> B -m-> C -m-> D -m-> E
That is essentially a variation on the theme. Since we are not given an exhaustive list, we just have to make logical inferences to compensate for the long list of possible combinations.
All bald people are hairless and all bald people have experienced hair loss. If that is the case then have all hairless people experienced hair loss?
Lawgic:
Bald -> Hairless
Bald-> Hair loss
Therefore, Bald -> Hairless -> Hair loss
Which gives us, Hairless -> Hair loss
Now suppose the total number of bald people are 100 in that case what our conclusion means is all 100 people are hairless and have experienced hair loss.
As some has a lower limit of at least 1 of the hundred but no upper limit yes this does also imply, Hairless Hair loss
While I have come to appreciate "Lawgic", it was a bit tricky to follow in the beginning. I wish the curriculum was structured so that we got this entire list (in this format) prior to learning about each type of valid argument so that as we were learning we could see how each component contributes to the overall argument, rather than the structure of the argument being revealed in Lawgic after the fact.
Not that it would make a huge difference to the overall learning, but I think it would have helped me follow along a bit easier.
Visualizing these arguments is helping me learn them:
The Conditional Argument:
Imagine a conveyor belt (→) where if A items come in, B items come out. If you put something on the conveyor belt, you will get the corresponding outcome.
The Contrapositive Argument:
Visualize a seesaw with A on one end and B on the other. If B goes down, then A must go up to balance it out.
Conditional Chaining:
Picture a train moving from A to B to C. If the train goes from A to B and then from B to C, logically, it's like it directly went from A to C.
Some Before All:
Imagine a group of people entering a room. If some people (A) are in a room with all of B and some of B are in a room with all of C, then logically, some of A must also be in the room with all of C.
Most Before All:
Think of a river flowing from A to B to C. If most of the water flows from A to B and then from B to C, it means that there's a significant portion of the water from A that ends up in C.
Two Mosts:
Visualize two buckets, one labeled B and the other labeled C. If most of the items in bucket A can fill both buckets B and C, then logically, there must be some overlap between the contents of buckets B and C.
One of the things that helped me with visualizing conditionals with quantifiers will be below. IDK if I am right so I welcome feedback. This is based on my understanding from the lessons on here and two videos I saw on YouTube.
With chaining the quantifiers is that when it is a quantifier before all, we are trying to see if there is a relationship between the two supersets. These must be connected by a common subset. And in all cases but one, one of the conditional relationships must be all:
X's → Y's
(all/most/some) X's → Z's
the inference will always be X's ←s→ Z's because we do not how big Y or Z are.
The one exception is most
Most X's → Y's
Most X's → Z's
Inference: Some Y's
The other one is a straight chain like A→B→C. Like we learned about the subset → superset lessons, I will call B the sub-superset because it acts as both and IDK a proper word for it. The reason why the quantifier needs to be before the all in the chain ( how its actually written does not matter) is because if we say all of A is B and most of B is C, we do not know how big A or B are.
However, since all of B is in C, even if A is just 0.1% in B, A is the same amount in C. Therefore, the relationship between the subset and the superset will be the same as the subset sub=superset. So, the sub-superset needs to be all in in the superset. If this relation unfolds, any relationship between A and B will be the same between A and C so long as all of B is in C.
Nope, i think it helps to think about it like this: 2 scoops of half plus 1 of A letters are being poured into bucket B and bucket C.
However, this isn't possible, I can't put half plus 1 of A into both bucket B and C. This means there must be some overlap between B and C, where they have some of A.
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.
47 comments
Verbally it reads
Conditional: If A then B, Thing of A is Thing of B
Contrapositive: If A then B, Thing is not B then Thing is not A
Conditional Chaining: If A then B then C, If A then C
Some before all: Some of A is in B and B is in C, therefore Some of A is in C
Most before All: Most of A is in B and B is in C, therefore Most of A is in C
Two Mosts: Most of A is in B | Most of A is in C, therefore some of B is in C
Is this correct? Wording might not be correct, lmk.
The fact that I could not understand a single letter here one week ago and now I can read it out loud without even checking my notes speaks volumes, even create quick examples on my mind that are valid when translated to lawgic. I'm gonna destroy this exam and you will also do it <3
@Fernabkn1 what helped you?
@dancingqueen138 translating each question and each of the possible answer to lawgic. It truly gets to the point in which u start mapping things out on your mind as you read. When u get to that point u truly feel that u are becoming faster without necessarily loosing accuracy. Which I think is the goal.
@Fernabkn1 Seriously!!! The fact I just read this out loud was such a confidence booster.
definitely will be coming back to review!
Definitely book marking!
Simple examples for anyone struggling:
1. Conditional:
Basketball players are athletes
Tom plays basketball
---
Tom is an athlete
2. Contrapositive:
Basketball players are athletes
Tom is not an athlete
---
Tom does not play basketball
3. Conditional Chaining:
Kittens are cats. Cats are cute
---
Kittens are cute
4. Some before all:
Some cats are pets. (All) Pets are kind
---
Some cats are kind
5. Most before all:
Most cats are pets. (All) Pets are kind
---
Most cats are kind
6. Two Mosts:
Most cats are pets. Most cats are kind.
---
Some pets are kind
@BenjaminBrady Thank you!!!
Love this breakdown summary probably most important page in this section
I'm still a bit confused about when we use all these formal arguments
#help for formal argument #6 Two Mosts:
Is it correct to say that if most of A are B, and most of A are C, that (instead of saying most of B are C) some of A are B and C?
I don't know how to comprehend this argument without thinking of it like that. Here's an example I've been using:
Most dogs have leashes.
Most dogs have collars.
Therefore, some dogs have leashes and collars.
But (correct me if I'm wrong) this^ argument in lawgic form would look more like:
Premise: A -m-> B
Premise: A -m-> C
Conclusion: A B and C
Am I just overcomplicating things? Or is my version incorrect?? Idk I'm getting confused if anyone can help me on this.
@lauren0683 Hello! I understood the formal argument #6 with the "floating" letters example. Most As are Bs. Most As are Cs. Therefore, it would be valid to say that some Bs are Cs (Remember the "range" that is covered by the word "some"- The range could be from at least one, which would be the minimum, to all of the members of the set).
Assume that the population of As is 5. If we know that most As are Bs, then it would mean that at least three of the As are Bs. If we also say that most As are Cs, then it would also mean that at least three of the As are Bs. This would cause an overlap of an A to have both a B and a C.
Visually:
A B
A B
A B C
A C
A C
Hopefully it helps!
Whoever is studying and reading this keep crushing it.
Stick with it you owe it to yourself.
over and outttieeee.
for "Some Before All" and "Most Before All" does it have to be before? can't it be an all statement, and then a some/most statement and it still works??
Some before all example
Some dogs love kibble. All dogs that love kibble are energetic.
dogs ←s→ love kibble → energetic
therefore we can conclude......
some dogs are energetic
Most before all example
Most people who drink water are healthy. All people who are healthy are athletic.
drink water ‑m→ healthy → athletic
therefore we can conclude...
Most people who drink water are athletic.
Two mosts example
Most students are smart. Most students have planners.
therefore we can conclude....
Some smart students have planners.
Correct me if any of these are incorrect.
If your answers are incorrect, then I will correct it. Inc→C, contrapositve that, /C→/Inc,
If I don't correct any of your answer, then they are correct.
I don't correct any of them, therefore they are correct.
If all implies most and most implies some, cant we just convert all these most arrows to some arrows making it easier for us to conclude that some A are E or whatever else we want to conclude?: A --> B -m-> C -m-> D -m-> E
In other words, wouldn't it be much easier to simplify all and most arrows to some arrows?
Not quite, because if you turn everything into Some, that will make you think you can't make certain inferences when you actually can.
All As are B.
Some As are C.
You can conclude from these that Some Bs are C.
"All As are B" does imply "Some As are B." But if you thought that all we know was
Some As are B
Some As are C
Then you would think there's no inference to be made from the combination of the two statements we started with.
#feedback in the skill builders, this form appears quite a lot:
All As are Bs
All As are Cs
Therefore, some Bs are Cs.
Shouldn't this also be part of the list then?
That is essentially a variation on the theme. Since we are not given an exhaustive list, we just have to make logical inferences to compensate for the long list of possible combinations.
#help Is this valid? A→B, A→C, therefore B←s→C?
Yes it is valid because "all" implies "most", and this would follow the formal argument structure of "two mosts"
Yes I think so and this is why:
example:
All bald people are hairless and all bald people have experienced hair loss. If that is the case then have all hairless people experienced hair loss?
Lawgic:
Bald -> Hairless
Bald-> Hair loss
Therefore, Bald -> Hairless -> Hair loss
Which gives us, Hairless -> Hair loss
Now suppose the total number of bald people are 100 in that case what our conclusion means is all 100 people are hairless and have experienced hair loss.
As some has a lower limit of at least 1 of the hundred but no upper limit yes this does also imply, Hairless Hair loss
yes its correct!
Is there a form that follows:
A‑m→B
A←s→C
--------
A←s→C?
I don't think this is a working argument coz you're repeating A-s-C.
I think the conclusion would be
B←s→C
Correct me if I'm wrong!!!!
While I have come to appreciate "Lawgic", it was a bit tricky to follow in the beginning. I wish the curriculum was structured so that we got this entire list (in this format) prior to learning about each type of valid argument so that as we were learning we could see how each component contributes to the overall argument, rather than the structure of the argument being revealed in Lawgic after the fact.
Not that it would make a huge difference to the overall learning, but I think it would have helped me follow along a bit easier.
completely agree i feel like my understanding moment came literally at this last stage
#feedback
Can we get this in a diagram?
what does the (x)/B mean?
X is a member of "not B" or X is not a member of B.
Visualizing these arguments is helping me learn them:
The Conditional Argument:
Imagine a conveyor belt (→) where if A items come in, B items come out. If you put something on the conveyor belt, you will get the corresponding outcome.
The Contrapositive Argument:
Visualize a seesaw with A on one end and B on the other. If B goes down, then A must go up to balance it out.
Conditional Chaining:
Picture a train moving from A to B to C. If the train goes from A to B and then from B to C, logically, it's like it directly went from A to C.
Some Before All:
Imagine a group of people entering a room. If some people (A) are in a room with all of B and some of B are in a room with all of C, then logically, some of A must also be in the room with all of C.
Most Before All:
Think of a river flowing from A to B to C. If most of the water flows from A to B and then from B to C, it means that there's a significant portion of the water from A that ends up in C.
Two Mosts:
Visualize two buckets, one labeled B and the other labeled C. If most of the items in bucket A can fill both buckets B and C, then logically, there must be some overlap between the contents of buckets B and C.
thank you!!
Should I memorize this or work on being familiar with the language surrounding these arguments?
How should I be learning this concept?
One of the things that helped me with visualizing conditionals with quantifiers will be below. IDK if I am right so I welcome feedback. This is based on my understanding from the lessons on here and two videos I saw on YouTube.
With chaining the quantifiers is that when it is a quantifier before all, we are trying to see if there is a relationship between the two supersets. These must be connected by a common subset. And in all cases but one, one of the conditional relationships must be all:
X's → Y's
(all/most/some) X's → Z's
the inference will always be X's ←s→ Z's because we do not how big Y or Z are.
The one exception is most
Most X's → Y's
Most X's → Z's
Inference: Some Y's
The other one is a straight chain like A→B→C. Like we learned about the subset → superset lessons, I will call B the sub-superset because it acts as both and IDK a proper word for it. The reason why the quantifier needs to be before the all in the chain ( how its actually written does not matter) is because if we say all of A is B and most of B is C, we do not know how big A or B are.
However, since all of B is in C, even if A is just 0.1% in B, A is the same amount in C. Therefore, the relationship between the subset and the superset will be the same as the subset sub=superset. So, the sub-superset needs to be all in in the superset. If this relation unfolds, any relationship between A and B will be the same between A and C so long as all of B is in C.
Just a general comment... I come from a mathematics background and all of this reminds me of calculus but with the English language.
Very complex structuring. Making sure we understand English very well, I can see, will help immensely on the exam!
Best of luck everyone!
Also (previous lesson):
A → B
A —m→ C
B ←s→ C
#help#feedback should the very last argument here have a conclusion of "B ←s→ C"?
Isn't the last one (Two Mosts) wrong? it should be B←s→C
#help (Added by Admin)
Nope, i think it helps to think about it like this: 2 scoops of half plus 1 of A letters are being poured into bucket B and bucket C.
However, this isn't possible, I can't put half plus 1 of A into both bucket B and C. This means there must be some overlap between B and C, where they have some of A.
I hope this helps!!!
I believe that's what gianfrancop15 is saying. The last answer should be B some C, which is the overlap you're talking about.