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Setup and Rule Diagrams

eober
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Hi,

I just want to make sure if my inferences are correct in this game.


I made the inferences:
U :dblline: W (from W :dbl: S and S :dbl: U)

R :dblline: T (from T :dblline: W and R :dblline: W)

W :dblline: Y (from R :dblline: W and R not :arrow: Y) because if W is there, R is not, if R is not then Y is.

Y :dblline: R (from Y not :arrow: R)

Does the inferences I made make sense? I get confused when combining a rule with a double arrow. Could you explain as a general rule how we make different inferences from double arrows and double not arrows?

Thank you!
David Boyle
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eober wrote:Hi,

I just want to make sure if my inferences are correct in this game.


I made the inferences:
U :dblline: W (from W :dbl: S and S :dbl: U)

R :dblline: T (from T :dblline: W and R :dblline: W)

W :dblline: Y (from R :dblline: W and R not :arrow: Y) because if W is there, R is not, if R is not then Y is.

Y :dblline: R (from Y not :arrow: R)

Does the inferences I made make sense? I get confused when combining a rule with a double arrow. Could you explain as a general rule how we make different inferences from double arrows and double not arrows?

Thank you!


Hello,

One might do them differently, something like u :arrow: s :arrow: w :arrow: slash t, slash r, with the slash r :arrow: y. (Because we know that slash y :arrow: r)

As for "different inferences from double arrows and double not arrows", the first is about "love", the second is about "hate". Make sure you correctly track any arrows coming in or out of double arrows, e.g., if A :dblline: B, and C :arrow: A, then you can assume that C :dblline: B. But if there were A :arrow: C, you can't automatically assume that C :dblline: B.

David
Dave Killoran
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This morning I received the following question:

I know this is a dumb question but I need some clarification. On Dec 2004 LSAT LG 3: a rule states Raimundo appears in every photograph that Yakira does not appear in. So I diagrammed it as not R then Y. wouldn't the contrapositive be Not Y then R? I reviewed a video lesson and it said that If not R then Y is free variable? which the #1 on the game was B which include R and Y. I just don't understand how the rule effects the game. Help!



First, there are never any dumb questions! I know people say that a lot, but it is still true. One small misunderstanding can sink you on a game, so if you ever are even a bit uncertain, ask! Second, this is a really tricky game, and they intentionally worded the rules in an unusual way in order to get students messed up. Basically, the first two rules look like "B appears when A appears" (A :arrow: B), and the third rule looks like "B appears when A does not appear" (A :arrow: B). So, the necessary condition is being presented first, and the sufficient condition appears later in the sentence (which is very normal and acceptable). Thus, the first two rules would be diagrammed as:


    S :arrow: W

    U :arrow: S

    (and those two can be combined to form the chain diagram: U :arrow: S :arrow: W )


With the third rule, we have a negative on the sufficient condition, leading to the following diagram:


    Y :arrow: R

So, if Y is NOT in a photograph, then R must be in that photograph. The contrapositive both reverses and negates the terms, and appears as follows:


    R :arrow: Y

When those two rules are considered together, the mean that if one of R or Y is absent from the photograph, then the other must be in the photograph. In other words, at least one of R/Y is always in a photograph(and possibly both). Rules like this,w here the sufficient condition is negated, are extremely tricky. I wrote an article about them that I think you would find worthwhile. It's called The Most Dangerous Conditional Rule on the LSAT and it explains exactly how rules like this work.

Ok, so when we look at your diagrams, it looks like you did great! You did it in reverse order from me, but that doesn't matter at all. A statement and its contrapositive are identical in meaning, so whichever is "first" is irrelevant.

As for the video lesson you reviewed, it definitely wasn't one of ours because we wouldn't say that if Y occurs then R is "free." In fact, it's not free at all—if Y, then R must be in that photograph. Could they have meant that if Y was in the photograph, then R was free? Because that would be true.

Hopefully that clears this rule up. It's a tough one, so make sure you are 100% comfortable with it. Please let me know what you think. Thanks!
Dave Killoran
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Zestor
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Hello, I am practicing this game in the LG bible book, and I seem a bit confused regarding the relationship between u>s>w>slash r > y relationship. I am able to infer that w>y, s>y, u>y, and slash y>slash w.

But, wouldn't it be correct to also infer that slash y> slash s, slash y> slash u? I am visually trying to make these inferences so i am looking at the u>s>w>slash r > y relationship, and taking the contrapositive of it to arrive at inferences. I am not sure if my process is icorrect.