Setup and Rule Diagrams

[Ladan Soleimani]

Hi Echx73!

I'm not completely sure how you are diagraming the rule, but I think I can help clarify a few things for you. I would hesitate to use subscripts for this because I think it makes it more confusing. I would stick to using a slash to represent the variable not being reduced, meaning it is in the out group.

Here is how that rule would be written:
G G
+ W contrapositive: W or
S S

So if W is not reduced then either G or S or both are not reduced. Because of the or, you cannot place both G and S in the out group. At most you could place G/S in a spot on the out group because at least one of them will not be reduced.

If there were more than two groups it is a little bit tricky to see how it works with the current wording. So lets switch the wording a little bit to groups X, Y, and Z and make the rule, If G and S are in group X, then W is in group X. All we would know from the contrapositve is that if W is not in X then either G or S couldn't be in X. We couldn't place G or S anywhere because they could be in either of the remaining groups. Does that make sense?

Ladan

[Echx73]

Ladan,

Thank you for walking me through this. You confirmed exactly what I was thinking. I do, though, have one more question. On this very same game, Question #9 has us introduce the a local rule of " L and S are both reduced, which one of the following could be a pair of areas of expenditure both of which ate reduced?"

I put L in the reduced group and I place P in the not reduced group. I then go to the 2 of LMR Rule and I place a MR block in the reduced group and place a MR block in the not reduced group. Now I turn to S. The only rule I have with S in it is the contrapositive the second rule of the game: If N is reduced, neither R nor S is reduced. The contrapositive is as follows: If both R and S are reduced, then N is not reduced.

For me to place N in the not reduced group, both R an S have to be reduced. The explanation of the correct answer for Question #9 on Page 300 states you can place N in the not reduced group solely on S being reduced...
Why are they saying we can solely put N in the not reduced group by just having S in the reduced group? The contrapositives says both R and S have to be reduced to put N in the not reduced group. Let me know what your thoughts are! Thank you so much for your time.

Respectfully,

Eric

[Anthony Esposito]

Hi Eric,

Great question! I'm glad that you're not satisfied with just getting the correct answer, but with also understanding fully why the correct answer is correct (and how to get there most efficiently).

So, everything was chugging along perfectly until you tried to use that R + S N contrapositive you made. (And, it was correct for you to try to make it during your set-up.)

The problem seems to be that the contrapositive of N R + S is R OR S N, not the R + S N you had. Remember, when you make a contrapositive with an "and" or "or", the "and" becomes "or" and the "or" becomes "and."

This means that by just having S in the reduced group, that is sufficient for us to place N in the not reduced group.

Hope this helps,
Anthony

[Echx73]

Hey Anthony,

The original statement is: If N is reduced, neither R nor S is reduced. That means the contrapositive would have an 'and'. What am I missing!? Or does Neither, nor me .....and? Sorry to be argumentative,,,,, Just trying to get on the same page! Let me know what your thoughts!

[Nikki Siclunov]

Hi Echx73,

You're correct in that "and" negates to "or" (and vice versa). Anthony's explanation does not contradict that. When two propositions A and B are joined using the conjunction "neither/or" (If N is reduced, then neither R nor S is reduced), this suggests that R is not reduced and S is not reduced. In the contrapositive form, it would be sufficient to show that either R or S is reduced to conclude that N cannot be reduced.

Does that make sense?

[aec2j]

In the analysis of this game you don't include the inference r s? Since we know n r and n s can't we infer that r s?

Thanks for your help

[Dave Killoran]

Hi AEC,

It's not included because it's not an inference

Yes, N won't go with R or S, but does that mean R and S can't both be selected? It does not. Think about it as if you had two enemies—conveniently named Russ and Shari—and that you won't go to any movie that they go to. Does that then mean that Russ and Shari can't go to a movie together? No, they could be best of friends and attend together (or maybe they don't like each other either—we don't know).

Does that help? Please let me know. Thanks!

[aec2j]

Can you clarify further as I think i'm missing something/confusing concepts. I thought that if N is sufficient condition in n r and a necessary in s n you could link to be s n r and therefore s r

Im guessing you can't link the chains with a but I just want to make sure! Thanks!

[Adam Tyson]

You're correct that you cannot link the chains in that way with the double not arrows. Also, N is Sufficient in both cases - it is sufficient to knock R out and also sufficient to knock S out. R and S are each sufficient to knock N out. That double not arrow is helpfully read as "I can't have both". If N is in, I cannot have R because I cannot have both N and R. Same thing happens with S - if N is in, I cannot have S because I cannot have both. If R is in, then N is out, but what does that tell me about S? Not a darn thing - maybe it's in, maybe not. I know for sure I won't have a problem with "both" because N is already out!

Hope that helps.

[Stephanie Oswalt]

We recently received the following question:

I'm having issues understanding the N-R-S connection from the scenario on page 324. Game 1 Oct 1996 Q 6-12. As well as the Diagram and explanation from pg. 326 [PowerScore Logic Games Bible 2017 edition].

The part about when N is reduced neither R nor S is reduced I think I understand. I believe if I were to say "when N IS reduced S&R are not reduced" ?

My confusion is to the counter positive and conditions it creates. From the written scenario I understand the counter positive should be "when R or S is reduced N is not reduced" The way I'm reading the Diagram it appears N-R and N-S are listed as two separate double not arrow rules. From the way it appears in the diagram if N is in the (Not Reduced) group, both R and S have to be reduced. But this would create issues with question 9's correct answer. How can R and N both be in the same group? Sincerely, Aaron

An instructor will respond to this question below. Thanks!