68 comments

  • Edited 4 days ago

    this confused me so i put it into something my brain could comprehend:

    All flowers are plants. (A -> B)

    Some plants grow strawberries. (B <-s-> C)

    Therefore, some flowers grow strawberries. (A <-s-> C)

    Think:

    1. What makes you think flowers and strawberry plants overlap in the plant community? Is there proof in a premise that tells us there are strawberry-growing flowers? NO. So the LSAT has no reason to think that and it's invalid. Sometimes it helps to imagine you are arguing with the LSAT prompt. Like "What makes you think that? Where's your proof? No proof? Invalid."

    2. You cannot infer that flowers are in the subset of plants that grow strawberries when there is no premise that provides that information. Your job is to only believe what the premises are telling you.

    3. This argument has flawed logic because it is making an assumption about flowers based off of it's membership in the plant group. You know flowers are plants right? You know some plants grow strawberries? Then why are you assuming flowers grow strawberries just because both are plants!!

    4. In my case, the word "all" in the first premise make it hard to think of flowers as a subset of plants. It helps in this stage to draw it out so your brain genuinely pictures how absurd this flawed logic is. Big circle is plants. Inside Big circle: Little circle #1: flowers and Little circle #2: strawberry growing plants. There is no info in the argument that says the two overlap so why is the conclusion assuming that?

    2
  • Wednesday, Jun 24

    Because I'm a watch nerd:

    Premises:

    All Submariner owners own Rolexes.

    Some Rolex owners own Grand Seikos.

    Can we conclude: Some Submariner owners own Grand Seikos?

    NO

    11
  • Wednesday, May 20

    ALL must always be secondary........ In order to include the SOME AND MOST if they are placed first the smaller pools trickle into the larger ones. .

    10
  • Monday, May 4

    I appreciate these explanations in written format as we already had videos to watch on this! Sometimes it's easier to understand when watching the videos, so reading it through helps me synthesize and think through the different arguments.

    6
  • Tuesday, Feb 24

    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.

    4
  • Wednesday, Feb 18

    Add a video for these flaw explanations please.

    42
    Saturday, Jul 11

    @ChristinaCorbo !!!

    1
  • Monday, Feb 9

    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.

    8
  • Thursday, Jan 22

    DRAW OR THINK OF VENN DIAGRAMS! It helps a lot :D

    8
  • 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!

    3
  • Saturday, Dec 6, 2025

    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!!

    27
    Tuesday, Mar 10

    @SheridanMcGadden thank you so much. This helped me understand it way better.

    2
  • Monday, Nov 24, 2025

    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?

    1
  • Wednesday, Nov 19, 2025

    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!

    35
  • Thursday, Oct 30, 2025

    who else is confused :p

    6
    Monday, Nov 24, 2025

    @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.

    2
  • Monday, Oct 6, 2025

    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.

    6
    Wednesday, Apr 8

    @kyorofan20 This was helpful thank you!

    1
  • Wednesday, Sep 24, 2025

    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.

    1
  • Tuesday, Sep 16, 2025

    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.

    1
  • Monday, Jul 21, 2025

    why is there no video for the important subjects :'(

    32
  • Wednesday, Jun 25, 2025

    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.

    4
  • Monday, Jun 9, 2025

    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.

    3
  • Friday, Apr 18, 2025

    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?

    1
    Sunday, Jun 1, 2025

    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

    2
  • Wednesday, Feb 12, 2025

    So what about the following argument? Is this valid?

    A→B

    A ←s→ C

    B←s→ C

    0
    Sunday, Feb 16, 2025

    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.

    2
  • Wednesday, Feb 5, 2025

    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.

    67
    Friday, Feb 28, 2025

    Hands down the best explanation I've found on here. Way better than what is in the lessons. Thank you !!

    9
    Monday, Feb 10, 2025

    Very helpful, thank you!

    1
    Tuesday, Jul 29, 2025

    @A.Belle you are a BEAST. 7Sage should update this with your formula.

    #YouRock

    1
    Tuesday, Jan 20

    @A.Belle great explanation. I like this more than the "scoop" example 7sage has been using.

    1
  • Tuesday, Jan 7, 2025

    I found that drawing the circles for the subsets and supersets make it much easier to understand.

    1
    Friday, Feb 7, 2025

    Same!

    0
  • Saturday, Dec 21, 2024

    im going to need a video explanation on this one chief im starting to dive myself insane

    22
    Tuesday, Jan 7, 2025

    Same! I'm so lost with these last 3 logic flaws.

    3
  • Monday, Nov 11, 2024

    So basically, the smaller amount needs to come before the larger amount in order for an argument to be true

    3

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