64 Game 2
Comments
On number 8, I failed to see that J could be out without K being out, is there a symbology that would help me keep that in mind?
I strongly recommend using conditional symbols for all conditional relationships, for exactly this purpose! I find that it's clear and accessible this way.
Hi Dave, I think in the games catalog (by template type) you have this listed as standard ordering, and you may want to update that! Thank you!
Thanks for pointing that out! It's fixed now.
Re ans. choice D: at first, I didn't get how you tried to make J->K false, but eventually I got it - you couldn't get to that step of trying to make it false due to having followed the first rule, which by default made J->K true. As simple as that was, I couldn't get past that. So while I was stuck, I tried the following on ans. choices C and D:
c) ~O->K->~N
N->~K->O
d) J->~N->K (resulting in J->K)
~K->N->~J (contrapositive ~K->~J)
For 12 answer choice D, why is it not possible to have both J and N not assigned (since it just says they can't be assigned together, but not that one of the two has to be assigned)? I assumed that was an option when doing the question.
Ah, but that's not what (D) says.
You're right that it is possible for both J and N to not be assigned (see, for example, Question 8).
But (D) is saying that the new rule is J and N cannot both be assigned.
This is the right answer, because it both (1) must be true under our old rules, and also (2) it guarantees our old rule is true.
That is, if you have J, you have K, which means you cannot have N. That's the same as saying if you have J you cannot have N. And if you have N, you cannot have K, which means you cannot have J. If you have N, you have no J.
J and N are not—they must not be—both assigned.
Is that clearer?
Let me know if you need more help!