If all the new cars in the exhibition are research models, then which one of the following statements must be true?

idonije on October 15, 2021

If R-->N and if R-->F then how could A be incorrect while D is correct? The rule is basically stating R-->N & F. So, how can we conclude R-->N-->F when no explicit order is stated?

Jay-Etter on January 23, 2022

1: F/S, N/U, P/R
2: F/S, N/U, P/R
3: F/S, N/U, P/R
(there are multiple ways to diagram this one but I found this diagram to work).
After we put our rules in we get
1: F/S, N, P/R
2: F/S, N/U, P/R
3: F/S, U, P
(based on 1 being N, 3 being U, and U not with R).

The question is saying if new cars in the exhibition are research (aka. RN always paired), then which must be true:
We can pair the R and the N we already have in 1 to get
1: F, N, R (Not R because RS cannot go together)
2: F/S, N/U, P/R
3: F/S, U, P

Now, 2 is still pretty undetermined, so lets do two scenarios. 2 could either have New or Used, so lets draw those both out.
When 2 is new:
1: F, N, R
2: F, N, R
3: F/S, U, P.

When 2 is used:
1: F, N, R
2: F/S, U, P
3: F/S, U, P

Now we can look at our options.
A: Doesn't have to be true because we could have F and U in 3.
B: Doesn't have to be true because we could have FP in 3
C: Doesn't have to be true because we have FN in 1
D: Has to be true. Notice everywhere we have N we have F!
E: Doesn't have to be true, we could have PS in 3.

Hope this helps!