I’m happy to help. Let’s first take a look at the rule set to see if we can make some deductions.
We know that we will have two categories: S or R, and that each of six doctors is at exactly one. Then we have a bunch of rules:
If jS -> kR Contrapositive: kS -> jR (note that the negation automatically puts the doctor at the other clinic, since we know each doctor must be at one of two clinics, so “not kR†simply means that k is at S.)
If jR -> oS (note the link with the contrapositive in the above chain! ...kS -> jR -> oS) Contrapositive: oR -> jS -> kR
If lS -> nR AND pR Contrapositive: If nS OR pS -> lR
If nR -> oR Contrapositive: If oS -> nS
If pR -> kS AND oS Contrapositive: If kR OR oR -> pS.
Whew! Now on to the problem. We are trying to determine the minimum number of doctors that could be at Souderton, so let’s start with the minimum and see if it could be true.
Could it be zero? No. Rule 1 tells us that if jS -> kR, so in the world in which j is at S, we have at least one S. Combine that with rule 2, which tells us about the world in which j is at R — then o is at S! So we know that there can never be zero at S, because j has to be in either S or R and there will be a different S doctor if j is at R.
What about one? Let’s keep working through our previous world options.
1: jR -> oS 2: oR -> jS
Now let’s work these worlds through by adding in our other rules.
1. .jR -> oS -> NS. Could everyone else go to R? No — if P is at R, then K must be S. So we would have at least 3 in this world. 2: oR -> jS + oR also means pS. Can everyone else go to R? Yes! J and P are our only doctors at s, and therefore there can be a minimum of two doctors at S.
3: Note that nothing precludes oS and jS. In this world we would have a minimum of two at S, so we don’t have to keep working it through.
So c is the right answer! A tough problem, but hopefully you see how we worked through the options and rule set. Please let us know if you have further questions.
