Making Formal Logic Inferences #7

[andwer123]

In the course Dave examines the example:

A B C

When I analyzed this before watching the video I got this:

(we don't start with a double not arrow sooo,)
B C therefore ~C ~B

Then we would have:
~C ~B ~A
.: ~C ~A or A C
Which is obviously not correct because it conflicts with the original chain. I seem to be generally confused and all turned around in this logical chain. I believe my confusion stems from not understanding how you can go from:
B C to C B.

I thought the only way to go the opposite direction of an arrow was to find the contra positive.

Thanks for the help

[Dave Killoran]

Hi Andwer123,

Thanks for the question! The step where you made the error is right here:

Then we would have:
~C ~B ~A
.: ~C ~A or A C

You recognized after the fact that there was a problem, so let's dive into what the issue was.

First, taking the contrapositive of B C was fine, and it resulted in the statement you correctly posted: C B.

But, you fell into a trap when interpreting the relationship between A and B in the diagram of A B. The meaning of that statement is that A and B can't go together. It is produced by either of the following two statements:

• A B
  
  B A

You mistakenly turned that into:

• B A

Which, if you take the contrapositive is the same as A B, which is exactly the opposite of what the original statement was saying


With that identified, let's go back and look again at B C. In this case, you didn't need a contrapositive since the terms are already linked. Essentially, once variables are connected, you can at that point get the inference you needed. Taking the CP in a case like this is only useful when you want to connect variables when they otherwise wouldn't connect easily. For example, if you were given this pair of statements, you'd want to take the contrapositive of the second term so you could connect through B:

• A B
  
  C B

So, since you don't need to take the contrapositive here, you need to rely on the inherent inference that resides within all arrow relationships (this is in another video). When you have B C, you automatically know that every B is a C. That means you also know the most Bs are Cs, and some Bs, are Cs. Well, some is a reversible terms, so you also know that some Cs are Bs, and that what we use to go "backward" from C to B.

Please let me know if that helps!

[andwer123]

...

• A B
  
  B A

You mistakenly turned that into:

• B A

...
Please let me know if that helps!

Thanks, Dave. This does help. I need to go back and review the double arrows again. I was unclear on exactly what those implied.