#12 - Local, Could Be True

[Adam Tyson]

Exactly right. This could be (but need not be) represented by a biconditional with one of the variables negated, or "out", and the other being positive, or "in".

J S

[T.B.Justin]

Hey LSAT and Adam,

Your discussion helped me to better understand the biconditional relationship between 'Not S' 'J,' with consideration of the original relationship between 'Not J' 'S,' where at least one of S or J, but not both, must be selected in this situation.

I have one related and unrelated question:

How to make this symbol "Not Variable" but with the negation line through the Variable, and with this circumstance is it correct that, if J is selected, H must be selected, G,W,S cannot be selected, and M could be selected or or could not be selected.

[Jay Donnell]

Hi T.B.!

The possible confusion between these two rules are hugely responsible for trouble in Grouping games:

1) If A is selected, B is not.

2) If C is not selected, D is.


As always, conditional rules need to have their contrapositives taken as an instinctual reflex. So, those rules also imply:

1cp) If B is selected, A is not.

2cp) If D is not selected, C is.


Rule 1 then involves the idea that we cannot select both A and B, so the symbol of ( A <-|-> B ) helps to reflect that we can select either just A, just B, or neither.

Rule 2 presents an opposite relationship. We MUST select at least either C or D into the game, so our options are to select just C, just D, or both. There is no official PS symbol for this relationship, so I want to offer two ideas:

First: ( C or D ) which like all uses of the word or on the LSAT is inclusive, should be taken as "OR" = "at least one, maybe both".

Second: (~C <-|-> ~D ) (Burdened by the format of the forum I'm using "~" where I would normally slash through the letter). This represents the idea that we cannot leave both C and D as unselected.

Another way to help clear up the two rules is this mathematical thought process:

1) A --> ~B

Since the negation is on the right side of the arrow, cross off the right side number of (0, 1, 2), which means we cannot select both.

2) ~C --> D

Since the negation is on the left side of the arrow, cross off the left side number of (0, 1, 2), which means we cannot select none.


For the latter part of your question, this is what the conditional chain should look like for this game:

~S --> J --> H --> ~G --> ~W
M-->

(That is meant to symbolize that either J or M is individually sufficient to bring in H, but otherwise M is unconnected to J and S.)

The contrapositive of that chain implies:

W --> G --> ~H --> ~J --> S
--> ~ M

(H kicks out both J and M, but again M is otherwise unrelated to J and S)

All MBT deductions must follow along the linear chain directly, so if we select J, the only things we can guarantee are:

H is selected
G and W are not selected

Since (S) is not directly affected down the one way linear line from the inclusion of J, it may be selected or not, and has no forced deduction.


Hope that helps!

[leslie7]

Hi,

the way I came to this answer was via my pre-made linked conditional .

I got

W->G->/H->/M &/J->S->/H then Contra (Which I didn't end up needing) H->/S->M or J -> H ->/ G ->/W

This was enough for me to eliminate the other AC and see that B could occur (both W and S could be in the forest)

but throughout this thread I'm seeing a lot of mention of a bi-conditional and it's effect on J and other variables I think?

Based on my diagramming can someone help me to understand/capture what I am missing to understand the effect of the new rule in its breadth/in an exhaustive way?

(I feel like I'm missing some key pieces here)

[Robert Carroll]

leslie,

The new rule makes S sufficient for H to be out. H is a requirement of J, so now, if S is in, J is out. But a rule already says that if J is out, S is in. So now "J out" and "S in" are sufficient and necessary for each other. The end of your master conditional has "H out"; you could recycle that back in your conditional to get that J is out, so...J being out makes S in, and S being in makes J out.

Robert Carroll