From the fourth rule we know that if N is in (r) then O is also in (r):
N(r) --> O(r)
The contrapositive of this is if O is in (s) then N is in (s):
O(s) --> N(s)
The question presents us with
both N and O being in different clinics which would mean we cannot place N in (r) or O in (s) as that would result in violating what we are being asked. As such, we know that
N must be in (s) and O in (r) as part of our set up.
Does N being in (s) and O in (r) trigger any other chains??
It does!
We know that if N is in (s), L is in (r).
and
if O is in (r), J is in (s), K is in (r) and P is in (s). Therefore, answer choice (A) must be true.
The last trigger was formed by combining the contrapositives of the the first two rules and the last rule. Here are how I wrote my rules out in the order presented:
Conditional Rules:1. J(s)---> K(r)
2. J(r) ---> O(s)
3. L(s) ---> N(r) and P(r)
4. N(r) ---> O(r)
5. P(r)---> K(s) and O(s)
Contrapositives:1. K(s)---> J(r)
2. O(r) ---> J(s)
3. N(s)
OR P(s) ---> L(r)
4. O(s) ---> N(s)
5. K(r)
OR O(r)---> P(s)
Transitive (Combined) Conditionals1. J(s)---> K(r)---> P(s)
2. J(r) ---> O(s)---> N(s)
3. O(r) ---> J(s)---> K(r)---> P(s)
4. P(r)---> K(s) and O(s)---> N(s)
5. L(s) ---> N(r) and P(r) ---> O(r), K(s) and O(s)
For the last transitive property conditionals, we can note a violation. If we have both N and P in (r), that will result in O having to be in both (r) and (s). As such, we know for certain N and P cannot both be in (r). This helps us answer question (23) along the process.
I hope this helps!!!