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ChantraS on November 21, 2017

Transitive Property Slide - C exists

You stated that P: C exists C: D exists C: E exists We are not sure about B or A. But is the following correct N/S logic? P: not C -> not B C: not B -> not A C: not C -> not A

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Mehran on November 21, 2017

@Chantra not sure I am understanding your question here.

Can you please clarify?

ChantraS on November 21, 2017

If we negate C, is the contrapositve in the other direction (to B&A) correct?

Mehran on November 24, 2017

@ChantraS yes that would be correct.

P1: A ==> B
not B ==> not A

P2: B ==> C
not C ==> not B

We can combined these two using the transitive chain as follows:

A ==> B ==> C
not C ==> not B ==> not A

Hope that helps! Please let us know if you have any other questions.

tyler.channell7@gmail.com on September 30, 2018

I had a question. I feel like I am lost with the homework.

With Transitive Property, it doesn't have to be:

A=>B
B=>C
Therefore, A=>C

I can be:
not A=>B
B=>C
Therefore, not A=>C

Correct? Or am I just totally missing transitive property?

tyler.channell7@gmail.com on September 30, 2018

I also have another question on the homework.

I was going through the flash cards and I was wondering if there has to be a correct order to how transitive property works.

For example,

For the missing premise flash card it states:

P:D => A
P:C = A
P:
C: not X => A

The missing conclusion that I arrived at was:
not D => X (or Contrapositive: not X => D)

But the answer was
not C => X

Couldn't my answer achieve the same goal? I am terribly confused at this. Please help.

Thanks.

Skylar on January 26, 2020

@tyler.channell7@gmail.com,

That is correct.

If we are given...:
P: not A -> B
not B -> A
P: B -> C
not C -> not B

...we can validly conclude:
not C -> not B -> A, or simplified: not C -> A
not A -> B -> C, or simplified: A -> not C

Therefore, the way the transitive property works remains the same regardless of what the variables are.

Skylar on January 26, 2020

@tyler.channell7@gmail.com,

Your answer is an equally valid possibility (assuming you meant to write the second given premise as "C->A" rather than "C=A").

We know
P: D -> A
not A -> not D
P: C -> A
not A -> not C
P: ?
C: not X -> A
not A -> X

The correct answer could be any of the four following answers:
1) not X -> D
2) not D -> X
3) not X -> C
4) not C -> X

Missing Premise Drills are unlike real LSAT section questions in their open-endedness, but this is because the drills are designed to acclimate you to the logic patterns behind more complex questions. In this case, either one of the first two premises can be used to come to the final conclusion, while the unused premise can be largely ignored. Moreover, the contrapositive of the missing premise has the same meaning as the missing premise, so it does not matter which you identify as the answer. Overall, these questions are meant to familiarize you with making logical connections, which you seem to be doing well.