All and Most Inferences

[HarmonRabb]

I understand we can make an inference when we have:

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A-----m----->B----->C

But can we deduce an inference from:

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A----->B-----m----->C

In the LR workbook, there is a problem:

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T<-----|----->U----->V-----m----->W

I thought you could make the inferences

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U-----m----->W
and
W<-----s----->/T

But apparently U-----m----->W is not a valid inference.

[Dave Killoran]

Hi Harmon,

With this statement: A B C, there is NO inference that can be made. It may help to imagine that there is only one A, and thousands of B, and then a lot fo Cs too. That one A doesn't have to intersect with the C (because not all Bs have to be Cs).

Applying that to the second problem, you can see it won't then create an inference.

Please let me know if that helps. Thanks!

[HarmonRabb]

Thanks for the reply Dave, really appreciate it, and on a Friday end of day none the less! I'm curious why in this case we assume there is only one A yet many of B? The way I read it to myself is "All A's are B's and most B's are C's" So I thought the inference made sense logically.

[Dave Killoran]

Haha, thanks! Logic doesn't care what day it is

As for assuming one A, that's just for the purposes of the example. But what that example does is prove that A and C don't have to overlap. Once that possibility exists, no inference can be made .