Necessary/Sufficient and Chain Relationships

[alluv001]

Hello everyone,

I'm having a hard time combining the sufficient/necessary conditions (For example, written in "if, then") statements into chain relationships, especially when it comes to diagramming them from the stimulus into the actual chain.

Can someone provide some insight/elaboration on this?

Thanks!

[Emily Haney-Caron]

Hi alluv001,

That can definitely be a tricky thing to get the hang of at first. Can you give me an example that was difficult for you, or help me understand what about the chain relationship you're struggling with? That way, I can tailor my answer to your experience.

Thanks!

[alluv001]

Sure.

For example, on Lesson Two of the Powerscore LSAT Course Books, pg. 2-11.

#4 creates the chain relationship from the passage in one manner, however #5 inverts the manner in which the chain link is created. The explanations available online states that it is done due to the presence of "only if" in the stimulus of #5, but I'm still a little confused as to how and when to diagram chain relationships and their association to necessary/sufficient conditions (I've noticed sometimes the necessary -------> sufficient, and at others it is flipped?).

Thanks!

[Emily Haney-Caron]

Hi alluv001,

Thanks for providing an example. I think I understand where you're getting tripped up. Sufficient will always go on the left of the arrow, and necessary will always go to the right, when you are diagramming a stimulus. I think the tricky part for you here is making chain relationships when that requires you to generate a contrapositive. So in #4 on 2-11, we get the following relationships:
Sufficient Necessary
RLN NBU
NBU TSIO

From this, we can see that the common piece is NBU, and so we can create a chain around that element. The chain is easy on this one; we just put the two together, and nothing needs to change direction:
RLN NBU TSIO

For #5, it is a little bit trickier. We get these relationships:
Sufficient Necessary
UM C
L PC
I notC

We can see that the common element here is C, but in one case it is negated, and in the other it isn't. Additionally, unlike in #4 where the common element was sufficient in one statement and necessary in the other, here the C is necessary in each statement. This means that in order to connect the two conditional statements, we'll need to take the contrapositive of one. So here, we'll take the contrapositive for the first statement, to negate the C to create a common element and to get the common element on different sides of the arrow:
notC notUM

Now we can combine the common elements from the first and third conditional statements, using this new contrapositive. It's the same thing as in #4 - just connect the common element, starting with the conditional statement that has the common element as the necessary condition:
I notC notUM

Does that help? The trick is recognizing when you need to take the contrapositive of one of the statements. Once you do that, it should be straightforward to create the chain.

[alluv001]

Hi alluv001,

Thanks for providing an example. I think I understand where you're getting tripped up. Sufficient will always go on the left of the arrow, and necessary will always go to the right, when you are diagramming a stimulus. I think the tricky part for you here is making chain relationships when that requires you to generate a contrapositive. So in #4 on 2-11, we get the following relationships:
Sufficient Necessary
RLN NBU
NBU TSIO

From this, we can see that the common piece is NBU, and so we can create a chain around that element. The chain is easy on this one; we just put the two together, and nothing needs to change direction:
RLN NBU TSIO

For #5, it is a little bit trickier. We get these relationships:
Sufficient Necessary
UM C
L PC
I notC

We can see that the common element here is C, but in one case it is negated, and in the other it isn't. Additionally, unlike in #4 where the common element was sufficient in one statement and necessary in the other, here the C is necessary in each statement. This means that in order to connect the two conditional statements, we'll need to take the contrapositive of one. So here, we'll take the contrapositive for the first statement, to negate the C to create a common element and to get the common element on different sides of the arrow:
notC notUM

Now we can combine the common elements from the first and third conditional statements, using this new contrapositive. It's the same thing as in #4 - just connect the common element, starting with the conditional statement that has the common element as the necessary condition:
I notC notUM

Does that help? The trick is recognizing when you need to take the contrapositive of one of the statements. Once you do that, it should be straightforward to create the chain.

Certainly does help, your elaboration combined with further study really opened up the concept for me. I was always be to go and do both problems from scratch without any issues understanding why each step was made.

My main issue was thinking that at times there would be cases where the diagrams would have the necessary on the left and then the sufficient on the right. ( N -----> S). Now I see that this is never the case (...right?) and its helped tremendously!

Thank you!

[Dave Killoran]

My main issue was thinking that at times there would be cases where the diagrams would have the necessary on the left and then the sufficient on the right. ( N -----> S). Now I see that this is never the case (...right?) and its helped tremendously!

Hey Alluv, you are indeed correct about this, but it may help to think about it as the sufficient is at the start of the arrow, and the necessary is at the end of the arrow, as opposed to left and right. This is because later on, we'll start point the arrow in different directions at times (it will make sense then, and make solving problems easier). So, from that perspective, each of these diagrams is identical in meaning (ignore the badly drawn arrows—best I could do using just keystrokes!):

• S ---> N
  
  
  N <--- S
  
  
  S
  
  |
  |
  v
  
  N
  
  
  N
  
  ^
  |
  |
  
  S

So, when you are working with single arrows, it doesn't matter which way they point: the sufficient is always at the start or beginning of the arrow, and the necessary is always at the end.

Thanks!

[alluv001]

Dave,

Thanks for your elaboration. As I progress through the lessons I see why it is that my previous view was incorrect, and your explanation definitely helped cement the proper way of viewing it!

Thank you!