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 Stephanie Oswalt
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#44387
We recently received the following question from a student. An instructor will respond below. Thanks!
Hello! On page 321 I would like to know how the first inference can be combined with the super rule to create C<-/->G
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 Jonathan Evans
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#44405
Hi,

Good question! Check out the illustration below followed by a brief explanation:

Image

Consider three scenarios:
  1. What if B is selected? If B is selected, G must be selected (Rule 2) and C must not be selected (Rules 1 & 3 combined).
  2. What if C is selected? If C is selected, B is not selected (Rules 1 & 3) and therefore G is not selected (Rule 4).
  3. What if G is selected? If G is selected, B is selected (Rule 4) and C is not selected (Rules 1 & 3).
Notice in all three scenarios, there is no possible way C and G can be selected together, therefore: C :dblline: G.

In the illustration above, you can see this link between the element in common between both statements, B.

I hope this helps!
 Diliabarron
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#44415
Thank you! Makes sense now.
 rwraulynaitis
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#75501
Hello,

I am working through the 6th question in the Grouping Setup Practice Drill. I got the answer correctly on my own, but in reading through the answer key on p. 341, I am confused about how the single arrow and the double not-arrow were combined.

I understand how we got: ~H -> L <-/-> M

but I do not see how that leads to the inference: ~H -> ~M

I would appreciate it if you would be able to clear this up!
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 Dave Killoran
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#75663
Hi R,

Thanks for the question! So, we are looking at this relationship: H :arrow: L :dblline: M

Let's first start with H :arrow: L :dblline: M to see how it works. In this diagram, what you get is if you have H, then you have L, and if you have L then you do not have M. So, you get H :arrow: M.

Now, look at when we start with: H :arrow: L :dblline: M. The only difference is that H has a slash on it now, for H. So, when we have H, then you have L, and if you have L then you do not have M. So, you get H :arrow: M. Same exact process but this time we accounted for the slash on H.

Of course, the contrapositive of H :arrow: M is in fact M :arrow: H, which how we get that tricky inference.

Please let me know if that helps. Thanks!
 rwraulynaitis
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#75701
Got it, thanks Dave!

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