I've gone through about 50-60 games so far (PT 1-35). I'm having a lot of trouble of creating the initial master board for never seen before games. That's my main issue, once I make the proper boards, I can clear the questions quickly.

Foolproofing has been "working" so far. For games that initially took me 15 minutes just to get half of the questions wrong because I don't know what the hell I was doing when making the masterboard, now takes me 5-7 minutes because I foolproofed them so many times.

It looks like my biggest fear is coming true. Foolproofing is making me faster at memorizing the games itself. Rather than memorizing the important inferences needed to be made when attacking never seen before games. To note, when I do foolproof, I do the whole game from scratch without memorizing the game rules, I try my best not to make any "cheat" inferences which are completely unintentional, but rather memorized because I foolproofed it (doing the game so many times over and over again)

Any tips?

I understand why A is correct, but why is E wrong? Is it because E does the opposite of weaken and just strengthens the argument?

"Unless Shiming's car is fixed, which cannot happen without Steve's help, there will be no road trip next week."

From the lesson, https://classic.7sage.com/lesson/quiz-complex-conditional-translations-4-w-answers/

I'm very confused in here because of the prior rules of mixing group 3 and group 4 conditional indicators. And in this sentence there's a bunch of them.

I understand the "final" necessary condition being /road trip next week. So the first conditional statement written in logic would be /car fixed > /road trip. This part I understand as we are using group 3 logic, negate sufficient. Simple enough.

Now to add the second part, "Unless Shiming's car is fixed, which cannot happen without Steve's help....."

In this sentence we have two group 3 indicators and one group 4 indicator. What do we do in this situation? In a previous lesson when we run into this situation, we pick one side, follow that group's negation rules, then treat the other "logical indicator" as a negation.

So for example, I want to follow group 3's rule on the indicator "unless". Which is choose a side, make it the sufficient condition, and negate. Okay. Let's pick car fixed.

So this puts me at /car fixed > which cannot happen without Steve's help.....

Now on the necessary side we have a group 3 and a group 4 logical indicator in there :DDDD.

What do we do here? If I treat both as a negation which cancels each other out, that turns into /car fixed > Steve's help. Which makes absolutely no sense at all. And even the other way around, /car fixed > /steve's help, which makes so much more sense, it doesn't fit in the final answer. This is because the contrapositive of /car fixed > /steve's help is steve's help > car fixed.

The final answer being /steve's help > /car fixed > /road trip. The contrapositive being road trip > car fixed > steve's help.

Help me please~~~

M → F

/J → /M

/M

I took the contrapositive of M > F which got me /F > /M

Then I inferred that /F or /J would net me /M. Which means that the arrows can split in the sufficient condition.

Which means the sufficient assumption for /M would either be to negate F (/F), or to make it that /F > /J (/F > /J > /M)

Is this correct? Or am I missing some other possible sufficient assumption.

Also, is the actual LSAT done like the program we use on 7sage when drilling/prep testing? Where we can highlight, underline, "X" out wrong answers, etc?

I'm looking for a strategy of writing/making notes because I cannot remember the whole passage on one go...

I'm trying to understand why did JY break the "contrapositive rule"when mapping out the first two rules?

Rule 1: If 2 C's and Min1 L same vote --> both moderates same vote

Then as his contrapositive, he just negated the sufficient and necessary without swapping their positions? I thought we are never allowed to do that.

For example, the typical contra positive is (A -> B) and (/B -> A)

But here he did (CC L same --> MM same) and (/CC L same --> /M same)

I understand the rule itself. If 2C and Min 1L are in, both M is in as well. And if either there is no 2C or min 1 L together, then the two Ms are floaters. Is this correct?

Admin Note: Edited title. Please use the format: "PT#.S#.Q# (G#) - brief description of question"

Explanation Video: https://classic.7sage.com/lsat_explanations/lsat-7-section-2-game-3/