Need Help Please! Not sure if my logic is correct on a question.

165LSATDUDE

Someone tell me if I'm correct in this logic. This is for question #2. Here is the link.

https://7sage.com/lesson/quiz-on-finding-sufficient-assumptions-with-intersection-statements-2-answers/?ss_completed_lesson=11895

G –m→ N

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G –m→ /A

G-m->N-m->/A = I’m not able to draw it out like how JY does so this is what I put however, on paper I draw it out like JY does.

N—>/A = Answer

So in regards to N-m->/A not being the answer, let me explain.

/A
/A
GN
GN
GN/A
GN/A
G/A
G/A
N/A
N/A

Now the reason N-m->/A is incorrect is because this answer would effect the conclusion. Say you had more A’s and more N’s then it wouldn’t be the case that G-m->/A is the answer because look, I have more A’s and N’s that are not G’s so how could G –m→ /A? This would make the conclusion invalid. It has to be certain that most G’s are /A and by getting ride of N-m->/A statement were getting ride of the extra two N’s that carry A with it as well as the other extra two /A’s. If N-m->/A then how does G-m->/A work? Out of the 6 N’s, two are A which pulls from the conclusion by saying, “No, G-m->/A invalid because we have two extra A’s therefore out of all the A’s which is 8 (100%) only 4 G’s are A’s and it needs to be 5 to count for most. The N-m->/A is satisfied but doesn’t allow the conclusion to be valid so the sufficient assumption that N—>/A makes it 100% true that yes G –m→ /A.

GN
GN
GN/A
GN/A
G/A
G/A

The2ndSage

If I am understanding you correctly, I think you are pretty much right here. Like you say, if your second premise was N-m->/A instead of N—>/A then you could not conclude G –m→ /A because it would then be a "some" relationship between G and /A rather than a most relationship.

There are two parts I did not understand if you can explain further:
For the first list you make (granted, I might be misunderstanding it), why do you have the two G/A's? Those then make a total of 4 out of 6 G's that have /A with it. So that would mean G-m->/A is true and I thought you were trying to show how it does not work with this list, right?

For the second list (again, I might be misunderstanding it), it looks like you are illustrating the SA premise of N—>/A, and if so, you need "/A" with the first two "GN"s. That is because for every N, you must have a /A, so you cannot have a GN without a GN/A.

165LSATDUDE

Hi, thank you for taking the time out to reply. I think I was way over thinking this big time and it didn't help that I was exhausted. I was trying to show that if N-m->/A, then the conclusion is invalid because as you stated there would be a some relationship in the conclusion rather then a most. Now in regards to the second list, I meant to put / next to the A. Bottom line is that N-m->/A doesn't allow the conclusion to be true and N—>/A does. Is this correct?