by ohthatpatrick Tue Nov 13, 2018 2:34 pm
R S T V Y Z
L: __+
M: __ __
P: __+
(ST)
~(VY)
YZ both in P or both not in P
Sure thing.
The condition is that R is alone is some group. He could either be alone in L or in P (can't be alone in M, since M has to have two members. So I would start by framing those two worlds. Since we know that M has exactly two members, if R is alone in a group, then the remaining group has the other three people.
L: r
M: __ __
P: __ __ __
or
L: __ __ __
M: __ __
P: r
From there, we start asking ourselves about the ST chunk, splitting up the VY enemies, and whether YZ are both in P or both not in P.
The most definitive thing we can do, when we're down to two available groups and have an enemy rule, is to split up the enemies.
L: r
M: v/y __
P: y/v __ __
or
L: v/y __ __
M: y/v __
P: r
This helps to clarify where the ST chunk could fit.
L: r
M: v/y __
P: y/v, s, t
or
L: v/y, s, t
M: y/v __
P: r
In the first frame, there's no way for Y and Z to both be in P, so they have to both not be in P. Thus, that frame looks like
L: r
M: y, z
P: v, s, t
for the other one, we just know that Z is the final member of group M, but it doesn't matter whether we have put V or Y into groups L or M.
L: v/y, s, t
M: y/v, z
P: r
Since the condition of this problem led to two worlds, and they're asking what must be true (i.e. "what is always true"), we can look at our two worlds for a common denominator.
It looks like in both cases Z is in Markets. So that should be the answer. Scan for it and pick (D).
(A) frame 1 is a counterexample
(B) frame 1 is a counterexample
(C) in frame 2, Y could be assigned to Labor
(E) frame 2 is a counterexample
Hope this helps.