[carmenstclaire]

Hi Mike,

I’m back at Minor Questions on p. 397 and I am not sure about q.6. Answer is C. But I’m not clear why A. isn’t correct. Explanation says that it doesn’t put P&H on opposite shifts. But if P&N are always together, wouldn’t H2 make N go into Shift 1?

Thanks again!

Carmen

[Mike Kim]

Hey Carmen — hope you don’t mind but I moved your Q to a new thread so that if others have the same question they can more easily find this in the future —

Per answer (A), which says that If N works 1, H works 2 —

We know that it two things must be true: If N is in 1, H is in 2 and the contrapositive: If H is in 1, N is in 2.

Per these rules, there are 3 possibilities for how N and H could be arranged: (N1, H2) (H1, N2) (N2, H2).

It is the third of these possibilities that makes it so that an “if…then” statement has a different consequence than a straight-up “P and H will work different shifts,” and that’s why (A) doesn’t have the same meaning that (C) does — (C), which is a biconditional (discussed on pg. 191), does not allow for them to work either shift together — (C) only allows for two options: (N1, H2) and (H1, N2) just like the original rule only allowed for two possibilities —

I hope that clears it up — good luck studying this week — Mike

 

[carmenstclaire]

Hi Mike, yes, that cleared up the confusion, the biconditional information on p, 191 was exactly on the money for this question. This is starting to come together, slowly. Thanks again!!!