6. Arrows and double-not arrows Question

[MordecaiSmith]

In the 6th principle it claims that "Any combination of an arrow and double-not arrow in succession will yield an inference" and it uses the example "A B C" to prove this point with its inference being "A C". This makes complete sense to me.

To test the theory that any combination of an arrow and double-not arrow will yield an inference I wrote out an alternative to the example statement as "A B C" and the inference, "C A", can't be true. For example there's one possibility that "C D A"

To me this disproves that "Any combination of an arrow and double-not arrow in succession will yield an inference".
Is there something I'm missing here?

[Dave Killoran]

Yes, you are missing something! I think you expected another power arrow inference like the one you cited, but in this case, let's walk through it and watch what happens:

• We start with: A B C
  
  Here we want to start with either A or C to begin making an inference, and it's easier to start with A since the C/B relationship is a double-not arrow.
  
  Starting at A then, how do we "go backward" against the arrow to get to B? Well, if all Bs are As, then we know for sure that at least some As are B, and so what "goes backwards" is A B.
  
  We can then attach that to the B/C relationship to get: A B C.
  
  Following the Some Train then, we can start at A and arrive at C, leading to the following inference:
  
  A C

This is actually an inference that we've seen before on the LSAT, so it's one you want to know (and you'll see more of it--that's a long chapter ).

Thanks!

[MordecaiSmith]

That does help thanks!

[MordecaiSmith]

Yes, you are missing something! I think you expected another power arrow inference like the one you cited, but in this case, let's walk through it and watch what happens:

• We start with: A B C
  
  Here we want to start with either A or C to begin making an inference, and it's easier to start with A since the C/B relationship is a double-not arrow.
  
  Starting at A then, how do we "go backward" against the arrow to get to B? Well, if all Bs are As, then we know for sure that at least some As are B, and so what "goes backwards" is A B.
  
  We can then attach that to the B/C relationship to get: A B C.
  
  Following the Some Train then, we can start at A and arrive at C, leading to the following inference:
  
  A C

This is actually an inference that we've seen before on the LSAT, so it's one you want to know (and you'll see more of it--that's a long chapter ).

Thanks!

Just realized principle number 7 explains this lol, thanks again