Those aren't in the rules, but are inferences based on the interaction of the rules. From the top post in this thread:
The key inference to this game is that at least one L must vote for Datalog because all three cannot vote against Datalog. If all three L’s attempted to vote against Datalog, then both C’s would have to vote for Datalog, which is impossible according to the given rules.
This comes from the last rule of the game, that at least one conservative voted against Datalog. Our two groups should be the For group and the Against group, and so in your base you must have those two groups, each with a minimum of two slots, and there has to be a C in the Against group. From there, you can infer that there must be at least one L in the For group, per the inference described above.
The diagram showing CCL with an A subscript is just conditional. IF the two conservatives voted the same way as each other, that has to be in the Against group. If a liberal joins them, then both the moderates must also join them, giving you CCLMM in the Against group. That leaves only two liberals left, and since the For group needs a minimum of two member, LL has to go in the For group. This doesn't mean that there is always a CCL in the Against group, but only shows what would have to happen if that were to occur.
Also, the two numerical distributions are flexible. They only mean that there is either a group of 5 and a group of 2, or else there is a group of 4 and a group of 3. It's not absolutely predetermined which group must be the larger of the two. For example, while the grouping described above has a 2-5 distribution, with 2 in the For group and 5 in the Against group, there is a possible 5-2 distribution, with the For group consisting of LLCMM and the Against group having CL. The main diagram shows only the thing that must occur in every solution - at least one L in the For group and at least one C in the Against group, with a minimum of two slots per group.
Adam M. Tyson
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