Help with PT.33.S4.Q12 (G2) - Bird-watchers explores a forest

BuddySmithIII

Good Evening, Everyone.

Can someone please share some knowledge with me revolving around question 12 of PT.33? My specific question pertains to understanding the written logic, which is used to solve the question.

The lawgic is /S --> J --> H --> /G --> /W. There is an “M” below the J pointing to the H as well; however, what is confusing to me is drawing an arched arrow backward from the H to the /S. Once I have done that, how the lawgic plays out is confusing to me; and therefore, how the correct answer is reached. I have watched the video on this question more than once and I am still confused.

If anyone can share anything regarding this question, I would appreciated it. I hope my question makes sense.

Thank you!

Admin Note: Edited title. Please use the format: "PT#.S#.Q# - brief description of the question"
https://7sage.com/lsat_explanations/lsat-33-section-4-game-2/

claremont

Try adding the new condition "S → ~H~" redundantly back into the master chain instead of just drawing an arrow the way JY did: (I omitted M → H because I don't know how to clearly represent that using this medium, but on your paper make sure to add it) and then make a mental note that the harrier, i.e., H, repeats.

The contrapositive of the new condition H → ~S~ fits at the beginning of the master chain:

H → ~S~ → J → H → ~G~ → ~W~

AC A: Can J and S both be in? No, if J is in then H is in, but if you look at the beginning of the chain if H is in then S is out.
AC B: Yes, contrapose back starting at W.
AC C: Can M and S both be in? No, if M is in then H is in. But look at the beginning of the chain if H is in S is out.
AC D : Can J be out where M is in? No, if M is in then H is in. If J is out then S is in and H is out.
AC E: No, H and S now have a "not both" relationship.

I hope that helps.

BuddySmithIII

claremont

This does help. Thank you very much!