Setup and Rule Diagrams

[cd1010]

Hi! I'm getting confused in the set-up for when O is in 5:
"Of course, from the fourth rule, if O cannot be chosen for stops 1 or 2, then N cannot be chosen for stops 2 or 3 (and then from the third rule N cannot be chosen for stop 4)".

I don't get what I'm not seeing here! Help!

I see that if O is in 5, then O can't be in 6 (which means that it can't be in 1). O also can't be in 4 because, similarly, that would break the rule about items not being allowed to be consecutive. So O can't be in 4 which means O can't be in 2. I can't see how you go from there and then deduce Not Laws for N for 2 and 3?

[cd1010]

Hi! I'm getting confused in the set-up for when O is in 5:
"Of course, from the fourth rule, if O cannot be chosen for stops 1 or 2, then N cannot be chosen for stops 2 or 3 (and then from the third rule N cannot be chosen for stop 4)".

I don't get what I'm not seeing here! Help!

I see that if O is in 5, then O can't be in 6 (which means that it can't be in 1). O also can't be in 4 because, similarly, that would break the rule about items not being allowed to be consecutive. So O can't be in 4 which means O can't be in 2. I can't see how you go from there and then deduce Not Laws for N for 2 and 3?

Actually, nvm, I figured it out! N is really restricted because LN can't be together, and MN can't be together. That means that if N shows up, the only option that can go before it is O. So an inference is:
N --> ON

When you can't have an O in 1, then that means you can't have an N. The contrapositive is why we have Not laws for N for 2 and 4:

not O1 --> not N2
not O2 --> not N3

Is that right?

[Luke Haqq]

Hi cd1010!

I'd adjust one thing you mention. You comment,

So an inference is: N ON

This is close. However, note the possibility that N could be the first stop. As the administrator notes, "if N appears in stops 2-6, then O must precede it." Given that N could appear as the first stop, we can't infer that whenever N occurs, it must be preceded by O.