PT34.S4.G4 - each of exactly six doctors

TheAnxious0L

I've watched the explanation for this video...but one of the conditional statements still has me confused. Basically, J.Y combined a /J and a J, to make a conditional chain...can someone take a look at this video and break it down?

https://7sage.com/lsat_explanations/lsat-34-section-4-game-4/

Achen165

Hi,

I am starting with a basic breakdown for foundation, which I build upon later to provide the most complete explanation possible. I hope that you find this useful and I am sorry if this first portion is too repetitive.

What he is doing is applying the conditional logic rules for contrapositives (where with a contrapositive, you flip the elements and retain ordered the placement of the negation-- only the variables move). The original conditional statement and its contrapositive have logically equivalent meanings. At their most basic level, recall that conditionals are "if, then" statements (where the former is the sufficient condition and the latter is the necessary condition).

In PT34.S4.Q#. (G4),
the first two rules (both of which refer to 'J') translate to***:

1. Kudrow is a Randsborough if Juarez is at Souderton.
   Conditional: J (s) ----> K (r)
2. Onawa is at Souderton if Juarez is at Randborough.
   Conditional: J (r) -----> O (s)
   ***the numbers used here and in section below correspond to rules (1) and (2), respectively.

To represent the grouping, J.Y. sets up the game in an analogous fashion, representing either/or as an In/Out game.

Note that in the game, which is a grouping game, all elements are in either one of two groups ('S'ouderton or 'R'andsborough). If not in 'S', then in 'J;' If not in 'J' then in 'S'....etc. They cannot simultaneously be in both groups, and they cannot be in neither group, since that they are either at exactly one of two clinics.

To illustrate that (J, K, L, N, O, and P) are in either the 'r' or 's' group, J.Y. sets up the game as an In/Out game where 's' (Sourderton) represents "in" and 'r' (Randsborough) represents "out." He does so to imply if not at 's' then at 'r'; if not at 'r', then at 's'.

1. (see above) J (s) ---> K (r) becomes
   J ---> /K
   to represent: "If J is in (s) then, K is not in (s)..." [because K is in (r)]...the only other place for K to go. So:

                 ** J ---> /K**
   

2. (see above) J (r) ---> O (s) becomes
   ****/J ----> O****

to represent: "If J is not in (s), then O is in (s)..." [because J is in (r)]...If J is not in (s), the only other place for J to go is (r). (Equivalent to rule 2: "Onawa is at Sourderton is Juarez is at Randsborough").

To combine the rules (** --> **), take the contrapositive of rule 2...

Conditional (original, corresponding to 2.)

              /J -----> O   becomes /O ----> J

From here, we create a chain....combining the conditional for Rule 1, and the contrapositive of the conditional for Rule 2.

1) J ---> /K {conditional, derived from rule 1]
2) /O ----> J {contrapositive, derived from rule 2}

The chain is born!

 /O ---> J ---> /K

He has two conditionals on the screen at once, one the CP, before deleting the original. The chain only includes the un-negated form of 'J', not both.