Setup and Rule Diagrams

[Adam Tyson]

Exactly right! Needs no further explanation - you got it. Just be sure to test inferences, though - the rules might not say it explicitly, but maybe the way they interact would prevent S from running the 1st or 3rd race, or maybe he has to run one of the races. You can't read that rule as meaning that he is DEFINITELY allowed to run in 1st or 3rd or that he could DEFINITELY be left out, just that the rules don't explicitly prohibit those things.

[srcline@noctrl.edu]

Hello

I am not sure what I did wrong on this game. Here is my setup and inferences.

_ _ _ _ _(out)
1 2 3 4 5

Rule 1:
Q QT
contra (+): (not) QT (not) Q

Rule 3:
not U R2
(+) not R2 U

Rule 4
R2 not U
(+) U not R2

Inference:
U R2

Thankyou
Sarah

[David Boyle]

Hello

I am not sure what I did wrong on this game. Here is my setup and inferences.

_ _ _ _ _(out)
1 2 3 4 5

Rule 1:
Q QT
contra (+): (not) QT (not) Q

Rule 3:
not U R2
(+) not R2 U

Rule 4
R2 not U
(+) U not R2

Inference:
U R2

Thankyou
Sarah

Hello Sarah,

I'm not sure if you did anything wrong, actually! (Though with U and R2, it may be more of a biconditional, "U slash R2", than a double not-arrow, maybe.) --Maybe you can do more: e.g., can Q be last? Do some of the rules interconnect, e.g., if S can't be 2 or 4, does that relate at all to the last two rules? But you already have some valuable observations!

Hope this helps,
David

[srcline@noctrl.edu]

Hello David

Why would it be a double arrow and not a double not arrow becasue the terms are negated on the necessessay side?

Thankyou
Sarah

[Robert Carroll]

Sarah,

Although you know that the things on either side of your Double Not-Arrow (U, R2) can't both happen, you know more than that - the last two rules together tell you that U's absence and R's being second (so, U, R2) are each of them sufficient to make the other necessary. Thus, U is out if and only if R is second, so U R2.

[srcline@noctrl.edu]

Hello Robert

Okay that makes more sense.

Thankyou
Sarah

[deck1134]

I'm really not sure how the biconditional plays a role in the game, as opposed to just griding rule 3 and 4. Any tips? There did not appear to be any inferences at all in this game.

[LSAT2018]

Biconditional Reasoning
Uzoma → Ramirez (Second)
Ramirez (Second) → Uzoma

Ramirez (Second) → Uzoma
Uzoma → Ramirez (Second)

So given the third and fourth rules, they mean that either Uzoma is in or Ramirez is second, but not both. It excludes the possibility from the third rule that both are in effect (Uzoma is in and Ramirez is second) and also excludes the possibility from the fourth rule that both are not in effect (Uzoma is out and Ramirez is not second). Is my understanding correct?

[Dave Killoran]

Biconditional Reasoning
Uzoma → Ramirez (Second)
Ramirez (Second) → Uzoma

Ramirez (Second) → Uzoma
Uzoma → Ramirez (Second)

So given the third and fourth rules, they mean that either Uzoma is in or Ramirez is second, but not both. It excludes the possibility from the third rule that both are in effect (Uzoma is in and Ramirez is second) and also excludes the possibility from the fourth rule that both are not in effect (Uzoma is out and Ramirez is not second). Is my understanding correct?

Yes, I believe that is what Robert stated above. Thanks!

[tfab]

So I read through the explanations given and I just wanted to clarify it for myself. Basically, due to the 3rd and 4th rules, "U" cannot be out at the same time that R is in the 2nd slot? I was very confused when I first read the rules for this game because I had never seen a game with rules similar to the 3rd and 4th ones here. This ultimately led to me getting Q14 incorrect which turned out to be one of the easiest for other test takers. Do you guys know of any games with rules that are similar to the 3rd and 4th one here? I'd like to eliminate this confusion from my mind for future tests, thanks!